{"year": "1959", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (POL) For every integer $n$ prove that the fraction $\\frac{21 n+4}{14 n+3}$ cannot be reduced any further.", "solution": "1. The desired result $(14 n+3,21 n+4)=1$ follows from  $$ 3(14 n+3)-2(21 n+4)=1 $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1959", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (ROM) For which real numbers $x$ do the following equations hold: (a) $\\sqrt{x+\\sqrt{2 x-1}}+\\sqrt{x+\\sqrt{2 x-1}}=\\sqrt{2}$, (b) $\\sqrt{x+\\sqrt{2 x-1}}+\\sqrt{x+\\sqrt{2 x-1}}=1$, (c) $\\sqrt{x+\\sqrt{2 x-1}}+\\sqrt{x+\\sqrt{2 x-1}}=2$ ?", "solution": "2. For the square roots to be real we must have $2 x-1 \\geq 0 \\Rightarrow x \\geq 1 / 2$ and $x \\geq \\sqrt{2 x-1} \\Rightarrow x^{2} \\geq 2 x-1 \\Rightarrow(x-1)^{2} \\geq 0$, which always holds. Then we have $\\sqrt{x+\\sqrt{2 x-1}}+\\sqrt{x-\\sqrt{2 x-1}}=c \\Longleftrightarrow$  $$ c^{2}=2 x+2 \\sqrt{x^{2}-\\sqrt{2 x-1}^{2}}=2 x+2|x-1|= \\begin{cases}2, & 1 / 2 \\leq x \\leq 1 \\\\ 4 x-2, & x \\geq 1\\end{cases} $$  (a) $c^{2}=2$. The equation holds for $1 / 2 \\leq x \\leq 1$. (b) $c^{2}=1$. The equation has no solution. (c) $c^{2}=4$. The equation holds for $4 x-2=4 \\Rightarrow x=3 / 2$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1959", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (HUN) Let $x$ be an angle and let the real numbers $a, b, c, \\cos x$ satisfy the following equation:  $$ a \\cos ^{2} x+b \\cos x+c=0 . $$  Write the analogous quadratic equation for $a, b, c, \\cos 2 x$. Compare the given and the obtained equality for $a=4, b=2, c=-1$.  Second Day", "solution": "3. Multiplying the equality by $4\\left(a \\cos ^{2} x-b \\cos x+c\\right)$, we obtain $4 a^{2} \\cos ^{4} x+$ $2\\left(4 a c-2 b^{2}\\right) \\cos ^{2} x+4 c^{2}=0$. Plugging in $2 \\cos ^{2} x=1+\\cos 2 x$ we obtain (after quite a bit of manipulation):  $$ a^{2} \\cos ^{2} 2 x+\\left(2 a^{2}+4 a c-2 b^{2}\\right) \\cos 2 x+\\left(a^{2}+4 a c-2 b^{2}+4 c^{2}\\right)=0 $$  For $a=4, b=2$, and $c=-1$ we get $4 \\cos ^{2} x+2 \\cos x-1=0$ and $16 \\cos ^{2} 2 x+8 \\cos 2 x-4=0 \\Rightarrow 4 \\cos ^{2} 2 x+2 \\cos 2 x-1=0$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1959", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (HUN) Construct a right-angled triangle whose hypotenuse $c$ is given if it is known that the median from the right angle equals the geometric mean of the remaining two sides of the triangle.", "solution": "4. Analysis. Let $a$ and $b$ be the other two sides of the triangle. From the conditions of the problem we have $c^{2}=a^{2}+b^{2}$ and $c / 2=\\sqrt{a b} \\Leftrightarrow 3 / 2 c^{2}=$ $a^{2}+b^{2}+2 a b=(a+b)^{2} \\Leftrightarrow \\sqrt{3 / 2} c=a+b$. Given a desired $\\triangle A B C$ let $D$ be a point on $(A C$ such that $C D=C B$. In that case, $A D=a+b=\\sqrt{3 / 2} c$, and also, since $B C=C D$, it follows that $\\angle A D B=45^{\\circ}$. Construction. From a segment of length $c$ we elementarily construct a segment $A D$ of length $\\sqrt{3 / 2} c$. We then construct a ray ( $D X$ such that $\\angle A D X=45^{\\circ}$ and a circle $k(A, c)$ that intersects the ray at point $B$. Finally, we construct the perpendicular from $B$ to $A D$; point $C$ is the foot of that perpendicular. Proof. It holds that $A B=c$, and, since $C B=C D$, it also holds that $A C+$ $C B=A C+C D=A D=\\sqrt{3 / 2} c$. From this it follows that $\\sqrt{A C \\cdot C B}=$ $c / 2$. Since $B C$ is perpendicular to $A D$, it follows that $\\measuredangle B C A=90^{\\circ}$. Thus $A B C$ is the desired triangle. Discussion. Since $A B \\sqrt{2}=\\sqrt{2} c>\\sqrt{3 / 2} c=A D>A B$, the circle $k$ intersects the ray $D X$ in exactly two points, which correspond to two symmetric solutions.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1959", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (ROM) A segment $A B$ is given and on it a point $M$. On the same side of $A B$ squares $A M C D$ and $B M F E$ are constructed. The circumcircles of the two squares, whose centers are $P$ and $Q$, intersect in $M$ and another point $N$. (a) Prove that lines $F A$ and $B C$ intersect at $N$. (b) Prove that all such constructed lines $M N$ pass through the same point $S$, regardless of the selection of $M$. (c) Find the locus of the midpoints of all segments $P Q$, as $M$ varies along the segment $A B$.", "solution": "5. (a) It suffices to prove that $A F \\perp B C$, since then for the intersection point $X$ we have $\\angle A X C=\\angle B X F=90^{\\circ}$, implying that $X$ belongs to the circumcircles of both squares and thus that $X=N$. The relation $A F \\perp B C$ holds because from $M A=M C, M F=M B$, and $\\angle A M C=\\angle F M B$ it follows that $\\triangle A M F$ is obtained by rotating $\\triangle B M C$ by $90^{\\circ}$ around $M$. (b) Since $N$ is on the circumcircle of $B M F E$, it follows that $\\angle A N M=$ $\\angle M N B=45^{\\circ}$. Hence $M N$ is the bisector of $\\angle A N B$. It follows that $M N$ passes through the midpoint of the $\\operatorname{arc} \\widehat{A B}$ of the circle with diameter $A B$ (i.e., the circumcircle of $\\triangle A B N$ ) not containing $N$. (c) Let us introduce a coordinate system such that $A=(0,0), B=(b, 0)$, and $M=(m, 0)$. Setting in general $W=\\left(x_{W}, y_{W}\\right)$ for an arbitrary point $W$ and denoting by $R$ the midpoint of $P Q$, we have $y_{R}=\\left(y_{P}+\\right.$ $\\left.y_{Q}\\right) / 2=(m+b-m) / 4=b / 4$ and $x_{R}=\\left(x_{P}+x_{Q}\\right) / 2=(m+m+b) / 4=$ $(2 m+b) / 4$, the parameter $m$ varying from 0 to $b$. Thus the locus of all points $R$ is the closed segment $R_{1} R_{2}$ where $R_{1}=(b / 4, b / 4)$ and $R_{2}=(b / 4,3 b / 4)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1959", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (CZS) Let $\\alpha$ and $\\beta$ be two planes intersecting at a line $p$. In $\\alpha$ a point $A$ is given and in $\\beta$ a point $C$ is given, neither of which lies on $p$. Construct $B$ in $\\alpha$ and $D$ in $\\beta$ such that $A B C D$ is an equilateral trapezoid, $A B \\| C D$, in which a circle can be inscribed.", "solution": "6. Analysis. For $A B \\| C D$ to hold evidently neither must intersect $p$ and hence constructing lines $r$ in $\\alpha$ through $A$ and $s$ in $\\beta$ through $C$, both being parallel to $p$, we get that $B \\in r$ and $D \\in s$. Hence the problem reduces to a planar problem in $\\gamma$, determined by $r$ and $s$. Denote by $A^{\\prime}$ the foot of the perpendicular from $A$ to $s$. Since $A B C D$ is isosceles and has an incircle, it follows $A D=B C=(A B+C D) / 2=A^{\\prime} C$. The remaining parts of the problem are now obvious.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1960", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (BUL) Find all the three-digit numbers for which one obtains, when dividing the number by 11 , the sum of the squares of the digits of the initial number.", "solution": "1. Given the number $\\overline{a c b}$, since $11 \\mid \\overline{a c b}$, it follows that $c=a+b$ or $c=$ $a+b-11$. In the first case, $a^{2}+b^{2}+(a+b)^{2}=10 a+b$, and in the second case, $a^{2}+b^{2}+(a+b-11)^{2}=10(a-1)+b$. In the first case the LHS is even, and hence $b \\in\\{0,2,4,6,8\\}$, while in the second case it is odd, and hence $b \\in\\{1,3,5,7,9\\}$. Analyzing the 10 quadratic equations for $a$ we obtain that the only valid solutions are 550 and 803.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1960", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (HUN) For which real numbers $x$ does the following inequality hold:  $$ \\frac{4 x^{2}}{(1-\\sqrt{1+2 x})^{2}}<2 x+9 ? $$", "solution": "2. The LHS term is well-defined for $x \\geq-1 / 2$ and $x \\neq 0$. Furthermore, $4 x^{2} /(1-\\sqrt{1+2 x})^{2}=(1+\\sqrt{1+2 x})^{2}$. Since $f(x)=(1+\\sqrt{1+2 x})^{2}-2 x-$ $9=2 \\sqrt{1+2 x}-7$ is increasing and since $f(45 / 8)=0$, it follows that the inequality holds precisely for $-1 / 2 \\leq x<45 / 8$ and $x \\neq 0$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1960", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (ROM) A right-angled triangle $A B C$ is given for which the hypotenuse $B C$ has length $a$ and is divided into $n$ equal segments, where $n$ is odd. Let $\\alpha$ be the angle with which the point $A$ sees the segment containing the middle of the hypotenuse. Prove that  $$ \\tan \\alpha=\\frac{4 n h}{\\left(n^{2}-1\\right) a}, $$  where $h$ is the height of the triangle.  ## Second Day", "solution": "3. Let $B^{\\prime} C^{\\prime}$ be the middle of the $n=2 k+1$ segments and let $D$ be the foot of the perpendicular from $A$ to the hypotenuse. Let us assume $\\mathcal{B}\\left(C, D, C^{\\prime}, B^{\\prime}, B\\right)$. Then from $C D<B D, C D+B D=a$, and $C D \\cdot B D=$ $h^{2}$ we have $C D^{2}-a \\cdot C D+h^{2}=0 \\Longrightarrow C D=\\left(a-\\sqrt{a^{2}-4 h^{2}}\\right) / 2$. Let us define $\\measuredangle D A C^{\\prime}=\\gamma$ and $\\measuredangle D A B^{\\prime}=\\beta$; then $\\tan \\beta=D B^{\\prime} / h$ and $\\tan \\gamma=$ $D C^{\\prime} / h$. Since $D B^{\\prime}=C B^{\\prime}-C D=(k+1) a /(2 k+1)-\\left(c-\\sqrt{c^{2}-4 h^{2}}\\right) / 2$ and $D C^{\\prime}=k a /(2 k+1)-\\left(c-\\sqrt{c^{2}-4 h^{2}}\\right) / 2$, we have  $$ \\begin{aligned} \\tan \\alpha=\\tan (\\beta-\\gamma) & =\\frac{\\tan \\beta-\\tan \\gamma}{1+\\tan \\beta \\cdot \\tan \\gamma}=\\frac{\\frac{a}{(2 k+1) h}}{1+\\frac{a^{2}-4 h^{2}}{4 h^{2}}-\\frac{a^{2}}{4 h^{2}(2 k+1)^{2}}} \\\\ & =\\frac{4 h(2 k+1)}{4 a k(k+1)}=\\frac{4 n h}{\\left(n^{2}-1\\right) a} . \\end{aligned} $$  The case $\\mathcal{B}\\left(C, C^{\\prime}, D, B^{\\prime}, B\\right)$ is similar.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1960", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (HUN) Construct a triangle $A B C$ whose lengths of heights $h_{a}$ and $h_{b}$ (from $A$ and $B$, respectively) and length of median $m_{a}$ (from $A$ ) are given.", "solution": "4. Analysis. Let $A^{\\prime}$ and $B^{\\prime}$ be the feet of the perpendiculars from $A$ and $B$, respectively, to the opposite sides, $A_{1}$ the midpoint of $B C$, and let $D^{\\prime}$ be the foot of the perpendicular from $A_{1}$ to $A C$. We then have $A A_{1}=m_{a}$, $A A^{\\prime}=h_{a}, \\angle A A^{\\prime} A_{1}=90^{\\circ}, A_{1} D^{\\prime}=h_{b} / 2$, and $\\angle A D^{\\prime} A_{1}=90^{\\circ}$. Construction. We construct the quadrilateral $A D^{\\prime} A_{1} A^{\\prime}$ (starting from the circle with diameter $A A_{1}$ ). Then $C$ is the intersection of $A^{\\prime} A_{1}$ and $A D^{\\prime}$, and $B$ is on the line $A_{1} C$ such that $C A_{1}=A_{1} B$ and $\\mathcal{B}\\left(B, A_{1}, C\\right)$. Discussion. We must have $m_{a} \\geq h_{a}$ and $m_{a} \\geq h_{b} / 2$. The number of solutions is 0 if $m_{a}=h_{a}=h_{b} / 2,1$ if two of $m_{a}, h_{a}, h_{b} / 2$ are equal, and 2 otherwise.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1960", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (CZS) A cube $A B C D A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime}$ is given. (a) Find the locus of all midpoints of segments $X Y$, where $X$ is any point on segment $A C$ and $Y$ any point on segment $B^{\\prime} D^{\\prime}$. (b) Find the locus of all points $Z$ on segments $X Y$ such that $\\overrightarrow{Z Y}=2 \\overrightarrow{X Z}$.", "solution": "5. (a) The locus of the points is the square $E F G H$ where these four points are the centers of the faces $A B B^{\\prime} A^{\\prime}, B C C^{\\prime} B^{\\prime}, C D D^{\\prime} C^{\\prime}$ and $D A A^{\\prime} D^{\\prime}$. (b) The locus of the points is the rectangle $I J K L$ where these points are on $A B^{\\prime}, C B^{\\prime}, C D^{\\prime}$, and $A D^{\\prime}$ at a distance of $A A^{\\prime} / 3$ with respect to the plane $A B C D$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1960", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (BUL) An isosceles trapezoid with bases $a$ and $b$ and height $h$ is given. (a) On the line of symmetry construct the point $P$ such that both (nonbase) sides are seen from $P$ with an angle of $90^{\\circ}$. (b) Find the distance of $P$ from one of the bases of the trapezoid. (c) Under what conditions for $a, b$, and $h$ can the point $P$ be constructed (analyze all possible cases)?", "solution": "6. Let $E, F$ respectively be the midpoints of the bases $A B, C D$ of the isosceles trapezoid $A B C D$. (a) The point $P$ is on the intersection of $E F$ and the circle with diameter $B C$. (b) Let $x=E P$. Since $\\triangle B E P \\sim \\triangle P F C$, we have $x(h-x)=a b / 4 \\Rightarrow$ $x_{1,2}=\\left(h \\pm \\sqrt{h^{2}-a b}\\right) / 2$. (c) If $h^{2}>a b$ there are two solutions, if $h^{2}=a b$ there is only one solution, and if $h^{2}<a b$ there are no solutions.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1960", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (GDR) A sphere is inscribed in a regular cone. Around the sphere a cylinder is circumscribed so that its base is in the same plane as the base of the cone. Let $V_{1}$ be the volume of the cone and $V_{2}$ the volume of the cylinder. (a) Prove that $V_{1}=V_{2}$ is impossible. (b) Find the smallest $k$ for which $V_{1}=k V_{2}$, and in this case construct the angle at the vertex of the cone.", "solution": "7. Let $A$ be the vertex of the cone, $O$ the center of the sphere, $S$ the center of the base of the cone, $B$ a point on the base circle, and $r$ the radius of the sphere. Let $\\angle S A B=\\alpha$. We easily obtain $A S=r(1+\\sin \\alpha) / \\sin \\alpha$ and $S B=r(1+\\sin \\alpha) \\tan \\alpha / \\sin \\alpha$ and hence $V_{1}=\\pi S B^{2} \\cdot S A / 3=\\pi r^{3}(1+$ $\\sin \\alpha)^{2} /[3 \\sin \\alpha(1-\\sin \\alpha)]$. We also have $V_{2}=2 \\pi r^{3}$ and hence  $$ k=\\frac{(1+\\sin \\alpha)^{2}}{6 \\sin \\alpha(1-\\sin \\alpha)} \\Rightarrow(1+6 k) \\sin ^{2} \\alpha+2(1-3 k) \\sin \\alpha+1=0 $$  The discriminant of this quadratic must be nonnegative: $(1-3 k)^{2}-(1+$ $6 k) \\geq 0 \\Rightarrow k \\geq 4 / 3$. Hence we cannot have $k=1$. For $k=4 / 3$ we have $\\sin \\alpha=1 / 3$, whose construction is elementary.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1961", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (HUN) Solve the following system of equations:  $$ \\begin{aligned} x+y+z & =a, \\\\ x^{2}+y^{2}+z^{2} & =b^{2}, \\\\ x y & =z^{2} \\end{aligned} $$  where $a$ and $b$ are given real numbers. What conditions must hold on $a$ and $b$ for the solutions to be positive and distinct?", "solution": "1. This is a problem solvable using elementary manipulations, so we shall state only the final solutions. For $a=0$ we get $(x, y, z)=(0,0,0)$. For $a \\neq 0$ we get $(x, y, z) \\in\\left\\{\\left(t_{1}, t_{2}, z_{0}\\right),\\left(t_{2}, t_{1}, z_{0}\\right)\\right\\}$, where  $$ z_{0}=\\frac{a^{2}-b^{2}}{2 a} \\quad \\text { and } \\quad t_{1,2}=\\frac{a^{2}+b^{2} \\pm \\sqrt{\\left(3 a^{2}-b^{2}\\right)\\left(3 b^{2}-a^{2}\\right)}}{4 a} $$  For the solutions to be positive and distinct the following conditions are necessary and sufficient: $3 b^{2}>a^{2}>b^{2}$ and $a>0$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1961", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (POL) Let $a, b$, and $c$ be the lengths of a triangle whose area is $S$. Prove that  $$ a^{2}+b^{2}+c^{2} \\geq 4 S \\sqrt{3} $$  In what case does equality hold?", "solution": "2. Using $S=b c \\sin \\alpha / 2, a^{2}=b^{2}+c^{2}-2 b c \\cos \\alpha$ and $(\\sqrt{3} \\sin \\alpha+\\cos \\alpha) / 2=$ $\\cos \\left(\\alpha-60^{\\circ}\\right)$ we have  $$ \\begin{gathered} a^{2}+b^{2}+c^{2} \\geq 4 S \\sqrt{3} \\Leftrightarrow b^{2}+c^{2} \\geq b c(\\sqrt{3} \\sin \\alpha+\\cos \\alpha) \\Leftrightarrow \\\\ \\Leftrightarrow(b-c)^{2}+2 b c\\left(1-\\cos \\left(\\alpha-60^{\\circ}\\right)\\right) \\geq 0, \\end{gathered} $$  where equality holds if and only if $b=c$ and $\\alpha=60^{\\circ}$, i.e., if the triangle is equilateral.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1961", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (BUL) Solve the equation $\\cos ^{n} x-\\sin ^{n} x=1$, where $n$ is a given positive integer.  ## Second Day", "solution": "3. For $n \\geq 2$ we have  $$ \\begin{aligned} 1 & =\\cos ^{n} x-\\sin ^{n} x \\leq\\left|\\cos ^{n} x-\\sin ^{n} x\\right| \\\\ & \\leq\\left|\\cos ^{n} x\\right|+\\left|\\sin ^{n} x\\right| \\leq \\cos ^{2} x+\\sin ^{2} x=1 \\end{aligned} $$  Hence $\\sin ^{2} x=\\left|\\sin ^{n} x\\right|$ and $\\cos ^{2} x=\\left|\\cos ^{n} x\\right|$, from which it follows that $\\sin x, \\cos x \\in\\{1,0,-1\\} \\Rightarrow x \\in \\pi \\mathbb{Z} / 2$. By inspection one obtains the set of solutions $\\{m \\pi \\mid m \\in \\mathbb{Z}\\}$ for even $n$ and $\\{2 m \\pi, 2 m \\pi-\\pi / 2 \\mid m \\in \\mathbb{Z}\\}$ for odd $n$. For $n=1$ we have $1=\\cos x-\\sin x=-\\sqrt{2} \\sin (x-\\pi / 4)$, which yields the set of solutions  $$ \\{2 m \\pi, 2 m \\pi-\\pi / 2 \\mid m \\in \\mathbb{Z}\\} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1961", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (GDR) In the interior of $\\triangle P_{1} P_{2} P_{3}$ a point $P$ is given. Let $Q_{1}, Q_{2}$, and $Q_{3}$ respectively be the intersections of $P P_{1}, P P_{2}$, and $P P_{3}$ with the opposing edges of $\\triangle P_{1} P_{2} P_{3}$. Prove that among the ratios $P P_{1} / P Q_{1}, P P_{2} / P Q_{2}$, and $P P_{3} / P Q_{3}$ there exists at least one not larger than 2 and at least one not smaller than 2.", "solution": "4. Let $x_{i}=P P_{i} / P Q_{i}$ for $i=1,2,3$. For all $i$ we have  $$ \\frac{1}{x_{i}+1}=\\frac{P Q_{i}}{P_{i} Q_{i}}=\\frac{S_{P P_{j} P_{k}}}{S_{P_{1} P_{2} P_{3}}} $$  where the indices $j$ and $k$ are distinct and different from $i$. Hence we have  $$ \\begin{aligned} f\\left(x_{1}, x_{2}, x_{3}\\right) & =\\frac{1}{x_{1}+1}+\\frac{1}{x_{2}+1}+\\frac{1}{x_{3}+1} \\\\ & =\\frac{S\\left(P P_{2} P_{3}\\right)+S\\left(P P_{1} P_{3}\\right)+S\\left(P P_{2} P_{3}\\right)}{S\\left(P_{1} P_{2} P_{3}\\right)}=1 \\end{aligned} $$  It follows that $1 /\\left(x_{i}+1\\right) \\geq 1 / 3$ for some $i$ and $1 /\\left(x_{j}+1\\right) \\leq 1 / 3$ for some $j$. Consequently, $x_{i} \\leq 2$ and $x_{j} \\geq 2$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1961", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (CZS) Construct a triangle $A B C$ if the following elements are given: $A C=b, A B=c$, and $\\measuredangle A M B=\\omega\\left(\\omega<90^{\\circ}\\right)$, where $M$ is the midpoint of $B C$. Prove that the construction has a solution if and only if  $$ b \\tan \\frac{\\omega}{2} \\leq c<b . $$  In what case does equality hold?", "solution": "5. Analysis. Let $C_{1}$ be the midpoint of $A B$. In $\\triangle A M B$ we have $M C_{1}=b / 2$, $A B=c$, and $\\angle A M B=\\omega$. Thus, given $A B=c$, the point $M$ is at the intersection of the circle $k\\left(C^{\\prime}, b / 2\\right)$ and the set of points $e$ that view $A B$ at an angle of $\\omega$. The construction of $A B C$ is now obvious. Discussion. It suffices to establish the conditions for which $k$ and $e$ intersect. Let $E$ be the midpoint of one of the arcs that make up $e$. A necessary and sufficient condition for $k$ to intersect $e$ is  $$ \\frac{c}{2}=C^{\\prime} A \\leq \\frac{b}{2} \\leq C^{\\prime} E=\\frac{c}{2} \\cot \\frac{\\omega}{2} \\Leftrightarrow b \\tan \\frac{\\omega}{2} \\leq c<b . $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1961", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (ROM) A plane $\\epsilon$ is given and on one side of the plane three noncollinear points $A, B$, and $C$ such that the plane determined by them is not parallel to $\\epsilon$. Three arbitrary points $A^{\\prime}, B^{\\prime}$, and $C^{\\prime}$ in $\\epsilon$ are selected. Let $L, M$, and $N$ be the midpoints of $A A^{\\prime}, B B^{\\prime}$, and $C C^{\\prime}$, and $G$ the centroid of $\\triangle L M N$. Find the locus of all points obtained for $G$ as $A^{\\prime}, B^{\\prime}$, and $C^{\\prime}$ are varied (independently of each other) across $\\epsilon$.", "solution": "6. Let $h(X)$ denote the distance of a point $X$ from $\\epsilon, X$ restricted to being on the same side of $\\epsilon$ as $A, B$, and $C$. Let $G_{1}$ be the (fixed) centroid of $\\triangle A B C$ and $G_{1}^{\\prime}$ the centroid of $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$. It is trivial to prove that $G$ is the midpoint of $G_{1} G_{1}^{\\prime}$. Hence varying $G_{1}^{\\prime}$ across $\\epsilon$, we get that the locus of $G$ is the plane $\\alpha$ parallel to $\\epsilon$ such that  $$ X \\in \\alpha \\Leftrightarrow h(X)=\\frac{h\\left(G_{1}\\right)}{2}=\\frac{h(A)+h(B)+h(C)}{6} . $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1962", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (POL) Find the smallest natural number $n$ with the following properties: (a) In decimal representation it ends with 6. (b) If we move this digit to the front of the number, we get a number 4 times larger.", "solution": "1. From the conditions of the problem we have $n=10 x+6$ and $4 n=$ $6 \\cdot 10^{m}+x$ for some integer $x$. Eliminating $x$ from these two equations, we get $40 n=6 \\cdot 10^{m+1}+n-6 \\Rightarrow n=2\\left(10^{m+1}-1\\right) / 13$. Hence we must find the smallest $m$ such that this fraction is an integer. By inspection, this happens for $m=6$, and for this $m$ we obtain $n=153846$, which indeed satisfies the conditions of the problem.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1962", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (HUN) Find all real numbers $x$ for which  $$ \\sqrt{3-x}-\\sqrt{x+1}>\\frac{1}{2} . $$", "solution": "2. We note that $f(x)=\\sqrt{3-x}-\\sqrt{x+1}$ is well-defined only for $-1 \\leq x \\leq 3$ and is decreasing (and obviously continuous) on this interval. We also note that $f(-1)=2>1 / 2$ and $f(1-\\sqrt{31} / 8)=\\sqrt{(1 / 4+\\sqrt{31} / 4)^{2}}-$ $\\sqrt{(1 / 4-\\sqrt{31} / 4)^{2}}=1 / 2$. Hence the inequality is satisfied for $-1 \\leq x<$ $1-\\sqrt{31} / 8$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1962", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (CZS) A cube $A B C D A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime}$ is given. The point $X$ is moving at a constant speed along the square $A B C D$ in the direction from $A$ to $B$. The point $Y$ is moving with the same constant speed along the square $B C C^{\\prime} B^{\\prime}$ in the direction from $B^{\\prime}$ to $C^{\\prime}$. Initially, $X$ and $Y$ start out from $A$ and $B^{\\prime}$ respectively. Find the locus of all the midpoints of $X Y$.  Second Day", "solution": "3. By inspecting the four different stages of this periodic motion we easily obtain that the locus of the midpoints of $X Y$ is the edges of $M N C Q$, where $M, N$, and $Q$ are the centers of $A B B^{\\prime} A^{\\prime}, B C C^{\\prime} B^{\\prime}$, and $A B C D$, respectively.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1962", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (ROM) Solve the equation  $$ \\cos ^{2} x+\\cos ^{2} 2 x+\\cos ^{2} 3 x=1 . $$", "solution": "4. Since $\\cos 2 x=1+\\cos ^{2} x$ and $\\cos \\alpha+\\cos \\beta=2 \\cos \\left(\\frac{\\alpha+\\beta}{2}\\right) \\cos \\left(\\frac{\\alpha-\\beta}{2}\\right)$, we have $\\cos ^{2} x+\\cos ^{2} 2 x+\\cos ^{2} 3 x=1 \\Leftrightarrow \\cos 2 x+\\cos 4 x+2 \\cos ^{2} 3 x=$ $2 \\cos 3 x(\\cos x+\\cos 3 x)=0 \\Leftrightarrow 4 \\cos 3 x \\cos 2 x \\cos x=0$. Hence the solutions are $x \\in\\{\\pi / 2+m \\pi, \\pi / 4+m \\pi / 2, \\pi / 6+m \\pi / 3 \\mid m \\in \\mathbb{Z}\\}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1962", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (BUL) On the circle $k$ three points $A, B$, and $C$ are given. Construct the fourth point on the circle $D$ such that one can inscribe a circle in $A B C D$.", "solution": "5. Analysis. Let $A B C D$ be the desired quadrilateral. Let us assume w.l.o.g. that $A B>B C$ (for $A B=B C$ the construction is trivial). For a tangent quadrilateral we have $A D-D C=A B-B C$. Let $X$ be a point on $A D$ such that $D X=D C$. We then have $A X=A B-B C$ and $\\measuredangle A X C=$ $\\measuredangle A D C+\\measuredangle C D X=180^{\\circ}-\\angle A B C / 2$. Constructing $X$ and hence $D$ is now obvious.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1962", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (GDR) Let $A B C$ be an isosceles triangle with circumradius $r$ and inradius $\\rho$. Prove that the distance $d$ between the circumcenter and incenter is given by  $$ d=\\sqrt{r(r-2 \\rho)} . $$", "solution": "6. This problem is a special case, when the triangle is isosceles, of Euler's formula, which holds for all triangles.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1962", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (USS) Prove that a tetrahedron $S A B C$ has five different spheres that touch all six lines determined by its edges if and only if it is regular.", "solution": "7. The spheres are arranged in a similar manner as in the planar case where we have one incircle and three excircles. Here we have one \"insphere\" and four \"exspheres\" corresponding to each of the four sides. Each vertex of the tetrahedron effectively has three tangent lines drawn from it to each of the five spheres. Repeatedly using the equality of the three tangent segments from a vertex (in the same vein as for tangent planar quadrilaterals) we obtain $S A+B C=S B+C A=S C+A B$ from the insphere. From the exsphere opposite of $S$ we obtain $S A-B C=S B-C A=S C-A B$, hence $S A=S B=S C$ and $A B=B C=C A$. By symmetry, we also have $A B=A C=A S$. Hence indeed, all the edges of the tetrahedron are equal in length and thus we have shown that the tetrahedron is regular.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1963", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (CZS) Determine all real solutions of the equation $\\sqrt{x^{2}-p}+2 \\sqrt{x^{2}-1}=$ $x$, where $p$ is a real number.", "solution": "1. Obviously, $x \\geq 0$; hence squaring the given equation yields an equivalent equation $5 x^{2}-p-4+4 \\sqrt{\\left(x^{2}-1\\right)\\left(x^{2}-p\\right)}=x^{2}$, i.e., $4 \\sqrt{\\left(x^{2}-1\\right)\\left(x^{2}-p\\right)}=$ $(p+4)-4 x^{2}$. If $4 x^{2} \\leq(p+4)$, we may square the equation once again to get $-16(p+1) x^{2}+16 p=-8(p+4) x^{2}+(p+4)^{2}$, which is equivalent to $x^{2}=(4-p)^{2} /[4(4-2 p)]$, i.e., $x=(4-p) /(2 \\sqrt{4-2 p})$. For this to be a solution we must have $p \\leq 2$ and $(4-p)^{2} /(4-2 p)=4 x^{2} \\leq(p+4)$. Hence $4 / 3 \\leq p \\leq 2$. Otherwise there is no solution.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1963", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (USS) Find the locus of points in space that are vertices of right angles of which one ray passes through a given point and the other intersects a given segment.", "solution": "2. Let $A$ be the given point, $B C$ the given segment, and $\\mathcal{B}_{1}, \\mathcal{B}_{2}$ the closed balls with the diameters $A B$ and $A C$ respectively. Consider one right angle $\\angle A O K$ with $K \\in[B C]$. If $B^{\\prime}, C^{\\prime}$ are the feet of the perpendiculars from $B, C$ to $A O$ respectively, then $O$ lies on the segment $B^{\\prime} C^{\\prime}$, which implies that it lies on exactly one of the segments $A B^{\\prime}, A C^{\\prime}$. Hence $O$ belongs to exactly one of the balls $\\mathcal{B}_{1}, \\mathcal{B}_{2}$; i.e., $O \\in \\mathcal{B}_{1} \\Delta \\mathcal{B}_{2}$. This is obviously the required locus.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1963", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (HUN) Prove that if all the angles of a convex $n$-gon are equal and the lengths of consecutive edges $a_{1}, \\ldots, a_{n}$ satisfy $a_{1} \\geq a_{2} \\geq \\cdots \\geq a_{n}$, then $a_{1}=a_{2}=\\cdots=a_{n}$.  Second Day", "solution": "3. Let $\\overrightarrow{O A_{1}}, \\overrightarrow{O A_{2}}, \\ldots, \\overrightarrow{O A_{n}}$ be the vectors corresponding respectively to the edges $a_{1}, a_{2}, \\ldots, a_{n}$ of the polygon. By the conditions of the problem, these vectors satisfy $\\overrightarrow{O A_{1}}+\\cdots+\\overrightarrow{O A_{n}}=\\overrightarrow{0}, \\angle A_{1} O A_{2}=\\angle A_{2} O A_{3}=\\cdots=$ $\\angle A_{n} O A_{1}=2 \\pi / n$ and $O A_{1} \\geq O A_{2} \\geq \\cdots \\geq O A_{n}$. Our task is to prove that $O A_{1}=\\cdots=O A_{n}$. Let $l$ be the line through $O$ perpendicular to $O A_{n}$, and $B_{1}, \\ldots, B_{n-1}$ the projections of $A_{1}, \\ldots, A_{n-1}$ onto $l$ respectively. By the assumptions, the sum of the $\\overrightarrow{O B_{i}}$ 's is $\\overrightarrow{0}$. On the other hand, since $O B_{i} \\leq O B_{n-i}$ for all $i \\leq n / 2$, all the sums $\\overrightarrow{O B_{i}}+\\overrightarrow{O B_{n-i}}$ lie on the same side of the point $O$. Hence all these sums must be equal to $\\overrightarrow{0}$. Consequently, $O A_{i}=O A_{n-i}$, from which the result immediately follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1963", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (USS) Find all solutions $x_{1}, \\ldots, x_{5}$ to the system of equations  $$ \\left\\{\\begin{array}{l} x_{5}+x_{2}=y x_{1}, \\\\ x_{1}+x_{3}=y x_{2}, \\\\ x_{2}+x_{4}=y x_{3}, \\\\ x_{3}+x_{5}=y x_{4}, \\\\ x_{4}+x_{1}=y x_{5}, \\end{array}\\right. $$  where $y$ is a real parameter.", "solution": "4. Summing up all the equations yields $2\\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\\right)=y\\left(x_{1}+\\right.$ $x_{2}+x_{3}+x_{4}+x_{5}$ ). If $y=2$, then the given equations imply $x_{1}-x_{2}=$ $x_{2}-x_{3}=\\cdots=x_{5}-x_{1}$; hence $x_{1}=x_{2}=\\cdots=x_{5}$, which is clearly a solution. If $y \\neq 2$, then $x_{1}+\\cdots+x_{5}=0$, and summing the first three equalities gives $x_{2}=y\\left(x_{1}+x_{2}+x_{3}\\right)$. Using that $x_{1}+x_{3}=y x_{2}$ we obtain $x_{2}=\\left(y^{2}+y\\right) x_{2}$, i.e., $\\left(y^{2}+y-1\\right) x_{2}=0$. If $y^{2}+y-1 \\neq 0$, then $x_{2}=0$, and similarly $x_{1}=\\cdots=x_{5}=0$. If $y^{2}+y-1=0$, it is easy to prove that the last two equations are the consequence of the first three. Thus choosing any values for $x_{1}$ and $x_{5}$ will give exactly one solution for $x_{2}, x_{3}, x_{4}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1963", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (GDR) Prove that $\\cos \\frac{\\pi}{7}-\\cos \\frac{2 \\pi}{7}+\\cos \\frac{3 \\pi}{7}=\\frac{1}{2}$.", "solution": "5. The LHS of the desired identity equals $S=\\cos (\\pi / 7)+\\cos (3 \\pi / 7)+$ $\\cos (5 \\pi / 7)$. Now  $$ S \\sin \\frac{\\pi}{7}=\\frac{\\sin \\frac{2 \\pi}{7}}{2}+\\frac{\\sin \\frac{4 \\pi}{7}-\\sin \\frac{2 \\pi}{7}}{2}+\\frac{\\sin \\frac{6 \\pi}{7}-\\sin \\frac{4 \\pi}{7}}{2}=\\frac{\\sin \\frac{6 \\pi}{7}}{2} \\Rightarrow S=\\frac{1}{2} . $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1963", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (HUN) Five students $A, B, C, D$, and $E$ have taken part in a certain competition. Before the competition, two persons $X$ and $Y$ tried to guess the rankings. $X$ thought that the ranking would be $A, B, C, D, E$; and $Y$ thought that the ranking would be $D, A, E, C, B$. At the end, it was revealed that $X$ didn't guess correctly any rankings of the participants, and moreover, didn't guess any of the orderings of pairs of consecutive participants. On the other hand, $Y$ guessed the correct rankings of two participants and the correct ordering of two pairs of consecutive participants. Determine the rankings of the competition.", "solution": "6. The result is $E D A C B$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1964", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (CZS) (a) Find all natural numbers $n$ such that the number $2^{n}-1$ is divisible by 7 . (b) Prove that for all natural numbers $n$ the number $2^{n}+1$ is not divisible by 7 .", "solution": "1. Let $n=3 k+r$, where $0 \\leq r<2$. Then $2^{n}=2^{3 k+r}=8^{k} \\cdot 2^{r} \\equiv 2^{r}(\\bmod 7)$. Thus the remainder of $2^{n}$ modulo 7 is $1,2,4$ if $n \\equiv 0,1,2(\\bmod 3)$. Hence $2^{n}-1$ is divisible by 7 if and only if $3 \\mid n$, while $2^{n}+1$ is never divisible by 7 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1964", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (HUN) Denote by $a, b, c$ the lengths of the sides of a triangle. Prove that  $$ a^{2}(b+c-a)+b^{2}(c+a-b)+c^{2}(a+b-c) \\leq 3 a b c $$", "solution": "2. By substituting $a=x+y, b=y+z$, and $c=z+x(x, y, z>0)$ the given inequality becomes  $$ 6 x y z \\leq x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x+z x^{2}, $$  which follows immediately by the AM-GM inequality applied to $x^{2} y, x y^{2}$, $x^{2} z, x z^{2}, y^{2} z, y z^{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1964", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (YUG) The incircle is inscribed in a triangle $A B C$ with sides $a, b, c$. Three tangents to the incircle are drawn, each of which is parallel to one side of the triangle $A B C$. These tangents form three smaller triangles (internal to $\\triangle A B C$ ) with the sides of $\\triangle A B C$. In each of these triangles an incircle is inscribed. Determine the sum of areas of all four incircles.  Second Day", "solution": "3. Let $r$ be the radius of the incircle of $\\triangle A B C, r_{a}, r_{b}, r_{c}$ the radii of the smaller circles corresponding to $A, B, C$, and $h_{a}, h_{b}, h_{c}$ the altitudes from $A, B, C$ respectively. The coefficient of similarity between the smaller triangle at $A$ and the triangle $A B C$ is $1-2 r / h_{a}$, from which we easily obtain $r_{a}=\\left(h_{a}-2 r\\right) r / h_{a}=(s-a) r / s$. Similarly, $r_{b}=(s-b) r / s$ and $r_{c}=(s-c) r / s$. Now a straightforward computation gives that the sum of areas of the four circles is given by  $$ \\Sigma=\\frac{(b+c-a)(c+a-b)(a+b-c)\\left(a^{2}+b^{2}+c^{2}\\right) \\pi}{(a+b+c)^{3}} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1964", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (HUN) Each of 17 students talked with every other student. They all talked about three different topics. Each pair of students talked about one topic. Prove that there are three students that talked about the same topic among themselves.", "solution": "4. Let us call the topics $T_{1}, T_{2}, T_{3}$. Consider an arbitrary student $A$. By the pigeonhole principle there is a topic, say $T_{3}$, he discussed with at least 6 other students. If two of these 6 students discussed $T_{3}$, then we are done. Suppose now that the 6 students discussed only $T_{1}$ and $T_{2}$ and choose one of them, say $B$. By the pigeonhole principle he discussed one of the topics, say $T_{2}$, with three of these students. If two of these three students also discussed $T_{2}$, then we are done. Otherwise, all the three students discussed only $T_{1}$, which completes the task.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1964", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (ROM) Five points are given in the plane. Among the lines that connect these five points, no two coincide and no two are parallel or perpendicular. Through each point we construct an altitude to each of the other lines. What is the maximal number of intersection points of these altitudes (excluding the initial five points)?", "solution": "5. Let us first compute the number of intersection points of the perpendiculars passing through two distinct points $B$ and $C$. The perpendiculars from $B$ to the lines through $C$ other than $B C$ meet all perpendiculars from $C$, which counts to $3 \\cdot 6=18$ intersection points. Each perpendicular from $B$ to the 3 lines not containing $C$ can intersect at most 5 of the perpendiculars passing through $C$, which counts to another $3 \\cdot 5=15$ intersection points. Thus there are $18+15=33$ intersection points corresponding to $B, C$. It follows that the required total number is at most $10 \\cdot 33=330$. But some of these points, namely the orthocenters of the triangles with vertices at the given points, are counted thrice. There are 10 such points. Hence the maximal number of intersection points is $330-2 \\cdot 10=310$.  Remark. The jury considered only the combinatorial part of the problem and didn't require an example in which 310 points appear. However, it is \"easily\" verified that, for instance, the set of points $A(1,1), B(e, \\pi)$, $C\\left(e^{2}, \\pi^{2}\\right), D\\left(e^{3}, \\pi^{3}\\right), E\\left(e^{4}, \\pi^{4}\\right)$ works.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1964", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (POL) Given a tetrahedron $A B C D$, let $D_{1}$ be the centroid of the triangle $A B C$ and let $A_{1}, B_{1}, C_{1}$ be the intersection points of the lines parallel to $D D_{1}$ and passing through the points $A, B, C$ with the opposite faces of the tetrahedron. Prove that the volume of the tetrahedron $A B C D$ is onethird the volume of the tetrahedron $A_{1} B_{1} C_{1} D_{1}$. Does the result remain true if the point $D_{1}$ is replaced with any point inside the triangle $A B C$ ?", "solution": "6. We shall prove that the statement is valid in the general case, for an arbitrary point $D_{1}$ inside $\\triangle A B C$. Since $D_{1}$ belongs to the plane $A B C$, there are real numbers $a, b, c$ such that $(a+b+c) \\overrightarrow{D D_{1}}=a \\overrightarrow{D A}+b \\overrightarrow{D B}+c \\overrightarrow{D C}$. Since $A A_{1} \\| D D_{1}$, it holds that $\\overrightarrow{A A_{1}}=k \\overrightarrow{D D_{1}}$ for some $k \\in \\mathbb{R}$. Now it is easy to get $\\overrightarrow{D A_{1}}=-(b \\overrightarrow{D B}+c \\overrightarrow{D C}) / a, \\overrightarrow{D B_{1}}=-(a \\overrightarrow{D A}+c \\overrightarrow{D C}) / b$, and $\\overrightarrow{D C_{1}}=-(a \\overrightarrow{D A}+b \\overrightarrow{D B}) / c$. This implies  $$ \\begin{aligned} & \\overrightarrow{D_{1} A_{1}}=-\\frac{a^{2} \\overrightarrow{D A}+b(a+2 b+c) \\overrightarrow{D B}+c(a+b+2 c) \\overrightarrow{D C}}{a(a+b+c)} \\\\ & \\overrightarrow{D_{1} B_{1}}=-\\frac{a(2 a+b+c) \\overrightarrow{D A}+b^{2} \\overrightarrow{D B}+c(a+b+2 c) \\overrightarrow{D C}}{b(a+b+c)}, \\text { and } \\\\ & \\overrightarrow{D_{1} C_{1}}=-\\frac{a(2 a+b+c) \\overrightarrow{D A}+b(a+2 b+c) \\overrightarrow{D B}+c^{2} \\overrightarrow{D C}}{c(a+b+c)} \\end{aligned} $$  By using  $$ 6 V_{D_{1} A_{1} B_{1} C_{1}}=\\left|\\left[\\overrightarrow{D_{1} A_{1}}, \\overrightarrow{D_{1} B_{1}}, \\overrightarrow{D_{1} C_{1}}\\right]\\right| \\text { and } 6 V_{D A B C}=|[\\overrightarrow{D A}, \\overrightarrow{D B}, \\overrightarrow{D C}]| $$  we get  $$ V_{D_{1} A_{1} B_{1} C_{1}}=\\frac{\\left\\|\\begin{array}{ccc} a^{2} & b(a+2 b+c) & c(a+b+2 c) \\\\ a(2 a+b+c) & b^{2} & c(a+b+2 c) \\\\ a(2 a+b+c) b(a+2 b+c) & c^{2} \\end{array}\\right\\|}{6 a b c(a+b+c)^{3}}=3 V_{D A B C} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1965", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (YUG) Find all real numbers $x \\in[0,2 \\pi]$ such that  $$ 2 \\cos x \\leq|\\sqrt{1+\\sin 2 x}-\\sqrt{1-\\sin 2 x}| \\leq \\sqrt{2} $$", "solution": "1. Let us set $S=|\\sqrt{1+\\sin 2 x}-\\sqrt{1-\\sin 2 x}|$. Observe that $S^{2}=2-$ $2 \\sqrt{1-\\sin ^{2} 2 x}=2-2|\\cos 2 x| \\leq 2$, implying $S \\leq \\sqrt{2}$. Thus the righthand inequality holds for all $x$. It remains to investigate the left-hand inequality. If $\\pi / 2 \\leq x \\leq 3 \\pi / 2$, then $\\cos x \\leq 0$ and the inequality trivially holds. Assume now that $\\cos x>$ 0 . Then the inequality is equivalent to $2+2 \\cos 2 x=4 \\cos ^{2} x \\leq S^{2}=$ $2-2|\\cos 2 x|$, which is equivalent to $\\cos 2 x \\leq 0$, i.e., to $x \\in[\\pi / 4, \\pi / 2] \\cup$ $[3 \\pi / 2,7 \\pi / 4]$. Hence the solution set is $\\pi / 4 \\leq x \\leq 7 \\pi / 4$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1965", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (POL) Consider the system of equations  $$ \\left\\{\\begin{array}{l} a_{11} x_{1}+a_{12} x_{2}+a_{13} x_{3}=0 \\\\ a_{21} x_{1}+a_{22} x_{2}+a_{23} x_{3}=0 \\\\ a_{31} x_{1}+a_{32} x_{2}+a_{33} x_{3}=0 \\end{array}\\right. $$  whose coefficients satisfy the following conditions: (a) $a_{11}, a_{22}, a_{33}$ are positive real numbers; (b) all other coefficients are negative; (c) in each of the equations the sum of the coefficients is positive. Prove that $x_{1}=x_{2}=x_{3}=0$ is the only solution to the system.", "solution": "2. Suppose that $\\left(x_{1}, x_{2}, x_{3}\\right)$ is a solution. We may assume w.l.o.g. that $\\left|x_{1}\\right| \\geq$ $\\left|x_{2}\\right| \\geq\\left|x_{3}\\right|$. Suppose that $\\left|x_{1}\\right|>0$. From the first equation we obtain that  $$ 0=\\left|x_{1}\\right| \\cdot\\left|a_{11}+a_{12} \\frac{x_{2}}{x_{1}}+a_{13} \\frac{x_{3}}{x_{1}}\\right| \\geq\\left|x_{1}\\right| \\cdot\\left(a_{11}-\\left|a_{12}\\right|-\\left|a_{13}\\right|\\right)>0 $$  which is a contradiction. Hence $\\left|x_{1}\\right|=0$ and consequently $x_{1}=x_{2}=x_{3}=$ 0.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1965", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (CZS) A tetrahedron $A B C D$ is given. The lengths of the edges $A B$ and $C D$ are $a$ and $b$, respectively, the distance between the lines $A B$ and $C D$ is $d$, and the angle between them is equal to $\\omega$. The tetrahedron is divided into two parts by the plane $\\pi$ parallel to the lines $A B$ and $C D$. Calculate the ratio of the volumes of the parts if the ratio between the distances of the plane $\\pi$ from $A B$ and $C D$ is equal to $k$.  Second Day", "solution": "3. Let $d$ denote the distance between the lines $A B$ and $C D$. Being parallel to $A B$ and $C D$, the plane $\\pi$ intersects the faces of the tetrahedron in a parallelogram $E F G H$. Let $X \\in A B$ be a points such that $H X \\| D B$. Clearly $V_{A E H B F G}=V_{A X E H}+$ $V_{X E H B F G}$. Let $M N$ be the common perpendicular to lines $A B$ and $C D(M \\in A B, N \\in C D)$ and let $M N, B N$ meet the plane $\\pi$ at $Q$ and $R$ respectively. Then it holds that $B R / R N=M Q / Q N=k$ and consequently $A X / X B=A E / E C=$ $A H / H D=B F / F C=B G / G D=$ $k$. Now we have $V_{A X E H} / V_{A B C D}=$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-352.jpg?height=417&width=535&top_left_y=1138&top_left_x=819) $k^{3} /(k+1)^{3}$ 。 Furthermore, if $h=3 V_{A B C D} / S_{A B C}$ is the height of $A B C D$ from $D$, then  $$ \\begin{aligned} V_{X E H B F G} & =\\frac{1}{2} S_{X B F E} \\frac{k}{k+1} h \\text { and } \\\\ S_{X B F E} & =S_{A B C}-S_{A X E}-S_{E F C}=\\frac{(k+1)^{2}-1-k^{2}}{(k+1)^{2}}=\\frac{2 k}{(1+k)^{2}} \\end{aligned} $$  These relations give us $V_{X E H B F G} / V_{A B C D}=3 k^{2} /(1+k)^{3}$. Finally,  $$ \\frac{V_{A E H B F G}}{V_{A B C D}}=\\frac{k^{3}+3 k^{2}}{(k+1)^{3}} $$  Similarly, $V_{C E F D H G} / V_{A B C D}=(3 k+1) /(k+1)^{3}$, and hence the required ratio is $\\left(k^{3}+3 k^{2}\\right) /(3 k+1)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1965", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (USS) Find four real numbers $x_{1}, x_{2}, x_{3}, x_{4}$ such that the sum of any of the numbers and the product of other three is equal to 2 .", "solution": "4. It is easy to see that all $x_{i}$ are nonzero. Let $x_{1} x_{2} x_{3} x_{4}=p$. The given system of equations can be rewritten as $x_{i}+p / x_{i}=2, i=1,2,3,4$. The equation $x+p / x=2$ has at most two real solutions, say $y$ and $z$. Then each $x_{i}$ is equal either to $y$ or to $z$. There are three cases: (i) $x_{1}=x_{2}=x_{3}=x_{4}=y$. Then $y+y^{3}=2$ and hence $y=1$. (ii) $x_{1}=x_{2}=x_{3}=y, x_{4}=z$. Then $z+y^{3}=y+y^{2} z=2$. It is easy to obtain that the only possibilities for $(y, z)$ are $(-1,3)$ and $(1,1)$. (iii) $x_{1}=x_{2}=y, x_{3}=x_{4}$. In this case the only possibility is $y=z=1$.  Hence the solutions for $\\left(x_{1}, x_{2}, x_{3}, x_{4}\\right)$ are $(1,1,1,1),(-1,-1,-1,3)$, and the cyclic permutations.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1965", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (ROM) Given a triangle $O A B$ such that $\\angle A O B=\\alpha<90^{\\circ}$, let $M$ be an arbitrary point of the triangle different from $O$. Denote by $P$ and $Q$ the feet of the perpendiculars from $M$ to $O A$ and $O B$, respectively. Let $H$ be the orthocenter of the triangle $O P Q$. Find the locus of points $H$ when: (a) $M$ belongs to the segment $A B$; (b) $M$ belongs to the interior of $\\triangle O A B$.", "solution": "5. (a) Let $A^{\\prime}$ and $B^{\\prime}$ denote the feet of the perpendiculars from $A$ and $B$ to $O B$ and $O A$ respectively. We claim that $H \\in A^{\\prime} B^{\\prime}$. Indeed, since $M P H Q$ is a parallelogram, we have $B^{\\prime} P / B^{\\prime} A=B M / B A=$ $M Q / A A^{\\prime}=P H / A A^{\\prime}$, which implies by Thales's theorem that $H \\in$ $A^{\\prime} B^{\\prime}$. It is easy to see that the locus of $H$ is the whole segment $A^{\\prime} B^{\\prime}$. (b) In this case the locus of points $H$ is obviously the interior of the triangle $O A^{\\prime} B^{\\prime}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1965", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (POL) We are given $n \\geq 3$ points in the plane. Let $d$ be the maximal distance between two of the given points. Prove that the number of pairs of points whose distance is equal to $d$ is less than or equal to $n$.", "solution": "6. We recall the simple statement that every two diameters of a set must have a common point. Consider any point $B$ that is an endpoint of $k \\geq 2$ diameters $B C_{1}, B C_{2}$, $\\ldots, B C_{k}$. We may assume w.l.o.g. that all the points $C_{1}, \\ldots, C_{k}$ lie on the $\\operatorname{arc} C_{1} C_{k}$, whose center is $B$ and measure does not exceed $60^{\\circ}$. We observe that for $1<i<k$ any diameter with the endpoint $C_{i}$ has to intersect both the diameters $C_{1} B$ and $C_{l} B$. Hence $C_{i} B$ is the only diameter with an endpoint at $C_{i}$ if $i=2, \\ldots, k-1$. In other words, with each point that is an endpoint of $k \\geq 2$ we can associate $k-2$ points that are endpoints of exactly one diameter. We now assume that each $A_{i}$ is an endpoint of exactly $k_{i} \\geq 0$ diameters, and that $k_{1}, \\ldots, k_{s} \\geq 2$, while $k_{s+1}, \\ldots, k_{n} \\leq 1$. The total number $D$ of diameters satisfies the inequality $2 D \\leq k_{1}+k_{2}+\\cdots+k_{s}+(n-s)$. On the other hand, by the above consideration we have $\\left(k_{1}-2\\right)+\\cdots+\\left(k_{s}-2\\right) \\leq$ $n-s$, i.e., $k_{1}+\\cdots+k_{s} \\leq n+s$. Hence $2 D \\leq(n+s)+(n-s)=2 n$, which proves the result.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1966", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (USS) Three problems $A, B$, and $C$ were given on a mathematics olympiad. All 25 students solved at least one of these problems. The number of students who solved $B$ and not $A$ is twice the number of students who solved $C$ and not $A$. The number of students who solved only $A$ is greater by 1 than the number of students who along with $A$ solved at least one other problem. Among the students who solved only one problem, half solved $A$. How many students solved only $B$ ?", "solution": "1. Let $N_{a}, N_{b}, N_{c}, N_{a b}, N_{a c}, N_{b c}, N_{a b c}$ denote the number of students who solved exactly the problems whose letters are stated in the index of the variable. From the conditions of the problem we have  $$ \\begin{gathered} N_{a}+N_{b}+N_{c}+N_{a b}+N_{b c}+N_{a c}+N_{a b c}=25 \\\\ N_{b}+N_{b c}=2\\left(N_{c}+N_{b c}\\right), \\quad N_{a}-1=N_{a c}+N_{a b c}+N_{a b}, \\quad N_{a}=N_{b}+N_{c} . \\end{gathered} $$  From the first and third equations we get $2 N_{a}+N_{b}+N_{c}+N_{b c}=26$, and from the second and fourth we get $4 N_{b}+N_{c}=26$ and thus $N_{b} \\leq 6$. On the other hand, we have from the second equation $N_{b}=2 N_{c}+N_{b c} \\Rightarrow$ $N_{c} \\leq N_{b} / 2 \\Rightarrow 26 \\leq 9 N_{b} / 2 \\Rightarrow N_{b} \\geq 6$; hence $N_{b}=6$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1966", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (HUN) If $a, b$, and $c$ are the sides and $\\alpha, \\beta$, and $\\gamma$ the respective angles of the triangle for which $a+b=\\tan \\frac{\\gamma}{2}(a \\tan \\alpha+b \\tan \\beta)$, prove that the triangle is isosceles.", "solution": "2. Angles $\\alpha$ and $\\beta$ are less than $90^{\\circ}$, otherwise if w.l.o.g. $\\alpha \\geq 90^{\\circ}$ we have $\\tan (\\gamma / 2) \\cdot(a \\tan \\alpha+b \\tan \\beta)<b \\tan (\\gamma / 2) \\tan \\beta \\leq b \\tan (\\gamma / 2) \\cot (\\gamma / 2)=$ $b<a+b$. Since $a \\geq b \\Leftrightarrow \\tan a \\geq \\tan b$, Chebyshev's inequality gives $a \\tan \\alpha+b \\tan \\beta \\geq(a+b)(\\tan \\alpha+\\tan \\beta) / 2$. Due to the convexity of the $\\tan$ function we also have $(\\tan \\alpha+\\tan \\beta) / 2 \\geq \\tan [(\\alpha+\\beta) / 2]=\\cot (\\gamma / 2)$. Hence we have  $$ \\begin{aligned} \\tan \\frac{\\gamma}{2}(a \\tan \\alpha+b \\tan \\beta) & \\geq \\frac{1}{2} \\tan \\frac{\\gamma}{2}(a+b)(\\tan \\alpha+\\tan \\beta) \\\\ & \\geq \\tan \\frac{\\gamma}{2}(a+b) \\cot \\frac{\\gamma}{2}=a+b \\end{aligned} $$  The equalities can hold only if $a=b$. Thus the triangle is isosceles.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1966", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (BUL) Prove that the sum of distances from the center of the circumsphere of the regular tetrahedron to its four vertices is less than the sum of distances from any other point to the four vertices.  Second Day", "solution": "3. Consider a coordinate system in which the points of the regular tetrahedron are placed at $A(-a,-a,-a), B(-a, a, a), C(a,-a, a)$ and $D(a, a$, $-a)$. Then the center of the tetrahedron is at $O(0,0,0)$. For a point $X(x, y, z)$ the sum $X A+X B+X C+X D$ by the QM-AM inequality does not exceed $2 \\sqrt{X A^{2}+X B^{2}+X C^{2}+X D^{2}}$. Now, since $X A^{2}=$ $(x+a)^{2}+(y+a)^{2}+(z+a)^{2}$ etc., we easily obtain  $$ \\begin{aligned} X A^{2}+X B^{2}+X C^{2}+X D^{2} & =4\\left(x^{2}+y^{2}+z^{2}\\right)+12 a^{2} \\\\ & \\geq 12 a^{2}=O A^{2}+O B^{2}+O C^{2}+O D^{2} \\end{aligned} $$  Hence $X A+X B+X C+X D \\geq 2 \\sqrt{O A^{2}+O B^{2}+O C^{2}+O D^{2}}=O A+$ $O B+O C+O D$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1966", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (YUG) Prove the following equality:  $$ \\frac{1}{\\sin 2 x}+\\frac{1}{\\sin 4 x}+\\frac{1}{\\sin 8 x}+\\cdots+\\frac{1}{\\sin 2^{n} x}=\\cot x-\\cot 2^{n} x $$  where $n \\in \\mathbb{N}$ and $x \\notin \\pi \\mathbb{Z} / 2^{k}$ for every $k \\in \\mathbb{N}$.", "solution": "4. It suffices to prove $1 / \\sin 2^{k} x=\\cot 2^{k-1} x-\\cot 2^{k} x$ for any integer $k$ and real $x$, i.e., $1 / \\sin 2 x=\\cot x-\\cot 2 x$ for all real $x$. We indeed have $\\cot x-\\cot 2 x=\\cot x-\\frac{\\cot ^{2} x-1}{2 \\cot x}=\\frac{\\left(\\frac{\\cos x}{\\sin x}\\right)^{2}+1}{2 \\frac{\\cos x}{\\sin x}}=\\frac{1}{2 \\sin x \\cos x}=\\frac{1}{\\sin 2 x}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1966", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (CZS) Solve the following system of equations:  $$ \\begin{aligned} & \\left|a_{1}-a_{2}\\right| x_{2}+\\left|a_{1}-a_{3}\\right| x_{3}+\\left|a_{1}-a_{4}\\right| x_{4}=1, \\\\ & \\left|a_{2}-a_{1}\\right| x_{1}+\\left|a_{2}-a_{3}\\right| x_{3}+\\left|a_{2}-a_{4}\\right| x_{4}=1, \\\\ & \\left|a_{3}-a_{1}\\right| x_{1}+\\left|a_{3}-a_{2}\\right| x_{2}+\\left|a_{3}-a_{4}\\right| x_{4}=1, \\\\ & \\left|a_{4}-a_{1}\\right| x_{1}+\\left|a_{4}-a_{2}\\right| x_{2}+\\left|a_{4}-a_{3}\\right| x_{3}=1, \\end{aligned} $$  where $a_{1}, a_{2}, a_{3}$, and $a_{4}$ are mutually distinct real numbers.", "solution": "5. We define $L_{1}=\\left|a_{1}-a_{2}\\right| x_{2}+\\left|a_{1}-a_{3}\\right| x_{3}+\\left|a_{1}-a_{4}\\right| x_{4}$ and analogously $L_{2}$, $L_{3}$, and $L_{4}$. Let us assume w.l.o.g. that $a_{1}<a_{2}<a_{3}<a_{4}$. In that case,  $$ \\begin{aligned} 2\\left|a_{1}-a_{2}\\right|\\left|a_{2}-a_{3}\\right| x_{2} & =\\left|a_{3}-a_{2}\\right| L_{1}-\\left|a_{1}-a_{3}\\right| L_{2}+\\left|a_{1}-a_{2}\\right| L_{3} \\\\ & =\\left|a_{3}-a_{2}\\right|-\\left|a_{1}-a_{3}\\right|+\\left|a_{1}-a_{2}\\right|=0 \\\\ 2\\left|a_{2}-a_{3}\\right|\\left|a_{3}-a_{4}\\right| x_{3} & =\\left|a_{4}-a_{3}\\right| L_{2}-\\left|a_{2}-a_{4}\\right| L_{3}+\\left|a_{2}-a_{3}\\right| L_{4} \\\\ & =\\left|a_{4}-a_{3}\\right|-\\left|a_{2}-a_{4}\\right|+\\left|a_{2}-a_{3}\\right|=0 . \\end{aligned} $$  Hence it follows that $x_{2}=x_{3}=0$ and consequently $x_{1}=x_{4}=1 /\\left|a_{1}-a_{4}\\right|$. This solution set indeed satisfies the starting equations. It is easy to generalize this result to any ordering of $a_{1}, a_{2}, a_{3}, a_{4}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1966", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (POL) Let $M, K$, and $L$ be points on $(A B),(B C)$, and $(C A)$, respectively. Prove that the area of at least one of the three triangles $\\triangle M A L$, $\\triangle K B M$, and $\\triangle L C K$ is less than or equal to one-fourth the area of $\\triangle A B C$.", "solution": "6. Let $S$ denote the area of $\\triangle A B C$. Let $A_{1}, B_{1}, C_{1}$ be the midpoints of $B C, A C, A B$ respectively. We note that $S_{A_{1} B_{1} C}=S_{A_{1} B C_{1}}=S_{A B_{1} C_{1}}=$ $S_{A_{1} B_{1} C_{1}}=S / 4$. Let us assume w.l.o.g. that $M \\in\\left[A C_{1}\\right]$. We then must have $K \\in\\left[B A_{1}\\right]$ and $L \\in\\left[C B_{1}\\right]$. However, we then have $S(K L M)>$ $S\\left(K L C_{1}\\right)>S\\left(K B_{1} C_{1}\\right)=S\\left(A_{1} B_{1} C_{1}\\right)=S / 4$. Hence, by the pigeonhole principle one of the remaining three triangles $\\triangle M A L, \\triangle K B M$, and $\\triangle L C K$ must have an area less than or equal to $S / 4$. This completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (BUL 1) Prove that all numbers in the sequence  $$ \\frac{107811}{3}, \\frac{110778111}{3}, \\frac{111077781111}{3}, \\ldots $$  are perfect cubes.", "solution": "1. Let us denote the $n$th term of the given sequence by $a_{n}$. Then  $$ \\begin{aligned} a_{n} & =\\frac{1}{3}\\left(\\frac{10^{3 n+3}-10^{2 n+3}}{9}+7 \\frac{10^{2 n+2}-10^{n+1}}{9}+\\frac{10^{n+2}-1}{9}\\right) \\\\ & =\\frac{1}{27}\\left(10^{3 n+3}-3 \\cdot 10^{2 n+2}+3 \\cdot 10^{n+1}-1\\right)=\\left(\\frac{10^{n+1}-1}{3}\\right)^{3} . \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (CZS 4) The square $A B C D$ is to be decomposed into $n$ triangles (nonoverlapping) all of whose angles are acute. Find the smallest integer $n$ for which there exists a solution to this problem and construct at least one decomposition for this $n$. Answer whether it is possible to ask additionally that (at least) one of these triangles has a perimeter less than an arbitrarily given positive number.", "solution": "10. Let $n$ be the number of triangles and let $b$ and $i$ be the numbers of vertices on the boundary and in the interior of the square, respectively. Since all the triangles are acute, each of the vertices of the square belongs to at least two triangles. Additionally, every vertex on the boundary belongs to at least three, and every vertex in the interior belongs to at least five triangles. Therefore  $$ 3 n \\geq 8+3 b+5 i $$  Moreover, the sum of angles at any vertex that lies in the interior, on the boundary, or at a vertex of the square is equal to $2 \\pi, \\pi, \\pi / 2$ respectively. The sum of all angles of the triangles equals $n \\pi$, which gives us $n \\pi=4 \\cdot \\pi / 2+b \\pi+2 i \\pi$, i.e., $n=$ $2+b+2 i$. This relation together with (1) easily yields that $i \\geq 2$. Since each of the vertices inside the square belongs to at least five triangles, and at most two contain both, it follows that $n \\geq 8$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-357.jpg?height=525&width=489&top_left_y=1547&top_left_x=813)  It is shown in the figure that the square can be decomposed into eight acute triangles. Obviously one of them can have an arbitrarily small perimeter.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (CZS 5) Let $n$ be a positive integer. Find the maximal number of noncongruent triangles whose side lengths are integers less than or equal to $n$.", "solution": "11. We have to find the number $p_{n}$ of triples of positive integers $(a, b, c)$ satisfying $a \\leq b \\leq c \\leq n$ and $a+b>c$. Let us denote by $p_{n}(k)$ the number of such triples with $c=k, k=1,2, \\ldots, n$. For $k$ even, $p_{n}(k)=k+(k-2)+(k-4)+\\cdots+2=\\left(k^{2}+2 k\\right) / 4$, and for $k$ odd, $p_{n}(k)=\\left(k^{2}+2 k+1\\right) / 4$. Hence $p_{n}=p_{n}(1)+p_{n}(2)+\\cdots+p_{n}(n)= \\begin{cases}n(n+2)(2 n+5) / 24, & \\text { for } 2 \\mid n, \\\\ (n+1)(n+3)(2 n+1) / 24, & \\text { for } 2 \\nmid n .\\end{cases}$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (CZS 6) Given a segment $A B$ of the length 1, define the set $M$ of points in the following way: it contains the two points $A, B$, and also all points obtained from $A, B$ by iterating the following rule: $(*)$ for every pair of points $X, Y$ in $M$, the set $M$ also contains the point $Z$ of the segment $X Y$ for which $Y Z=3 X Z$. (a) Prove that the set $M$ consists of points $X$ from the segment $A B$ for which the distance from the point $A$ is either  $$ A X=\\frac{3 k}{4^{n}} \\quad \\text { or } \\quad A X=\\frac{3 k-2}{4^{n}} $$  where $n, k$ are nonnegative integers. (b) Prove that the point $X_{0}$ for which $A X_{0}=1 / 2=X_{0} B$ does not belong to the set $M$.", "solution": "12. Let us denote by $M_{n}$ the set of points of the segment $A B$ obtained from $A$ and $B$ by not more than $n$ iterations of $(*)$. It can be proved by induction that  $$ M_{n}=\\left\\{X \\in A B \\left\\lvert\\, A X=\\frac{3 k}{4^{n}}\\right. \\text { or } \\frac{3 k-2}{4^{n}} \\text { for some } k \\in \\mathbb{N}\\right\\} $$  Thus (a) immediately follows from $M=\\bigcup M_{n}$. It also follows that if $a, b \\in \\mathbb{N}$ and $a / b \\in M$, then $3 \\mid a(b-a)$. Therefore $1 / 2 \\notin M$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (GDR 1) Find whether among all quadrilaterals whose interiors lie inside a semicircle of radius $r$ there exists one (or more) with maximal area. If so, determine their shape and area.", "solution": "13. The maximum area is $3 \\sqrt{3} r^{2} / 4$ (where $r$ is the radius of the semicircle) and is attained in the case of a trapezoid with two vertices at the endpoints of the diameter of the semicircle and the other two vertices dividing the semicircle into three equal arcs.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (GDR 2) Which fraction $p / q$, where $p, q$ are positive integers less than 100 , is closest to $\\sqrt{2}$ ? Find all digits after the decimal point in the decimal representation of this fraction that coincide with digits in the decimal representation of $\\sqrt{2}$ (without using any tables).", "solution": "14. We have that  $$ \\left|\\frac{p}{q}-\\sqrt{2}\\right|=\\frac{|p-q \\sqrt{2}|}{q}=\\frac{\\left|p^{2}-2 q^{2}\\right|}{q(p+q \\sqrt{2})} \\geq \\frac{1}{q(p+q \\sqrt{2})} $$  because $\\left|p^{2}-2 q^{2}\\right| \\geq 1$. The greatest solution to the equation $\\left|p^{2}-2 q^{2}\\right|=1$ with $p, q \\leq 100$ is $(p, q)=(99,70)$. It is easy to verify using (1) that $\\frac{99}{70}$ best approximates $\\sqrt{2}$ among the fractions $p / q$ with $p, q \\leq 100$. Second solution. By using some basic facts about Farey sequences one can find that $\\frac{41}{29}<\\sqrt{2}<\\frac{99}{70}$ and that $\\frac{41}{29}<\\frac{p}{q}<\\frac{99}{70}$ implies $p \\geq 41+99>100$ because $99 \\cdot 29-41 \\cdot 70=1$. Of the two fractions $41 / 29$ and $99 / 70$, the latter is closer to $\\sqrt{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (GDR 3) Suppose $\\tan \\alpha=p / q$, where $p$ and $q$ are integers and $q \\neq 0$. Prove that the number $\\tan \\beta$ for which $\\tan 2 \\beta=\\tan 3 \\alpha$ is rational only when $p^{2}+q^{2}$ is the square of an integer.", "solution": "15. Given that $\\tan \\alpha \\in \\mathbb{Q}$, we have that $\\tan \\beta$ is rational if and only if $\\tan \\gamma$ is rational, where $\\gamma=\\beta-\\alpha$ and $2 \\gamma=\\alpha$. Putting $t=\\tan \\gamma$ we obtain $\\frac{p}{q}=\\tan 2 \\gamma=\\frac{2 t}{1-t^{2}}$, which leads to the quadratic equation $p t^{2}+2 q t-p=0$. This equation has rational solutions if and only if its discriminant $4\\left(p^{2}+q^{2}\\right)$ is a perfect square, and the result follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (GDR 4) Prove the following statement: If $r_{1}$ and $r_{2}$ are real numbers whose quotient is irrational, then any real number $x$ can be approximated arbitrarily well by numbers of the form $z_{k_{1}, k_{2}}=k_{1} r_{1}+k_{2} r_{2}, k_{1}, k_{2}$ integers; i.e., for every real number $x$ and every positive real number $p$ two integers $k_{1}$ and $k_{2}$ can be found such that $\\left|x-\\left(k_{1} r_{1}+k_{2} r_{2}\\right)\\right|<p$.", "solution": "16. First let us notice that all the numbers $z_{m_{1}, m_{2}}=m_{1} r_{1}+m_{2} r_{2}$ ( $m_{1}, m_{2} \\in$ $\\mathbb{Z})$ are distinct, since $r_{1} / r_{2}$ is irrational. Thus for any $n \\in \\mathbb{N}$ the interval $\\left[-n\\left(\\left|r_{1}\\right|+\\left|r_{2}\\right|\\right), n\\left(\\left|r_{1}\\right|+\\left|r_{2}\\right|\\right)\\right]$ contains $(2 n+1)^{2}$ numbers $z_{m_{1}, m_{2}}$, where $\\left|m_{1}\\right|,\\left|m_{2}\\right| \\leq n$. Therefore some two of these $(2 n+1)^{2}$ numbers, say $z_{m_{1}, m_{2}}, z_{n_{1}, n_{2}}$, differ by at most $\\frac{2 n\\left(\\left|r_{1}\\right|+\\left|r_{2}\\right|\\right)}{(2 n+1)^{2}-1}=\\frac{\\left(\\left|r_{1}\\right|+\\left|r_{2}\\right|\\right)}{2(n+1)}$. By taking $n$ large enough we can achieve that  $$ z_{q_{1}, q_{2}}=\\left|z_{m_{1}, m_{2}}-z_{n_{1}, n_{2}}\\right| \\leq p $$  If now $k$ is the integer such that $k z_{q_{1}, q_{2}} \\leq x<(k+1) z_{q_{1}, q_{2}}$, then $z_{k q_{1}, k q_{2}}=$ $k z_{q_{1}, q_{2}}$ differs from $x$ by at most $p$, as desired.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (GBR 1) ${ }^{\\mathrm{IMO} 3}$ Let $k, m$, and $n$ be positive integers such that $m+k+1$ is a prime number greater than $n+1$. Write $c_{s}$ for $s(s+1)$. Prove that the product $\\left(c_{m+1}-c_{k}\\right)\\left(c_{m+2}-c_{k}\\right) \\cdots\\left(c_{m+n}-c_{k}\\right)$ is divisible by the product $c_{1} c_{2} \\cdots c_{n}$.", "solution": "17. Using $c_{r}-c_{s}=(r-s)(r+s+1)$ we can easily get  $$ \\frac{\\left(c_{m+1}-c_{k}\\right) \\cdots\\left(c_{m+n}-c_{k}\\right)}{c_{1} c_{2} \\cdots c_{n}}=\\frac{(m-k+n)!}{(m-k)!n!} \\cdot \\frac{(m+k+n+1)!}{(m+k+1)!(n+1)!} $$  The first factor $\\frac{(m-k+n)!}{(m-k)!n!}=\\binom{m-k+n}{n}$ is clearly an integer. The second factor is also an integer because by the assumption, $m+k+1$ and ( $m+$ $k)!(n+1)$ ! are coprime, and $(m+k+n+1)$ ! is divisible by both; hence it is also divisible by their product.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (GBR 5) If $x$ is a positive rational number, show that $x$ can be uniquely expressed in the form  $$ x=a_{1}+\\frac{a_{2}}{2!}+\\frac{a_{3}}{3!}+\\cdots, $$  where $a_{1}, a_{2}, \\ldots$ are integers, $0 \\leq a_{n} \\leq n-1$ for $n>1$, and the series terminates. Show also that $x$ can be expressed as the sum of reciprocals of different integers, each of which is greater than $10^{6}$.", "solution": "18. In the first part, it is sufficient to show that each rational number of the form $m / n!, m, n \\in \\mathbb{N}$, can be written uniquely in the required form. We prove this by induction on $n$. The statement is trivial for $n=1$. Let us assume it holds for $n-1$, and let there be given a rational number $m / n$ !. Let us take $a_{n} \\in\\{0, \\ldots, n-1\\}$ such that $m-a_{n}=n m_{1}$ for some $m_{1} \\in \\mathbb{N}$. By the inductive hypothesis, there are unique $a_{1} \\in \\mathbb{N}_{0}, a_{i} \\in\\{0, \\ldots, i-1\\}(i=1, \\ldots, n-1)$ such that $m_{1} /(n-1)!=\\sum_{i=1}^{n-1} a_{i} / i$ !, and then  $$ \\frac{m}{n!}=\\frac{m_{1}}{(n-1)!}+\\frac{a_{n}}{n!}=\\sum_{i=1}^{n} \\frac{a_{i}}{i!} $$  as desired. On the other hand, if $m / n!=\\sum_{i=1}^{n} a_{i} / i$ !, multiplying by $n$ ! we see that $m-a_{n}$ must be a multiple of $n$, so the choice of $a_{n}$ was unique and therefore the representation itself. This completes the induction. In particular, since $a_{i} \\mid i!$ and $i!/ a_{i}>(i-1)!\\geq(i-1)!/ a_{i-1}$, we conclude that each rational $q, 0<q<1$, can be written as the sum of different reciprocals. Now we prove the second part. Let $x>0$ be a rational number. For any integer $m>10^{6}$, let $n>m$ be the greatest integer such that $y=$ $x-\\frac{1}{m}-\\frac{1}{m+1}-\\cdots-\\frac{1}{n}>0$. Then $y$ can be written as the sum of reciprocals of different positive integers, which all must be greater than $n$. The result follows immediately.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (GBR 6) The $n$ points $P_{1}, P_{2}, \\ldots, P_{n}$ are placed inside or on the boundary of a disk of radius 1 in such a way that the minimum distance $d_{n}$ between any two of these points has its largest possible value $D_{n}$. Calculate $D_{n}$ for $n=2$ to 7 and justify your answer.", "solution": "19. Suppose $n \\leq 6$. Let us decompose the disk by its radii into $n$ congruent regions, so that one of the points $P_{j}$ lies on the boundaries of two of these regions. Then one of these regions contains two of the $n$ given points. Since the diameter of each of these regions is $2 \\sin \\frac{\\pi}{n}$, we have $d_{n} \\leq 2 \\sin \\frac{\\pi}{n}$. This value is attained if $P_{i}$ are the vertices of a regular $n$-gon inscribed in the boundary circle. Hence $D_{n}=2 \\sin \\frac{\\pi}{n}$. For $n=7$ we have $D_{7} \\leq D_{6}=1$. This value is attained if six of the seven points form a regular hexagon inscribed in the boundary circle and the seventh is at the center. Hence $D_{7}=1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (BUL 2) Prove that $\\frac{1}{3} n^{2}+\\frac{1}{2} n+\\frac{1}{6} \\geq(n!)^{2 / n}$ ( $n$ is a positive integer) and that equality is possible only in the case $n=1$.", "solution": "2. $(n!)^{2 / n}=\\left((1 \\cdot 2 \\cdots n)^{1 / n}\\right)^{2} \\leq\\left(\\frac{1+2+\\cdots+n}{n}\\right)^{2}=\\left(\\frac{n+1}{2}\\right)^{2} \\leq \\frac{1}{3} n^{2}+\\frac{1}{2} n+\\frac{1}{6}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. (HUN 1) In space, $n$ points $(n \\geq 3)$ are given. Every pair of points determines some distance. Suppose all distances are different. Connect every point with the nearest point. Prove that it is impossible to obtain a polygonal line in such a way. ${ }^{1}$", "solution": "20. The statement so formulated is false. It would be true under the additional assumption that the polygonal line is closed. However, from the offered solution, which is not clear, it does not seem that the proposer had this in mind.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. (HUN 2) Without using any tables, find the exact value of the product  $$ P=\\cos \\frac{\\pi}{15} \\cos \\frac{2 \\pi}{15} \\cos \\frac{3 \\pi}{15} \\cos \\frac{4 \\pi}{15} \\cos \\frac{5 \\pi}{15} \\cos \\frac{6 \\pi}{15} \\cos \\frac{7 \\pi}{15} $$", "solution": "21. Using the formula  $$ \\cos x \\cos 2 x \\cos 4 x \\cdots \\cos 2^{n-1} x=\\frac{\\sin 2^{n} x}{2^{n} \\sin x} $$  which is shown by simple induction, we obtain  $$ \\begin{gathered} \\cos \\frac{\\pi}{15} \\cos \\frac{2 \\pi}{15} \\cos \\frac{4 \\pi}{15} \\cos \\frac{7 \\pi}{15}=-\\cos \\frac{\\pi}{15} \\cos \\frac{2 \\pi}{15} \\cos \\frac{4 \\pi}{15} \\cos \\frac{8 \\pi}{15}=\\frac{1}{16} \\\\ \\cos \\frac{3 \\pi}{15} \\cos \\frac{6 \\pi}{15} \\end{gathered}=\\frac{1}{4}, \\quad \\cos \\frac{5 \\pi}{15}=\\frac{1}{2} . $$  Multiplying these equalities, we get that the required product $P$ equals 1/128.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "22", "problem_type": null, "exam": "IMO", "problem": "22. (HUN 3) The distance between the centers of the circles $k_{1}$ and $k_{2}$ with radii $r$ is equal to $r$. Points $A$ and $B$ are on the circle $k_{1}$, symmetric with respect to the line connecting the centers of the circles. Point $P$ is an arbitrary point on $k_{2}$. Prove that  $$ P A^{2}+P B^{2} \\geq 2 r^{2} $$  When does equality hold?", "solution": "22. Let $O_{1}$ and $O_{2}$ be the centers of circles $k_{1}$ and $k_{2}$ and let $C$ be the midpoint of the segment $A B$. Using the well-known relation for elements of a triangle, we obtain  $$ P A^{2}+P B^{2}=2 P C^{2}+2 C A^{2} \\geq 2 O_{1} C^{2}+2 C A^{2}=2 O_{1} A^{2}=2 r^{2} $$  Equality holds if $P$ coincides with $O_{1}$ or if $A$ and $B$ coincide with $O_{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "23", "problem_type": null, "exam": "IMO", "problem": "23. (HUN 4) Prove that for an arbitrary pair of vectors $f$ and $g$ in the plane, the inequality  $$ a f^{2}+b f g+c g^{2} \\geq 0 $$  holds if and only if the following conditions are fulfilled: $a \\geq 0, c \\geq 0$, $4 a c \\geq b^{2}$.", "solution": "23. Suppose that $a \\geq 0, c \\geq 0,4 a c \\geq b^{2}$. If $a=0$, then $b=0$, and the inequality reduces to the obvious $c g^{2} \\geq 0$. Also, if $a>0$, then  $$ a f^{2}+b f g+c g^{2}=a\\left(f+\\frac{b}{2 a} g\\right)^{2}+\\frac{4 a c-b^{2}}{4 a} g^{2} \\geq 0 $$  Suppose now that $a f^{2}+b f g+c g^{2} \\geq 0$ holds for an arbitrary pair of vectors $f, g$. Substituting $f$ by $t g(t \\in \\mathbb{R})$ we get that $\\left(a t^{2}+b t+c\\right) g^{2} \\geq 0$ holds for any real number $t$. Therefore $a \\geq 0, c \\geq 0,4 a c \\geq b^{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "24", "problem_type": null, "exam": "IMO", "problem": "24. (HUN 5) ${ }^{\\text {IMO6 }}$ Father has left to his children several identical gold coins. According to his will, the oldest child receives one coin and one-seventh of the remaining coins, the next child receives two coins and one-seventh of the remaining coins, the third child receives three coins and one-seventh of the remaining coins, and so on through the youngest child. If every child inherits an integer number of coins, find the number of children and the number of coins.", "solution": "24. Let the $k$ th child receive $x_{k}$ coins. By the condition of the problem, the number of coins that remain after him was $6\\left(x_{k}-k\\right)$. This gives us a recurrence relation  $$ x_{k+1}=k+1+\\frac{6\\left(x_{k}-k\\right)-k-1}{7}=\\frac{6}{7} x_{k}+\\frac{6}{7}, $$  which, together with the condition $x_{1}=1+(m-1) / 7$, yields  $$ x_{k}=\\frac{6^{k-1}}{7^{k}}(m-36)+6 \\text { for } 1 \\leq k \\leq n . $$  Since we are given $x_{n}=n$, we obtain $6^{n-1}(m-36)=7^{n}(n-6)$. It follows that $6^{n-1} \\mid n-6$, which is possible only for $n=6$. Hence, $n=6$ and $m=36$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "25", "problem_type": null, "exam": "IMO", "problem": "25. (HUN 6) Three disks of diameter $d$ are touching a sphere at their centers. Moreover, each disk touches the other two disks. How do we choose the radius $R$ of the sphere so that the axis of the whole figure makes an angle  [^0]of $60^{\\circ}$ with the line connecting the center of the sphere with the point on the disks that is at the largest distance from the axis? (The axis of the figure is the line having the property that rotation of the figure through $120^{\\circ}$ about that line brings the figure to its initial position. The disks are all on one side of the plane, pass through the center of the sphere, and are orthogonal to the axes.)", "solution": "25. The answer is $R=(4+\\sqrt{3}) d / 6$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "26", "problem_type": null, "exam": "IMO", "problem": "26. (ITA 1) Let $A B C D$ be a regular tetrahedron. To an arbitrary point $M$ on one edge, say $C D$, corresponds the point $P=P(M)$, which is the intersection of two lines $A H$ and $B K$, drawn from $A$ orthogonally to $B M$ and from $B$ orthogonally to $A M$. What is the locus of $P$ as $M$ varies?", "solution": "26. Let $L$ be the midpoint of the edge $A B$. Since $P$ is the orthocenter of $\\triangle A B M$ and $M L$ is its altitude, $P$ lies on $M L$ and therefore belongs to the triangular area $L C D$. Moreover, from the similarity of triangles $A L P$ and $M L B$ we have $L P \\cdot L M=L A \\cdot L B=a^{2} / 4$, where $a$ is the side length of tetrahedron $A B C D$. It easily follows that the locus of $P$ is the image of the segment $C D$ under the inversion of the plane $L C D$ with center $L$ and radius $a / 2$. This locus is the arc of a circle with center $L$ and endpoints at the orthocenters of triangles $A B C$ and $A B D$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "27", "problem_type": null, "exam": "IMO", "problem": "27. (ITA 2) Which regular polygons can be obtained (and how) by cutting a cube with a plane?", "solution": "27. Regular polygons with 3,4 , and 6 sides can be obtained by cutting a cube with a plane, as shown in the figure. A polygon with more than 6 sides cannot be obtained in such a way, for a cube has 6 faces. Also, if a pentagon is obtained by cutting a ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-361.jpg?height=299&width=312&top_left_y=991&top_left_x=935) cube with a plane, then its sides lying on opposite faces are parallel; hence it cannot be regular.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "28", "problem_type": null, "exam": "IMO", "problem": "28. (ITA 3) Find values of the parameter $u$ for which the expression $$ y=\\frac{\\tan (x-u)+\\tan x+\\tan (x+u)}{\\tan (x-u) \\tan x \\tan (x+u)} $$ does not depend on $x$.", "solution": "28. The given expression can be transformed into  $$ y=\\frac{4 \\cos 2 u+2}{\\cos 2 u-\\cos 2 x}-3 . $$  It does not depend on $x$ if and only if $\\cos 2 u=-1 / 2$, i.e., $u= \\pm \\pi / 3+k \\pi$ for some $k \\in \\mathbb{Z}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "29", "problem_type": null, "exam": "IMO", "problem": "29. (ITA 4) ${ }^{\\mathrm{IMO} 4}$ The triangles $A_{0} B_{0} C_{0}$ and $A^{\\prime} B^{\\prime} C^{\\prime}$ have all their angles acute. Describe how to construct one of the triangles $A B C$, similar to $A^{\\prime} B^{\\prime} C^{\\prime}$ and circumscribing $A_{0} B_{0} C_{0}$ (so that $A, B, C$ correspond to $A^{\\prime}$, $B^{\\prime}, C^{\\prime}$, and $A B$ passes through $C_{0}, B C$ through $A_{0}$, and $C A$ through $B_{0}$ ). Among these triangles $A B C$, describe, and prove, how to construct the triangle with the maximum area.", "solution": "29. Let arc $l_{a}$ be the locus of points $A$ lying on the opposite side from $A_{0}$ with respect to the line $B_{0} C_{0}$ such that $\\angle B_{0} A C_{0}=\\angle A^{\\prime}$. Let $k_{a}$ be the circle containing $l_{a}$, and let $S_{a}$ be the center of $k_{a}$. We similarly define $l_{b}, l_{c}, k_{b}, k_{c}, S_{b}, S_{c}$. It is easy to show that circles $k_{a}, k_{b}, k_{c}$ have a common point $S$ inside $\\triangle A B C$. Let $A_{1}, B_{1}, C_{1}$ be the points on the arcs $l_{a}, l_{b}, l_{c}$ diametrically opposite to $S$ with respect to $S_{a}, S_{b}, S_{c}$ respectively. Then $A_{0} \\in B_{1} C_{1}$ because $\\angle B_{1} A_{0} S=\\angle C_{1} A_{0} S=90^{\\circ}$; similarly, $B_{0} \\in A_{1} C_{1}$ and $C_{0} \\in A_{1} B_{1}$. Hence the triangle $A_{1} B_{1} C_{1}$ is circumscribed about $\\triangle A_{0} B_{0} C_{0}$ and similar to $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$. Moreover, we claim that $\\triangle A_{1} B_{1} C_{1}$ is the triangle $A B C$ with the desired properties having the maximum side $B C$ and hence the maximum area.  Indeed, if $A B C$ is any other such triangle and $S_{b}^{\\prime}, S_{c}^{\\prime}$ are the projections of $S_{b}$ and $S_{c}$ onto the line $B C$, it holds that $B C=2 S_{b}^{\\prime} S_{c}^{\\prime} \\leq 2 S_{b} S_{c}=B_{1} C_{1}$, which proves the maximality of $B_{1} C_{1}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (BUL 3) Prove the trigonometric inequality $\\cos x<1-\\frac{x^{2}}{2}+\\frac{x^{4}}{16}$, where $x \\in(0, \\pi / 2)$.", "solution": "3. Consider the function $f:[0, \\pi / 2] \\rightarrow \\mathbb{R}$ defined by $f(x)=1-x^{2} / 2+$ $x^{4} / 16-\\cos x$. It is easy to calculate that $f^{\\prime}(0)=f^{\\prime \\prime}(0)=f^{\\prime \\prime \\prime}(0)=0$ and $f^{\\prime \\prime \\prime \\prime}(x)=$ $3 / 2-\\cos x$. Since $f^{\\prime \\prime \\prime \\prime}(x)>0, f^{\\prime \\prime \\prime}(x)$ is increasing. Together with $f^{\\prime \\prime \\prime}(0)=0$, this gives $f^{\\prime \\prime \\prime}(x)>0$ for $x>0$; hence $f^{\\prime \\prime}(x)$ is increasing, etc. Continuing in the same way we easily conclude that $f(x)>0$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "30", "problem_type": null, "exam": "IMO", "problem": "30. (MON 1) Given $m+n$ numbers $a_{i}(i=1,2, \\ldots, m), b_{j}(j=1,2, \\ldots, n)$, determine the number of pairs $\\left(a_{i}, b_{j}\\right)$ for which $|i-j| \\geq k$, where $k$ is a nonnegative integer.", "solution": "30. We assume w.l.o.g. that $m \\leq n$. Let $r$ and $s$ be the numbers of pairs for which $i-j \\geq k$ and of those for which $j-i \\geq k$. The desired number is $r+s$. We easily find that  $$ \\begin{aligned} & r= \\begin{cases}(m-k)(m-k+1) / 2, & k<m \\\\ 0, & k \\geq m\\end{cases} \\\\ & s= \\begin{cases}m(2 n-2 k-m+1) / 2, & k<n-m \\\\ (n-k)(n-k+1) / 2, & n-m \\leq k<n \\\\ 0, & k \\geq n\\end{cases} \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "31", "problem_type": null, "exam": "IMO", "problem": "31. (MON 2) An urn contains balls of $k$ different colors; there are $n_{i}$ balls of the $i$ th color. Balls are drawn at random from the urn, one by one, without replacement. Find the smallest number of draws necessary for getting $m$ balls of the same color.", "solution": "31. Suppose that $n_{1} \\leq n_{2} \\leq \\cdots \\leq n_{k}$. If $n_{k}<m$, there is no solution. Otherwise, the solution is $1+(m-1)(k-s+1)+\\sum_{i<s} n_{i}$, where $s$ is the smallest $i$ for which $m \\leq n_{i}$ holds.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "32", "problem_type": null, "exam": "IMO", "problem": "32. (MON 3) Determine the volume of the body obtained by cutting the ball of radius $R$ by the trihedron with vertex in the center of that ball if its dihedral angles are $\\alpha, \\beta, \\gamma$.", "solution": "32. Let us denote by $V$ the volume of the given body, and by $V_{a}, V_{b}, V_{c}$ the volumes of the parts of the given ball that lie inside the dihedra of the given trihedron. It holds that $V_{a}=2 R^{3} \\alpha / 3, V_{b}=2 R^{3} \\beta / 3$, $V_{c}=2 R^{3} \\gamma / 3$. It is easy to see that $2\\left(V_{a}+V_{b}+V_{c}\\right)=4 V+4 \\pi R^{3} / 3$, from ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-362.jpg?height=344&width=356&top_left_y=939&top_left_x=897) which it follows that  $$ V=\\frac{1}{3} R^{3}(\\alpha+\\beta+\\gamma-\\pi) $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "33", "problem_type": null, "exam": "IMO", "problem": "33. (MON 4) In what case does the system \\[  $$ \\begin{aligned} & x+y+m z=a, \\\\ & x+m y+z=b, \\\\ & m x+y+z=c, \\end{aligned} $$ \\]  have a solution? Find the conditions under which the unique solution of the above system is an arithmetic progression.", "solution": "33. If $m \\notin\\{-2,1\\}$, the system has the unique solution  $$ x=\\frac{b+a-(1+m) c}{(2+m)(1-m)}, \\quad y=\\frac{a+c-(1+m) b}{(2+m)(1-m)}, \\quad z=\\frac{b+c-(1+m) a}{(2+m)(1-m)} $$  The numbers $x, y, z$ form an arithmetic progression if and only if $a, b, c$ do so. For $m=1$ the system has a solution if and only if $a=b=c$, while for $m=-2$ it has a solution if and only if $a+b+c=0$. In both these cases it has infinitely many solutions.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "34", "problem_type": null, "exam": "IMO", "problem": "34. (MON 5) The faces of a convex polyhedron are six squares and eight equilateral triangles, and each edge is a common side for one triangle and one square. All dihedral angles obtained from the triangle and square with a common edge are equal. Prove that it is possible to circumscribe a sphere around this polyhedron and compute the ratio of the squares of the volumes of the polyhedron and of the ball whose boundary is the circumscribed sphere.", "solution": "34. Each vertex of the polyhedron is a vertex of exactly two squares and triangles (more than two is not possible; otherwise, the sum of angles at a vertex exceeds $360^{\\circ}$ ). By using the condition that the trihedral angles are equal it is easy to see that such a polyhedron is uniquely determined by its side length.  The polyhedron obtained from a cube by \"cutting\" its vertices, as shown in the figure, satisfies the conditions. Now it is easy to calculate that the ratio of the squares of volumes of that polyhedron and of the ball whose boundary is the circum- ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-363.jpg?height=333&width=344&top_left_y=260&top_left_x=908) scribed sphere is equal to $25 /\\left(8 \\pi^{2}\\right)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "35", "problem_type": null, "exam": "IMO", "problem": "35. (MON 6) Prove the identity  $$ \\sum_{k=0}^{n}\\binom{n}{k}\\left(\\tan \\frac{x}{2}\\right)^{2 k}\\left[1+2^{k} \\frac{1}{\\left(1-\\tan ^{2}(x / 2)\\right)^{k}}\\right]=\\sec ^{2 n} \\frac{x}{2}+\\sec ^{n} x $$", "solution": "35. The given sum can be rewritten as  $$ \\sum_{k=0}^{n}\\binom{n}{k}\\left(\\tan ^{2} \\frac{x}{2}\\right)^{k}+\\sum_{k=0}^{n}\\binom{n}{k}\\left(\\frac{2 \\tan ^{2} \\frac{x}{2}}{1-\\tan ^{2} \\frac{x}{2}}\\right)^{k} $$  Since $\\frac{2 \\tan ^{2}(x / 2)}{1-\\tan ^{2}(x / 2)}=\\frac{1-\\cos x}{\\cos x}$, the above sum is transformed using the binomial formula into  $$ \\left(1+\\tan ^{2} \\frac{x}{2}\\right)^{n}+\\left(1+\\frac{1-\\cos x}{\\cos x}\\right)^{n}=\\sec ^{2 n} \\frac{x}{2}+\\sec ^{n} x $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "36", "problem_type": null, "exam": "IMO", "problem": "36. (POL 1) Prove that the center of the sphere circumscribed around a tetrahedron $A B C D$ coincides with the center of a sphere inscribed in that tetrahedron if and only if $A B=C D, A C=B D$, and $A D=B C$.", "solution": "36. Suppose that the skew edges of the tetrahedron $A B C D$ are equal. Let $K$, $L, M, P, Q, R$ be the midpoints of edges $A B, A C, A D, C D, D B, B C$ respectively. Segments $K P, L Q, M R$ have the common midpoint $T$. We claim that the lines $K P, L Q$ and $M R$ are axes of symmetry of the tetrahedron $A B C D$. From $L M\\|C D\\| R Q$ and similarly $L R \\| M Q$ and $L M=C D / 2=$ $A B / 2=L R$ it follows that $L M Q R$ is a rhombus and therefore $L Q \\perp$ $M R$. We similarly show that $K P$ is perpendicular to $L Q$ and $M R$, and ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-363.jpg?height=390&width=442&top_left_y=1219&top_left_x=897) thus it is perpendicular to the plane $L M Q R$. Since the lines $A B$ and $C D$ are parallel to the plane $L M Q R$, they are perpendicular to $K P$. Hence the points $A$ and $C$ are symmetric to $B$ and $D$ with respect to the line $K P$, which means that $K P$ is an axis of symmetry of the tetrahedron $A B C D$. Similarly, so are the lines $L Q$ and $M R$. The centers of circumscribed and inscribed spheres of tetrahedron $A B C D$ must lie on every axis of symmetry of the tetrahedron, and hence both coincide with $T$. Conversely, suppose that the centers of circumscribed and inscribed spheres of the tetrahedron $A B C D$ coincide with some point $T$. Then the orthogonal projections of $T$ onto the faces $A B C$ and $A B D$ are the circumcenters $O_{1}$ and $O_{2}$ of these two triangles, and moreover, $T O_{1}=T O_{2}$.  Pythagoras's theorem gives $A O_{1}=A O_{2}$, which by the law of sines implies $\\angle A C B=\\angle A D B$. Now it easily follows that the sum of the angles at one vertex of the tetrahedron is equal to $180^{\\circ}$. Let $D^{\\prime}, D^{\\prime \\prime}$, and $D^{\\prime \\prime \\prime}$ be the points in the plane $A B C$ lying outside $\\triangle A B C$ such that $\\triangle D^{\\prime} B C \\cong \\triangle D B C, \\triangle D^{\\prime \\prime} C A \\cong \\triangle D C A$, and $\\triangle D^{\\prime \\prime \\prime} A B \\cong \\triangle D A B$. The angle $D^{\\prime \\prime} A D^{\\prime \\prime \\prime}$ is then straight, and hence $A, B, C$ are midpoints of the segments $D^{\\prime \\prime} D^{\\prime \\prime \\prime}, D^{\\prime \\prime \\prime} D^{\\prime}, D^{\\prime} D^{\\prime \\prime}$ respectively. Hence $A D=D^{\\prime \\prime} D^{\\prime \\prime \\prime} / 2=B C$, and analogously $A B=C D$ and $A C=B D$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "37", "problem_type": null, "exam": "IMO", "problem": "37. (POL 2) Prove that for arbitrary positive numbers the following inequality holds:  $$ \\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\leq \\frac{a^{8}+b^{8}+c^{8}}{a^{3} b^{3} c^{3}} $$", "solution": "37. Using the $\\mathrm{A}-\\mathrm{G}$ mean inequality we obtain  $$ \\begin{aligned} & 8 a^{2} b^{3} c^{3} \\leq 2 a^{8}+3 b^{8}+3 c^{8} \\\\ & 8 a^{3} b^{2} c^{3} \\leq 3 a^{8}+2 b^{8}+3 c^{8} \\\\ & 8 a^{3} b^{3} c^{2} \\leq 3 a^{8}+3 b^{8}+2 c^{8} \\end{aligned} $$  By adding these inequalities and dividing by $3 a^{3} b^{3} c^{3}$ we obtain the desired one.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "38", "problem_type": null, "exam": "IMO", "problem": "38. (POL 3) Does there exist an integer such that its cube is equal to $3 n^{2}+3 n+7$, where $n$ is integer?", "solution": "38. Suppose that there exist integers $n$ and $m$ such that $m^{3}=3 n^{2}+3 n+7$. Then from $m^{3} \\equiv 1(\\bmod 3)$ it follows that $m=3 k+1$ for some $k \\in \\mathbb{Z}$. Substituting into the initial equation we obtain $3 k\\left(3 k^{2}+3 k+1\\right)=n^{2}+$ $n+2$. It is easy to check that $n^{2}+n+2$ cannot be divisible by 3 , and so this equality cannot be true. Therefore our equation has no solutions in integers.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "39", "problem_type": null, "exam": "IMO", "problem": "39. (POL 4) Show that the triangle whose angles satisfy the equality  $$ \\frac{\\sin ^{2} A+\\sin ^{2} B+\\sin ^{2} C}{\\cos ^{2} A+\\cos ^{2} B+\\cos ^{2} C}=2 $$  is a right-angled triangle.", "solution": "39. Since $\\sin ^{2} A+\\sin ^{2} B+\\sin ^{2} C+\\cos ^{2} A+\\cos ^{2} B+\\cos ^{2} C=3$, the given equality is equivalent to $\\cos ^{2} A+\\cos ^{2} B+\\cos ^{2} C=1$, which by multiplying by 2 is transformed into  $$ \\begin{aligned} 0 & =\\cos 2 A+\\cos 2 B+2 \\cos ^{2} C=2 \\cos (A+B) \\cos (A-B)+2 \\cos ^{2} C \\\\ & =2 \\cos C(\\cos (A-B)-\\cos C) \\end{aligned} $$  It follows that either $\\cos C=0$ or $\\cos (A-B)=\\cos C$. In both cases the triangle is right-angled.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (BUL 4) Suppose medians $m_{a}$ and $m_{b}$ of a triangle are orthogonal. Prove that: (a) The medians of that triangle correspond to the sides of a right-angled triangle. (b) The inequality  $$ 5\\left(a^{2}+b^{2}-c^{2}\\right) \\geq 8 a b $$  is valid, where $a, b$, and $c$ are side lengths of the given triangle.", "solution": "4. (a) Let $A B C D$ be a parallelogram, and $K, L$ the midpoints of segments $B C$ and $C D$ respectively. The sides of $\\triangle A K L$ are equal and parallel to the medians of $\\triangle A B C$. (b) Using the formulas $4 m_{a}^{2}=2 b^{2}+2 c^{2}-a^{2}$ etc., it is easy to obtain that $m_{a}^{2}+m_{b}^{2}=m_{c}^{2}$ is equivalent to $a^{2}+b^{2}=5 c^{2}$. Then  $$ 5\\left(a^{2}+b^{2}-c^{2}\\right)=4\\left(a^{2}+b^{2}\\right) \\geq 8 a b $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "40", "problem_type": null, "exam": "IMO", "problem": "40. (POL 5) ${ }^{\\mathrm{IMO} 2}$ Exactly one side of a tetrahedron is of length greater than 1. Show that its volume is less than or equal to $1 / 8$.", "solution": "40. Suppose $C D$ is the longest edge of the tetrahedron $A B C D, A B=a, C K$ and $D L$ are the altitudes of the triangles $A B C$ and $A B D$ respectively, and $D M$ is the altitude of the tetrahedron $A B C D$. Then $C K^{2} \\leq 1-a^{2} / 4$, since $C K$ is a leg of the right triangle whose other leg has length not less than $a / 2$ and whose hypotenuse has length not greater than 1 (AKC or $B K C)$. In the similar way we can show that $D L^{2} \\leq 1-a^{2} / 4$. Since $D M \\leq D L$, then $D M^{2} \\leq 1-a^{2} / 4$. It follows that  $$ \\begin{aligned} V & =\\frac{1}{3}\\left(\\frac{a}{2} C K\\right) D M \\leq \\frac{1}{6} a\\left(1-\\frac{a^{2}}{4}\\right)=\\frac{1}{24} a(2-a)(2+a) \\\\ & =\\frac{1}{24}\\left[1-(a-1)^{2}\\right](2+a) \\leq \\frac{1}{24} \\cdot 1 \\cdot 3=\\frac{1}{8} . \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "41", "problem_type": null, "exam": "IMO", "problem": "41. (POL 6) A line $l$ is drawn through the intersection point $H$ of the altitudes of an acute-angled triangle. Prove that the symmetric images $l_{a}$, $l_{b}, l_{c}$ of $l$ with respect to sides $B C, C A, A B$ have one point in common, which lies on the circumcircle of $A B C$.", "solution": "41. It is well known that the points $K, L, M$, symmetric to $H$ with respect to $B C, C A, A B$ respectively, lie on the circumcircle $k$ of the triangle $A B C$. For $K$, this follows from an elementary calculation of angles of triangles $H B C$ and noting that $\\measuredangle K B C=\\measuredangle H B C=\\measuredangle K A C$. For other points the proof is analogous. Since the lines $l_{a}, l_{b}$ pass through $K$ and $L$ and $l_{b}$ is obtained from $l_{a}$ by rotation about $C$ for an angle $2 \\gamma=\\angle L C K$, it follows that the intersection point $P$ of $l_{a}$ and $l_{b}$ is at the circumcircle of $K L C$, that is, $k$. Similarly, $l_{b}$ and $l_{c}$ meet at a point on $k$; hence they must pass through the same point $P$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-365.jpg?height=387&width=399&top_left_y=430&top_left_x=869)", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "42", "problem_type": null, "exam": "IMO", "problem": "42. (ROM 1) Decompose into real factors the expression $1-\\sin ^{5} x-\\cos ^{5} x$.", "solution": "42. $E=(1-\\sin x)(1-\\cos x)[3+2(\\sin x+\\cos x)+2 \\sin x \\cos x+\\sin x \\cos x(\\sin x+$ $\\cos x)]$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "43", "problem_type": null, "exam": "IMO", "problem": "43. (ROM 2) The equation  $$ x^{5}+5 \\lambda x^{4}-x^{3}+(\\lambda \\alpha-4) x^{2}-(8 \\lambda+3) x+\\lambda \\alpha-2=0 $$  is given. (a) Determine $\\alpha$ such that the given equation has exactly one root independent of $\\lambda$. (b) Determine $\\alpha$ such that the given equation has exactly two roots independent of $\\lambda$.", "solution": "43. We can write the given equation in the form  $$ x^{5}-x^{3}-4 x^{2}-3 x-2+\\lambda\\left(5 x^{4}+\\alpha x^{2}-8 x+\\alpha\\right)=0 . $$  A root of this equation is independent of $\\lambda$ if and only if it is a common root of the equations  $$ x^{5}-x^{3}-4 x^{2}-3 x-2=0 \\quad \\text { and } \\quad 5 x^{4}+\\alpha x^{2}-8 x+\\alpha=0 . $$  The first of these two equations is equivalent to $(x-2)\\left(x^{2}+x+1\\right)^{2}=0$ and has three different roots: $x_{1}=2, x_{2,3}=(-1 \\pm i \\sqrt{3}) / 2$. (a) For $\\alpha=-64 / 5, x_{1}=2$ is the unique root independent of $\\lambda$. (b) For $\\alpha=-3$ there are two roots independent of $\\lambda$ : $x_{1}=\\omega$ and $x_{2}=\\omega^{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "44", "problem_type": null, "exam": "IMO", "problem": "44. (ROM 3) Suppose $p$ and $q$ are two different positive integers and $x$ is a real number. Form the product $(x+p)(x+q)$. (a) Find the sum $S(x, n)=\\sum(x+p)(x+q)$, where $p$ and $q$ take values from 1 to $n$. (b) Do there exist integer values of $x$ for which $S(x, n)=0$ ?", "solution": "44. (a) $S(x, n)=n(n-1)\\left[x^{2}+(n+1) x+(n+1)(3 n+2) / 12\\right]$. (b) It is easy to see that the equation $S(x, n)=0$ has two roots $x_{1,2}=$ $(-(n+1) \\pm \\sqrt{(n+1) / 3}) / 2$. They are integers if and only if $n=$ $3 k^{2}-1$ for some $k \\in \\mathbb{N}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "45", "problem_type": null, "exam": "IMO", "problem": "45. (ROM 4) (a) Solve the equation  $$ \\sin ^{3} x+\\sin ^{3}\\left(\\frac{2 \\pi}{3}+x\\right)+\\sin ^{3}\\left(\\frac{4 \\pi}{3}+x\\right)+\\frac{3}{4} \\cos 2 x=0 . $$  (b) Suppose the solutions are in the form of arcs $A B$ of the trigonometric circle (where $A$ is the beginning of arcs of the trigonometric circle), and $P$ is a regular $n$-gon inscribed in the circle with one vertex at $A$. (1) Find the subset of arcs with the endpoint $B$ at a vertex of the regular dodecagon. (2) Prove that the endpoint $B$ cannot be at a vertex of $P$ if $2,3 \\nmid n$ or $n$ is prime.", "solution": "45. (a) Using the formula $4 \\sin ^{3} x=3 \\sin x-\\sin 3 x$ one can easily reduce the given equation to $\\sin 3 x=\\cos 2 x$. Its solutions are given by $x=$ $(4 k+1) \\pi / 10, k \\in \\mathbb{Z}$. (b) (1) The point $B$ corresponding to the solution $x=(4 k+1) \\pi / 10$ is a vertex of the regular dodecagon if and only if $(4 k+1) \\pi / 10=$ $2 m \\pi / 12$, i.e., $3(4 k+1)=5 m$ for some $m \\in \\mathbb{Z}$. This is possible if and only if $5 \\mid 4 k+1$, i.e., $k \\equiv 1(\\bmod 5)$. (2) Similarly, if the point $B$ corresponding to $x=(4 k+1) \\pi / 10$ is a vertex of a polygon $P$, then $(4 k+1) n=20 m$ for some $m \\in \\mathbb{N}$, which implies that $4 \\mid n$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "46", "problem_type": null, "exam": "IMO", "problem": "46. (ROM 5) If $x, y, z$ are real numbers satisfying the relations $x+y+z=1$ and $\\arctan x+\\arctan y+\\arctan z=\\pi / 4$, prove that  $$ x^{2 n+1}+y^{2 n+1}+z^{2 n+1}=1 $$  for all positive integers $n$.", "solution": "46. Let us set $\\arctan x=a, \\arctan y=b, \\arctan z=c$. Then $\\tan (a+b)=\\frac{x+y}{1-x y}$ and $\\tan (a+b+c)=\\frac{x+y+z-x y z}{1-y z-z x-x y}=1$, which implies that  $$ (x-1)(y-1)(z-1)=x y z-x y-y z-z x+x+y+z-1=0 . $$  One of $x, y, z$ is equal to 1 , say $z=1$, and consequently $x+y=0$. Therefore  $$ x^{2 n+1}+y^{2 n+1}+z^{2 n+1}=x^{2 n+1}+(-x)^{2 n+1}+1^{2 n+1}=1 $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "47", "problem_type": null, "exam": "IMO", "problem": "47. (ROM 6) Prove the inequality $x_{1} x_{2} \\cdots x_{k}\\left(x_{1}^{n-1}+x_{2}^{n-1}+\\cdots+x_{k}^{n-1}\\right) \\leq x_{1}^{n+k-1}+x_{2}^{n+k-1}+\\cdots+x_{k}^{n+k-1}$, where $x_{i}>0(i=1,2, \\ldots, k), k \\in N, n \\in N$.", "solution": "47. Using the $\\mathrm{A}-\\mathrm{G}$ mean inequality we get  $$ \\begin{gathered} (n+k-1) x_{1}^{n} x_{2} \\cdots x_{k} \\leq n x_{1}^{n+k-1}+x_{2}^{n+k-1}+\\cdots+x_{k}^{n+k-1} \\\\ (n+k-1) x_{1} x_{2}^{n} \\cdots x_{k} \\leq x_{1}^{n+k-1}+n x_{2}^{n+k-1}+\\cdots+x_{k}^{n+k-1} \\\\ \\cdots \\cdots \\cdots \\\\ \\cdots \\cdots \\cdots \\\\ (n+k-1) x_{1} x_{2} \\cdots x_{k}^{n} \\leq x_{1}^{n+k-1}+x_{2}^{n+k-1}+\\cdots+n x_{k}^{n+k-1} \\end{gathered} $$  By adding these inequalities and dividing by $n+k-1$ we obtain the desired one. Remark. This is also an immediate consequence of Muirhead's inequality.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "48", "problem_type": null, "exam": "IMO", "problem": "48. (SWE 1) Determine all positive roots of the equation $x^{x}=1 / \\sqrt{2}$.", "solution": "48. Put $f(x)=x \\ln x$. The given equation is equivalent to $f(x)=f(1 / 2)$, which has the solutions $x_{1}=1 / 2$ and $x_{2}=1 / 4$. Since the function $f$ is decreasing on $(0,1 / e)$, and increasing on $(1 / e,+\\infty)$, this equation has no other solutions.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "49", "problem_type": null, "exam": "IMO", "problem": "49. (SWE 2) Let $n$ and $k$ be positive integers such that $1 \\leq n \\leq N+1$, $1 \\leq k \\leq N+1$. Show that  $$ \\min _{n \\neq k}|\\sin n-\\sin k|<\\frac{2}{N} $$", "solution": "49. Since $\\sin 1, \\sin 2, \\ldots, \\sin (N+1) \\in(-1,1)$, two of these $N+1$ numbers have distance less than $2 / N$. Therefore $|\\sin n-\\sin k|<2 / N$ for some integers $1 \\leq k, n \\leq N+1, n \\neq k$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (BUL 5) Solve the system  $$ \\begin{aligned} & x^{2}+x-1=y \\\\ & y^{2}+y-1=z \\\\ & z^{2}+z-1=x \\end{aligned} $$", "solution": "5. If one of $x, y, z$ is equal to 1 or -1 , then we obtain solutions $(-1,-1,-1)$ and $(1,1,1)$. We claim that these are the only solutions to the system. Let $f(t)=t^{2}+t-1$. If among $x, y, z$ one is greater than 1 , say $x>1$, we have $x<f(x)=y<f(y)=z<f(z)=x$, which is impossible. It follows that $x, y, z \\leq 1$. Suppose now that one of $x, y, z$, say $x$, is less than -1 . Since $\\min _{t} f(t)=$ $-5 / 4$, we have $x=f(z) \\in[-5 / 4,-1)$. Also, since $f([-5 / 4,-1))=$ $(-1,-11 / 16) \\subseteq(-1,0)$ and $f((-1,0))=[-5 / 4,-1)$, it follows that $y=f(x) \\in(-1,0), z=f(y) \\in[-5 / 4,-1)$, and $x=f(z) \\in(-1,0)$, which is a contradiction. Therefore $-1 \\leq x, y, z \\leq 1$. If $-1<x, y, z<1$, then $x>f(x)=y>f(y)=z>f(z)=x$, a contradiction. This proves our claim.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "50", "problem_type": null, "exam": "IMO", "problem": "50. (SWE 3) The function $\\varphi(x, y, z)$, defined for all triples $(x, y, z)$ of real numbers, is such that there are two functions $f$ and $g$ defined for all pairs of real numbers such that  $$ \\varphi(x, y, z)=f(x+y, z)=g(x, y+z) $$  for all real $x, y$, and $z$. Show that there is a function $h$ of one real variable such that  $$ \\varphi(x, y, z)=h(x+y+z) $$  for all real $x, y$, and $z$.", "solution": "50. Since $\\varphi(x, y, z)=f(x+y, z)=\\varphi(0, x+y, z)=g(0, x+y+z)$, it is enough to put $h(t)=g(0, t)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "51", "problem_type": null, "exam": "IMO", "problem": "51. (SWE 4) A subset $S$ of the set of integers $0, \\ldots, 99$ is said to have property A if it is impossible to fill a crossword puzzle with 2 rows and 2 columns with numbers in $S$ ( 0 is written as 00,1 as 01 , and so on). Determine the maximal number of elements in sets $S$ with property A.", "solution": "51. If there exist two numbers $\\overline{a b}, \\overline{b c} \\in S$, then one can fill a crossword puzzle as $\\left(\\begin{array}{ll}a & b \\\\ b & c\\end{array}\\right)$. The converse is obvious. Hence the set $S$ has property $A$ if and only if the set of first digits and the set of second digits of numbers in $S$ are disjoint. Thus the maximum size of $S$ is 25 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "52", "problem_type": null, "exam": "IMO", "problem": "52. (SWE 5) In the plane a point $O$ and a sequence of points $P_{1}, P_{2}, P_{3}, \\ldots$ are given. The distances $O P_{1}, O P_{2}, O P_{3}, \\ldots$ are $r_{1}, r_{2}, r_{3}, \\ldots$, where $r_{1} \\leq$ $r_{2} \\leq r_{3} \\leq \\cdots$. Let $\\alpha$ satisfy $0<\\alpha<1$. Suppose that for every $n$ the distance from the point $P_{n}$ to any other point of the sequence is greater than or equal to $r_{n}^{\\alpha}$. Determine the exponent $\\beta$, as large as possible, such that for some $C$ independent of $n,{ }^{2}$  $$ r_{n} \\geq C n^{\\beta}, \\quad n=1,2, \\ldots $$", "solution": "52. This problem is not elementary. The solution offered by the proposer was not quite clear and complete (the existence was not proved).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "53", "problem_type": null, "exam": "IMO", "problem": "53. (SWE 6) In making Euclidean constructions in geometry it is permitted to use a straightedge and compass. In the constructions considered in this question, no compasses are permitted, but the straightedge is assumed to have two parallel edges, which can be used for constructing two parallel lines through two given points whose distance is at least equal to the breadth of the ruler. Then the distance between the parallel lines is equal to the breadth of the straightedge. Carry through the following constructions with such a straightedge. Construct: (a) The bisector of a given angle. (b) The midpoint of a given rectilinear segment. (c) The center of a circle through three given noncollinear points. (d) A line through a given point parallel to a given line.", "solution": "53. (a) We can construct two lines parallel to the rays of the angle, at equal distances from the rays. The intersection of these two lines lies on the bisector of the angle. (b) If the length of a segment $A B$ exceeds the breadth of the ruler, we can construct parallel lines through $A$ and $B$ in two different ways. The diagonal in the resulting rhombus is the perpendicular bisector of the segment $A B$.  If the segment $A B$ is too short, we can construct a line $l$ parallel to $A B$ and centrally project $A B$ onto $l$ from a point $C$ chosen sufficiently close to the segment, thus obtaining an arbitrarily long segment $A^{\\prime} B^{\\prime} \\|$ $A B$. Then we construct the midpoint $D^{\\prime}$ of $A^{\\prime} B^{\\prime}$ as above. The line $D^{\\prime} C$ intersects the segment $A B$ at its midpoint $D$. By means of lines parallel to $D C$ the segment $A B$ can be prolonged symmetrically, and then the perpendicular bisector can be found as above. (c) follows immediately from part (b). (d) Let there be given a point $P$ and a line $l$. We draw an arbitrary line through $P$ that intersects $l$ at $A$, and two lines $l_{1}$ and $l_{2}$ parallel to $A P$, at equal distances from $A P$ and on either side of $A P$. Line $l_{1}$ intersects $l$ at $B$. We can construct the midpoint $C$ of $A P$. If $B C$ intersects $l_{2}$ at $D$, then $P D$ is parallel to $l$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "54", "problem_type": null, "exam": "IMO", "problem": "54. (USS 1) Is it possible to put 100 (or 200) points on a wooden cube such that by all rotations of the cube the points map into themselves? Justify your answer.", "solution": "54. Let $S$ be the given set of points on the cube. Let $x, y, z$ denote the numbers of points from $S$ lying at a vertex, at the midpoint of an edge, at the midpoint of a face of the cube, respectively, and let $u$ be the number of all other points from $S$. Either there are no points from $S$ at the vertices of the cube, or there is a point from $S$ at each vertex. Hence $x$ is either 0 or 8 . Similarly, $y$ is either 0 or 12 , and $z$ is either 0 or 6 . Any other point of $S$ has 24 possible images under rotations of the cube. Hence $u$ is divisible by 24 . Since $n=x+y+z+u$ and $6 \\mid y, z, u$, it follows that either $6 \\mid n$ or $6 \\mid n-8$, i.e., $n \\equiv 0$ or $n \\equiv 2(\\bmod 6)$. Thus $n=200$ is possible, while $n=100$ is not, because $n \\equiv 4(\\bmod 6)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "55", "problem_type": null, "exam": "IMO", "problem": "55. (USS 2) Find all $x$ for which for all $n$,  $$ \\sin x+\\sin 2 x+\\sin 3 x+\\cdots+\\sin n x \\leq \\frac{\\sqrt{3}}{2} $$", "solution": "55. It is enough to find all $x$ from $(0,2 \\pi]$ such that the given inequality holds for all $n$. Suppose $0<x<2 \\pi / 3$. If $n$ is the maximum integer for which $n x \\leq$ $2 \\pi / 3$, we have $\\pi / 3<n x \\leq 2 \\pi / 3$, and consequently $\\sin n x \\geq \\sqrt{3} / 2$. Thus $\\sin x+\\sin 2 x+\\cdots+\\sin n x>\\sqrt{3} / 2$. Suppose now that $2 \\pi / 3 \\leq x<2 \\pi$. We have  $$ \\sin x+\\cdots+\\sin n x=\\frac{\\cos \\frac{x}{2}-\\cos \\frac{2 n+1}{2} x}{2 \\sin \\frac{x}{2}} \\leq \\frac{\\cos \\frac{x}{2}+1}{2 \\sin \\frac{x}{2}}=\\frac{\\cot \\frac{x}{4}}{2} \\leq \\frac{\\sqrt{3}}{2} . $$  For $x=2 \\pi$ the given inequality clearly holds for all $n$. Hence, the inequality holds for all $n$ if and only if $2 \\pi / 3+2 k \\pi \\leq x \\leq 2 \\pi+2 k \\pi$ for some integer $k$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "56", "problem_type": null, "exam": "IMO", "problem": "56. (USS 3) In a group of interpreters each one speaks one or several foreign languages; 24 of them speak Japanese, 24 Malay, 24 Farsi. Prove that it is possible to select a subgroup in which exactly 12 interpreters speak Japanese, exactly 12 speak Malay, and exactly 12 speak Farsi.", "solution": "56. We shall prove by induction on $n$ the following statement: If in some group of interpreters exactly $n$ persons, $n \\geq 2$, speak each of the three languages, then it is possible to select a subgroup in which each language is spoken by exactly two persons. The statement of the problem easily follows from this: it suffices to select six such groups.  The case $n=2$ is trivial. Let us assume $n \\geq 2$, and let $N_{j}, N_{m}, N_{f}, N_{j m}$, $N_{j f}, N_{m f}, N_{j m f}$ be the sets of those interpreters who speak only Japanese, only Malay, only Farsi, only Japanese and Malay, only Japanese and Farsi, only Malay and Farsi, and all the three languages, respectively, and $n_{j}, n_{m}$, $n_{f}, n_{j m}, n_{j f}, n_{m f}, n_{j m f}$ the cardinalities of these sets, respectively. By the condition of the problem, $n_{j}+n_{j m}+n_{j f}+n_{j m f}=n_{m}+n_{j m}+n_{m f}+n_{j m f}=$ $n_{f}+n_{j f}+n_{m f}+n_{j m f}=24$, and consequently  $$ n_{j}-n_{m f}=n_{m}-n_{j f}=n_{f}-n_{j m}=c $$  Now if $c<0$, then $n_{j m}, n_{j f}, n_{m f}>0$, and it is enough to select one interpreter from each of the sets $N_{j m}, N_{j f}, N_{m f}$. If $c>0$, then $n_{j}, n_{m}, n_{f}>0$, and it is enough to select one interpreter from each of the sets $N_{j}, N_{m}, N_{f}$ and then use the inductive assumption. Also, if $c=0$, then w.l.o.g. $n_{j}=n_{m f}>0$, and it is enough to select one interpreter from each of the sets $N_{j}, N_{m f}$ and then use the inductive hypothesis. This completes the induction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "57", "problem_type": null, "exam": "IMO", "problem": "57. (USS 4) ${ }^{\\mathrm{IMO} 5}$ Consider the sequence $\\left(c_{n}\\right)$ :  $$ \\begin{gathered} c_{1}=a_{1}+a_{2}+\\cdots+a_{8}, \\\\ c_{2}=a_{1}^{2}+a_{2}^{2}+\\cdots+a_{8}^{2}, \\\\ \\cdots \\\\ \\cdots \\cdots \\cdots \\\\ c_{n}=a_{1}^{n}+a_{2}^{n}+\\cdots+a_{8}^{n}, \\end{gathered} $$  [^1]where $a_{1}, a_{2}, \\ldots, a_{8}$ are real numbers, not all equal to zero. Given that among the numbers of the sequence $\\left(c_{n}\\right)$ there are infinitely many equal to zero, determine all the values of $n$ for which $c_{n}=0$.", "solution": "57. Obviously $c_{n}>0$ for all even $n$. Thus $c_{n}=0$ is possible only for an odd $n$. Let us assume $a_{1} \\leq a_{2} \\leq \\cdots \\leq a_{8}$ : in particular, $a_{1} \\leq 0 \\leq a_{8}$. If $\\left|a_{1}\\right|<\\left|a_{8}\\right|$, then there exists $n_{0}$ such that for every odd $n>n_{0}, 7\\left|a_{1}\\right|^{n}<$ $a_{8}^{n} \\Rightarrow a_{1}^{n}+\\cdots+a_{7}^{n}+a_{8}^{n}>7 a_{1}^{n}+a_{8}^{n}>0$, contradicting the condition that $c_{n}=0$ for infinitely many $n$. Similarly $\\left|a_{1}\\right|>\\left|a_{8}\\right|$ is impossible, and we conclude that $a_{1}=-a_{8}$. Continuing in the same manner we can show that $a_{2}=-a_{7}, a_{3}=-a_{6}$ and $a_{4}=-a_{5}$. Hence $c_{n}=0$ for every odd $n$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "58", "problem_type": null, "exam": "IMO", "problem": "58. (USS 5) A linear binomial $l(z)=A z+B$ with complex coefficients $A$ and $B$ is given. It is known that the maximal value of $|l(z)|$ on the segment $-1 \\leq x \\leq 1(y=0)$ of the real line in the complex plane $(z=x+i y)$ is equal to $M$. Prove that for every $z$ $$ |l(z)| \\leq M \\rho, $$ where $\\rho$ is the sum of distances from the point $P=z$ to the points $Q_{1}$ : $z=1$ and $Q_{3}: z=-1$.", "solution": "58. The following sequence of equalities and inequalities gives an even stronger estimate than needed.  $$ \\begin{aligned} |l(z)| & =|A z+B|=\\frac{1}{2}|(z+1)(A+B)+(z-1)(A-B)| \\\\ & =\\frac{1}{2}|(z+1) f(1)+(z-1) f(-1)| \\\\ & \\leq \\frac{1}{2}(|z+1| \\cdot|f(1)|+|z-1| \\cdot|f(-1)|) \\\\ & \\leq \\frac{1}{2}(|z+1|+|z-1|) M=\\frac{1}{2} \\rho M . \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "59", "problem_type": null, "exam": "IMO", "problem": "59. (USS 6) On the circle with center $O$ and radius 1 the point $A_{0}$ is fixed and points $A_{1}, A_{2}, \\ldots, A_{999}, A_{1000}$ are distributed in such a way that $\\angle A_{0} O A_{k}=k$ (in radians). Cut the circle at points $A_{0}, A_{1}, \\ldots, A_{1000}$. How many arcs with different lengths are obtained?  ### 3.10 The Tenth IMO", "solution": "59. By the $\\operatorname{arc} A B$ we shall always mean the positive $\\operatorname{arc} A B$. We denote by $|A B|$ the length of arc $A B$. Let a basic arc be one of the $n+1$ arcs into which the circle is partitioned by the points $A_{0}, A_{1}, \\ldots, A_{n}$, where $n \\in \\mathbb{N}$. Suppose that $A_{p} A_{0}$ and $A_{0} A_{q}$ are the basic arcs with an endpoint at $A_{0}$, and that $x_{n}, y_{n}$ are their lengths, respectively. We show by induction on $n$ that for each $n$ the length of a basic arc is equal to $x_{n}, y_{n}$ or $x_{n}+y_{n}$. The statement is trivial for $n=1$. Assume that it holds for $n$, and let $A_{i} A_{n+1}, A_{n+1} A_{j}$ be basic arcs. We shall prove that these two arcs have lengths $x_{n}, y_{n}$, or $x_{n}+y_{n}$. If $i, j$ are both strictly positive, then $\\left|A_{i} A_{n+1}\\right|=$ $\\left|A_{i-1} A_{n}\\right|$ and $\\left|A_{n+1} A_{j}\\right|=\\left|A_{n} A_{j-1}\\right|$ are equal to $x_{n}, y_{n}$, or $x_{n}+y_{n}$ by the inductive hypothesis. Let us assume now that $i=0$, i.e., that $A_{p} A_{n+1}$ and $A_{n+1} A_{0}$ are basic arcs. Then $\\left|A_{p} A_{n+1}\\right|=\\left|A_{0} A_{n+1-p}\\right| \\geq\\left|A_{0} A_{q}\\right|=y_{n}$ and similarly $\\left|A_{n+1} A_{q}\\right| \\geq x_{n}$, but $\\left|A_{p} A_{q}\\right|=x_{n}+y_{n}$, from which it follows that $\\left|A_{p} A_{n+1}\\right|=\\left|A_{0} A_{q}\\right|=y_{n}$ and consequently $n+1=p+q$. Also, $x_{n+1}=\\left|A_{n+1} A_{0}\\right|=y_{n}-x_{n}$ and $y_{n+1}=y_{n}$. Now, all basic arcs have lengths $y_{n}-x_{n}, x_{n}, y_{n}, x_{n}+y_{n}$. A presence of a basic arc of length $x_{n}+y_{n}$ would spoil our inductive step. However, if any basic arc $A_{k} A_{l}$ has length $x_{n}+y_{n}$, then we must have $l-q=k-p$ because $2 \\pi$ is irrational, and therefore the arc $A_{k} A_{l}$ contains either the point $A_{k-p}$ (if $k \\geq p$ ) or the point $A_{k+q}$ (if $k<p$ ), which is impossible; hence, the proof is complete for $i=0$. The proof for $j=0$ is analogous. This completes the induction. It can be also seen from the above considerations that the basic arcs take only two distinct lengths if and only if $n=p+q-1$. If we denote by $n_{k}$ the sequence of $n$ 's for which this holds, and by $p_{k}, q_{k}$ the sequences of the corresponding $p, q$, we have $p_{1}=q_{1}=1$ and  $$ \\left(p_{k+1}, q_{k+1}\\right)=\\left\\{\\begin{array}{l} \\left(p_{k}+q_{k}, q_{k}\\right), \\text { if }\\left\\{p_{k} /(2 \\pi)\\right\\}+\\left\\{q_{k} /(2 \\pi)\\right\\}>1 \\\\ \\left(p_{k}, p_{k}+q_{k}\\right), \\text { if }\\left\\{p_{k} /(2 \\pi)\\right\\}+\\left\\{q_{k} /(2 \\pi)\\right\\}<1 \\end{array}\\right. $$  It is now \"easy\" to calculate that $p_{19}=p_{20}=333, q_{19}=377, q_{20}=710$, and thus $n_{19}=709<1000<1042=n_{20}$. It follows that the lengths of the basic arcs for $n=1000$ take exactly three different values.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (BUL 6) Solve the system  $$ \\begin{aligned} |x+y|+|1-x| & =6 \\\\ |x+y+1|+|1-y| & =4 \\end{aligned} $$", "solution": "6. The given system has two solutions: $(-2,-1)$ and $(-14 / 3,13 / 3)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (CZS 1) Find all real solutions of the system of equations  $$ \\begin{aligned} x_{1}+x_{2}+\\cdots+x_{n} & =a, \\\\ x_{1}^{2}+x_{2}^{2}+\\cdots+x_{n}^{2} & =a^{2}, \\\\ \\ldots \\cdots \\cdots \\cdots+x_{n}^{n} & =a^{n} . \\end{aligned} $$", "solution": "7. Let $S_{k}=x_{1}^{k}+x_{2}^{k}+\\cdots+x_{n}^{k}$ and let $\\sigma_{k}, k=1,2, \\ldots, n$ denote the $k$ th elementary symmetric polynomial in $x_{1}, \\ldots, x_{n}$. The given system can be written as $S_{k}=a^{k}, k=1, \\ldots, n$. Using Newton's formulas  $$ k \\sigma_{k}=S_{1} \\sigma_{k-1}-S_{2} \\sigma_{k-2}+\\cdots+(-1)^{k} S_{k-1} \\sigma_{1}+(-1)^{k-1} S_{k}, \\quad k=1,2, \\ldots, n $$  the system easily leads to $\\sigma_{1}=a$ and $\\sigma_{k}=0$ for $k=2, \\ldots, n$. By Vieta's formulas, $x_{1}, x_{2}, \\ldots, x_{n}$ are the roots of the polynomial $x^{n}-a x^{n-1}$, i.e., $a, 0,0, \\ldots, 0$ in some order. Remark. This solution does not use the assumption that the $x_{j}$ 's are real.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. $(\\mathbf{C Z S} 2)^{\\mathrm{IMO1}} A B C D$ is a parallelogram; $A B=a, A D=1, \\alpha$ is the size of $\\angle D A B$, and the three angles of the triangle $A B D$ are acute. Prove that the four circles $K_{A}, K_{B}, K_{C}, K_{D}$, each of radius 1, whose centers are the vertices $A, B, C, D$, cover the parallelogram if and only if $a \\leq$ $\\cos \\alpha+\\sqrt{3} \\sin \\alpha$.", "solution": "8. The circles $K_{A}, K_{B}, K_{C}, K_{D}$ cover the parallelogram if and only if for every point $X$ inside the parallelogram, the length of one of the segments $X A, X B, X C, X D$ does not exceed 1. Let $O$ and $r$ be the center and radius of the circumcircle of $\\triangle A B D$. For every point $X$ inside $\\triangle A B D$, it holds that $X A \\leq r$ or $X B \\leq r$ or $X D \\leq r$. Similarly, for $X$ inside $\\triangle B C D, X B \\leq r$ or $X C \\leq r$ or $X D \\leq r$. Hence $K_{A}, K_{B}, K_{C}, K_{D}$ cover the parallelogram if and only if $r \\leq 1$, which is equivalent to $\\angle A B D \\geq 30^{\\circ}$. However, this last is exactly equivalent to $a=A B=2 r \\sin \\angle A D B \\leq 2 \\sin \\left(\\alpha+30^{\\circ}\\right)=\\sqrt{3} \\sin \\alpha+\\cos \\alpha$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1967", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (CZS 3) The circle $k$ and its diameter $A B$ are given. Find the locus of the centers of circles inscribed in the triangles having one vertex on $A B$ and two other vertices on $k$.", "solution": "9. The incenter of any such triangle lies inside the circle $k$. We shall show that every point $S$ interior to the circle $S$ is the incenter of one such triangle. If $S$ lies on the segment $A B$, then it is obviously the incenter of an isosceles triangle inscribed in $k$ that has $A B$ as an axis of symmetry. Let us now suppose $S$ does not lie on $A B$. Let $X$ and $Y$ be the intersection points of lines $A S$ and $B S$ with $k$, and let $Z$ be the foot of the perpendicular from $S$ to $A B$. Since the quadrilateral $B Z S X$ is cyclic, we have $\\angle Z X S=$ $\\angle A B S=\\angle S X Y$ and analogously $\\angle Z Y S=\\angle S Y X$, which implies that $S$ is the incenter of $\\triangle X Y Z$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (SWE 2) Two ships sail on the sea with constant speeds and fixed directions. It is known that at 9:00 the distance between them was 20 miles; at 9:35, 15 miles; and at 9:55, 13 miles. At what moment were the ships the smallest distance from each other, and what was that distance?", "solution": "1. Since the ships are sailing with constant speeds and directions, the second ship is sailing at a constant speed and direction in reference to the first ship. Let $A$ be the constant position of the first ship in this frame. Let $B_{1}$, $B_{2}, B_{3}$, and $B$ on line $b$ defining the trajectory of the ship be positions of the second ship with respect to the first ship at 9:00, 9:35, 9:55, and at the moment the two ships were closest. Then we have the following equations for distances (in miles):  $$ \\begin{gathered} A B_{1}=20, \\quad A B_{2}=15, \\quad A B_{3}=13 \\\\ B_{1} B_{2}: B_{2} B_{3}=7: 4, \\quad A B_{i}^{2}=A B^{2}+B B_{i}^{2} \\end{gathered} $$  Since $B B_{1}>B B_{2}>B B_{3}$, it follows that $\\mathcal{B}\\left(B_{3}, B, B_{2}, B_{1}\\right)$ or $\\mathcal{B}\\left(B, B_{3}, B_{2}\\right.$, $B_{1}$ ). We get a system of three quadratic equations with three unknowns: $A B, B B_{3}$ and $B_{3} B_{2}$ ( $B B_{3}$ being negative if $\\mathcal{B}\\left(B_{3}, B, B_{1}, B_{2}\\right)$, positive otherwise). This can be solved by eliminating $A B$ and then $B B_{3}$. The unique solution ends up being  $$ A B=12, \\quad B B_{3}=5, \\quad B_{3} B_{2}=4 $$  and consequently, the two ships are closest at 10:20 when they are at a distance of 12 miles.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (ROM 4) Consider two segments of length $a, b(a>b)$ and a segment of length $c=\\sqrt{a b}$. (a) For what values of $a / b$ can these segments be sides of a triangle? (b) For what values of $a / b$ is this triangle right-angled, obtuse-angled, or acute-angled?", "solution": "10. (a) Let us set $k=a / b>1$. Then $a=k b$ and $c=\\sqrt{k} b$, and $a>c>b$. The segments $a, b, c$ form a triangle if and only if $k<\\sqrt{k}+1$, which holds if and only if $1<k<\\frac{3+\\sqrt{5}}{2}$. (b) The triangle is right-angled if and only if $a^{2}=b^{2}+c^{2} \\Leftrightarrow k^{2}=k+1 \\Leftrightarrow$ $k=\\frac{1+\\sqrt{5}}{2}$. Also, it is acute-angled if and only if $k^{2}<k+1 \\Leftrightarrow 1<$ $k<\\frac{1+\\sqrt{5}}{2}$ and obtuse-angled if $\\frac{1+\\sqrt{5}}{2}<k<\\frac{3+\\sqrt{5}}{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (ROM 6) Find all solutions $\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)$ of the equation $1+\\frac{1}{x_{1}}+\\frac{x_{1}+1}{x_{1} x_{2}}+\\frac{\\left(x_{1}+1\\right)\\left(x_{2}+1\\right)}{x_{1} x_{2} x_{3}}+\\cdots+\\frac{\\left(x_{1}+1\\right) \\cdots\\left(x_{n-1}+1\\right)}{x_{1} x_{2} \\cdots x_{n}}=0$.", "solution": "11. Introducing $y_{i}=\\frac{1}{x_{i}}$, we transform our equation to  $$ \\begin{aligned} 0 & =1+y_{1}+\\left(1+y_{1}\\right) y_{2}+\\cdots+\\left(1+y_{1}\\right) \\cdots\\left(1+y_{n-1}\\right) y_{n} \\\\ & =\\left(1+y_{1}\\right)\\left(1+y_{2}\\right) \\cdots\\left(1+y_{n}\\right) \\end{aligned} $$  The solutions are $n$-tuples $\\left(y_{1}, \\ldots, y_{n}\\right)$ with $y_{i} \\neq 0$ for all $i$ and $y_{j}=-1$ for at least one index $j$. Returning to $x_{i}$, we conclude that the solutions are all the $n$-tuples $\\left(x_{1}, \\ldots, x_{n}\\right)$ with $x_{i} \\neq 0$ for all $i$, and $x_{j}=-1$ for at least one index $j$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (POL 1) If $a$ and $b$ are arbitrary positive real numbers and $m$ an integer, prove that  $$ \\left(1+\\frac{a}{b}\\right)^{m}+\\left(1+\\frac{b}{a}\\right)^{m} \\geq 2^{m+1} $$", "solution": "12. The given inequality is equivalent to $(a+b)^{m} / b^{m}+(a+b)^{m} / a^{m} \\geq 2^{m+1}$, which can be rewritten as  $$ \\frac{1}{2}\\left(\\frac{1}{a^{m}}+\\frac{1}{b^{m}}\\right) \\geq\\left(\\frac{2}{a+b}\\right)^{m} $$  Since $f(x)=1 / x^{m}$ is a convex function for every $m \\in \\mathbb{Z}$, the last inequality immediately follows from Jensen's inequality $(f(a)+f(b)) / 2 \\geq$ $f((a+b) / 2)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (POL 5) Given two congruent triangles $A_{1} A_{2} A_{3}$ and $B_{1} B_{2} B_{3}\\left(A_{i} A_{k}=\\right.$ $B_{i} B_{k}$ ), prove that there exists a plane such that the orthogonal projections of these triangles onto it are congruent and equally oriented.", "solution": "13. Translating one of the triangles if necessary, we may assume w.l.o.g. that $B_{1} \\equiv A_{1}$. We also assume that $B_{2} \\not \\equiv A_{2}$ and $B_{3} \\not \\equiv A_{3}$, since the result is obvious otherwise. There exists a plane $\\pi$ through $A_{1}$ that is parallel to both $A_{2} B_{2}$ and $A_{3} B_{3}$. Let $A_{2}^{\\prime}, A_{3}^{\\prime}, B_{2}^{\\prime}, B_{3}^{\\prime}$ denote the orthogonal projections of $A_{2}, A_{3}, B_{2}, B_{3}$ onto $\\pi$, and let $h_{2}, h_{3}$ denote the distances of $A_{2}, B_{2}$ and of $A_{3}, B_{3}$ from $\\pi$. By the Pythagorean theorem, $A_{2}^{\\prime} A_{3}^{\\prime 2}=A_{2} A_{3}^{2}-\\left(h_{2}+h_{3}\\right)^{2}=B_{2} B_{3}^{2}-$ $\\left(h_{2}+h_{3}\\right)^{2}=B_{2}^{\\prime}{B_{3}^{\\prime}}^{2}$, and similarly $A_{1} A_{2}^{\\prime}=A_{1} B_{2}^{\\prime}$ and $A_{1} A_{3}^{\\prime}=A_{1} B_{3}^{\\prime}$; hence $\\triangle A_{1} A_{2}^{\\prime} A_{3}^{\\prime}$ and $\\triangle A_{1} B_{2}^{\\prime} B_{3}^{\\prime}$ are congruent. If these two triangles are equally oriented, then we have finished. Otherwise, they are symmetric with respect to some line $a$ passing through $A_{1}$, and consequently the projections of the triangles $A_{1} A_{2} A_{3}$ and $A_{1} B_{2} B_{3}$ onto the plane through $a$ perpendicular to $\\pi$ coincide.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (BUL 5) A line in the plane of a triangle $A B C$ intersects the sides $A B$ and $A C$ respectively at points $X$ and $Y$ such that $B X=C Y$. Find the locus of the center of the circumcircle of triangle $X A Y$.", "solution": "14. Let $O, D, E$ be the circumcenter of $\\triangle A B C$ and the midpoints of $A B$ and $A C$, and given arbitrary $X \\in A B$ and $Y \\in A C$ such that $B X=C Y$, let $O_{1}, D_{1}, E_{1}$ be the circumcenter of $\\triangle A X Y$ and the midpoints of $A X$ and $A Y$, respectively. Since $A D=A B / 2$ and $A D_{1}=A X / 2$, it follows that $D D_{1}=B X / 2$ and similarly $E E_{1}=C Y / 2$. Hence $O_{1}$ is at the same distance $B X / 2=C Y / 2$ from the lines $O D$ and $O E$ and lies on the halfline bisector $l$ of $\\angle D O E$. If we let $X, Y$ vary along the segments $A B$ and $A C$, we obtain that the locus of $O_{1}$ is the segment $O P$, where $P \\in l$ is a point at distance $\\min (A B, A C) / 2$ from $O D$ and $O E$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (GBR 1) ${ }^{\\mathrm{IMO} 6}$ Let $[x]$ denote the integer part of $x$, i.e., the greatest integer not exceeding $x$. If $n$ is a positive integer, express as a simple function of $n$ the sum  $$ \\left[\\frac{n+1}{2}\\right]+\\left[\\frac{n+2}{4}\\right]+\\cdots+\\left[\\frac{n+2^{i}}{2^{i+1}}\\right]+\\cdots $$", "solution": "15. Set  $$ f(n)=\\left[\\frac{n+1}{2}\\right]+\\left[\\frac{n+2}{4}\\right]+\\cdots+\\left[\\frac{n+2^{i}}{2^{i+1}}\\right]+\\ldots $$  We prove by induction that $f(n)=n$. This obviously holds for $n=1$. Let us assume that $f(n-1)=n-1$. Define  $$ g(i, n)=\\left[\\frac{n+2^{i}}{2^{i+1}}\\right]-\\left[\\frac{n-1+2^{i}}{2^{i+1}}\\right] . $$  We have that $f(n)-f(n+1)=\\sum_{i=0}^{\\infty} g(i, n)$. We also note that $g(i, n)=1$ if and only if $2^{i+1} \\mid n+2^{i}$; otherwise, $g(i, n)=0$. The divisibility $2^{i+1} \\mid$ $n+2^{i}$ is equivalent to $2^{i} \\mid n$ and $2^{i+1} \\nmid n$, which for a given $n$ holds for exactly one $i \\in \\mathbb{N}_{0}$. Thus it follows that $f(n)-f(n-1)=1 \\Rightarrow f(n)=n$. The proof by induction is now complete. Second solution. It is easy to show that $[x+1 / 2]=[2 x]-[x]$ for $x \\in \\mathbb{R}$. Now $f(x)=([x]-[x / 2])+([x / 2]-[x / 4])+\\cdots=[x]$. Hence, $f(n)=n$ for all $n \\in \\mathbb{N}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (GBR 3) A polynomial $p(x)=a_{0} x^{k}+a_{1} x^{k-1}+\\cdots+a_{k}$ with integer coefficients is said to be divisible by an integer $m$ if $p(x)$ is divisible by $m$ for all integers $x$. Prove that if $p(x)$ is divisible by $m$, then $k!a_{0}$ is also divisible by $m$. Also prove that if $a_{0}, k, m$ are nonnegative integers for which $k!a_{0}$ is divisible by $m$, there exists a polynomial $p(x)=a_{0} x^{k}+\\cdots+$ $a_{k}$ divisible by $m$.", "solution": "16. We shall prove the result by induction on $k$. It trivially holds for $k=0$. Assume that the statement is true for some $k-1$, and let $p(x)$ be a polynomial of degree $k$. Let us set $p_{1}(x)=p(x+1)-p(x)$. Then $p_{1}(x)$ is a polynomial of degree $k-1$ with leading coefficient $k a_{0}$. Also, $m \\mid p_{1}(x)$ for all $x \\in \\mathbb{Z}$ and hence by the inductive assumption $m \\mid(k-1)!\\cdot k a_{0}=k!a_{0}$, which completes the induction. On the other hand, for any $a_{0}, k$ and $m \\mid k!a_{0}, p(x)=k!a_{0}\\binom{x}{k}$ is a polynomial with leading coefficient $a_{0}$ that is divisible by $m$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (GBR 4) Given a point $O$ and lengths $x, y, z$, prove that there exists an equilateral triangle $A B C$ for which $O A=x, O B=y, O C=z$, if and only if $x+y \\geq z, y+z \\geq x, z+x \\geq y$ (the points $O, A, B, C$ are coplanar).", "solution": "17. Let there be given an equilateral triangle $A B C$ and a point $O$ such that $O A=x, O B=y, O C=z$. Let $X$ be the point in the plane such that $\\triangle C X B$ and $\\triangle C O A$ are congruent and equally oriented. Then $B X=x$ and the triangle $X O C$ is equilateral, which implies $O X=z$. Thus we have a triangle $O B X$ with $B X=x, B O=y$, and $O X=z$. Conversely, given a triangle $O B X$ with $B X=x, B O=y$ and $O X=z$ it is easy to construct the triangle $A B C$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (ITA 2) If an acute-angled triangle $A B C$ is given, construct an equilateral triangle $A^{\\prime} B^{\\prime} C^{\\prime}$ in space such that lines $A A^{\\prime}, B B^{\\prime}, C C^{\\prime}$ pass through a given point.", "solution": "18. The required construction is not feasible. In fact, let us consider the special case $\\angle B O C=135^{\\circ}, \\angle A O C=120^{\\circ}, \\angle A O B=90^{\\circ}$, where $A A^{\\prime} \\cap B B^{\\prime} \\cap$ $C C^{\\prime}=\\{O\\}$. Denoting $O A^{\\prime}, O B^{\\prime}, O C^{\\prime}$ by $a, b, c$ respectively we obtain the system of equations $a^{2}+b^{2}=a^{2}+c^{2}+a c=b^{2}+c^{2}+\\sqrt{2} b c$. Assuming w.l.o.g. $c=1$ we easily obtain $a^{3}-a^{2}-a-1=0$, which is an irreducible equation of third degree. By a known theorem, its solution $a$ is not constructible by ruler and compass.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (ITA 5) We are given a fixed point on the circle of radius 1, and going from this point along the circumference in the positive direction on curved distances $0,1,2, \\ldots$ from it we obtain points with abscisas $n=0,1,2, \\ldots$ respectively. How many points among them should we take to ensure that some two of them are less than the distance $1 / 5$ apart?", "solution": "19. We shall denote by $d_{n}$ the shortest curved distance from the initial point to the $n$th point in the positive direction. The sequence $d_{n}$ goes as follows: $0,1,2,3,4,5,6,0.72,1.72, \\ldots, 5.72,0.43,1.43, \\ldots, 5.43,0.15=d_{19}$. Hence the required number of points is 20 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (ROM 5) ${ }^{\\mathrm{IMO} 1}$ Prove that there exists a unique triangle whose side lengths are consecutive natural numbers and one of whose angles is twice the measure of one of the others.", "solution": "2. The sides $a, b, c$ of a triangle $A B C$ with $\\angle A B C=2 \\angle B A C$ satisfy $b^{2}=$ $a(a+c)$ (this statement is the lemma in (SL98-7)). Taking into account the remaining condition that $a, b, c$ are consecutive integers with $a<b$, we obtain three cases: (i) $a=n, b=n+1, c=n+2$. We get the equation $(n+1)^{2}=n(2 n+2)$, giving us $(a, b, c)=(1,2,3)$, which is not a valid triangle. (ii) $a=n, b=n+2, c=n+1$. We get $(n+2)^{2}=n(2 n+1) \\Rightarrow$ $(n-4)(n+1)=0$, giving us the triangle $(a, b, c)=(4,6,5)$. (iii) $a=n+1, b=n+2, c=n$. We get $(n+2)^{2}=(n+1)(2 n+1) \\Rightarrow$ $n^{2}-n-3=0$, which has no positive integer solutions for $n$. Hence, the only solution is the triangle with sides of lengths 4,5 , and 6 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. (CZS 1) Given $n(n \\geq 3)$ points in space such that every three of them form a triangle with one angle greater than or equal to $120^{\\circ}$, prove that these points can be denoted by $A_{1}, A_{2}, \\ldots, A_{n}$ in such a way that for each $i, j, k, 1 \\leq i<j<k \\leq n$, angle $A_{i} A_{j} A_{k}$ is greater than or equal to $120^{\\circ}$.", "solution": "20. Let us denote the points $A_{1}, A_{2}, \\ldots, A_{n}$ in such a manner that $A_{1} A_{n}$ is a diameter of the set of given points, and $A_{1} A_{2} \\leq A_{1} A_{3} \\leq \\cdots \\leq A_{1} A_{n}$. Since for each $1<i<n$ it holds that $A_{1} A_{i}<A_{1} A_{n}$, we have $\\angle A_{i} A_{1} A_{n}<120^{\\circ}$ and hence $\\angle A_{i} A_{1} A_{n}<60^{\\circ}$ (otherwise, all angles in $\\triangle A_{1} A_{i} A_{n}$ are less than $120^{\\circ}$ ). It follows that for all $1<i<j \\leq n$, $\\angle A_{i} A_{1} A_{j}<120^{\\circ}$. Consequently, the angle in the triangle $A_{1} A_{i} A_{j}$ that is at least $120^{\\circ}$ must be $\\angle A_{1} A_{i} A_{j}$. Moreover, for any $1<i<j<k \\leq n$ it holds that $\\angle A_{i} A_{j} A_{k} \\geq \\angle A_{1} A_{j} A_{k}-\\angle A_{1} A_{j} A_{i}>120^{\\circ}-60^{\\circ}=60^{\\circ}$ (because $\\angle A_{1} A_{j} A_{i}<60^{\\circ}$ ); hence $\\angle A_{i} A_{j} A_{k} \\geq 120^{\\circ}$. This proves that the denotation is correct. Remark. It is easy to show that the diameter is unique. Hence the denotation is also unique.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. (CZS 2) Let $a_{0}, a_{1}, \\ldots, a_{k}(k \\geq 1)$ be positive integers. Find all positive integers $y$ such that  $$ a_{0}\\left|y ; \\quad\\left(a_{0}+a_{1}\\right)\\right|\\left(y+a_{1}\\right) ; \\ldots ; \\quad\\left(a_{0}+a_{n}\\right) \\mid\\left(y+a_{n}\\right) . $$", "solution": "21. The given conditions are equivalent to $y-a_{0}$ being divisible by $a_{0}, a_{0}+$ $a_{1}, a_{0}+a_{2}, \\ldots, a_{0}+a_{n}$, i.e., to $y=k\\left[a_{0}, a_{0}+a_{1}, \\ldots, a_{0}+a_{n}\\right]+a_{0}, k \\in \\mathbb{N}_{0}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "22", "problem_type": null, "exam": "IMO", "problem": "22. (CZS 3) ${ }^{\\mathrm{IMO} 2}$ Find all positive integers $x$ for which $p(x)=x^{2}-10 x-22$, where $p(x)$ denotes the product of the digits of $x$.", "solution": "22. It can be shown by induction on the number of digits of $x$ that $p(x) \\leq x$ for all $x \\in \\mathbb{N}$. It follows that $x^{2}-10 x-22 \\leq x$, which implies $x \\leq 12$.  Since $0<x^{2}-10 x-22=(x-12)(x+2)+2$, one easily obtains $x \\geq 12$. Now one can directly check that $x=12$ is indeed a solution, and thus the only one.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "23", "problem_type": null, "exam": "IMO", "problem": "23. (CZS 4) Find all complex numbers $m$ such that polynomial  $$ x^{3}+y^{3}+z^{3}+m x y z $$  can be represented as the product of three linear trinomials.", "solution": "23. We may assume w.l.o.g. that in all the factors the coefficient of $x$ is 1 . Suppose that $x+a y+b z$ is one of the linear factors of $p(x, y, z)=x^{3}+$ $y^{3}+z^{3}+m x y z$. Then $p(x)$ is 0 at every point $(x, y, z)$ with $z=-a x-b y$. Hence $x^{3}+y^{3}+(-a x-b y)^{3}+m x y(-a x-b y)=\\left(1-a^{3}\\right) x^{3}-(3 a b+$ $m)(a x+b y) x y+\\left(1-b^{3}\\right) y^{3} \\equiv 0$. This is obviously equivalent to $a^{3}=b^{3}=1$ and $m=-3 a b$, from which it follows that $m \\in\\left\\{-3,-3 \\omega,-3 \\omega^{2}\\right\\}$, where $\\omega=\\frac{-1+i \\sqrt{3}}{2}$. Conversely, for each of the three possible values for $m$ there are exactly three possibilities $(a, b)$. Hence $-3,-3 \\omega,-3 \\omega^{2}$ are the desired values.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "24", "problem_type": null, "exam": "IMO", "problem": "24. (MON 1) Find the number of all $n$-digit numbers for which some fixed digit stands only in the $i$ th $(1<i<n)$ place and the last $j$ digits are distinct. ${ }^{3}$", "solution": "24. If the $i$ th digit is 0 , then the result is $9^{k-j} 9$ !/ $(10-j)$ ! if $i>k-j$ and $9^{k-j-1} 9$ !/ $(9-j)$ ! otherwise. If the $i$ th digit is not 0 , then the above results are multiplied by 8 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "25", "problem_type": null, "exam": "IMO", "problem": "25. (MON 2) Given $k$ parallel lines and a few points on each of them, find the number of all possible triangles with vertices at these given points. ${ }^{4}$", "solution": "25. The answer is  $$ \\sum_{1 \\leq p<q<r \\leq k} n_{p} n_{q} n_{r}+\\sum_{1 \\leq p<q \\leq k}\\left[n_{p}\\binom{n_{q}}{2}+n_{q}\\binom{n_{p}}{2}\\right] . $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "26", "problem_type": null, "exam": "IMO", "problem": "26. (GDR) ${ }^{\\text {IMO5 }}$ Let $a>0$ be a real number and $f(x)$ a real function defined on all of $\\mathbb{R}$, satisfying for all $x \\in \\mathbb{R}$,  $$ f(x+a)=\\frac{1}{2}+\\sqrt{f(x)-f(x)^{2}} . $$  (a) Prove that the function $f$ is periodic; i.e., there exists $b>0$ such that for all $x, f(x+b)=f(x)$. (b) Give an example of such a nonconstant function for $a=1$.  [^2]", "solution": "26. (a) We shall show that the period of $f$ is $2 a$. From $(f(x+a)-1 / 2)^{2}=$ $f(x)-f(x)^{2}$ we obtain  $$ \\left(f(x)-f(x)^{2}\\right)+\\left(f(x+a)-f(x+a)^{2}\\right)=\\frac{1}{4} $$  Subtracting the above relation for $x+a$ in place of $x$ we get $f(x)-$ $f(x)^{2}=f(x+2 a)-f(x+2 a)^{2}$, which implies $(f(x)-1 / 2)^{2}=$ $(f(x+2 a)-1 / 2)^{2}$. Since $f(x) \\geq 1 / 2$ holds for all $x$ by the condition of the problem, we conclude that $f(x+2 a)=f(x)$. (b) The following function, as is directly verified, satisfies the conditions:  $$ f(x)=\\left\\{\\begin{array}{cl} 1 / 2 & \\text { if } 2 n \\leq x<2 n+1, \\\\ 1 & \\text { if } 2 n+1 \\leq x<2 n+2, \\end{array} \\text { for } n=0,1,2, \\ldots\\right. $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (POL 4) ${ }^{\\mathrm{IMO} 4}$ Prove that in any tetrahedron there is a vertex such that the lengths of its sides through that vertex are sides of a triangle.", "solution": "3. A triangle cannot be formed out of three lengths if and only if one of them is larger than the sum of the other two. Let us assume this is the case for all triplets of edges out of each vertex in a tetrahedron $A B C D$. Let w.l.o.g. $A B$ be the largest edge of the tetrahedron. Then $A B \\geq A C+A D$ and $A B \\geq B C+B D$, from which it follows that $2 A B \\geq A C+A D+B C+B D$. This implies that either $A B \\geq A C+B C$ or $A B \\geq A D+B D$, contradicting the triangle inequality. Hence the three edges coming out of at least one of the vertices $A$ and $B$ form a triangle. Remark. The proof can be generalized to prove that in a polyhedron with only triangular surfaces there is a vertex such that the edges coming out of this vertex form a triangle.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (BUL 2) ${ }^{\\mathrm{IMO} 3}$ Let $a, b, c$ be real numbers. Prove that the system of equations  $$ \\left\\{\\begin{array}{r} a x_{1}^{2}+b x_{1}+c=x_{2} \\\\ a x_{2}^{2}+b x_{2}+c=x_{3} \\\\ \\cdots \\cdots \\cdots \\cdots \\\\ a x_{n-1}^{2}+b x_{n-1}+c=x_{n} \\\\ a x_{n}^{2}+b x_{n}+c=x_{1} \\end{array}\\right. $$  has a unique real solution if and only if $(b-1)^{2}-4 a c=0$. Remark. It is assumed that $a \\neq 0$.", "solution": "4. We will prove the equivalence in the two directions separately: $(\\Rightarrow)$ Suppose $\\left\\{x_{1}, \\ldots, x_{n}\\right\\}$ is the unique solution of the equation. Since $\\left\\{x_{n}, x_{1}, x_{2} \\ldots, x_{n-1}\\right\\}$ is also a solution, it follows that $x_{1}=x_{2}=\\cdots=$ $x_{n}=x$ and the system of equations reduces to a single equation $a x^{2}+$ $(b-1) x+c=0$. For the solution for $x$ to be unique the discriminant $(b-1)^{2}-4 a c$ of this quadratic equation must be 0 . $(\\Leftarrow)$ Assume $(b-1)^{2}-4 a c=0$. Adding up the equations, we get  $$ \\sum_{i=1}^{n} f\\left(x_{i}\\right)=0, \\quad \\text { where } \\quad f(x)=a x^{2}+(b-1) x+c $$  But by the assumed condition, $f(x)=a\\left(x+\\frac{b-1}{2 a}\\right)^{2}$. Hence we must have $f\\left(x_{i}\\right)=0$ for all $i$, and $x_{i}=-\\frac{b-1}{2 a}$, which is indeed a solution.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (BUL 5) Let $h_{n}$ be the apothem (distance from the center to one of the sides) of a regular $n$-gon $(n \\geq 3)$ inscribed in a circle of radius $r$. Prove the inequality  $$ (n+1) h_{n+1}-n h_{n}>r . $$  Also prove that if $r$ on the right side is replaced with a greater number, the inequality will not remain true for all $n \\geq 3$.", "solution": "5. We have $h_{k}=r \\cos (\\pi / k)$ for all $k \\in \\mathbb{N}$. Using $\\cos x=1-2 \\sin ^{2}(x / 2)$ and $\\cos x=2 /\\left(1+\\tan ^{2}(x / 2)\\right)-1$ and $\\tan x>x>\\sin x$ for all $0<x<\\pi / 2$, it suffices to prove  $$ \\begin{aligned} & (n+1)\\left(1-2 \\frac{\\pi^{2}}{4(n+1)^{2}}\\right)-n\\left(\\frac{2}{1+\\pi^{2} /\\left(4 n^{2}\\right)}-1\\right)>1 \\\\ \\Leftrightarrow & 1+2 n\\left(1-\\frac{1}{1+\\pi^{2} /\\left(4 n^{2}\\right)}\\right)-\\frac{\\pi^{2}}{2(n+1)}>1 \\\\ \\Leftrightarrow & 1+\\frac{\\pi^{2}}{2}\\left(\\frac{1}{n+\\pi^{2} /(4 n)}-\\frac{1}{n+1}\\right)>1, \\end{aligned} $$  where the last inequality holds because $\\pi^{2}<4 n$. It is also apparent that as $n$ tends to infinity the term in parentheses tends to 0 , and hence it is not possible to strengthen the bound. This completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (HUN 1) If $a_{i}(i=1,2, \\ldots, n)$ are distinct non-zero real numbers, prove that the equation  $$ \\frac{a_{1}}{a_{1}-x}+\\frac{a_{2}}{a_{2}-x}+\\cdots+\\frac{a_{n}}{a_{n}-x}=n $$  has at least $n-1$ real roots.", "solution": "6. We define $f(x)=\\frac{a_{1}}{a_{1}-x}+\\frac{a_{2}}{a_{2}-x}+\\cdots+\\frac{a_{n}}{a_{n}-x}$. Let us assume w.l.o.g. $a_{1}<a_{2}<\\cdots<a_{n}$. We note that for all $1 \\leq i<n$ the function $f$ is continuous in the interval $\\left(a_{i}, a_{i+1}\\right)$ and satisfies $\\lim _{x \\rightarrow a_{i}} f(x)=-\\infty$ and $\\lim _{x \\rightarrow a_{i+1}} f(x)=\\infty$. Hence the equation $f(x)=n$ will have a real solution in each of the $n-1$ intervals $\\left(a_{i}, a_{i+1}\\right)$. Remark. In fact, this equation has exactly $n$ solutions, and hence they are all real. Moreover, the solutions are distinct if all $a_{i}$ are of the same sign, since $x=0$ is an evident solution.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (HUN 5) Prove that the product of the radii of three circles exscribed to a given triangle does not exceed $\\frac{3 \\sqrt{3}}{8}$ times the product of the side lengths of the triangle. When does equality hold?", "solution": "7. Let $r_{a}, r_{b}, r_{c}$ denote the radii of the exscribed circles corresponding to the sides of lengths $a, b, c$ respectively, and $R, p$ and $S$ denote the circumradius, semiperimeter, and area of the given triangle. It is well-known that $r_{a}(p-a)=r_{b}(p-b)=r_{c}(p-c)=S=\\sqrt{p(p-a)(p-b)(p-c)}=\\frac{a b c}{4 R}$. Hence, the desired inequality $r_{a} r_{b} r_{c} \\leq \\frac{3 \\sqrt{3}}{8} a b c$ reduces to $p \\leq \\frac{3 \\sqrt{3}}{2} R$, which is by the law of sines equivalent to  $$ \\sin \\alpha+\\sin \\beta+\\sin \\gamma \\leq \\frac{3 \\sqrt{3}}{2} $$  This inequality immediately follows from Jensen's inequality, since the sine is concave on $[0, \\pi]$. Equality holds if and only if the triangle is equilateral.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (ROM 2) Given an oriented line $\\Delta$ and a fixed point $A$ on it, consider all trapezoids $A B C D$ one of whose bases $A B$ lies on $\\Delta$, in the positive direction. Let $E, F$ be the midpoints of $A B$ and $C D$ respectively. Find the loci of vertices $B, C, D$ of trapezoids that satisfy the following: (i) $|A B| \\leq a \\quad(a$ fixed); (ii) $|E F|=l \\quad(l$ fixed); (iii) the sum of squares of the nonparallel sides of the trapezoid is constant. Remark. The constants are chosen so that such trapezoids exist.", "solution": "8. Let $G$ be the point such that $B C D G$ is a parallelogram and let $H$ be the midpoint of $A G$. Obviously $H E F D$ is also a parallelogram, and thus $D H=E F=l$. If $A D^{2}+B C^{2}=m^{2}$ is fixed, then from the Stewart theorem we have  $$ D H^{2}=\\frac{2 D A^{2}+2 D G^{2}-A G^{2}}{4}=\\frac{2 m^{2}-A G^{2}}{4} $$  which is fixed. Thus $G$ and $H$ are fixed points, and from here the locus of $D$ is a circle with center $H$ and radius $l$. The locus of $B$ is the segment (GI], where $I \\in \\Delta$ is a point in the positive direction such that $A I=a$. Finally, the locus of $C$ is a region of the plane consisting of a rectangle sandwiched between two semicircles of radius $l$ centered at points $H$ and $H^{\\prime}$, where $H^{\\prime}$ is a point such that $\\overrightarrow{I H^{\\prime}}=\\overrightarrow{G H}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1968", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (ROM 3) Let $A B C$ be an arbitrary triangle and $M$ a point inside it. Let $d_{a}, d_{b}, d_{c}$ be the distances from $M$ to sides $B C, C A, A B ; a, b, c$ the lengths of the sides respectively, and $S$ the area of the triangle $A B C$. Prove the inequality  $$ a b d_{a} d_{b}+b c d_{b} d_{c}+c a d_{c} d_{a} \\leq \\frac{4 S^{2}}{3} $$  Prove that the left-hand side attains its maximum when $M$ is the centroid of the triangle.", "solution": "9. We note that $S_{a}=a d_{a} / 2, S_{b}=b d_{b} / 2$, and $S_{c}=c d_{c} / 2$ are the areas of the triangles $M B C, M C A$, and $M A B$ respectively. The desired inequality now follows from  $$ S_{a} S_{b}+S_{b} S_{c}+S_{c} S_{a} \\leq \\frac{1}{3}\\left(S_{a}+S_{b}+S_{c}\\right)^{2}=\\frac{S^{2}}{3} $$  Equality holds if and only if $S_{a}=S_{b}=S_{c}$, which is equivalent to $M$ being the centroid of the triangle.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1969", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. Prove that there exist infinitely many natural numbers $a$ with the following property: the number $z=n^{4}+a$ is not prime for any natural number $n$.", "solution": "1. Set $a=4 m^{4}$, where $m \\in \\mathbb{N}$ and $m>1$. We then have $z=n^{4}+4 m^{4}=$ $\\left(n^{2}+2 m^{2}\\right)^{2}-(2 m n)^{2}=\\left(n^{2}+2 m^{2}+2 m n\\right)\\left(n^{2}+2 m^{2}-2 m n\\right)$. Since $n^{2}+2 m^{2}-2 m n=(n-m)^{2}+m^{2} \\geq m^{2}>1$, it follows that $z$ must be composite. Thus we have found infinitely many $a$ that satisfy the condition of the problem.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1969", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. Let $a_{1}, a_{2}, \\ldots, a_{n}$ be real constants and  $$ y(x)=\\cos \\left(a_{1}+x\\right)+\\frac{\\cos \\left(a_{2}+x\\right)}{2}+\\frac{\\cos \\left(a_{3}+x\\right)}{2^{2}}+\\cdots+\\frac{\\cos \\left(a_{n}+x\\right)}{2^{n-1}} . $$  If $x_{1}, x_{2}$ are real and $y\\left(x_{1}\\right)=y\\left(x_{2}\\right)=0$, prove that $x_{1}-x_{2}=m \\pi$ for some integer $m$.", "solution": "2. Using $\\cos (a+x)=\\cos a \\cos x-\\sin a \\sin x$, we obtain $f(x)=A \\sin x+$ $B \\cos x$ where $A=-\\sin a_{1}-\\sin a_{2} / 2-\\cdots-\\sin a_{n} / 2^{n-1}$ and $B=\\cos a_{1}+$ $\\cos a_{2} / 2+\\cdots+\\cos a_{n} / 2^{n-1}$. Numbers $A$ and $B$ cannot both be equal to 0 , for otherwise $f$ would be identically equal to 0 , while on the other hand, we have $f\\left(-a_{1}\\right)=\\cos \\left(a_{1}-a_{1}\\right)+\\cos \\left(a_{2}-a_{1}\\right) / 2+\\cdots+\\cos \\left(a_{n}-a_{1}\\right) / 2^{n-1} \\geq$ $1-1 / 2-\\cdots-1 / 2^{n-1}=1 / 2^{n-1}>0$. Setting $A=C \\cos \\phi$ and $B=C \\sin \\phi$, where $C \\neq 0$ (such $C$ and $\\phi$ always exist), we get $f(x)=C \\sin (x+\\phi)$. It follows that the zeros of $f$ are of the form $x_{0} \\in-\\phi+\\pi \\mathbb{Z}$, from which $f\\left(x_{1}\\right)=f\\left(x_{2}\\right) \\Rightarrow x_{1}-x_{2}=m \\pi$ immediately follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1969", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. Find conditions on the positive real number $a$ such that there exists a tetrahedron $k$ of whose edges $(k=1,2,3,4,5)$ have length $a$, and the other $6-k$ edges have length 1 .  Second Day (July 11)", "solution": "3. We have several cases: $1^{\\circ} k=1$. W.l.o.g. let $A B=a$ and the remaining segments have length 1. Let $M$ be the midpoint of $C D$. Then $A M=B M=\\sqrt{3} / 2(\\triangle C D A$ and $\\triangle C D B$ are equilateral) and $0<A B<A M+B M=\\sqrt{3}$, i.e., $0<a<\\sqrt{3}$. It is evident that all values of $a$ within this interval are realizable. $2^{\\circ} k=2$. We have two subcases. First, let $A C=A D=a$. Let $M$ be the midpoint of $C D$. We have $C D=1, A M=\\sqrt{a^{2}-1 / 4}$, and $B M=\\sqrt{3} / 2$. Then we have $1-$ $\\sqrt{3} / 2=A B-B M<A M<A B+B M=1+\\sqrt{3} / 2$, which gives us $\\sqrt{2-\\sqrt{3}}<a<\\sqrt{2+\\sqrt{3}}$. Second, let $A B=C D=a$. Let $M$ be the midpoint of $C D$. From $\\triangle M A B$ we get $a<\\sqrt{2}$. Thus, from $\\sqrt{2-\\sqrt{3}}<\\sqrt{2}<\\sqrt{2+\\sqrt{3}}$ it follows that the required condition in this case is $0<a<\\sqrt{2+\\sqrt{3}}$. All values for $a$ in this range are realizable. $3^{\\circ} k=3$. We show that such a tetrahedron exists for all $a$. Assume $a>1$. Assume $A B=A C=A D=a$. Varying $A$ along the line perpendicular to the plane $B C D$ and through the center of $\\triangle B C D$ we achieve all values of $a>1 / \\sqrt{3}$. For $a<1 / \\sqrt{3}$ we can observe a similar tetrahedron with three edges of length $1 / a$ and three of length 1 and proceed as before. $4^{\\circ} k=4$. By observing the similar tetrahedron we reduce this case to $k=2$ with length $1 / a$ instead of $a$. Thus we get $a>\\sqrt{2-\\sqrt{3}}$. $5^{\\circ} k=5$. We reduce to $k=1$ and get $a>1 / \\sqrt{3}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1969", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. Let $A B$ be a diameter of a circle $\\gamma$. A point $C$ different from $A$ and $B$ is on the circle $\\gamma$. Let $D$ be the projection of the point $C$ onto the line $A B$. Consider three other circles $\\gamma_{1}, \\gamma_{2}$, and $\\gamma_{3}$ with the common tangent $A B: \\gamma_{1}$ inscribed in the triangle $A B C$, and $\\gamma_{2}$ and $\\gamma_{3}$ tangent to both (the segment) $C D$ and $\\gamma$. Prove that $\\gamma_{1}, \\gamma_{2}$, and $\\gamma_{3}$ have two common tangents.", "solution": "4. Let $O$ be the midpoint of $A B$, i.e., the center of $\\gamma$. Let $O_{1}, O_{2}$, and $O_{3}$ respectively be the centers of $\\gamma_{1}, \\gamma_{2}$, and $\\gamma_{3}$ and let $r_{1}, r_{2}, r_{3}$ respectively be the radii of $\\gamma_{1}, \\gamma_{2}$ and $\\gamma_{3}$. Let $C_{1}, C_{2}$, and $C_{3}$ respectively be the points of tangency of $\\gamma_{1}, \\gamma_{2}$ and $\\gamma_{3}$ with $A B$. Let $D_{2}$ and $D_{3}$ respectively be the points of tangency of $\\gamma_{2}$ and $\\gamma_{3}$ with $C D$. Finally, let $G_{2}$ and $G_{3}$ respectively be the points of tangency of $\\gamma_{2}$ and $\\gamma_{3}$ with $\\gamma$. We have $\\mathcal{B}\\left(G_{2}, O_{2}, O\\right)$, $G_{2} O_{2}=O_{2} D_{2}$, and $G_{2} O=O B$. Hence, $G_{2}, D_{2}, B$ are collinear. Similarly, $G_{3}, D_{3}, A$ are collinear. It follows that $A G_{2} D_{2} D$ and $B G_{3} D_{3} D$ are cyclic, since $\\angle A G_{2} D_{2}=\\angle D_{2} D A=\\angle D_{3} D B=\\angle B G_{3} D_{3}=90^{\\circ}$. Hence $B C_{2}^{2}=B D_{2} \\cdot B G_{2}=B D \\cdot B A=B C^{2} \\Rightarrow B C_{2}=B C$ and hence $A C_{2}=A B-B C$. Similarly, $A C_{3}=A C$. We thus have $A C_{1}=(A C+A B-B C) / 2=\\left(A C_{3}+A C_{2}\\right) / 2$. Hence, $C_{1}$ is the midpoint of $C_{2} C_{3}$. We also have $r_{2}+r_{3}=C_{2} C_{3}=A C+B C-A B=2 r_{1}$, from which it follows that $O_{1}, O_{2}, O_{3}$ are collinear. Second solution. We shall prove the statement for arbitrary points $A, B, C$ on $\\gamma$. Let us apply the inversion $\\psi$ with respect to the circle $\\gamma_{1}$. We denote by $\\widehat{X}$ the image of an object $X$ under $\\psi$. Also, $\\psi$ maps lines $B C, C A, A B$ onto circles $\\widehat{a}, \\widehat{b}, \\widehat{c}$, respectively. Circles $\\widehat{a}, \\widehat{b}, \\widehat{c}$ pass through the center $O_{1}$ of $\\gamma_{1}$ and have radii equal to the radius of $\\widehat{\\gamma}$. Let $P, Q, R$ be the centers of $\\widehat{a}, \\widehat{b}, \\widehat{c}$ respectively. The line $C D$ maps onto a circle $k$ through $\\widehat{C}$ and $O_{1}$ that is perpendicular to $\\widehat{c}$. Therefore its center $K$ lies in the intersection of the tangent $t$ to $\\widehat{c}$ and the line $P Q$ (which bisects $\\left.\\widehat{C} O_{1}\\right)$. Let $O$ be a point such that $R O_{1} K O$ is a parallelogram and $\\gamma_{2}^{\\prime}, \\gamma_{3}^{\\prime}$ the circles centered at $O$ tangent to $k$. It is easy to see that $\\gamma_{2}^{\\prime}$ and $\\gamma_{3}^{\\prime}$ are also tangent to $\\widehat{c}$, since $O R$ and $O K$ have lengths equal to the radii of $k$ and $\\widehat{c}$. Hence $\\gamma_{2}^{\\prime}$ and $\\gamma_{3}^{\\prime}$ are the images of $\\gamma_{2}$ and $\\gamma_{3}$ under $\\psi$. Moreover, since $Q \\widehat{A} O K$ and $P \\widehat{B} O K$ are parallelograms and $Q, P, K$ are collinear, it follows that $\\widehat{A}, \\widehat{B}, O$ are also collinear. Hence the centers of $\\gamma_{1}, \\gamma_{2}, \\gamma_{3}$ are collinear, lying on the line $O_{1} O$, and the statement follows.  Third solution. Moreover, the statement holds for an arbitrary point $D \\in B C$. Let $E, F, G, H$ be the points of tangency of $\\gamma_{2}$ with $A B, C D$ and of $\\gamma_{3}$ with $A B, C D$, respectively. Let $O_{i}$ be the center of $\\gamma_{i}, i=1,2,3$. As is shown in the third solution of (SL93-3), $E F$ and $G H$ meet at $O_{1}$. Hence the problem of proving the collinearity of $O_{1}, O_{2}, O_{3}$ reduces to the following simple problem:  Let $D, E, F, G, H$ be points such that $D \\in E G, F \\in D H$ and $D E=D F, D G=D H$. Let $O_{1}, O_{2}, O_{3}$ be points such that $\\angle O_{2} E D=$ $\\angle O_{2} F D=90^{\\circ}, \\angle O_{3} G D=\\angle O_{3} H D=90^{\\circ}$, and $O_{1}=E F \\cap G H$. Then $O_{1}, O_{2}, O_{3}$ are collinear. Let $K_{2}=D O_{2} \\cap E F$ and $K_{3}=D O_{3} \\cap G H$. Then $O_{2} K_{2} / O_{2} D=$ $D K_{3} / D O_{3}=K_{2} O_{1} / D O_{3}$ and hence by Thales' theorem $O_{1} \\in O_{2} O_{3}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1969", "tier": "T0", "problem_label": "5", "problem_type": "Geometry", "exam": "IMO", "problem": "5. Given $n$ points in the plane such that no three of them are collinear, prove that one can find at least $\\binom{n-3}{2}$ convex quadrilaterals with their vertices at these points.", "solution": "5. We first prove the following lemma.  Lemma. If of five points in a plane no three belong to a single line, then there exist four that are the vertices of a convex quadrilateral.  Proof. If the convex hull of the five points $A, B, C, D, E$ is a pentagon or a quadrilateral, the statement automatically holds. If the convex hull is a triangle, then w.l.o.g. let $\\triangle A B C$ be that triangle and $D, E$ points in its interior. Let the line $D E$ w.l.o.g. intersect $[A B]$ and $[A C]$. Then $B, C, D, E$ form the desired quadrilateral. We now observe each quintuplet of points within the set. There are $\\binom{n}{5}$ such quintuplets, and for each of them there is at least one quadruplet of points forming a convex quadrilateral. Each quadruplet, however, will be counted up to $n-4$ times. Hence we have found at least $\\frac{1}{n-4}\\binom{n}{5}$ quadruplets. Since $\\frac{1}{n-4}\\binom{n}{5} \\geq\\binom{ n-3}{2} \\Leftrightarrow(n-5)(n-6)(n+8) \\geq 0$, which always holds, it follows that we have found at least $\\binom{n-3}{2}$ desired quadruplets of points.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1969", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. Under the conditions $x_{1}, x_{2}>0, x_{1} y_{1}>z_{1}^{2}$, and $x_{2} y_{2}>z_{2}^{2}$, prove the inequality  $$ \\frac{8}{\\left(x_{1}+x_{2}\\right)\\left(y_{1}+y_{2}\\right)-\\left(z_{1}+z_{2}\\right)^{2}} \\leq \\frac{1}{x_{1} y_{1}-z_{1}^{2}}+\\frac{1}{x_{2} y_{2}-z_{2}^{2}} $$", "solution": "6. Define $u_{1}=\\sqrt{x_{1} y_{1}}+z_{1}, u_{2}=\\sqrt{x_{2} y_{2}}+z_{2}, v_{1}=\\sqrt{x_{1} y_{1}}-z_{1}$, and $v_{2}=$ $\\sqrt{x_{2} y_{2}}-z_{2}$. By expanding both sides of the equation we can easily verify $\\left(x_{1}+x_{2}\\right)\\left(y_{1}+y_{2}\\right)-\\left(z_{1}+z_{2}\\right)^{2}=\\left(u_{1}+u_{2}\\right)\\left(v_{1}+v_{2}\\right)+\\left(\\sqrt{x_{1} y_{2}}-\\sqrt{x_{2} y_{1}}\\right)^{2} \\geq$ $\\left(u_{1}+u_{2}\\right)\\left(v_{1}+v_{2}\\right)$. Since $x_{i} y_{i}-z_{i}^{2}=u_{i} v_{i}$ for $i=1,2$, it suffices to prove  $$ \\begin{aligned} & \\frac{8}{\\left(u_{1}+u_{2}\\right)\\left(v_{1}+v_{2}\\right)} \\leq \\frac{1}{u_{1} v_{1}}+\\frac{1}{u_{2} v_{2}} \\\\ \\Leftrightarrow & 8 u_{1} u_{2} v_{1} v_{2} \\leq\\left(u_{1}+u_{2}\\right)\\left(v_{1}+v_{2}\\right)\\left(u_{1} v_{1}+u_{2} v_{2}\\right) \\end{aligned} $$  which trivially follows from the AM-GM inequalities $2 \\sqrt{u_{1} u_{2}} \\leq u_{1}+u_{2}$, $2 \\sqrt{v_{1} v_{2}} \\leq v_{1}+v_{2}$ and $2 \\sqrt{u_{1} v_{1} u_{2} v_{2}} \\leq u_{1} v_{1}+u_{2} v_{2}$. Equality holds if and only if $x_{1} y_{2}=x_{2} y_{1}, u_{1}=u_{2}$ and $v_{1}=v_{2}$, i.e. if and only if $x_{1}=x_{2}, y_{1}=y_{2}$ and $z_{1}=z_{2}$. Second solution. Let us define $f(x, y, z)=1 /\\left(x y-z^{2}\\right)$. The problem actually states that  $$ 2 f\\left(\\frac{x_{1}+x_{2}}{2}, \\frac{y_{1}+y_{2}}{2}, \\frac{z_{1}+z_{2}}{2}\\right) \\leq f\\left(x_{1}, y_{1}, z_{1}\\right)+f\\left(x_{2}, y_{2}, z_{2}\\right) $$  i.e., that the function $f$ is convex on the set $D=\\left\\{(x, y, z) \\in \\mathbb{R}^{2} \\mid x y-\\right.$ $\\left.z^{2}>0\\right\\}$. It is known that a twice continuously differentiable function $f\\left(t_{1}, t_{2}, \\ldots, t_{n}\\right)$ is convex if and only if its Hessian $\\left[f_{i j}^{\\prime \\prime}\\right]_{i, j=1}^{n}$ is positive semidefinite, or equivalently, if its principal minors $D_{k}=\\operatorname{det}\\left[f_{i j}^{\\prime \\prime}\\right]_{i, j=1}^{k}, k=$ $1,2, \\ldots, n$, are nonnegative. In the case of our $f$ this is directly verified: $D_{1}=2 y^{2} /\\left(x y-z^{2}\\right)^{3}, D_{2}=3 x y+z^{2} /\\left(x y-z^{2}\\right)^{5}, D_{3}=6 /\\left(x y-z^{2}\\right)^{6}$ are obviously positive.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1970", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (BEL 3) Consider a regular $2 n$-gon and the $n$ diagonals of it that pass through its center. Let $P$ be a point of the inscribed circle and let $a_{1}, a_{2}, \\ldots, a_{n}$ be the angles in which the diagonals mentioned are visible from the point $P$. Prove that  $$ \\sum_{i=1}^{n} \\tan ^{2} a_{i}=2 n \\frac{\\cos ^{2} \\frac{\\pi}{2 n}}{\\sin ^{4} \\frac{\\pi}{2 n}} $$", "solution": "1. Denote respectively by $R$ and $r$ the radii of the circumcircle and incircle, by $A_{1}, \\ldots, A_{n}, B_{1}, \\ldots, B_{n}$, the vertices of the $2 n$-gon and by $O$ its center. Let $P^{\\prime}$ be the point symmetric to $P$ with respect to $O$. Then $A_{i} P^{\\prime} B_{i} P$ is a parallelogram, and applying cosine theorem on triangles $A_{i} B_{i} P$ and $P P^{\\prime} B_{i}$ yields  $$ \\begin{aligned} 4 R^{2} & =P A_{i}^{2}+P B_{i}^{2}-2 P A_{i} \\cdot P B_{i} \\cos a_{i} \\\\ 4 r^{2} & =P B_{i}^{2}+P^{\\prime} B_{i}^{2}-2 P B_{i} \\cdot P^{\\prime} B_{i} \\cos \\angle P B_{i} P^{\\prime} \\end{aligned} $$  Since $A_{i} P^{\\prime} B_{i} P$ is a parallelogram, we have that $P^{\\prime} B_{i}=P A_{i}$ and $\\angle P B_{i} P^{\\prime}=\\pi-a_{i}$. Subtracting the expression for $4 r^{2}$ from the one for $4 R^{2}$ yields $4\\left(R^{2}-r^{2}\\right)=-4 P A_{i} \\cdot P B_{i} \\cos a_{i}=-8 S_{\\triangle A_{i} B_{i} P} \\cot a_{i}$, hence we conclude that  $$ \\tan ^{2} a_{i}=\\frac{4 S_{\\triangle A_{i} B_{i} P}^{2}}{\\left(R^{2}-r^{2}\\right)^{2}} $$  Denote by $M_{i}$ the foot of the perpendicular from $P$ to $A_{i} B_{i}$ and let $m_{i}=$ $P M_{i}$. Then $S_{\\triangle A_{i} B_{i} P}=R m_{i}$. Substituting this into (1) and adding up these relations for $i=1,2, \\ldots, n$, we obtain  $$ \\sum_{i=1}^{n} \\tan ^{2} a_{i}=\\frac{4 R^{2}}{\\left(R^{2}-r^{2}\\right)^{2}}\\left(\\sum_{i=1}^{n} m_{i}^{2}\\right) $$  Note that all the points $M_{i}$ lie on a circle with diameter $O P$ and form a regular $n$-gon. Denote its center by $F$. We have that $m_{i}^{2}=\\left\\|\\overrightarrow{P M_{i}}\\right\\|^{2}=$ $\\left\\|\\overrightarrow{F M_{i}}-\\overrightarrow{F P}\\right\\|^{2}=\\left\\|\\overrightarrow{F M}_{i}^{2}\\right\\|+\\left\\|\\overrightarrow{F P}^{2}\\right\\|-2\\left\\langle\\overrightarrow{F M_{i}}, \\overrightarrow{F P}\\right\\rangle=r^{2} / 2-2\\left\\langle\\overrightarrow{F M_{i}}, \\overrightarrow{F P}\\right\\rangle$. From this it follows that $\\sum_{i=1}^{n} m_{i}^{2}=2 n(r / 2)^{2}-2 \\sum_{i=1}^{n}\\left\\langle\\overrightarrow{F M_{i}}, \\overrightarrow{F P}\\right\\rangle=$ $2 n(r / 2)^{2}-2\\left\\langle\\sum_{i=1}^{n} \\overrightarrow{F M_{i}}, \\overrightarrow{F P}\\right\\rangle=2 n(r / 2)^{2}$, because $\\sum_{i=1}^{n} \\overrightarrow{F M_{i}}=\\overrightarrow{0}$. Thus  $$ \\sum_{i=1}^{n} \\tan ^{2} a_{i}=\\frac{4 R^{2}}{\\left(R^{2}-r^{2}\\right)^{2}} 2 n\\left(\\frac{r}{2}\\right)^{2}=2 n \\frac{(r / R)^{2}}{\\left(1-(r / R)^{2}\\right)^{2}}=2 n \\frac{\\cos ^{2} \\frac{\\pi}{2 n}}{\\sin ^{4} \\frac{\\pi}{2 n}} $$  Remark. For $n=1$ there is no regular 2-gon. However, if we think of a 2-gon as a line segment, the statement will remain true.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1970", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (SWE 4) ${ }^{\\mathrm{IMO} 3}$ Let $1=a_{0} \\leq a_{1} \\leq a_{2} \\leq \\cdots \\leq a_{n} \\leq \\cdots$ be a sequence of real numbers. Consider the sequence $b_{1}, b_{2}, \\ldots$ defined by:  $$ b_{n}=\\sum_{k=1}^{n}\\left(1-\\frac{a_{k-1}}{a_{k}}\\right) \\frac{1}{\\sqrt{a_{k}}} $$  Prove that: (a) For all natural numbers $n, 0 \\leq b_{n}<2$. (b) Given an arbitrary $0 \\leq b<2$, there is a sequence $a_{0}, a_{1}, \\ldots, a_{n}, \\ldots$ of the above type such that $b_{n}>b$ is true for infinitely many natural numbers $n$.", "solution": "10. (a) Since $a_{n-1}<a_{n}$, we have  $$ \\begin{aligned} \\left(1-\\frac{a_{k-1}}{a_{k}}\\right) \\frac{1}{\\sqrt{a_{k}}} & =\\frac{a_{k}-a_{k-1}}{a_{k}^{3 / 2}} \\\\ & \\leq \\frac{2\\left(\\sqrt{a_{k}}-\\sqrt{a_{k-1}}\\right) \\sqrt{a_{k}}}{a_{k} \\sqrt{a_{k-1}}}=2\\left(\\frac{1}{\\sqrt{a_{k-1}}}-\\frac{1}{\\sqrt{a_{k}}}\\right) . \\end{aligned} $$  Summing up all these inequalities for $k=1,2, \\ldots, n$ we obtain  $$ b_{n} \\leq 2\\left(\\frac{1}{\\sqrt{a_{0}}}-\\frac{1}{\\sqrt{a_{n}}}\\right)<2 $$  (b) Choose a real number $q>1$, and let $a_{k}=q^{k}, k=1,2, \\ldots$. Then $\\left(1-a_{k-1} / a_{k}\\right) / \\sqrt{a_{k}}=(1-1 / q) / q^{k / 2}$, and consequently  $$ b_{n}=\\left(1-\\frac{1}{q}\\right) \\sum_{k=1}^{n} \\frac{1}{q^{k / 2}}=\\frac{\\sqrt{q}+1}{q}\\left(1-\\frac{1}{q^{n / 2}}\\right) . $$  Since $(\\sqrt{q}+1) / q$ can be arbitrarily close to 2 , one can set $q$ such that $(\\sqrt{q}+1) / q>b$. Then $b_{n} \\geq b$ for all sufficiently large $n$. Second solution. (a) Note that  $$ b_{n}=\\sum_{k=1}^{n}\\left(1-\\frac{a_{k-1}}{a_{k}}\\right) \\frac{1}{\\sqrt{a_{k}}}=\\sum_{k=1}^{n}\\left(a_{k}-a_{k-1}\\right) \\cdot \\frac{1}{a_{k}^{3 / 2}} $$  hence $b_{n}$ represents exactly the lower Darboux sum for the function $f(x)=x^{-3 / 2}$ on the interval $\\left[a_{0}, a_{n}\\right]$. Then $b_{n} \\leq \\int_{a_{0}}^{a_{n}} x^{-3 / 2} d x<$ $\\int_{1}^{+\\infty} x^{-3 / 2} d x=2$. (b) For each $b<2$ there exists a number $\\alpha>1$ such that $\\int_{1}^{\\alpha} x^{-3 / 2} d x>$ $b+(2-b) / 2$. Now, by Darboux's theorem, there exists an array $1=$ $a_{0} \\leq a_{1} \\leq \\cdots \\leq a_{n}=\\alpha$ such that the corresponding Darboux sums are arbitrarily close to the value of the integral. In particular, there is an array $a_{0}, \\ldots, a_{n}$ with $b_{n}>b$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1970", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (SWE 6) Let $P, Q, R$ be polynomials and let $S(x)=P\\left(x^{3}\\right)+x Q\\left(x^{3}\\right)+$ $x^{2} R\\left(x^{3}\\right)$ be a polynomial of degree $n$ whose roots $x_{1}, \\ldots, x_{n}$ are distinct. Construct with the aid of the polynomials $P, Q, R$ a polynomial $T$ of degree $n$ that has the roots $x_{1}^{3}, x_{2}^{3}, \\ldots, x_{n}^{3}$.", "solution": "11. Let $S(x)=\\left(x-x_{1}\\right)\\left(x-x_{2}\\right) \\cdots\\left(x-x_{n}\\right)$. We have $x^{3}-x_{i}^{3}=\\left(x-x_{i}\\right)(\\omega x-$ $\\left.x_{i}\\right)\\left(\\omega^{2} x-x_{i}\\right)$, where $\\omega$ is a primitive third root of 1 . Multiplying these equalities for $i=1, \\ldots, n$ we obtain  $$ T\\left(x^{3}\\right)=\\left(x^{3}-x_{1}^{3}\\right)\\left(x^{3}-x_{2}^{3}\\right) \\cdots\\left(x^{3}-x_{n}^{3}\\right)=S(x) S(\\omega x) S\\left(\\omega^{2} x\\right) $$  Since $S(\\omega x)=P\\left(x^{3}\\right)+\\omega x Q\\left(x^{3}\\right)+\\omega^{2} x^{2} R\\left(x^{3}\\right)$ and $S\\left(\\omega^{2} x\\right)=P\\left(x^{3}\\right)+$ $\\omega^{2} x Q\\left(x^{3}\\right)+\\omega x^{2} R\\left(x^{3}\\right)$, the above expression reduces to  $$ T\\left(x^{3}\\right)=P^{3}\\left(x^{3}\\right)+x^{3} Q^{3}\\left(x^{3}\\right)+x^{6} R^{3}\\left(x^{3}\\right)-3 P\\left(x^{3}\\right) Q\\left(x^{3}\\right) R\\left(x^{3}\\right) $$  Therefore the zeros of the polynomial  $$ T(x)=P^{3}(x)+x Q^{3}(x)+x^{2} R^{3}(x)-3 P(x) Q(x) R(x) $$  are exactly $x_{1}^{3}, \\ldots, x_{n}^{3}$. It is easily verified that $\\operatorname{deg} T=\\operatorname{deg} S=n$, and hence $T$ is the desired polynomial.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1970", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (USS 4) ${ }^{\\mathrm{IMO} 6}$ We are given 100 points in the plane, no three of which are on the same line. Consider all triangles that have all vertices chosen from the 100 given points. Prove that at most $70 \\%$ of these triangles are acute angled.", "solution": "12. Lemma. Five points are given in the plane such that no three of them are collinear. Then there are at least three triangles with vertices at these points that are not acute-angled. Proof. We consider three cases, according to whether the convex hull of these points is a triangle, quadrilateral, or pentagon. (i) Let a triangle $A B C$ be the convex hull and two other points $D$ and $E$ lie inside the triangle. At least two of the triangles $A D B, B D C$ and $C D A$ have obtuse angles at the point $D$. Similarly, at least two of the triangles $A E B, B E C$ and $C E A$ are obtuse-angled. Thus there are at least four non-acute-angled triangles. (ii) Suppose that $A B C D$ is the convex hull and that $E$ is a point of its interior. At least one angle of the quadrilateral is not acute, determining one non-acute-angled triangle. Also, the point $E$ lies in the interior of either $\\triangle A B C$ or $\\triangle C D A$ hence, as in the previous case, it determines another two obtuse-angled triangles. (iii) It is easy to see that at least two of the angles of the pentagon are not acute. We may assume that these two angles are among the angles corresponding to vertices $A, B$, and $C$. Now consider the quadrilateral $A C D E$. At least one its angles is not acute. Hence, there are at least three triangles that are not acute-angled.  Now we consider all combinations of 5 points chosen from the given 100. There are $\\binom{100}{5}$ such combinations, and for each of them there are at least three non-acute-angled triangles with vertices in it. On the other hand, vertices of each of the triangles are counted $\\binom{97}{2}$ times. Hence there are at least $3\\binom{100}{5} /\\binom{97}{2}$ non-acute-angled triangles with vertices in the given 100 points. Since the number of all triangles with vertices in the given points is $\\binom{100}{3}$, the ratio between the number of acute-angled triangles and the number of all triangles cannot be greater than  $$ 1-\\frac{3\\binom{100}{5}}{\\binom{97}{2}\\binom{100}{3}}=0.7 $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1970", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (ROM 1) ${ }^{\\mathrm{IMO} 2}$ Let $a$ and $b$ be the bases of two number systems and let  $$ \\begin{array}{ll} A_{n}={\\overline{x_{1} x_{2} \\ldots x_{n}}}^{(a)}, & A_{n+1}={\\overline{x_{0} x_{1} x_{2} \\ldots x_{n}}}^{(a)}, \\\\ B_{n}={\\overline{x_{1} x_{2} \\ldots x_{n}}}^{(b)}, & B_{n+1}={\\overline{x_{0} x_{1} x_{2} \\ldots x_{n}}}^{(b)}, \\end{array} $$  be numbers in the number systems with respective bases $a$ and $b$, so that $x_{0}, x_{1}, x_{2}, \\ldots, x_{n}$ denote digits in the number system with base $a$ as well as in the number system with base $b$. Suppose that neither $x_{0}$ nor $x_{1}$ is zero. Prove that $a>b$ if and only if  $$ \\frac{A_{n}}{A_{n+1}}<\\frac{B_{n}}{B_{n+1}} . $$", "solution": "2. Suppose that $a>b$. Consider the polynomial $P(X)=x_{1} X^{n-1}+x_{2} X^{n-2}+$ $\\cdots+x_{n-1} X+x_{n}$. We have $A_{n}=P(a), B_{n}=P(b), A_{n+1}=x_{0} a^{n}+$ $P(a)$, and $B_{n+1}=x_{0} b^{n}+P(b)$. Now $A_{n} / A_{n+1}<B_{n} / B_{n+1}$ becomes $P(a) /\\left(x_{0} a^{n}+P(a)\\right)<P(b) /\\left(x_{0} b^{n}+P(b)\\right)$, i.e.,  $$ b^{n} P(a)<a^{n} P(b) $$  Since $a>b$, we have that $a^{i}>b^{i}$ and hence $x_{i} a^{n} b^{n-i} \\geq x_{i} b^{n} a^{n-i}$ (also, for $i \\geq 1$ the inequality is strict). Summing up all these inequalities for $i=1, \\ldots, n$ we get $a^{n} P(b)>b^{n} P(a)$, which completes the proof for $a>b$.  On the other hand, for $a<b$ we analogously obtain the opposite inequality $A_{n} / A_{n+1}>B_{n} / B_{n+1}$, while for $a=b$ we have equality. Thus $A_{n} / A_{n+1}<$ $B_{n} / B_{n+1} \\Leftrightarrow a>b$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1970", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (BUL 6) ${ }^{\\mathrm{IMO}}$ In the tetrahedron $S A B C$ the angle $B S C$ is a right angle, and the projection of the vertex $S$ to the plane $A B C$ is the intersection of the altitudes of the triangle $A B C$. Let $z$ be the radius of the inscribed circle of the triangle $A B C$. Prove that  $$ S A^{2}+S B^{2}+S C^{2} \\geq 18 z^{2} $$", "solution": "3 . We shall use the following lemma Lemma. If an altitude of a tetrahedron passes through the orthocenter of the opposite side, then each of the other altitudes possesses the same property. Proof. Denote the tetrahedron by $S A B C$ and let $a=B C, b=C A$, $c=A B, m=S A, n=S B, p=S C$. It is enough to prove that an altitude passes through the orthocenter of the opposite side if and only if $a^{2}+m^{2}=b^{2}+n^{2}=c^{2}+p^{2}$. Suppose that the foot $S^{\\prime}$ of the altitude from $S$ is the orthocenter of $A B C$. Then $S S^{\\prime} \\perp A B C \\Rightarrow S B^{2}-S C^{2}=S^{\\prime} B^{2}-S^{\\prime} C^{2}$. But from $A S^{\\prime} \\perp B C$ it follows that $A B^{2}-A C^{2}=S^{\\prime} B^{2}-S^{\\prime} C^{2}$. From these two equalities it can be concluded that $n^{2}-p^{2}=c^{2}-b^{2}$, or equivalently, $n^{2}+b^{2}=c^{2}+p^{2}$. Analogously, $a^{2}+m^{2}=n^{2}+b^{2}$, so we have proved the first part of the equivalence. Now suppose that $a^{2}+m^{2}=b^{2}+n^{2}=c^{2}+p^{2}$. Defining $S^{\\prime}$ as before, we get $n^{2}-p^{2}=S^{\\prime} B^{2}-S^{\\prime} C^{2}$. From the condition $n^{2}-p^{2}=c^{2}-b^{2}$ $\\left(\\Leftrightarrow b^{2}+n^{2}=c^{2}+p^{2}\\right.$ ) we conclude that $A S^{\\prime} \\perp B C$. In the same way $C S^{\\prime} \\perp A B$, which proves that $S^{\\prime}$ is the orthocenter of $\\triangle A B C$. The lemma is thus proven. Now using the lemma it is easy to see that if one of the angles at $S$ is right, than so are the others. Indeed, suppose that $\\angle A S B=\\pi / 2$. From the lemma we have that the altitude from $C$ passes through the orthocenter of $\\triangle A S B$, which is $S$, so $C S \\perp A S B$ and $\\angle C S A=\\angle C S B=\\pi / 2$. Therefore $m^{2}+n^{2}=c^{2}, n^{2}+p^{2}=a^{2}$, and $p^{2}+m^{2}=b^{2}$, so it follows that $m^{2}+n^{2}+p^{2}=\\left(a^{2}+b^{2}+c^{2}\\right) / 2$. By the inequality between the arithmetic and quadric means, we have that $\\left(a^{2}+b^{2}+c^{2}\\right) / 2 \\geq 2 s^{2} / 3$, where $s$ denotes the semiperimeter of $\\triangle A B C$. It remains to be shown that $2 s^{2} / 3 \\geq 18 r^{2}$. Since $S_{\\triangle A B C}=s r$, this is equivalent to $2 s^{4} / 3 \\geq$ $18 S_{A B C}^{2}=18 s(s-a)(s-b)(s-c)$ by Heron's formula. This reduces to $s^{3} \\geq 27(s-a)(s-b)(s-c)$, which is an obvious consequence of the AM-GM mean inequality. Remark. In the place of the lemma one could prove that the opposite edges of the tetrahedron are mutually perpendicular and proceed in the same way.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1970", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (CZS 1) $)^{\\mathrm{IMO} 4}$ For what natural numbers $n$ can the product of some of the numbers $n, n+1, n+2, n+3, n+4, n+5$ be equal to the product of the remaining ones?", "solution": "4. Suppose that $n$ is such a natural number. If a prime number $p$ divides any of the numbers $n, n+1, \\ldots, n+5$, then it must divide another one of them, so the only possibilities are $p=2,3,5$. Moreover, $n+1, n+2, n+3, n+4$ have no prime divisors other than 2 and 3 (if some prime number greater than 3 divides one of them, then none of the remaining numbers can have that divisor). Since two of these numbers are odd, they must be powers of  3 (greater than 1). However, there are no two powers of 3 whose difference is 2 . Therefore there is no such natural number $n$. Second solution. Obviously, none of $n, n+1, \\ldots, n+5$ is divisible by 7; hence they form a reduced system of residues. We deduce that $n(n+$ 1) $\\cdots(n+5) \\equiv 1 \\cdot 2 \\cdots 6 \\equiv-1(\\bmod 7)$. If $\\{n, \\ldots, n+5\\}$ can be partitioned into two subsets with the same products, both congruent to, say, $p$ modulo 7 , then $p^{2} \\equiv-1(\\bmod 7)$, which is impossible. Remark. Erdős has proved that a set $n, n+1, \\ldots, n+m$ of consecutive natural numbers can never be partitioned into two subsets with equal products of elements.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1970", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (CZS 3) Let $M$ be an interior point of the tetrahedron $A B C D$. Prove that  $$ \\begin{aligned} & \\overrightarrow{M A} \\operatorname{vol}(M B C D)+\\overrightarrow{M B} \\operatorname{vol}(M A C D) \\\\ & \\quad+\\overrightarrow{M C} \\operatorname{vol}(M A B D)+\\overrightarrow{M D} \\operatorname{vol}(M A B C)=0 \\end{aligned} $$  ( $\\operatorname{vol}(P Q R S)$ denotes the volume of the tetrahedron $P Q R S)$.", "solution": "5. Denote respectively by $A_{1}, B_{1}, C_{1}$ and $D_{1}$ the points of intersection of the lines $A M, B M, C M$, and $D M$ with the opposite sides of the tetrahedron. Since $\\operatorname{vol}(M B C D)=\\operatorname{vol}(A B C D) \\overrightarrow{M A_{1}} / \\overrightarrow{A A_{1}}$, the relation we have to prove is equivalent to  $$ \\overrightarrow{M A} \\cdot \\frac{\\overrightarrow{M A_{1}}}{\\overrightarrow{A A_{1}}}+\\overrightarrow{M B} \\cdot \\frac{\\overrightarrow{M B_{1}}}{\\overrightarrow{B B_{1}}}+\\overrightarrow{M C} \\cdot \\frac{\\overrightarrow{M C_{1}}}{\\overrightarrow{C C_{1}}}+\\overrightarrow{M D} \\cdot \\frac{\\overrightarrow{M D_{1}}}{\\overrightarrow{D D_{1}}}=0 $$  There exist unique real numbers $\\alpha, \\beta, \\gamma$, and $\\delta$ such that $\\alpha+\\beta+\\gamma+\\delta=1$ and for every point $O$ in space  $$ \\overrightarrow{O M}=\\alpha \\overrightarrow{O A}+\\beta \\overrightarrow{O B}+\\gamma \\overrightarrow{O C}+\\delta \\overrightarrow{O D} $$  (This follows easily from $\\overrightarrow{O M}=\\overrightarrow{O A}+\\overrightarrow{A M}=\\overrightarrow{O A}+k \\overrightarrow{A B}+l \\overrightarrow{A C}+m \\overrightarrow{A D}=$ $\\overrightarrow{A B}+k(\\overrightarrow{O B}-\\overrightarrow{O A})+l(\\overrightarrow{O C}-\\overrightarrow{O A})+m(\\overrightarrow{O D}-\\overrightarrow{O A})$ for some $k, l, m \\in \\mathbb{R}$.) Further, from the condition that $A_{1}$ belongs to the plane $B C D$ we obtain for every $O$ in space the following equality for some $\\beta^{\\prime}, \\gamma^{\\prime}, \\delta^{\\prime}$ :  $$ \\overrightarrow{O A_{1}}=\\beta^{\\prime} \\overrightarrow{O B}+\\gamma^{\\prime} \\overrightarrow{O C}+\\delta^{\\prime} \\overrightarrow{O D} $$  However, for $\\lambda=\\overrightarrow{M A_{1}} / \\overrightarrow{A A_{1}}, \\overrightarrow{O M}=\\lambda \\overrightarrow{O A}+(1-\\lambda) \\overrightarrow{O A_{1}}$; hence substituting (2) and (3) in this expression and equating coefficients for $\\overrightarrow{O A}$ we obtain $\\lambda=\\overrightarrow{M A_{1}} / \\overrightarrow{A A_{1}}=\\alpha$. Analogously, $\\beta=\\overrightarrow{M B_{1}} / \\overrightarrow{B B_{1}}, \\gamma=\\overrightarrow{M C_{1}} / \\overrightarrow{C C_{1}}$, and $\\delta=\\overrightarrow{M D_{1}} / \\overrightarrow{D D_{1}}$; hence (1) follows immediately for $O=M$. Remark. The statement of the problem actually follows from the fact that $M$ is the center of mass of the system with masses $\\operatorname{vol}(M B C D)$, $\\operatorname{vol}(M A C D), \\operatorname{vol}(M A B D), \\operatorname{vol}(M A B C)$ at $A, B, C, D$ respectively. Our proof is actually a formal verification of this fact.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1970", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (FRA 1) In the triangle $A B C$ let $B^{\\prime}$ and $C^{\\prime}$ be the midpoints of the sides $A C$ and $A B$ respectively and $H$ the foot of the altitude passing through the vertex $A$. Prove that the circumcircles of the triangles $A B^{\\prime} C^{\\prime}, B C^{\\prime} H$, and $B^{\\prime} C H$ have a common point $I$ and that the line $H I$ passes through the midpoint of the segment $B^{\\prime} C^{\\prime}$.", "solution": "6. Let $F$ be the midpoint of $B^{\\prime} C^{\\prime}, A^{\\prime}$ the midpoint of $B C$, and $I$ the intersection point of the line $H F$ and the circle circumscribed about $\\triangle B H C^{\\prime}$. Denote by $M$ the intersection point of the line $A A^{\\prime}$ with the circumscribed circle about the triangle $A B C$. Triangles $H B^{\\prime} C^{\\prime}$ and $A B C$ are similar. Since $\\angle C^{\\prime} I F=\\angle A B C=\\angle A^{\\prime} M C, \\angle C^{\\prime} F I=\\angle A A^{\\prime} B=\\angle M A^{\\prime} C$, $2 C^{\\prime} F=C^{\\prime} B^{\\prime}$, and $2 A^{\\prime} C=C B$, it follows that $\\triangle C^{\\prime} I B^{\\prime} \\sim \\triangle C M B$, hence $\\angle F I B^{\\prime}=\\angle A^{\\prime} M B=\\angle A C B$. Now one concludes that $I$ belongs to the circumscribed circles of $\\triangle A B^{\\prime} C^{\\prime}$ (since $\\left.\\angle C^{\\prime} I B^{\\prime}=180^{\\circ}-\\angle C^{\\prime} A B^{\\prime}\\right)$ and $\\triangle H C B^{\\prime}$. Second Solution. We denote the angles of $\\triangle A B C$ by $\\alpha, \\beta, \\gamma$. Evidently $\\triangle A B C \\sim \\triangle H C^{\\prime} B^{\\prime}$. Within $\\triangle H C^{\\prime} B^{\\prime}$ there exists a unique point $I$ such that $\\angle H I B^{\\prime}=180^{\\circ}-\\gamma, \\angle H I C^{\\prime}=180^{\\circ}-\\beta$, and $\\angle C^{\\prime} I B^{\\prime}=180^{\\circ}-\\alpha$, and all three circles must contain this point. Let $H I$ and $B^{\\prime} C^{\\prime}$ intersect in $F$. It remains to show that $F B^{\\prime}=F C^{\\prime}$. From $\\angle H I B^{\\prime}+\\angle H B^{\\prime} F=180^{\\circ}$ we obtain $\\angle I H B^{\\prime}=\\angle I B^{\\prime} F$. Similarly, $\\angle I H C^{\\prime}=\\angle I C^{\\prime} F$. Thus circles around $\\triangle I H C^{\\prime}$ and $\\triangle I H B^{\\prime}$ are both tangent to $B^{\\prime} C^{\\prime}$, giving us $F B^{\\prime 2}=$ $F I \\cdot F H=F C^{\\prime 2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1970", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (USS 5) For which digits $a$ do exist integers $n \\geq 4$ such that each digit of $\\frac{n(n+1)}{2}$ equals $a$ ?", "solution": "7. For $a=5$ one can take $n=10$, while for $a=6$ one takes $n=11$. Now assume $a \\notin\\{5,6\\}$. If there exists an integer $n$ such that each digit of $n(n+1) / 2$ is equal to $a$, then there is an integer $k$ such that $n(n+1) / 2=\\left(10^{k}-1\\right) a / 9$. After multiplying both sides of the equation by 72 , one obtains $36 n^{2}+36 n=$ $8 a \\cdot 10^{k}-8 a$, which is equivalent to  $$ 9(2 n+1)^{2}=8 a \\cdot 10^{k}-8 a+9 $$  So $8 a \\cdot 10^{k}-8 a+9$ is the square of some odd integer. This means that its last digit is 1,5 , or 9 . Therefore $a \\in\\{1,3,5,6,8\\}$. If $a=3$ or $a=8$, the number on the RHS of (1) is divisible by 5 , but not by 25 (for $k \\geq 2$ ), and thus cannot be a square. It remains to check the case $a=1$. In that case, (1) becomes $9(2 n+1)^{2}=8 \\cdot 10^{k}+1$, or equivalently $[3(2 n+1)-1][3(2 n+1)+1]=8 \\cdot 10^{k} \\Rightarrow(3 n+1)(3 n+2)=2 \\cdot 10^{k}$. Since the factors $3 n+1,3 n+2$ are relatively prime, this implies that one of them is $2^{k+1}$ and the other one is $5^{k}$. It is directly checked that their difference really equals 1 only for $k=1$ and $n=1$, which is excluded. Hence, the desired $n$ exists only for $a \\in\\{5,6\\}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1970", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (POL 2) ${ }^{\\mathrm{IMO} 1}$ Given a point $M$ on the side $A B$ of the triangle $A B C$, let $r_{1}$ and $r_{2}$ be the radii of the inscribed circles of the triangles $A C M$ and $B C M$ respectively and let $\\rho_{1}$ and $\\rho_{2}$ be the radii of the excircles of the triangles $A C M$ and $B C M$ at the sides $A M$ and $B M$ respectively. Let $r$ and $\\rho$ denote the radii of the inscribed circle and the excircle at the side $A B$ of the triangle $A B C$ respectively. Prove that  $$ \\frac{r_{1}}{\\rho_{1}} \\frac{r_{2}}{\\rho_{2}}=\\frac{r}{\\rho} $$", "solution": "8. Let $A C=b, B C=a, A M=x, B M=y, C M=l$. Denote by $I_{1}$ the incenter and by $S_{1}$ the center of the excircle of $\\triangle A M C$. Suppose that $P_{1}$ and $Q_{1}$ are feet of perpendiculars from $I_{1}$ and $S_{1}$, respectively, to the line $A C$. Then $\\triangle I_{1} C P_{1} \\sim \\triangle S_{1} C Q_{1}$, hence $r_{1} / \\rho_{1}=C P_{1} / C Q_{1}$. We have $C P_{1}=(A C+M C-A M) / 2=(b+l-x) / 2$ and $C Q_{1}=$ $(A C+M C+A M) / 2=(b+l+x) / 2$. Hence  $$ \\frac{r_{1}}{\\rho_{1}}=\\frac{b+l-x}{b+l+x} $$  We similarly obtain  $$ \\frac{r_{2}}{\\rho_{2}}=\\frac{b+l-y}{b+l+y} \\text { and } \\frac{r}{\\rho}=\\frac{a+b-x-y}{a+b+x+y} $$  What we have to prove is now equivalent to  $$ \\frac{(b+l-x)(a+l-y)}{(b+l+x)(a+l+y)}=\\frac{a+b-x-y}{a+b+x+y} $$  Multiplying both sides of (1) by $(a+l+y)(b+l+x)(a+b+x+y)$ we obtain an expression that reduces to $l^{2} x+l^{2} y+x^{2} y+x y^{2}=b^{2} y+a^{2} x$. Dividing both sides by $c=x+y$, we get that (1) is equivalent to $l^{2}=$ $b^{2} y /(x+y)+a^{2} x /(x+y)-x y$, which is exactly Stewart's theorem for $l$. This finally proves the desired result.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1970", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (GDR 3) Let $u_{1}, u_{2}, \\ldots, u_{n}, v_{1}, v_{2}, \\ldots, v_{n}$ be real numbers. Prove that  $$ 1+\\sum_{i=1}^{n}\\left(u_{i}+v_{i}\\right)^{2} \\leq \\frac{4}{3}\\left(1+\\sum_{i=1}^{n} u_{i}^{2}\\right)\\left(1+\\sum_{i=1}^{n} v_{i}^{2}\\right) $$  In what case does equality hold?", "solution": "9. Let us set $a=\\sqrt{\\sum_{i=1}^{n} u_{i}^{2}}$ and $b=\\sqrt{\\sum_{i=1}^{n} v_{i}^{2}}$. By Minkowski's inequality (for $p=2$ ) we have $\\sum_{i=1}^{n}\\left(u_{i}+v_{i}\\right)^{2} \\leq(a+b)^{2}$. Hence the LHS of the desired inequality is not greater than $1+(a+b)^{2}$, while the RHS is equal to $4\\left(1+a^{2}\\right)\\left(1+b^{2}\\right) / 3$. Now it is sufficient to prove that  $$ 3+3(a+b)^{2} \\leq 4\\left(1+a^{2}\\right)\\left(1+b^{2}\\right) $$  The last inequality can be reduced to the trivial $0 \\leq(a-b)^{2}+(2 a b-1)^{2}$. The equality in the initial inequality holds if and only if $u_{i} / v_{i}=c$ for some $c \\in \\mathbb{R}$ and $a=b=1 / \\sqrt{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (BUL 2) Consider a sequence of polynomials $P_{0}(x), P_{1}(x), P_{2}(x), \\ldots$, $P_{n}(x), \\ldots$, where $P_{0}(x)=2, P_{1}(x)=x$ and for every $n \\geq 1$ the following equality holds:  $$ P_{n+1}(x)+P_{n-1}(x)=x P_{n}(x) $$  Prove that there exist three real numbers $a, b, c$ such that for all $n \\geq 1$,  $$ \\left(x^{2}-4\\right)\\left[P_{n}^{2}(x)-4\\right]=\\left[a P_{n+1}(x)+b P_{n}(x)+c P_{n-1}(x)\\right]^{2} $$", "solution": "1. Assuming that $a, b, c$ in (1) exist, let us find what their values should be. Since $P_{2}(x)=x^{2}-2$, equation (1) for $n=1$ becomes $\\left(x^{2}-4\\right)^{2}=$ $\\left[a\\left(x^{2}-2\\right)+b x+2 c\\right]^{2}$. Therefore, there are two possibilities for $(a, b, c)$ : $(1,0,-1)$ and $(-1,0,1)$. In both cases we must prove that  $$ \\left(x^{2}-4\\right)\\left[P_{n}(x)^{2}-4\\right]=\\left[P_{n+1}(x)-P_{n-1}(x)\\right]^{2} $$  It suffices to prove (2) for all $x$ in the interval $[-2,2]$. In this interval we can set $x=2 \\cos t$ for some real $t$. We prove by induction that  $$ P_{n}(x)=2 \\cos n t \\quad \\text { for all } n $$  This is trivial for $n=0,1$. Assume (3) holds for some $n-1$ and $n$. Then $P_{n+1}(x)=4 \\cos t \\cos n t-2 \\cos (n-1) t=2 \\cos (n+1) t$ by the additive formula for the cosine. This completes the induction. Now (2) reduces to the obviously correct equality  $$ 16 \\sin ^{2} t \\sin ^{2} n t=(2 \\cos (n+1) t-2 \\cos (n-1) t)^{2} $$  Second solution. If $x$ is fixed, the linear recurrence relation $P_{n+1}(x)+$ $P_{n-1}(x)=x P_{n}(x)$ can be solved in the standard way. The characteristic polynomial $t^{2}-x t+1$ has zeros $t_{1,2}$ with $t_{1}+t_{2}=x$ and $t_{1} t_{2}=1$; hence, the general $P_{n}(x)$ has the form $a t_{1}^{n}+b t_{2}^{n}$ for some constants $a$, $b$. From $P_{0}=2$ and $P_{1}=x$ we obtain that  $$ P_{n}(x)=t_{1}^{n}+t_{2}^{n} . $$  Plugging in these values and using $t_{1} t_{2}=1$ one easily verifies (2).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (POL 2) ${ }^{\\mathrm{IMO} 3}$ Prove that the sequence $2^{n}-3(n>1)$ contains a subsequence of numbers relatively prime in pairs.", "solution": "10. We use induction. Suppose that every two of the numbers $a_{1}=2^{n_{1}}-$ $3, a_{2}=2^{n_{2}}-3, \\ldots, a_{k}=2^{n_{k}}-3$, where $2=n_{1}<n_{2}<\\cdots<n_{k}$, are coprime. Then one can construct $a_{k+1}=2^{n_{k+1}}-3$ in the following way:  Set $s=a_{1} a_{2} \\ldots a_{k}$. Among the numbers $2^{0}, 2^{1}, \\ldots, 2^{s}$, two give the same residue upon division by $s$, say $s \\mid 2^{\\alpha}-2^{\\beta}$. Since $s$ is odd, it can be assumed w.l.o.g. that $\\beta=0$ (this is actually a direct consequence of Euler's theorem). Let $2^{\\alpha}-1=q s, q \\in \\mathbb{N}$. Since $2^{\\alpha+2}-3=4 q s+1$ is then coprime to $s$, it is enough to take $n_{k+1}=\\alpha+2$. We obviously have $n_{k+1}>n_{k}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (POL 3) The matrix  $$ \\left(\\begin{array}{ccc} a_{11} & \\ldots & a_{1 n} \\\\ \\vdots & \\ldots & \\vdots \\\\ a_{n 1} & \\ldots & a_{n n} \\end{array}\\right) $$  satisfies the inequality $\\sum_{j=1}^{n}\\left|a_{j 1} x_{1}+\\cdots+a_{j n} x_{n}\\right| \\leq M$ for each choice of numbers $x_{i}$ equal to $\\pm 1$. Show that  $$ \\left|a_{11}+a_{22}+\\cdots+a_{n n}\\right| \\leq M $$", "solution": "11. We use induction. The statement for $n=1$ is trivial. Suppose that it holds for $n=k$ and consider $n=k+1$. From the given condition, we have  $$ \\begin{gathered} \\sum_{j=1}^{k}\\left|a_{j, 1} x_{1}+\\cdots+a_{j, k} x_{k}+a_{j, k+1}\\right| \\\\ +\\left|a_{k+1,1} x_{1}+\\cdots+a_{k+1, k} x_{k}+a_{k+1, k+1}\\right| \\leq M \\\\ \\sum_{j=1}^{k}\\left|a_{j, 1} x_{1}+\\cdots+a_{j, k} x_{k}-a_{j, k+1}\\right| \\\\ +\\left|a_{k+1,1} x_{1}+\\cdots+a_{k+1, k} x_{k}-a_{k+1, k+1}\\right| \\leq M \\end{gathered} $$  for each choice of $x_{i}= \\pm 1$. Since $|a+b|+|a-b| \\geq 2|a|$ for all $a, b$, we obtain  $$ \\begin{aligned} 2 \\sum_{j=1}^{k}\\left|a_{j 1} x_{1}+\\cdots+a_{j k} x_{k}\\right|+2\\left|a_{k+1, k+1}\\right| & \\leq 2 M, \\text { that is, } \\\\ \\sum_{j=1}^{k}\\left|a_{j 1} x_{1}+\\cdots+a_{j k} x_{k}\\right| & \\leq M-\\left|a_{k+1, k+1}\\right| \\end{aligned} $$  Now by the inductive assumption $\\sum_{j=1}^{k}\\left|a_{j j}\\right| \\leq M-\\left|a_{k+1, k+1}\\right|$, which is equivalent to the desired inequality.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (POL 6) Two congruent equilateral triangles $A B C$ and $A^{\\prime} B^{\\prime} C^{\\prime}$ in the plane are given. Show that the midpoints of the segments $A A^{\\prime}, B B^{\\prime}, C C^{\\prime}$ either are collinear or form an equilateral triangle.", "solution": "12. Let us start with the case $A=A^{\\prime}$. If the triangles $A B C$ and $A^{\\prime} B^{\\prime} C^{\\prime}$ are oppositely oriented, then they are symmetric with respect to some axis, and the statement is true. Suppose that they are equally oriented. There is a rotation around $A$ by $60^{\\circ}$ that maps $A B B^{\\prime}$ onto $A C C^{\\prime}$. This rotation also maps the midpoint $B_{0}$ of $B B^{\\prime}$ onto the midpoint $C_{0}$ of $C C^{\\prime}$, hence the triangle $A B_{0} C_{0}$ is equilateral. In the general case, when $A \\neq A^{\\prime}$, let us denote by $T$ the translation that maps $A$ onto $A^{\\prime}$. Let $X^{\\prime}$ be the image of a point $X$ under the (unique) isometry mapping $A B C$ onto $A^{\\prime} B^{\\prime} C^{\\prime}$, and $X^{\\prime \\prime}$ the image of $X$ under $T$. Furthermore, let $X_{0}, X_{0}^{\\prime}$ be the midpoints of segments $X X^{\\prime}, X^{\\prime} X^{\\prime \\prime}$. Then $X_{0}$ is the image of $X_{0}^{\\prime}$ under the translation $-(1 / 2) T$. However, since it has already been proven that the triangle $A_{0}^{\\prime} B_{0}^{\\prime} C_{0}^{\\prime}$ is equilateral, its image $A_{0} B_{0} C_{0}$ under $(1 / 2) T$ is also equilateral. The statement of the problem is thus proven.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (SWE 5) ${ }^{\\mathrm{IMO} 6}$ Consider the $n \\times n$ array of nonnegative integers  $$ \\left(\\begin{array}{cccc} a_{11} & a_{12} & \\ldots & a_{1 n} \\\\ a_{21} & a_{22} & \\ldots & a_{2 n} \\\\ \\vdots & \\vdots & & \\vdots \\\\ a_{n 1} & a_{n 2} & \\ldots & a_{n n} \\end{array}\\right) $$  with the following property: If an element $a_{i j}$ is zero, then the sum of the elements of the $i$ th row and the $j$ th column is greater than or equal to $n$. Prove that the sum of all the elements is greater than or equal to $\\frac{1}{2} n^{2}$.", "solution": "13. Let $p$ be the least of all the sums of elements in one row or column. If $p \\geq n / 2$, then the sum of all elements of the array is $s \\geq n p \\geq n^{2} / 2$.  Now suppose that $p<n / 2$. Without loss of generality, one can assume that the sum of elements in the first row is $p$, and that exactly the first $q$ elements of it are different from zero. Then the sum of elements in the last $n-q$ columns is greater than or equal to $(n-p)(n-q)$. Furthermore, the sum of elements in the first $q$ columns is greater than or equal to $p q$. This implies that the sum of all elements in the array is  $$ s \\geq(n-p)(n-q)+p q=\\frac{1}{2} n^{2}+\\frac{1}{2}(n-2 p)(n-2 q) \\geq \\frac{1}{2} n^{2} $$  since $n \\geq 2 p \\geq 2 q$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (USS 1) A broken line $A_{1} A_{2} \\ldots A_{n}$ is drawn in a $50 \\times 50$ square, so that the distance from any point of the square to the broken line is less than 1. Prove that its total length is greater than 1248.", "solution": "14. Denote by $V$ the figure made by a circle of radius 1 whose center moves along the broken line. From the condition of the problem, $V$ contains the whole $50 \\times 50$ square, and thus the area $S(V)$ of $V$ is not less than 2500 . Let $L$ be the length of the broken line. We shall show that $S(V) \\leq 2 L+\\pi$, from which it will follow that $L \\geq 1250-\\pi / 2>1248$. For each segment $l_{i}=A_{i} A_{i+1}$ of the broken line, consider the figure $V_{i}$ obtained by a circle of radius 1 whose center moves along it, and let $\\overline{V_{i}}$ be obtained by cutting off the circle of radius 1 with center at the starting point of $l_{i}$. The area of $\\overline{V_{i}}$ is equal to $2 A_{i} A_{i+1}$. It is clear that the union of all the figures $\\overline{V_{i}}$ together with a semicircle with center in $A_{1}$ and a semicircle with center in $A_{n}$ contains $V$ completely. Therefore  $$ S(V) \\leq \\pi+2 A_{1} A_{2}+2 A_{2} A_{3}+\\cdots+2 A_{n-1} A_{n}=\\pi+2 L . $$  This completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (USS 2) Natural numbers from 1 to 99 (not necessarily distinct) are written on 99 cards. It is given that the sum of the numbers on any subset of cards (including the set of all cards) is not divisible by 100. Show that all the cards contain the same number.", "solution": "15. Assume the opposite. Then one can numerate the cards 1 to 99, with a number $n_{i}$ written on the card $i$, so that $n_{98} \\neq n_{99}$. Denote by $x_{i}$ the remainder of $n_{1}+n_{2}+\\cdots+n_{i}$ upon division by 100 , for $i=1,2, \\ldots, 99$. All $x_{i}$ must be distinct: Indeed, if $x_{i}=x_{j}, i<j$, then $n_{i+1}+\\cdots+n_{j}$ is divisible by 100, which is impossible. Also, no $x_{i}$ can be equal to 0 . Thus, the numbers $x_{1}, x_{2}, \\ldots, x_{99}$ take exactly the values $1,2, \\ldots, 99$ in some order. Let $x$ be the remainder of $n_{1}+n_{2}+\\cdots+n_{97}+n_{99}$ upon division by 100 . It is not zero; hence it must be equal to $x_{k}$ for some $k \\in\\{1,2, \\ldots, 99\\}$. There are three cases: (i) $x=x_{k}, k \\leq 97$. Then $n_{k+1}+n_{k+2}+\\cdots+n_{97}+n_{99}$ is divisible by 100, a contradiction; (ii) $x=x_{98}$. Then $n_{98}=n_{99}$, a contradiction; (iii) $x=x_{99}$. Then $n_{98}$ is divisible by 100, a contradiction.  Therefore, all the cards contain the same number.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (USS 5) ${ }^{\\mathrm{IMO} 2}$ Given a convex polyhedron $P_{1}$ with 9 vertices $A_{1}, \\ldots, A_{9}$, let us denote by $P_{2}, P_{3}, \\ldots, P_{9}$ the images of $P_{1}$ under the translations mapping the vertex $A_{1}$ to $A_{2}, A_{3}, \\ldots, A_{9}$ respectively. Prove that among the polyhedra $P_{1}, \\ldots, P_{9}$ at least two have a common interior point.", "solution": "16. Denote by $P^{\\prime}$ the polyhedron defined as the image of $P$ under the homothety with center at $A_{1}$ and coefficient of similarity 2 . It is easy to see that all $P_{i}, i=1, \\ldots, 9$, are contained in $P^{\\prime}$ (indeed, if $M \\in P_{k}$, then $\\frac{1}{2} \\overrightarrow{A_{1} M}=\\frac{1}{2}\\left(\\overrightarrow{A_{1} A_{k}}+\\overrightarrow{A_{1} M^{\\prime}}\\right)$ for some $M^{\\prime} \\in P$, and the claim follows from the convexity of $P$ ). But the volume of $P^{\\prime}$ is exactly 8 times the volume of $P$, while the volumes of $P_{i}$ add up to 9 times that volume. We conclude that not all $P_{i}$ have disjoint interiors.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (YUG 1) Prove the inequality  $$ \\frac{a_{1}+a_{3}}{a_{1}+a_{2}}+\\frac{a_{2}+a_{4}}{a_{2}+a_{3}}+\\frac{a_{3}+a_{1}}{a_{3}+a_{4}}+\\frac{a_{4}+a_{2}}{a_{4}+a_{1}} \\geq 4 $$  where $a_{i}>0, i=1,2,3,4$.", "solution": "17. We use the following obvious consequences of $(a+b)^{2} \\geq 4 a b$ :  $$ \\begin{aligned} & \\frac{1}{\\left(a_{1}+a_{2}\\right)\\left(a_{3}+a_{4}\\right)} \\geq \\frac{4}{\\left(a_{1}+a_{2}+a_{3}+a_{4}\\right)^{2}} \\\\ & \\frac{1}{\\left(a_{1}+a_{4}\\right)\\left(a_{2}+a_{3}\\right)} \\geq \\frac{4}{\\left(a_{1}+a_{2}+a_{3}+a_{4}\\right)^{2}} \\end{aligned} $$  Now we have  $$ \\begin{aligned} & \\frac{a_{1}+a_{3}}{a_{1}+a_{2}}+\\frac{a_{2}+a_{4}}{a_{2}+a_{3}}+\\frac{a_{3}+a_{1}}{a_{3}+a_{4}}+\\frac{a_{4}+a_{2}}{a_{4}+a_{1}} \\\\ = & \\frac{\\left(a_{1}+a_{3}\\right)\\left(a_{1}+a_{2}+a_{3}+a_{4}\\right)}{\\left(a_{1}+a_{2}\\right)\\left(a_{3}+a_{4}\\right)}+\\frac{\\left(a_{2}+a_{4}\\right)\\left(a_{1}+a_{2}+a_{3}+a_{4}\\right)}{\\left(a_{1}+a_{4}\\right)\\left(a_{2}+a_{3}\\right)} \\\\ \\geq & \\frac{4\\left(a_{1}+a_{3}\\right)}{a_{1}+a_{2}+a_{3}+a_{4}}+\\frac{4\\left(a_{2}+a_{4}\\right)}{a_{1}+a_{2}+a_{3}+a_{4}}=4 . \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (BUL 5) ${ }^{\\mathrm{IMO} 5}$ Prove that for every natural number $m \\geq 1$ there exists a finite set $S_{m}$ of points in the plane satisfying the following condition: If $A$ is any point in $S_{m}$, then there are exactly $m$ points in $S_{m}$ whose distance to $A$ equals 1.", "solution": "2 . We will construct such a set $S_{m}$ of $2^{m}$ points. Take vectors $u_{1}, \\ldots, u_{m}$ in a given plane, such that $\\left|u_{i}\\right|=1 / 2$ and $0 \\neq\\left|c_{1} u_{1}+c_{2} u_{2}+\\cdots+c_{n} u_{n}\\right| \\neq 1 / 2$ for any choice of numbers $c_{i}$ equal to 0 or $\\pm 1$. Such vectors are easily constructed by induction on $m$ : For $u_{1}, \\ldots, u_{m-1}$ fixed, there are only finitely many vector values $u_{m}$ that violate the upper condition, and we may set $u_{m}$ to be any other vector of length $1 / 2$. Let $S_{m}$ be the set of all points $M_{0}+\\varepsilon_{1} u_{1}+\\varepsilon_{2} u_{2}+\\cdots+\\varepsilon_{m} u_{m}$, where $M_{0}$ is any fixed point in the plane and $\\varepsilon_{i}= \\pm 1$ for $i=1, \\ldots, m$. Then $S_{m}$ obviously satisfies the condition of the problem.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (GDR 1) Knowing that the system  $$ \\begin{aligned} x+y+z & =3, \\\\ x^{3}+y^{3}+z^{3} & =15, \\\\ x^{4}+y^{4}+z^{4} & =35, \\end{aligned} $$  has a real solution $x, y, z$ for which $x^{2}+y^{2}+z^{2}<10$, find the value of $x^{5}+y^{5}+z^{5}$ for that solution.", "solution": "3. Let $x, y, z$ be a solution of the given system with $x^{2}+y^{2}+z^{2}=\\alpha<10$. Then  $$ x y+y z+z x=\\frac{(x+y+z)^{2}-\\left(x^{2}+y^{2}+z^{2}\\right)}{2}=\\frac{9-\\alpha}{2} . $$  Furthermore, $3 x y z=x^{3}+y^{3}+z^{3}-(x+y+z)\\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\\right)$, which gives us $x y z=3(9-\\alpha) / 2-4$. We now have  $$ \\begin{aligned} 35= & x^{4}+y^{4}+z^{4}=\\left(x^{3}+y^{3}+z^{3}\\right)(x+y+z) \\\\ & -\\left(x^{2}+y^{2}+z^{2}\\right)(x y+y z+z x)+x y z(x+y+z) \\\\ = & 45-\\frac{\\alpha(9-\\alpha)}{2}+\\frac{9(9-\\alpha)}{2}-12 . \\end{aligned} $$  The solutions in $\\alpha$ are $\\alpha=7$ and $\\alpha=11$. Therefore $\\alpha=7$, xyz $=-1$, $x y+x z+y z=1$, and  $$ \\begin{aligned} x^{5}+y^{5}+z^{5}= & \\left(x^{4}+y^{4}+z^{4}\\right)(x+y+z) \\\\ & -\\left(x^{3}+y^{3}+z^{3}\\right)(x y+x z+y z)+x y z\\left(x^{2}+y^{2}+z^{2}\\right) \\\\ = & 35 \\cdot 3-15 \\cdot 1+7 \\cdot(-1)=83 . \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (GBR 3) We are given two mutually tangent circles in the plane, with radii $r_{1}, r_{2}$. A line intersects these circles in four points, determining three segments of equal length. Find this length as a function of $r_{1}$ and $r_{2}$ and the condition for the solvability of the problem.", "solution": "4. In the coordinate system in which the $x$-axis passes through the centers of the circles and the $y$-axis is their common tangent, the circles have equations  $$ x^{2}+y^{2}+2 r_{1} x=0, \\quad x^{2}+y^{2}-2 r_{2} x=0 . $$  Let $p$ be the desired line with equation $y=a x+b$. The abscissas of points of intersection of $p$ with both circles satisfy one of  $$ \\left(1+a^{2}\\right) x^{2}+2\\left(a b+r_{1}\\right) x+b^{2}=0, \\quad\\left(1+a^{2}\\right) x^{2}+2\\left(a b-r_{2}\\right) x+b^{2}=0 $$  Let us denote the lengths of the chords and their projections onto the $x$-axis by $d$ and $d_{1}$, respectively. From these equations it follows that  $$ d_{1}^{2}=\\frac{4\\left(a b+r_{1}\\right)^{2}}{\\left(1+a^{2}\\right)^{2}}-\\frac{4 b^{2}}{1+a^{2}}=\\frac{4\\left(a b-r_{2}\\right)^{2}}{\\left(1+a^{2}\\right)^{2}}-\\frac{4 b^{2}}{1+a^{2}} $$  Consider the point of intersection of $p$ with the $y$-axis. This point has equal powers with respect to both circles. Hence, if that point divides the segment determined on $p$ by the two circles on two segments of lengths $x$ and $y$, this power equals $x(x+d)=y(y+d)$, which implies $x=y=d / 2$. Thus each of the equations in (1) has two roots, one of which is thrice the other. This fact gives us $\\left(a b+r_{1}\\right)^{2}=4\\left(1+a^{2}\\right) b^{2} / 3$. From (1) and this we obtain  $$ \\begin{gathered} a b=\\frac{r_{2}-r_{1}}{2}, \\quad 4 b^{2}+a^{2} b^{2}=3\\left[\\left(a b+r_{1}\\right)^{2}-a^{2} b^{2}\\right]=3 r_{1} r_{2} \\\\ a^{2}=\\frac{4\\left(r_{2}-r_{1}\\right)^{2}}{14 r_{1} r_{2}-r_{1}^{2}-r_{2}^{2}}, \\quad b^{2}=\\frac{14 r_{1} r_{2}-r_{1}^{2}-r_{2}^{2}}{16} \\\\ d_{1}^{2}=\\frac{\\left(14 r_{1} r_{2}-r_{1}^{2}-r_{2}^{2}\\right)^{2}}{36\\left(r_{1}+r_{2}\\right)^{2}} \\end{gathered} $$  Finally, since $d^{2}=d_{1}^{2}\\left(1+a^{2}\\right)$, we conclude that  $$ d^{2}=\\frac{1}{12}\\left(14 r_{1} r_{2}-r_{1}^{2}-r_{2}^{2}\\right), $$  and that the problem is solvable if and only if $7-4 \\sqrt{3} \\leq \\frac{r_{1}}{r_{2}} \\leq 7+4 \\sqrt{3}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (HUN 1) ${ }^{\\mathrm{IMO}}$ Let $a, b, c, d, e$ be real numbers. Prove that the expression  $$ \\begin{gathered} (a-b)(a-c)(a-d)(a-e)+(b-a)(b-c)(b-d)(b-e)+(c-a)(c-b)(c-d)(c-e) \\\\ +(d-a)(d-b)(d-c)(d-e)+(e-a)(e-b)(e-c)(e-d) \\end{gathered} $$  is nonnegative.", "solution": "5. Without loss of generality, we may assume that $a \\geq b \\geq c \\geq d \\geq e$. Then $a-b=-(b-a) \\geq 0, a-c \\geq b-c \\geq 0, a-d \\geq b-d \\geq 0$ and $a-e \\geq b-e \\geq 0$, and hence  $$ (a-b)(a-c)(a-d)(a-e)+(b-a)(b-c)(b-d)(b-e) \\geq 0 $$  Analogously, $(d-a)(d-b)(d-c)(d-e)+(e-a)(e-b)(e-c)(e-d) \\geq 0$. Finally, $(c-a)(c-b)(c-d)(c-e) \\geq 0$ as a product of two nonnegative numbers, from which the inequality stated in the problem follows. Remark. The problem in an alternative formulation, accepted for the IMO, asked to prove that the analogous inequality  $$ \\begin{gathered} \\left(a_{1}-a_{2}\\right)\\left(a_{1}-a_{2}\\right) \\cdots\\left(a_{1}-a_{n}\\right)+\\left(a_{2}-a_{1}\\right)\\left(a_{2}-a_{3}\\right) \\cdots\\left(a_{2}-a_{n}\\right)+\\cdots \\\\ +\\left(a_{n}-a_{1}\\right)\\left(a_{n}-a_{2}\\right) \\cdots\\left(a_{n}-a_{n-1}\\right) \\geq 0 \\end{gathered} $$  holds for arbitrary real numbers $a_{i}$ if and only if $n=3$ or $n=5$. The case $n=3$ is analogous to $n=5$. For $n=4$, a counterexample is $a_{1}=0, a_{2}=a_{3}=a_{4}=1$, while for $n>5$ one can take $a_{1}=a_{2}=\\cdots=$ $a_{n-4}=0, a_{n-3}=a_{n-2}=a_{n-1}=2, a_{n}=1$ as a counterexample.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (HUN 7) Let $n \\geq 2$ be a natural number. Find a way to assign natural numbers to the vertices of a regular $2^{n}$-gon such that the following conditions are satisfied: (1) only digits 1 and 2 are used; (2) each number consists of exactly $n$ digits; (3) different numbers are assigned to different vertices; (4) the numbers assigned to two neighboring vertices differ at exactly one digit.", "solution": "6. The proof goes by induction on $n$. For $n=2$, the following numeration satisfies the conditions (a)-(d): $C_{1}=11, C_{2}=12, C_{3}=22, C_{4}=21$. Suppose that $n>2$, and that the numeration $C_{1}, C_{2}, \\ldots, C_{2^{n-1}}$ of a regular $2^{n-1}$-gon, in cyclical order, satisfies (i)-(iv). Then one can assign to the vertices of a $2^{n}$-gon cyclically the following numbers:  $$ \\overline{1 C_{1}}, \\overline{1 C_{2}}, \\ldots, \\overline{1 C_{2^{n-1}}}, \\overline{2 C_{2^{n-1}}}, \\ldots, \\overline{2 C_{2}}, \\overline{2 C_{1}} $$  The conditions (i), (ii) obviously hold, while (iii) and (iv) follow from the inductive assumption.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (NET 1) ${ }^{\\mathrm{IMO} /}$ Given a tetrahedron $A B C D$ whose all faces are acuteangled triangles, set  $$ \\sigma=\\measuredangle D A B+\\measuredangle B C D-\\measuredangle A B C-\\measuredangle C D A $$  Consider all closed broken lines $X Y Z T X$ whose vertices $X, Y, Z, T$ lie in the interior of segments $A B, B C, C D, D A$ respectively. Prove that: (a) if $\\sigma \\neq 0$, then there is no broken line $X Y Z T$ of minimal length; (b) if $\\sigma=0$, then there are infinitely many such broken lines of minimal length. That length equals $2 A C \\sin (\\alpha / 2)$, where  $$ \\alpha=\\measuredangle B A C+\\measuredangle C A D+\\measuredangle D A B $$", "solution": "7. (a) Suppose that $X, Y, Z$ are fixed on segments $A B, B C, C D$. It is proven in a standard way that if $\\angle A T X \\neq \\angle Z T D$, then $Z T+T X$ can be reduced. It follows that if there exists a broken line $X Y Z T X$ of minimal length, then the following conditions hold:  $$ \\begin{aligned} & \\angle D A B=\\pi-\\angle A T X-\\angle A X T \\\\ & \\angle A B C=\\pi-\\angle B X Y-\\angle B Y X=\\pi-\\angle A X T-\\angle C Y Z \\\\ & \\angle B C D=\\pi-\\angle C Y Z-\\angle C Z Y \\\\ & \\angle C D A=\\pi-\\angle D T Z-\\angle D Z T=\\pi-\\angle A T X-\\angle C Z Y . \\end{aligned} $$  Thus $\\sigma=0$. (b) Now let $\\sigma=0$. Let us cut the surface of the tetrahedron along the edges $A C, C D$, and $D B$ and set it down into a plane. Consider the plane figure $\\mathcal{S}=A C D^{\\prime} B D^{\\prime \\prime} C^{\\prime}$ thus obtained made up of triangles $B C D^{\\prime}, A B C, A B D^{\\prime \\prime}$, and $A C^{\\prime} D^{\\prime \\prime}$, with $Z^{\\prime}, T^{\\prime}, Z^{\\prime \\prime}$ respectively on $C D^{\\prime}, A D^{\\prime \\prime}, C^{\\prime} D^{\\prime \\prime}$ (here $C^{\\prime}$ corresponds to $C$, etc.). Since $\\angle C^{\\prime} D^{\\prime \\prime} A+\\angle D^{\\prime \\prime} A B+\\angle A B C+\\angle B C D^{\\prime}=0$ as an oriented angle (because $\\sigma=0$ ), the lines $C D^{\\prime}$ and $C^{\\prime} D^{\\prime \\prime}$ are parallel and equally oriented; i.e., $C D^{\\prime} D^{\\prime \\prime} C^{\\prime}$ is a parallelogram. The broken line $X Y Z T X$ has minimal length if and only if $Z^{\\prime \\prime}, T^{\\prime}, X$, $Y, Z^{\\prime}$ are collinear (where $Z^{\\prime} Z^{\\prime \\prime} \\|$ $C C^{\\prime}$ ), and then this length equals $Z^{\\prime} Z^{\\prime \\prime}=C C^{\\prime}=2 A C \\sin (\\alpha / 2)$. There is an infinity of such lines, one for every line $Z^{\\prime} Z^{\\prime \\prime}$ parallel to $C C^{\\prime}$ that meets the interiors of all the segments $C B, B A, A D^{\\prime \\prime}$. Such ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-389.jpg?height=300&width=516&top_left_y=408&top_left_x=835) $Z^{\\prime} Z^{\\prime \\prime}$ exist. Indeed, the triangles $C A B$ and $D^{\\prime \\prime} A B$ are acute-angled, and thus the segment $A B$ has a common interior point with the parallelogram $C D^{\\prime} D^{\\prime \\prime} C^{\\prime}$. Therefore the desired result follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (NET 4) Determine whether there exist distinct real numbers $a, b, c, t$ for which: (i) the equation $a x^{2}+b t x+c=0$ has two distinct real roots $x_{1}, x_{2}$, (ii) the equation $b x^{2}+c t x+a=0$ has two distinct real roots $x_{2}, x_{3}$, (iii) the equation $c x^{2}+a t x+b=0$ has two distinct real roots $x_{3}, x_{1}$.", "solution": "8. Suppose that $a, b, c, t$ satisfy all the conditions. Then $a b c \\neq 0$ and  $$ x_{1} x_{2}=\\frac{c}{a}, \\quad x_{2} x_{3}=\\frac{a}{b}, \\quad x_{3} x_{1}=\\frac{b}{c} . $$  Multiplying these equations, we obtain $x_{1}^{2} x_{2}^{2} x_{3}^{2}=1$, and hence $x_{1} x_{2} x_{3}=$ $\\varepsilon= \\pm 1$. From (1) we get $x_{1}=\\varepsilon b / a, x_{2}=\\varepsilon c / b, x_{3}=\\varepsilon a / c$. Substituting $x_{1}$ in the first equation, we get $a b^{2} / a^{2}+t \\varepsilon b^{2} / a+c=0$, which gives us  $$ b^{2}(1+t \\varepsilon)=-a c . $$  Analogously, $c^{2}(1+t \\varepsilon)=-a b$ and $a^{2}(1+t \\varepsilon)=-b c$, and therefore $(1+$ $t \\varepsilon)^{3}=-1$; i.e., $1+t \\varepsilon=-1$, since it is real. This also implies together with (1) that $b^{2}=a c, c^{2}=a b$, and $a^{2}=b c$, and consequently  $$ a=b=c \\text {. } $$  Thus the three equations in the problem are equal, which is impossible. Hence, such $a, b, c, t$ do not exist.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1971", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (POL 1) Let $T_{k}=k-1$ for $k=1,2,3,4$ and  $$ T_{2 k-1}=T_{2 k-2}+2^{k-2}, \\quad T_{2 k}=T_{2 k-5}+2^{k} \\quad(k \\geq 3) $$  Show that for all $k$,  $$ 1+T_{2 n-1}=\\left[\\frac{12}{7} 2^{n-1}\\right] \\quad \\text { and } \\quad 1+T_{2 n}=\\left[\\frac{17}{7} 2^{n-1}\\right] $$  where $[x]$ denotes the greatest integer not exceeding $x$.", "solution": "9. We use induction. Since $T_{1}=0, T_{2}=1, T_{3}=2, T_{4}=3, T_{5}=5, T_{6}=8$, the statement is true for $n=1,2,3$. Suppose that both formulas from the problem hold for some $n \\geq 3$. Then  $$ \\begin{aligned} & T_{2 n+1}=1+T_{2 n}+2^{n-1}=\\left[\\frac{17}{7} 2^{n-1}+2^{n-1}\\right]=\\left[\\frac{12}{7} 2^{n}\\right] \\\\ & T_{2 n+2}=1+T_{2 n-3}+2^{n+1}=\\left[\\frac{12}{7} 2^{n-2}+2^{n+1}\\right]=\\left[\\frac{17}{7} 2^{n}\\right] . \\end{aligned} $$  Therefore the formulas hold for $n+1$, which completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1972", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (BUL 7) ${ }^{\\mathrm{IMO} 5}$ Let $f$ and $\\varphi$ be real functions defined on the set $\\mathbb{R}$ satisfying the functional equation  $$ f(x+y)+f(x-y)=2 \\varphi(y) f(x) $$  for arbitrary real $x, y$ (give examples of such functions). Prove that if $f(x)$ is not identically 0 and $|f(x)| \\leq 1$ for all $x$, then $|\\varphi(x)| \\leq 1$ for all $x$.", "solution": "1. Suppose that $f\\left(x_{0}\\right) \\neq 0$ and for a given $y$ define the sequence $x_{k}$ by the formula  $$ x_{k+1}= \\begin{cases}x_{k}+y, & \\text { if }\\left|f\\left(x_{k}+y\\right)\\right| \\geq\\left|f\\left(x_{k}-y\\right)\\right| \\\\ x_{k}-y, & \\text { otherwise. }\\end{cases} $$  It follows from (1) that $\\left|f\\left(x_{k+1}\\right)\\right| \\geq|\\varphi(y)|\\left|f\\left(x_{k}\\right)\\right|$; hence by induction, $\\left|f\\left(x_{k}\\right)\\right| \\geq|\\varphi(y)|^{k}\\left|f\\left(x_{0}\\right)\\right|$. Since $\\left|f\\left(x_{k}\\right)\\right| \\leq 1$ for all $k$, we obtain $|\\varphi(y)| \\leq 1$. Second solution. Let $M=\\sup f(x) \\leq 1$, and $x_{k}$ any sequence, possibly constant, such that $f\\left(x_{k}\\right) \\rightarrow M, k \\rightarrow \\infty$. Then for all $k$,  $$ |\\varphi(y)|=\\frac{\\left|f\\left(x_{k}+y\\right)+f\\left(x_{k}-y\\right)\\right|}{2\\left|f\\left(x_{k}\\right)\\right|} \\leq \\frac{2 M}{2\\left|f\\left(x_{k}\\right)\\right|} \\rightarrow 1, \\quad k \\rightarrow \\infty $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1972", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (NET 3) ${ }^{\\text {IMO2 }}$ Prove that for each $n \\geq 4$ every cyclic quadrilateral can be decomposed into $n$ cyclic quadrilaterals.", "solution": "10. Consider first a triangle. It can be decomposed into $k=3$ cyclic quadrilaterals by perpendiculars from some interior point of it to the sides; also, it can be decomposed into a cyclic quadrilateral and a triangle, and it follows by induction that this decomposition is possible for every $k$. Since every triangle can be cut into two triangles, the required decomposition is possible for each $n \\geq 6$. It remains to treat the cases $n=4$ and $n=5$. $n=4$. If the center $O$ of the circumcircle is inside a cyclic quadrilateral $A B C D$, then the required decomposition is effected by perpendiculars from $O$ to the four sides. Otherwise, let $C$ and $D$ be the vertices of the obtuse angles of the quadrilateral. Draw the perpendiculars at $C$ and $D$ to the lines $B C$ and $A D$ respectively, and choose points $P$ and $Q$ on them such that $P Q \\| A B$. Then the required decomposition is effected by $C P, P Q, Q D$ and the perpendiculars from $P$ and $Q$ to $A B$. $n=5$. If $A B C D$ is an isosceles trapezoid with $A B \\| C D$ and $A D=B C$, then it is trivially decomposed by lines parallel to $A B$. Otherwise, $A B C D$ can be decomposed into a cyclic quadrilateral and a trapezoid; this trapezoid can be cut into an isosceles trapezoid and a triangle, which can further be cut into three cyclic quadrilaterals and an isosceles trapezoid.  Remark. It can be shown that the assertion is not true for $n=2$ and $n=3$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1972", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (NET 6) Consider a sequence of circles $K_{1}, K_{2}, K_{3}, K_{4}, \\ldots$ of radii $r_{1}, r_{2}, r_{3}, r_{4}, \\ldots$, respectively, situated inside a triangle $A B C$. The circle $K_{1}$ is tangent to $A B$ and $A C ; K_{2}$ is tangent to $K_{1}, B A$, and $B C ; K_{3}$ is tangent to $K_{2}, C A$, and $C B ; K_{4}$ is tangent to $K_{3}, A B$, and $A C$; etc. (a) Prove the relation  $$ r_{1} \\cot \\frac{1}{2} A+2 \\sqrt{r_{1} r_{2}}+r_{2} \\cot \\frac{1}{2} B=r\\left(\\cot \\frac{1}{2} A+\\cot \\frac{1}{2} B\\right), $$  where $r$ is the radius of the incircle of the triangle $A B C$. Deduce the existence of a $t_{1}$ such that  $$ r_{1}=r \\cot \\frac{1}{2} B \\cot \\frac{1}{2} C \\sin ^{2} t_{1} . $$  (b) Prove that the sequence of circles $K_{1}, K_{2}, \\ldots$ is periodic.", "solution": "11. Let $\\angle A=2 x, \\angle B=2 y, \\angle C=2 z$. (a) Denote by $M_{i}$ the center of $K_{i}, i=1,2, \\ldots$ If $N_{1}, N_{2}$ are the projections of $M_{1}, M_{2}$ onto $A B$, we have $A N_{1}=r_{1} \\cot x, N_{2} B=r_{2} \\cot y$, and $N_{1} N_{2}=\\sqrt{\\left(r_{1}+r_{2}\\right)^{2}-\\left(r_{1}-r_{2}\\right)^{2}}=2 \\sqrt{r_{1} r_{2}}$. The required relation between $r_{1}, r_{2}$ follows from $A B=A N_{1}+N_{1} N_{2}+N_{2} B$. If this relation is further considered as a quadratic equation in $\\sqrt{r_{2}}$, then its discriminant, which equals  $$ \\Delta=4\\left(r(\\cot x+\\cot y) \\cot y-r_{1}(\\cot x \\cot y-1)\\right), $$  must be nonnegative, and therefore $r_{1} \\leq r \\cot y \\cot z$. Then $t_{1}, t_{2}, \\ldots$ exist, and we can assume that $t_{i} \\in[0, \\pi / 2]$. (b) Substituting $r_{1}=r \\cot y \\cot z \\sin ^{2} t_{1}, r_{2}=r \\cot z \\cot x \\sin ^{2} t_{2}$ in the relation of (a) we obtain that $\\sin ^{2} t_{1}+\\sin ^{2} t_{2}+k^{2}+2 k \\sin t_{1} \\sin t_{2}=1$, where we set $k=\\sqrt{\\tan x \\tan y}$. It follows that $\\left(k+\\sin t_{1} \\sin t_{2}\\right)^{2}=$ $\\left(1-\\sin ^{2} t_{1}\\right)\\left(1-\\sin ^{2} t_{2}\\right)=\\cos ^{2} t_{1} \\cos ^{2} t_{2}$, and hence  $$ \\cos \\left(t_{1}+t_{2}\\right)=\\cos t_{1} \\cos t_{2}-\\sin t_{1} \\sin t_{2}=k=\\sqrt{\\tan x \\tan y} $$  which is constant. Writing the analogous relations for each $t_{i}, t_{i+1}$ we conclude that $t_{1}+t_{2}=t_{4}+t_{5}, t_{2}+t_{3}=t_{5}+t_{6}$, and $t_{3}+t_{4}=t_{6}+t_{7}$. It follows that $t_{1}=t_{7}$, i.e., $K_{1}=K_{7}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1972", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (USS 2) ${ }^{\\mathrm{IMO} 1} \\mathrm{~A}$ set of 10 positive integers is given such that the decimal expansion of each of them has two digits. Prove that there are two disjoint subsets of the set with equal sums of their elements.", "solution": "12. First we observe that it is not essential to require the subsets to be disjoint (if they aren't, one simply excludes their intersection). There are $2^{10}-1=$ 1023 different subsets and at most 990 different sums. By the pigeonhole principle there are two different subsets with equal sums.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1972", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (CZS 1) We are given $3 n$ points $A_{1}, A_{2}, \\ldots, A_{3 n}$ in the plane, no three of them collinear. Prove that one can construct $n$ disjoint triangles with vertices at the points $A_{i}$.", "solution": "2. We use induction. For $n=1$ the assertion is obvious. Assume that it is true for a positive integer $n$. Let $A_{1}, A_{2}, \\ldots, A_{3 n+3}$ be given $3 n+3$ points, and let w.l.o.g. $A_{1} A_{2} \\ldots A_{m}$ be their convex hull. Among all the points $A_{i}$ distinct from $A_{1}, A_{2}$, we choose the one, say $A_{k}$, for which the angle $\\angle A_{k} A_{1} A_{2}$ is minimal (this point is uniquely determined, since no three points are collinear). The line $A_{1} A_{k}$ separates the plane into two half-planes, one of which contains $A_{2}$ only, and the other one all the remaining $3 n$ points. By the inductive hypothesis, one can construct $n$ disjoint triangles with vertices in these $3 n$ points. Together with the triangle $A_{1} A_{2} A_{k}$, they form the required system of disjoint triangles.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1972", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (CZS 4) Let $x_{1}, x_{2}, \\ldots, x_{n}$ be real numbers satisfying $x_{1}+x_{2}+\\cdots+x_{n}=$ 0 . Let $m$ be the least and $M$ the greatest among them. Prove that  $$ x_{1}^{2}+x_{2}^{2}+\\cdots+x_{n}^{2} \\leq-n m M $$", "solution": "3. We have for each $k=1,2, \\ldots, n$ that $m \\leq x_{k} \\leq M$, which gives $(M-$ $\\left.x_{k}\\right)\\left(m-x_{k}\\right) \\leq 0$. It follows directly that  $$ 0 \\geq \\sum_{k=1}^{n}\\left(M-x_{k}\\right)\\left(m-x_{k}\\right)=n m M-(m+M) \\sum_{k=1}^{n} x_{k}+\\sum_{k=1}^{n} x_{k}^{2} $$  But $\\sum_{k=1}^{n} x_{k}=0$, implying the required inequality.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1972", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (GDR 1) Let $n_{1}, n_{2}$ be positive integers. Consider in a plane $E$ two disjoint sets of points $M_{1}$ and $M_{2}$ consisting of $2 n_{1}$ and $2 n_{2}$ points, respectively, and such that no three points of the union $M_{1} \\cup M_{2}$ are collinear. Prove that there exists a straightline $g$ with the following property: Each of the two half-planes determined by $g$ on $E$ ( $g$ not being included in either) contains exactly half of the points of $M_{1}$ and exactly half of the points of $M_{2}$.", "solution": "4. Choose in $E$ a half-line $s$ beginning at a point $O$. For every $\\alpha$ in the interval $\\left[0,180^{\\circ}\\right]$, denote by $s(\\alpha)$ the line obtained by rotation of $s$ about $O$ by $\\alpha$, and by $g(\\alpha)$ the oriented line containing $s(\\alpha)$ on which $s(\\alpha)$ defines the positive direction. For each $P$ in $M_{i}, i=1,2$, let $P(\\alpha)$ be the foot of the perpendicular from $P$ to $g(\\alpha)$, and $l_{P}(\\alpha)$ the oriented (positive, negative or zero) distance of $P(\\alpha)$ from $O$. Then for $i=1,2$ one can arrange the $l_{P}(\\alpha)\\left(P \\in M_{i}\\right)$ in ascending order, as $l_{1}(\\alpha), l_{2}(\\alpha), \\ldots, l_{2 n_{i}}(\\alpha)$. Call $J_{i}(\\alpha)$ the interval $\\left[l_{n_{i}}(\\alpha), l_{n_{i}+1}(\\alpha)\\right]$. It is easy to see that any line perpendicular to $g(\\alpha)$ and passing through the point with the distance $l$ in the interior of $J_{i}(\\alpha)$ from $O$, will divide the set $M_{i}$ into two subsets of equal cardinality. Therefore it remains to show that for some $\\alpha$, the interiors of intervals $J_{1}(\\alpha)$ and $J_{2}(\\alpha)$ have a common point. If this holds for $\\alpha=0$, then we have finished. Suppose w.l.o.g. that $J_{1}(0)$ lies on $g(0)$ to the left of $J_{2}(0)$; then $J_{1}\\left(180^{\\circ}\\right)$ lies to the right of $J_{2}\\left(180^{\\circ}\\right)$. Note that $J_{1}$ and $J_{2}$ cannot simultaneously degenerate to a point (otherwise, we would have four collinear points in $M_{1} \\cup M_{2}$ ); also, each of them degenerates to a point for only finitely many values of $\\alpha$. Since $J_{1}(\\alpha)$ and $J_{2}(\\alpha)$ move continuously, there exists a subinterval $I$ of $\\left[0,180^{\\circ}\\right]$ on which they are not disjoint. Thus, at some point of $I$, they are both nondegenerate and have a common interior point, as desired.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1972", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (GDR 2) Prove the following assertion: The four altitudes of a tetrahedron $A B C D$ intersect in a point if and only if  $$ A B^{2}+C D^{2}=B C^{2}+A D^{2}=C A^{2}+B D^{2} $$", "solution": "5. Lemma. If $X, Y, Z, T$ are points in space, then the lines $X Z$ and $Y T$ are perpendicular if and only if $X Y^{2}+Z T^{2}=Y Z^{2}+T X^{2}$. Proof. Consider the plane $\\pi$ through $X Z$ parallel to $Y T$. If $Y^{\\prime}, T^{\\prime}$ are the feet of the perpendiculars to $\\pi$ from $Y, T$ respectively, then  $$ \\begin{aligned} & \\\\ & X Y^{2}+Z T^{2}=X Y^{\\prime 2}+Z T^{\\prime 2}+2 Y Y^{\\prime 2} \\\\ & \\text { and } \\quad Y Z^{2}+T X^{2}=Y^{\\prime} Z^{2}+T^{\\prime} X^{2}+2 Y Y^{\\prime 2} \\end{aligned} $$  Since by the Pythagorean theorem $X Y^{\\prime 2}+Z T^{\\prime 2}=Y^{\\prime} Z^{2}+T^{\\prime} X^{2}$, i.e., $X Y^{\\prime 2}-Y^{\\prime} Z^{2}=X T^{\\prime 2}-T^{\\prime} Z^{2}$, if and only if $Y^{\\prime} T^{\\prime} \\perp X Z$, the statement follows. Assume that the four altitudes intersect in a point $P$. Then we have $D P \\perp$ $A B C \\Rightarrow D P \\perp A B$ and $C P \\perp A B D \\Rightarrow C P \\perp A B$, which implies that $C D P \\perp A B$, and $C D \\perp A B$. By the lemma, $A C^{2}+B D^{2}=A D^{2}+B C^{2}$. Using the same procedure we obtain the relation $A D^{2}+B C^{2}=A B^{2}+$ $C D^{2}$ 。 Conversely, assume that $A B^{2}+C D^{2}=A C^{2}+B D^{2}=A D^{2}+B C^{2}$. The lemma implies that $A B \\perp C D, A C \\perp B D, A D \\perp B C$. Let $\\pi$ be the plane containing $C D$ that is perpendicular to $A B$, and let $h_{D}$ be the altitude from $D$ to $A B C$. Since $\\pi \\perp A B$, we have $\\pi \\perp A B C \\Rightarrow h_{D} \\subset \\pi$ and $\\pi \\perp A B D \\Rightarrow h_{C} \\subset \\pi$. The altitudes $h_{D}$ and $h_{C}$ are not parallel; thus they have an intersection point $P_{C D}$. Analogously, $h_{B} \\cap h_{C}=\\left\\{P_{B C}\\right\\}$ and $h_{B} \\cap h_{D}=\\left\\{P_{B D}\\right\\}$, where both these points belong to $\\pi$. On the other hand, $h_{B}$ doesn't belong to $\\pi$; otherwise, it would be perpendicular to both $A C D$ and $A B \\subset \\pi$, i.e. $A B \\subset A C D$, which is impossible. Hence, $h_{B}$ can have at most one common point with $\\pi$, implying $P_{B D}=P_{C D}$. Analogously, $P_{A B}=P_{B D}=P_{C D}=P_{A B C D}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1972", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (GDR 3) Show that for any $n \\not \\equiv 0(\\bmod 10)$ there exists a multiple of $n$ not containing the digit 0 in its decimal expansion.", "solution": "6. Let $n=2^{\\alpha} 5^{\\beta} m$, where $\\alpha=0$ or $\\beta=0$. These two cases are analogous, and we treat only $\\alpha=0, n=5^{\\beta} m$. The case $m=1$ is settled by the following lemma. Lemma. For any integer $\\beta \\geq 1$ there exists a multiple $M_{\\beta}$ of $5^{\\beta}$ with $\\beta$ digits in decimal expansion, all different from 0. Proof. For $\\beta=1, M_{1}=5$ works. Assume that the lemma is true for $\\beta=k$. There is a positive integer $C_{k} \\leq 5$ such that $C_{k} 2^{k}+m_{k} \\equiv$ $0(\\bmod 5)$, where $5^{k} m_{k}=M_{k}$, i.e. $C_{k} 10^{k}+M_{k} \\equiv 0\\left(\\bmod 5^{k+1}\\right)$. Then $M_{k+1}=C_{k} 10^{k}+M_{k}$ satisfies the conditions, and proves the lemma.  In the general case, consider, the sequence $1,10^{\\beta}, 10^{2 \\beta}, \\ldots$ It contains two numbers congruent modulo $\\left(10^{\\beta}-1\\right) m$, and therefore for some $k>0$, $10^{k \\beta} \\equiv 1\\left(\\bmod \\left(10^{\\beta}-1\\right) m\\right)$ (this is in fact a consequence of Fermat's theorem). The number  $$ \\frac{10^{k \\beta}-1}{10^{\\beta}-1} M_{\\beta}=10^{(k-1) \\beta} M_{\\beta}+10^{(k-2) \\beta} M_{\\beta}+\\cdots+M_{\\beta} $$  is a multiple of $n=5^{\\beta} m$ with the required property.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1972", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. $(\\mathbf{G B R} 1)^{\\mathrm{IMO} 6}$ (a) A plane $\\pi$ passes through the vertex $O$ of the regular tetrahedron $O P Q R$. We define $p, q, r$ to be the signed distances of $P, Q, R$ from $\\pi$ measured along a directed normal to $\\pi$. Prove that  $$ p^{2}+q^{2}+r^{2}+(q-r)^{2}+(r-p)^{2}+(p-q)^{2}=2 a^{2} $$  where $a$ is the length of an edge of a tetrahedron. (b) Given four parallel planes not all of which are coincident, show that a regular tetrahedron exists with a vertex on each plane.", "solution": "7. (i) Consider the circumscribing cube $O Q_{1} P R_{1} O_{1} Q P_{1} R$ (that is, the cube in which the edges of the tetrahedron are small diagonals), of side $b=a \\sqrt{2} / 2$. The left-hand side is the sum of squares of the projections of the edges of the tetrahedron onto a perpendicular $l$ to $\\pi$. On the other hand, if $l$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-395.jpg?height=363&width=333&top_left_y=609&top_left_x=936) forms angles $\\varphi_{1}, \\varphi_{2}, \\varphi_{3}$ with $O O_{1}, O Q_{1}, O R_{1}$ respectively, then the projections of $O P$ and $Q R$ onto $l$ have lengths $b\\left(\\cos \\varphi_{2}+\\cos \\varphi_{3}\\right)$ and $b\\left|\\cos \\varphi_{2}-\\cos \\varphi_{3}\\right|$. Summing up all these expressions, we obtain  $$ 4 b^{2}\\left(\\cos ^{2} \\varphi_{1}+\\cos ^{2} \\varphi_{2}+\\cos ^{2} \\varphi_{3}\\right)=4 b^{2}=2 a^{2} $$  (ii) We construct a required tetrahedron of edge length $a$ given in (i). Take $O$ arbitrarily on $\\pi_{0}$, and let $p, q, r$ be the distances of $O$ from $\\pi_{1}, \\pi_{2}, \\pi_{3}$. Since $a>p, q, r,|p-q|$, we can choose $P$ on $\\pi_{1}$ anywhere at distance $a$ from $O$, and $Q$ at one of the two points on $\\pi_{2}$ at distance $a$ from both $O$ and $P$. Consider the fourth vertex of the tetrahedron: its distance from $\\pi_{0}$ will satisfy the equation from (i); i.e., there are two values for this distance; clearly, one of them is $r$, putting $R$ on $\\pi_{3}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1972", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (GBR 2) ${ }^{\\mathrm{IMO} 3}$ Let $m$ and $n$ be nonnegative integers. Prove that $m!n!(m+$ $n)$ ! divides $(2 m)!(2 n)!$.", "solution": "8. Let $f(m, n)=\\frac{(2 m)!(2 n)!}{m!n!(m+n)!}$. Then it is directly shown that  $$ f(m, n)=4 f(m, n-1)-f(m+1, n-1), $$  and thus $n$ may be successively reduced until one obtains $f(m, n)=$ $\\sum_{r} c_{r} f(r, 0)$. Now $f(r, 0)$ is a simple binomial coefficient, and the $c_{r}$ 's are integers. Second solution. For each prime $p$, the greatest exponents of $p$ that divide the numerator $(2 m)!(2 n)$ ! and denominator $m!n!(m+n)$ ! are respectively  $$ \\sum_{k>0}\\left(\\left[\\frac{2 m}{p^{k}}\\right]+\\left[\\frac{2 n}{p^{k}}\\right]\\right) \\quad \\text { and } \\quad \\sum_{k>0}\\left(\\left[\\frac{m}{p^{k}}\\right]+\\left[\\frac{n}{p^{k}}\\right]+\\left[\\frac{m+n}{p^{k}}\\right]\\right) $$  hence it suffices to show that the first exponent is not less than the second one for every $p$. This follows from the fact that for each real $x,[2 x]+[2 y] \\geq$ $[x]+[y]+[x+y]$, which is straightforward to prove (for example, using $[2 x]=[x]+[x+1 / 2])$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1972", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (NET 2) $)^{\\mathrm{IMO} 4}$ Find all solutions in positive real numbers $x_{i}(i=$ $1,2,3,4,5)$ of the following system of inequalities:  $$ \\begin{aligned} & \\left(x_{1}^{2}-x_{3} x_{5}\\right)\\left(x_{2}^{2}-x_{3} x_{5}\\right) \\leq 0, \\\\ & \\left(x_{2}^{2}-x_{4} x_{1}\\right)\\left(x_{3}^{2}-x_{4} x_{1}\\right) \\leq 0, \\\\ & \\left(x_{3}^{2}-x_{5} x_{2}\\right)\\left(x_{4}^{2}-x_{5} x_{2}\\right) \\leq 0, \\\\ & \\left(x_{4}^{2}-x_{1} x_{3}\\right)\\left(x_{5}^{2}-x_{1} x_{3}\\right) \\leq 0, \\\\ & \\left(x_{5}^{2}-x_{2} x_{4}\\right)\\left(x_{1}^{2}-x_{2} x_{4}\\right) \\leq 0 . \\end{aligned} $$", "solution": "9. Clearly $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}$ is a solution. We shall show that this describes all solutions. Suppose that not all $x_{i}$ are equal. Then among $x_{3}, x_{5}, x_{2}, x_{4}, x_{1}$ two consecutive are distinct: Assume w.l.o.g. that $x_{3} \\neq x_{5}$. Moreover, since $\\left(1 / x_{1}, \\ldots, 1 / x_{5}\\right)$ is a solution whenever $\\left(x_{1}, \\ldots, x_{5}\\right)$ is, we may assume that $x_{3}<x_{5}$. Consider first the case $x_{1} \\leq x_{2}$. We infer from (i) that $x_{1} \\leq \\sqrt{x_{3} x_{5}}<x_{5}$ and $x_{2} \\geq \\sqrt{x_{3} x_{5}}>x_{3}$. Then $x_{5}^{2}>x_{1} x_{3}$, which together with (iv) gives $x_{4}^{2} \\leq x_{1} x_{3}<x_{3} x_{5}$; but we also have $x_{3}^{2} \\leq x_{5} x_{2}$; hence by (iii), $x_{4}^{2} \\geq$ $x_{5} x_{2}>x_{5} x_{3}$, a contradiction. Consider next the case $x_{1}>x_{2}$. We infer from (i) that $x_{1} \\geq \\sqrt{x_{3} x_{5}}>x_{3}$ and $x_{2} \\leq \\sqrt{x_{3} x_{5}}<x_{5}$. Then by (ii) and (v),  $$ x_{1} x_{4} \\leq \\max \\left(x_{2}^{2}, x_{3}^{2}\\right) \\leq x_{3} x_{5} \\quad \\text { and } \\quad x_{2} x_{4} \\geq \\min \\left(x_{1}^{2}, x_{5}^{2}\\right) \\geq x_{3} x_{5} $$  which contradicts the assumption $x_{1}>x_{2}$. Second solution.  $$ \\begin{aligned} 0 & \\geq L_{1}=\\left(x_{1}^{2}-x_{3} x_{5}\\right)\\left(x_{2}^{2}-x_{3} x_{5}\\right)=x_{1}^{2} x_{2}^{2}+x_{3}^{2} x_{5}^{2}-\\left(x_{1}^{2}+x_{2}^{2}\\right) x_{3} x_{5} \\\\ & \\geq x_{1}^{2} x_{2}^{2}+x_{3}^{2} x_{5}^{2}-\\frac{1}{2}\\left(x_{1}^{2} x_{3}^{2}+x_{1}^{2} x_{5}^{2}+x_{2}^{2} x_{3}^{2}+x_{2}^{2} x_{5}^{2}\\right) \\end{aligned} $$  and analogously for $L_{2}, \\ldots, L_{5}$. Therefore $L_{1}+L_{2}+L_{3}+L_{4}+L_{5} \\geq 0$, with the only case of equality $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (BUL 6) Let a tetrahedron $A B C D$ be inscribed in a sphere $S$. Find the locus of points $P$ inside the sphere $S$ for which the equality  $$ \\frac{A P}{P A_{1}}+\\frac{B P}{P B_{1}}+\\frac{C P}{P C_{1}}+\\frac{D P}{P D_{1}}=4 $$  holds, where $A_{1}, B_{1}, C_{1}$, and $D_{1}$ are the intersection points of $S$ with the lines $A P, B P, C P$, and $D P$, respectively.", "solution": "1. The condition of the point $P$ can be written in the form $\\frac{A P^{2}}{A P \\cdot P A_{1}}+\\frac{B P^{2}}{B P \\cdot P B_{1}}+$ $\\frac{C P^{2}}{C P \\cdot P C_{1}}+\\frac{D P^{2}}{D P \\cdot P D_{1}}=4$. All the four denominators are equal to $R^{2}-O P^{2}$, i.e., to the power of $P$ with respect to $S$. Thus the condition becomes  $$ A P^{2}+B P^{2}+C P^{2}+D P^{2}=4\\left(R^{2}-O P^{2}\\right) $$  Let $M$ and $N$ be the midpoints of segments $A B$ and $C D$ respectively, and $G$ the midpoint of $M N$, or the centroid of $A B C D$. By Stewart's formula, an arbitrary point $P$ satisfies  $$ \\begin{aligned} A P^{2}+B P^{2}+C P^{2}+D P^{2} & =2 M P^{2}+2 N P^{2}+\\frac{1}{2} A B^{2}+\\frac{1}{2} C D^{2} \\\\ & =4 G P^{2}+M N^{2}+\\frac{1}{2}\\left(A B^{2}+C D^{2}\\right) \\end{aligned} $$  Particularly, for $P \\equiv O$ we get $4 R^{2}=4 O G^{2}+M N^{2}+\\frac{1}{2}\\left(A B^{2}+C D^{2}\\right)$, and the above equality becomes  $$ A P^{2}+B P^{2}+C P^{2}+D P^{2}=4 G P^{2}+4 R^{2}-4 O G^{2} $$  Therefore (1) is equivalent to $O G^{2}=O P^{2}+G P^{2} \\Leftrightarrow \\angle O P G=90^{\\circ}$. Hence the locus of points $P$ is the sphere with diameter $O G$. Now the converse is easy.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (SWE 3) ${ }^{\\mathrm{IMO} 6}$ Let $a_{1}, a_{2}, \\ldots, a_{n}$ be positive numbers and $q$ a given real number, $0<q<1$. Find $n$ real numbers $b_{1}, b_{2}, \\ldots, b_{n}$ that satisfy: (1) $a_{k}<b_{k}$ for all $k=1,2, \\ldots, n$; (2) $q<\\frac{b_{k+1}}{b_{k}}<\\frac{1}{q}$ for all $k=1,2, \\ldots, n-1$; (3) $b_{1}+b_{2}+\\cdots+b_{n}<\\frac{1+q}{1-q}\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)$.", "solution": "10. Let $b_{k}=a_{1} q^{k-1}+\\cdots+a_{k-1} q+a_{k}+a_{k+1} q+\\cdots+a_{n} q^{n-k}, \\quad k=1,2, \\ldots, n$. We show that these numbers satisfy the required conditions. Obviously $b_{k} \\geq a_{k}$. Further,  $$ b_{k+1}-q b_{k}=-\\left[\\left(q^{2}-1\\right) a_{k+1}+\\cdots+q^{n-k-1}\\left(q^{2}-1\\right) a_{n}\\right]>0 $$  we analogously obtain $q b_{k+1}-b_{k}<0$. Finally,  $$ \\begin{aligned} b_{1}+b_{2}+\\cdots+b_{n}= & a_{1}\\left(q^{n-1}+\\cdots+q+1\\right)+\\ldots \\\\ & +a_{k}\\left(q^{n-k}+\\cdots+q+1+q+\\cdots+q^{k-1}\\right)+\\ldots \\\\ \\leq & \\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)\\left(1+2 q+2 q^{2}+\\cdots+2 q^{n-1}\\right) \\\\ < & \\frac{1+q}{1-q}\\left(a_{1}+\\cdots+a_{n}\\right) \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (SWE 4) ${ }^{\\mathrm{IMO} 3}$ Determine the minimum of $a^{2}+b^{2}$ if $a$ and $b$ are real numbers for which the equation  $$ x^{4}+a x^{3}+b x^{2}+a x+1=0 $$  has at least one real solution.", "solution": "11. Putting $x+\\frac{1}{x}=t$ we also get $x^{2}+\\frac{1}{x^{2}}=t^{2}-2$, and the given equation reduces to $t^{2}+a t+b-2=0$. Since $x=\\frac{t \\pm \\sqrt{t^{2}-4}}{2}, x$ will be real if and only if $|t| \\geq 2, t \\in \\mathbb{R}$. Thus we need the minimum value of $a^{2}+b^{2}$ under the condition $a t+b=-\\left(t^{2}-2\\right),|t| \\geq 2$. However, by the Cauchy-Schwarz inequality we have  $$ \\left(a^{2}+b^{2}\\right)\\left(t^{2}+1\\right) \\geq(a t+b)^{2}=\\left(t^{2}-2\\right)^{2} $$  It follows that $a^{2}+b^{2} \\geq h(t)=\\frac{\\left(t^{2}-2\\right)^{2}}{t^{2}+1}$. Since $h(t)=\\left(t^{2}+1\\right)+\\frac{9}{t^{2}+1}-6$ is increasing for $t \\geq 2$, we conclude that $a^{2}+b^{2} \\geq h(2)=\\frac{4}{5}$. The cases of equality are easy to examine: These are $a= \\pm \\frac{4}{5}$ and $b=-\\frac{2}{5}$. Second solution. In fact, there was no need for considering $x=t+1 / t$. By the Cauchy-Schwarz inequality we have $\\left(a^{2}+2 b^{2}+a^{2}\\right)\\left(x^{6}+x^{4} / 2+x^{2}\\right) \\geq$ $\\left(a x^{3}+b x^{2}+a x\\right)^{2}=\\left(x^{4}+1\\right)^{2}$. Hence  $$ a^{2}+b^{2} \\geq \\frac{\\left(x^{4}+1\\right)^{2}}{2 x^{6}+x^{4}+2 x^{2}} \\geq \\frac{4}{5} $$  with equality for $x=1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (SWE 6) Consider the two square matrices  $$ A=\\left[\\begin{array}{rrrrr} 1 & 1 & 1 & 1 & 1 \\\\ 1 & 1 & 1 & -1 & -1 \\\\ 1 & -1 & -1 & 1 & 1 \\\\ 1 & -1 & -1 & -1 & 1 \\\\ 1 & 1 & -1 & 1 & -1 \\end{array}\\right] \\quad \\text { and } \\quad B=\\left[\\begin{array}{rrrrr} 1 & 1 & 1 & 1 & 1 \\\\ 1 & 1 & 1 & -1 & -1 \\\\ 1 & 1 & -1 & 1 & -1 \\\\ 1 & -1 & -1 & 1 & 1 \\\\ 1 & -1 & 1 & -1 & 1 \\end{array}\\right] $$  with entries 1 and -1 . The following operations will be called elementary: (1) Changing signs of all numbers in one row; (2) Changing signs of all numbers in one column; (3) Interchanging two rows (two rows exchange their positions); (4) Interchanging two columns.  Prove that the matrix $B$ cannot be obtained from the matrix $A$ using these operations.", "solution": "12. Observe that the absolute values of the determinants of the given matrices are invariant under all the admitted operations. The statement follows from $\\operatorname{det} A=16 \\neq \\operatorname{det} B=0$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (YUG 4) Find the sphere of maximal radius that can be placed inside every tetrahedron that has all altitudes of length greater than or equal to 1.", "solution": "13. Let $S_{1}, S_{2}, S_{3}, S_{4}$ denote the areas of the faces of the tetrahedron, $V$ its volume, $h_{1}, h_{2}, h_{3}, h_{4}$ its altitudes, and $r$ the radius of its inscribed sphere. Since  $$ 3 V=S_{1} h_{1}=S_{2} h_{2}=S_{3} h_{3}=S_{4} h_{4}=\\left(S_{1}+S_{2}+S_{3}+S_{4}\\right) r, $$  it follows that  $$ \\frac{1}{h_{1}}+\\frac{1}{h_{2}}+\\frac{1}{h_{3}}+\\frac{1}{h_{4}}=\\frac{1}{r} . $$  In our case, $h_{1}, h_{2}, h_{3}, h_{4} \\geq 1$, hence $r \\geq 1 / 4$. On the other hand, it is clear that a sphere of radius greater than $1 / 4$ cannot be inscribed in a tetrahedron all of whose altitudes have length equal to 1 . Thus the answer is $1 / 4$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (YUG 5) ${ }^{\\mathrm{IMO} 4} \\mathrm{~A}$ soldier has to investigate whether there are mines in an area that has the form of an equilateral triangle. The radius of his detector is equal to one-half of an altitude of the triangle. The soldier starts from one vertex of the triangle. Determine the shortest path that the soldier has to traverse in order to check the whole region.", "solution": "14. Suppose that the soldier starts at the vertex $A$ of the equilateral triangle $A B C$ of side length $a$. Let $\\varphi, \\psi$ be the arcs of circles with centers $B$ and $C$ and radii $a \\sqrt{3} / 4$ respectively, that lie inside the triangle. In order to check the vertices $B, C$, he must visit some points $D \\in \\varphi$ and $E \\in \\psi$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-401.jpg?height=324&width=394&top_left_y=1496&top_left_x=887)  Thus his path cannot be shorter than the path $A D E$ (or $A E D$ ) itself. The length of the path $A D E$ is $A D+D E \\geq A D+D C-a \\sqrt{3} / 4$. Let $F$ be the reflection of $C$ across the line $M N$, where $M, N$ are the midpoints of $A B$ and $B C$. Then $D C \\geq D F$ and hence $A D+D C \\geq A D+D F \\geq A F$. Consequently $A D+D E \\geq A F-a \\frac{\\sqrt{3}}{4}=a\\left(\\frac{\\sqrt{7}}{2}-\\frac{\\sqrt{3}}{4}\\right)$, with equality if and only if $D$ is the midpoint of $\\operatorname{arc} \\varphi$ and $E=(C D) \\cap \\psi$.  Moreover, it is easy to verify that, in following the path $A D E$, the soldier will check the whole region. Therefore this path (as well as the one symmetric to it) is shortest possible path that the soldier can take in order to check the entire field.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (CUB 1) Prove that for all $n \\in \\mathbb{N}$ the following is true:  $$ 2^{n} \\prod_{k=1}^{n} \\sin \\frac{k \\pi}{2 n+1}=\\sqrt{2 n+1} $$", "solution": "15. If $z=\\cos \\theta+i \\sin \\theta$, then $z-z^{-1}=2 i \\sin \\theta$. Now put $z=\\cos \\frac{\\pi}{2 n+1}+$ $i \\sin \\frac{\\pi}{2 n+1}$. Using de Moivre's formula we transform the required equality into  $$ A=\\prod_{k=1}^{n}\\left(z^{k}-z^{-k}\\right)=i^{n} \\sqrt{2 n+1} $$  On the other hand, the complex numbers $z^{2 k}(k=-n,-n+1, \\ldots, n)$ are the roots of $x^{2 n+1}-1$, and hence  $$ \\prod_{k=1}^{n}\\left(x-z^{2 k}\\right)\\left(x-z^{-2 k}\\right)=\\frac{x^{2 n+1}-1}{x-1}=x^{2 n}+\\cdots+x+1 $$  Now we go back to proving (1). We have  $$ (-1)^{n} z^{n(n+1) / 2} A=\\prod_{k=1}^{n}\\left(1-z^{2 k}\\right) \\quad \\text { and } \\quad z^{-n(n+1) / 2} A=\\prod_{k=1}^{n}\\left(1-z^{-2 k}\\right) $$  Multiplying these two equalities, we obtain $(-1)^{n} A^{2}=\\prod_{k=1}^{n}\\left(1-z^{2 k}\\right)(1-$ $\\left.z^{-2 k}\\right)=2 n+1$, by (2). Therefore $A= \\pm i^{-n} \\sqrt{2 n+1}$. This actually implies that the required product is $\\pm \\sqrt{2 n+1}$, but it must be positive, since all the sines are, and the result follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (CUB 2) Given $a, \\theta \\in \\mathbb{R}, m \\in \\mathbb{N}$, and $P(x)=x^{2 m}-2|a|^{m} x^{m} \\cos \\theta+a^{2 m}$, factorize $P(x)$ as a product of $m$ real quadratic polynomials.", "solution": "16. First, we have $P(x)=Q(x) R(x)$ for $Q(x)=x^{m}-|a|^{m} e^{i \\theta}$ and $R(x)=$ $x^{m}-|a|^{m} e^{-i \\theta}$, where $e^{i \\varphi}$ means of course $\\cos \\varphi+i \\sin \\varphi$. It remains to factor both $Q$ and $R$. Suppose that $Q(x)=\\left(x-q_{1}\\right) \\cdots\\left(x-q_{m}\\right)$ and $R(x)=\\left(x-r_{1}\\right) \\cdots\\left(x-r_{m}\\right)$. Considering $Q(x)$, we see that $\\left|q_{k}^{m}\\right|=|a|^{m}$ and also $\\left|q_{k}\\right|=|a|$ for $k=$ $1, \\ldots, m$. Thus we may put $q_{k}=|a| e^{i \\beta_{k}}$ and obtain by de Moivre's formula $q_{k}^{m}=|a|^{m} e^{i m \\beta_{k}}$. It follows that $m \\beta_{k}=\\theta+2 j \\pi$ for some $j \\in \\mathbb{Z}$, and we have exactly $m$ possibilities for $\\beta_{k}$ modulo $2 \\pi$ : $\\beta_{k}=\\frac{\\theta+2(k-1) \\pi}{m}$ for $k=1,2, \\ldots, m$. Thus $q_{k}=|a| e^{i \\beta_{k}}$; analogously we obtain for $R(x)$ that $r_{k}=|a| e^{-i \\beta_{k}}$. Consequently, $x^{m}-|a|^{m} e^{i \\theta}=\\prod_{k=1}^{m}\\left(x-|a| e^{i \\beta_{k}}\\right) \\quad$ and $\\quad x^{m}-|a|^{m} e^{-i \\theta}=\\prod_{k=1}^{m}\\left(x-|a| e^{-i \\beta_{k}}\\right)$. Finally, grouping the $k$ th factors of both polynomials, we get  $$ \\begin{aligned} P(x) & =\\prod_{k=1}^{m}\\left(x-|a| e^{i \\beta_{k}}\\right)\\left(x-|a| e^{-i \\beta_{k}}\\right)=\\prod_{k=1}^{m}\\left(x^{2}-2|a| x \\cos \\beta_{k}+a^{2}\\right) \\\\ & =\\prod_{k=1}^{m}\\left(x^{2}-2|a| x \\cos \\frac{\\theta+2(k-1) \\pi}{m}+a^{2}\\right) \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (POL 1) ${ }^{\\mathrm{IMO} 5}$ Let $\\mathcal{F}$ be a nonempty set of functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ of the form $f(x)=a x+b$, where $a$ and $b$ are real numbers and $a \\neq 0$. Suppose that $\\mathcal{F}$ satisfies the following conditions: (1) If $f, g \\in \\mathcal{F}$, then $g \\circ f \\in \\mathcal{F}$, where $(g \\circ f)(x)=g[f(x)]$. (2) If $f \\in \\mathcal{F}$ and $f(x)=a x+b$, then the inverse $f^{-1}$ of $f$ belongs to $\\mathcal{F}$ $\\left(f^{-1}(x)=(x-b) / a\\right)$. (3) None of the functions $f(x)=x+c$, for $c \\neq 0$, belong to $\\mathcal{F}$.  Prove that there exists $x_{0} \\in \\mathbb{R}$ such that $f\\left(x_{0}\\right)=x_{0}$ for all $f \\in \\mathcal{F}$.", "solution": "17. Let $f_{1}(x)=a x+b$ and $f_{2}(x)=c x+d$ be two functions from $\\mathcal{F}$. We define $g(x)=f_{1} \\circ f_{2}(x)=a c x+(a d+b)$ and $h(x)=f_{2} \\circ f_{1}(x)=a c x+(b c+d)$.  By the condition for $\\mathcal{F}$, both $g(x)$ and $h(x)$ belong to $\\mathcal{F}$. Moreover, there exists $h^{-1}(x)=\\frac{x-(b c+d)}{a c}$, and  $$ h^{-1} \\circ g(x)=\\frac{a c x+(a d+b)-(b c+d)}{a c}=x+\\frac{(a d+b)-(b c+d)}{a c} $$  belongs to $\\mathcal{F}$. Now it follows that we must have $a d+b=b c+d$ for every $f_{1}, f_{2} \\in \\mathcal{F}$, which is equivalent to $\\frac{b}{a-1}=\\frac{d}{c-1}=k$. But these formulas exactly describe the fixed points of $f_{1}$ and $f_{2}: f_{1}(x)=a x+b=x \\Rightarrow x=$ $\\frac{b}{a-1}$. Hence all the functions in $\\mathcal{F}$ fix the point $k$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (CZS 1) Given a circle $K$, find the locus of vertices $A$ of parallelograms $A B C D$ with diagonals $A C \\leq B D$, such that $B D$ is inside $K$.", "solution": "2. Let $D^{\\prime}$ be the reflection of $D$ across $A$. Since $B C A D^{\\prime}$ is then a parallelogram, the condition $B D \\geq A C$ is equivalent to $B D \\geq B D^{\\prime}$, which is in turn equivalent to $\\angle B A D \\geq \\angle B A D^{\\prime}$, i.e. to $\\angle B A D \\geq 90^{\\circ}$. Thus the needed locus is actually the locus of points $A$ for which there exist points $B, D$ inside $K$ with $\\angle B A D=90^{\\circ}$. Such points $B, D$ exist if and only if the two tangents from $A$ to $K$, say $A P$ and $A Q$, determine an obtuse angle. Then if $P, Q \\in K$, we have $\\angle P A O=\\angle Q A O=\\varphi>45^{\\circ}$; hence $O A=\\frac{O P}{\\sin \\varphi}<O P \\sqrt{2}$. Therefore the locus of $A$ is the interior of the circle $K^{\\prime}$ with center $O$ and radius $\\sqrt{2}$ times the radius of $K$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (CZS 6) ${ }^{\\mathrm{IMO}}$ Prove that the sum of an odd number of unit vectors passing through the same point $O$ and lying in the same half-plane whose border passes through $O$ has length greater than or equal to 1 .", "solution": "3 . We use induction on odd numbers $n$. For $n=1$ there is nothing to prove. Suppose that the result holds for $n-2$ vectors, and let us be given vectors $v_{1}, v_{2}, \\ldots, v_{n}$ arranged clockwise. Set $v^{\\prime}=v_{2}+v_{3}+\\cdots+v_{n-1}, u=v_{1}+v_{n}$, and $v=v_{1}+v_{2}+\\cdots+v_{n}=v^{\\prime}+u$. By the inductive hypothesis we have $\\left|v^{\\prime}\\right| \\geq 1$. Now if the angles between $v^{\\prime}$ and the vectors $v_{1}, v_{n}$ are $\\alpha$ and $\\beta$ respectively, then the angle between $u$ and $v^{\\prime}$ is $|\\alpha-\\beta| / 2 \\leq 90^{\\circ}$. Hence $\\left|v^{\\prime}+u\\right| \\geq\\left|v^{\\prime}\\right| \\geq 1$. Second solution. Again by induction, it can be easily shown that all possible values of the sum $v=v_{1}+v_{2}+\\cdots+v_{n}$, for $n$ vectors $v_{1}, \\ldots, v_{n}$ in the upper half-plane (with $y \\geq 0$ ), are those for which $|v| \\leq n$ and $|v-k e| \\geq 1$ for every integer $k$ for which $n-k$ is odd, where $e$ is the unit vector on the $x$ axis.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (GBR 1) Let $P$ be a set of 7 different prime numbers and $C$ a set of 28 different composite numbers each of which is a product of two (not necessarily different) numbers from $P$. The set $C$ is divided into 7 disjoint four-element subsets such that each of the numbers in one set has a common prime divisor with at least two other numbers in that set. How many such partitions of $C$ are there?", "solution": "4. Each of the subsets must be of the form $\\left\\{a^{2}, a b, a c, a d\\right\\}$ or $\\left\\{a^{2}, a b, a c, b c\\right\\}$. It is now easy to count up the partitions. The result is 26460 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (FRA 2) A circle of radius 1 is located in a right-angled trihedron and touches all its faces. Find the locus of centers of such circles.", "solution": "5. Let $O$ be the vertex of the trihedron, $Z$ the center of a circle $k$ inscribed in the trihedron, and $A, B, C$ points in which the plane of the circle meets the edges of the trihedron. We claim that the distance $O Z$ is constant. Set $O A=x, O B=y, O C=z, B C=a, C A=b, A B=c$, and let $S$ and $r=1$ be the area and inradius of $\\triangle A B C$. Since $Z$ is the incenter of $A B C$, we have $(a+b+c) \\overrightarrow{O Z}=a \\overrightarrow{O A}+b \\overrightarrow{O B}+c \\overrightarrow{O C}$. Hence  $$ (a+b+c)^{2} O Z^{2}=(a \\overrightarrow{O A}+b \\overrightarrow{O B}+c \\overrightarrow{O C})^{2}=a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2} $$  But since $y^{2}+z^{2}=a^{2}, z^{2}+x^{2}=b^{2}$ and $x^{2}+y^{2}=c^{2}$, we obtain $x^{2}=\\frac{-a^{2}+b^{2}+c^{2}}{2}, y^{2}=\\frac{a^{2}-b^{2}+c^{2}}{2}, z^{2}=\\frac{a^{2}+b^{2}-c^{2}}{2}$. Substituting these values in (1) yields  $$ \\begin{aligned} (a+b+c)^{2} O Z^{2} & =\\frac{2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2}-a^{4}-b^{4}-c^{4}}{2} \\\\ & =8 S^{2}=2(a+b+c)^{2} r^{2} \\end{aligned} $$  Hence $O Z=r \\sqrt{2}=\\sqrt{2}$, and $Z$ belongs to a sphere $\\sigma$ with center $O$ and radius $\\sqrt{2}$. Moreover, the distances of $Z$ from the faces of the trihedron do not exceed 1 ; hence $Z$ belongs to a part of $\\sigma$ that lies inside the unit cube with three faces lying on the faces of the trihedron. It is easy to see that this part of $\\sigma$ is exactly the required locus.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (POL 2) ${ }^{\\mathrm{IMO} 2}$ Does there exist a finite set $M$ of points in space, not all in the same plane, such that for each two points $A, B \\in M$ there exist two other points $C, D \\in M$ such that lines $A B$ and $C D$ are parallel?", "solution": "6. Yes. Take for $\\mathcal{M}$ the set of vertices of a cube $A B C D E F G H$ and two points $I, J$ symmetric to the center $O$ of the cube with respect to the laterals $A B C D$ and $E F G H$. Remark. We prove a stronger result: Given an arbitrary finite set of points $\\mathcal{S}$, then there is a finite set $\\mathcal{M} \\supset \\mathcal{S}$ with the described property. Choose a point $A \\in \\mathcal{S}$ and any point $O$ such that $A O \\| B C$ for some two points $B, C \\in \\mathcal{S}$. Now let $X^{\\prime}$ be the point symmetric to $X$ with respect to $\\underline{O}$, and $\\mathcal{S}^{\\prime}=\\left\\{X, X^{\\prime} \\mid X \\in S\\right\\}$. Finally, take $\\mathcal{M}=\\left\\{X, \\bar{X} \\mid X \\in S^{\\prime}\\right\\}$, where $\\bar{X}$ denotes the point symmetric to $X$ with respect to $A$. This $\\mathcal{M}$ has the desired property: If $X, Y \\in \\mathcal{M}$ and $Y \\neq \\bar{X}$, then $X Y \\| \\overline{X Y}$; otherwise, $X \\bar{X}$, i.e., $X A$ is parallel to $X^{\\prime} A^{\\prime}$ if $X \\neq A^{\\prime}$, or to $B C$ otherwise.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (POL 3) Given a tetrahedron $A B C D$, let $x=A B \\cdot C D, y=A C \\cdot B D$, and $z=A D \\cdot B C$. Prove that there exists a triangle with edges $x, y, z$.", "solution": "7. The result follows immediately from Ptolemy's inequality.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (ROM 1) Prove that there are exactly $\\binom{k}{[k / 2]}$ arrays $a_{1}, a_{2}, \\ldots, a_{k+1}$ of nonnegative integers such that $a_{1}=0$ and $\\left|a_{i}-a_{i+1}\\right|=1$ for $i=1,2, \\ldots, k$.", "solution": "8. Let $f_{n}$ be the required total number, and let $f_{n}(k)$ denote the number of sequences $a_{1}, \\ldots, a_{n}$ of nonnegative integers such that $a_{1}=0, a_{n}=k$, and $\\left|a_{i}-a_{i+1}\\right|=1$ for $i=1, \\ldots, n-1$. In particular, $f_{1}(0)=1$ and $f_{n}(k)=0$ if $k<0$ or $k \\geq n$. Since $a_{n-1}$ is either $k-1$ or $k+1$, we have  $$ f_{n}(k)=f_{n-1}(k+1)+f_{n-1}(k-1) \\quad \\text { for } k \\geq 1 $$  By successive application of (1) we obtain  $$ f_{n}(k)=\\sum_{i=0}^{r}\\left[\\binom{r}{i}-\\binom{r}{i-k-1}\\right] f_{n-r}(k+r-2 i) . $$  This can be verified by direct induction. Substituting $r=n-1$ in (2), we get at most one nonzero summand, namely the one for which $i=\\frac{k+n-1}{2}$. Therefore $f_{n}(n-1-2 j)=\\binom{n-1}{j}-\\binom{n-1}{n-1}$. Adding up these equalities for $j=0,1, \\ldots,\\left[\\frac{n-1}{2}\\right]$ we obtain $f_{n}=\\binom{n-1}{\\left[\\frac{n-1}{2}\\right]}$, as required.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1973", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (ROM 2) Let $O x, O y, O z$ be three rays, and $G$ a point inside the trihedron $O x y z$. Consider all planes passing through $G$ and cutting $O x, O y, O z$ at points $A, B, C$, respectively. How is the plane to be placed in order to yield a tetrahedron $O A B C$ with minimal perimeter?", "solution": "9. Let $a, b, c$ be vectors going along $O x, O y, O z$, respectively, such that $\\overrightarrow{O G}=$ $\\xrightarrow{a+b}+c$. Now let $A \\in O x, B \\in O y, C \\in O z$ and let $\\overrightarrow{O A}=\\alpha a, \\overrightarrow{O B}=\\beta b$, $\\overrightarrow{O C}=\\gamma c$, where $\\alpha, \\beta, \\gamma>0$. Point $G$ belongs to a plane $A B C$ with $A \\in O x, B \\in O y, C \\in O z$ if and only if there exist positive real numbers $\\lambda, \\mu, \\nu$ with sum 1 such that $\\lambda \\overrightarrow{O A}+\\mu \\overrightarrow{O B}+\\nu \\overrightarrow{O C}=\\overrightarrow{O G}$, which is equivalent to $\\lambda \\alpha=\\mu \\beta=\\nu \\gamma=1$. Such $\\lambda, \\mu, \\nu$ exist if and only if  $$ \\alpha, \\beta, \\gamma>0 \\quad \\text { and } \\quad \\frac{1}{\\alpha}+\\frac{1}{\\beta}+\\frac{1}{\\gamma}=1 $$  Since the volume of $O A B C$ is proportional to the product $\\alpha \\beta \\gamma$, it is minimized when $\\frac{1}{\\alpha} \\cdot \\frac{1}{\\beta} \\cdot \\frac{1}{\\gamma}$ is maximized, which occurs when $\\alpha=\\beta=\\gamma=3$ and $G$ is the centroid of $\\triangle A B C$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1974", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. I 1 (USA 4) ${ }^{\\mathrm{IMO}}$ Alice, Betty, and Carol took the same series of examinations. There was one grade of $A$, one grade of $B$, and one grade of $C$ for each examination, where $A, B, C$ are different positive integers. The final test scores were  | Alice | Betty | Carol | | :---: | :---: | :---: | | 20 | 10 | 9 |  If Betty placed first in the arithmetic examination, who placed second in the spelling examination?", "solution": "1. Denote by $n$ the number of exams. We have $n(A+B+C)=20+10+9=39$, and since $A, B, C$ are distinct, their sum is at least 6 ; therefore $n=3$ and $A+B+C=13$. Assume w.l.o.g. that $A>B>C$. Since Betty gained $A$ points in arithmetic, but fewer than 13 points in total, she had $C$ points in both remaining exams (in spelling as well). Furthermore, Carol also gained fewer than 13 points, but with at least $B$ points on two examinations (on which Betty scored $C$ ), including spelling. If she had $A$ in spelling, then she would have at least $A+B+C=13$ points in total, a contradiction. Hence, Carol scored $B$ and placed second in spelling. Remark. Moreover, it follows that Alice, Betty, and Carol scored $B+A+A$, $A+C+C$, and $C+B+B$ respectively, and that $A=8, B=4, C=1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1974", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. II 4 (FIN 3) ${ }^{\\mathrm{IMO} 2}$ Let $\\triangle A B C$ be a triangle. Prove that there exists a point $D$ on the side $A B$ such that $C D$ is the geometric mean of $A D$ and $B D$ if and only if $\\sqrt{\\sin A \\sin B} \\leq \\sin \\frac{C}{2}$.", "solution": "10. If we set $\\angle A C D=\\gamma_{1}$ and $\\angle B C D=\\gamma_{2}$ for a point $D$ on the segment $A B$, then by the sine theorem,  $$ f(D)=\\frac{C D^{2}}{A D \\cdot B D}=\\frac{C D}{A D} \\cdot \\frac{C D}{B D}=\\frac{\\sin \\alpha \\sin \\beta}{\\sin \\gamma_{1} \\sin \\gamma_{2}} . $$  The denominator of the last fraction is  $$ \\begin{aligned} \\sin \\gamma_{1} \\sin \\gamma_{2} & =\\frac{1}{2}\\left(\\cos \\left(\\gamma_{1}-\\gamma_{2}\\right)-\\cos \\left(\\gamma_{1}+\\gamma_{2}\\right)\\right) \\\\ & =\\frac{1}{2}\\left(\\cos \\left(\\gamma_{1}-\\gamma_{2}\\right)-\\cos \\gamma\\right) \\leq \\frac{1-\\cos \\gamma}{2}=\\sin ^{2} \\frac{\\gamma}{2} \\end{aligned} $$  from which we deduce that the set of values of $f(D)$ is the interval $\\left[\\frac{\\sin \\alpha \\sin \\beta}{\\sin ^{2} \\frac{1}{2}},+\\infty\\right.$ ). Hence $f(D)=1$ (equivalently, $\\left.C D^{2}=A D \\cdot B D\\right)$ is possible if and only if $\\sin \\alpha \\sin \\beta \\leq \\sin ^{2} \\frac{\\gamma}{2}$, i.e.,  $$ \\sqrt{\\sin \\alpha \\sin \\beta} \\leq \\sin \\frac{\\gamma}{2} $$  Second solution. Let $E$ be the second point of intersection of the line $C D$ with the circumcircle $k$ of $A B C$. Since $A D \\cdot B D=C D \\cdot E D$ (power of $D$ with respect to $k$ ), $C D^{2}=A D \\cdot B D$ ie equivalent to $E D \\geq C D$. Clearly the ratio $\\frac{E D}{C D}(D \\in A B)$ takes a minimal value when $E$ is the midpoint of the arc $A B$ not containing $C$. (This follows from $E D: C D=E^{\\prime} D: C^{\\prime} D$ when $C^{\\prime}$ and $E^{\\prime}$ are respectively projections from $C$ and $E$ onto $A B$.) On the other hand, it is directly shown that in this case  $$ \\frac{E D}{C D}=\\frac{\\sin ^{2} \\frac{\\gamma}{2}}{\\sin \\alpha \\sin \\beta} $$  and the assertion follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1974", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. II 5 (BUL 1) ${ }^{\\mathrm{IMO} 4}$ Consider a partition of an $8 \\times 8$ chessboard into $p$ rectangles whose interiors are disjoint such that each of them has an equal number of white and black cells. Assume that $a_{1}<a_{2}<\\cdots<a_{p}$, where $a_{i}$ denotes the number of white cells in the $i$ th rectangle. Find the maximal $p$ for which such a partition is possible and for that $p$ determine all possible corresponding sequences $a_{1}, a_{2}, \\ldots, a_{p}$.", "solution": "11. First, we notice that $a_{1}+a_{2}+\\cdots+a_{p}=32$. The numbers $a_{i}$ are distinct, and consequently $a_{i} \\geq i$ and $a_{1}+\\cdots+a_{p} \\geq p(p+1) / 2$. Therefore $p \\leq 7$. The number 32 can be represented as a sum of 7 mutually distinct positive integers in the following ways: (1) $32=1+2+3+4+5+6+11$; (2) $32=1+2+3+4+5+7+10$; (3) $32=1+2+3+4+5+8+9$; (4) $32=1+2+3+4+6+7+9$; (5) $32=1+2+3+5+6+7+8$.  The case (1) is eliminated because there is no rectangle with 22 cells on an $8 \\times 8$ chessboard. In the other cases the partitions are realized as below. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-408.jpg?height=239&width=250&top_left_y=1281&top_left_x=234)  Case (2) ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-408.jpg?height=239&width=243&top_left_y=1281&top_left_x=518)  Case (3) ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-408.jpg?height=236&width=241&top_left_y=1282&top_left_x=808)  Case (4) ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-408.jpg?height=238&width=243&top_left_y=1279&top_left_x=1098)  Case (5)", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1974", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. II 6 (USS 1) In a certain language words are formed using an alphabet of three letters. Some words of two or more letters are not allowed, and any two such distinct words are of different lengths. Prove that one can form a word of arbitrary length that does not contain any nonallowed word.", "solution": "12. We say that a word is good if it doesn't contain any nonallowed word. Let $a_{n}$ be the number of good words of length $n$. If we prolong any good word of length $n$ by adding one letter to its end (there are $3 a_{n}$ words that can be so obtained), we get either (i) a good word of length $n+1$, or (ii) an $(n+1)$-letter word of the form $X Y$, where $X$ is a good word and $Y$ a nonallowed word. The number of words of type (ii) with word $Y$ of length $k$ is exactly $a_{n+1-k}$; hence the total number of words of kind (ii) doesn't exceed $a_{n-1}+$ $\\cdots+a_{1}+a_{0}$ (where $a_{0}=1$ ). Hence  $$ a_{n+1} \\geq 3 a_{n}-\\left(a_{n-1}+\\cdots+a_{1}+a_{0}\\right), \\quad a_{0}=1, a_{1}=3 $$  We prove by induction that $a_{n+1}>2 a_{n}$ for all $n$. For $n=1$ the claim is trivial. If it holds for $i \\leq n$, then $a_{i} \\leq 2^{i-n} a_{n}$; thus we obtain from (1)  $$ a_{n+1}>a_{n}\\left(3-\\frac{1}{2}-\\frac{1}{2^{2}}-\\cdots-\\frac{1}{2^{n}}\\right)>2 a_{n} $$  Therefore $a_{n} \\geq 2^{n}$ for all $n$ (moreover, one can show from (1) that $a_{n} \\geq$ $(n+2) 2^{n-1}$ ); hence there exist good words of length $n$. Remark. If there are two nonallowed words (instead of one) of each length greater than 1, the statement of the problem need not remain true.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1974", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. I 2 (POL 1) Prove that the squares with sides $1 / 1,1 / 2,1 / 3, \\ldots$ may be put into the square with side $3 / 2$ in such a way that no two of them have any interior point in common.", "solution": "2. We denote by $q_{i}$ the square with side $\\frac{1}{i}$. Let us divide the big square into rectangles $r_{i}$ by parallel lines, where the size of $r_{i}$ is $\\frac{3}{2} \\times \\frac{1}{2^{i}}$ for $i=2,3, \\ldots$ and $\\frac{3}{2} \\times 1$ for $i=1$ (this can be done because $1+\\sum_{i=2}^{\\infty} \\frac{1}{2^{i}}=\\frac{3}{2}$ ). In rectangle $r_{1}$, one can put the squares $q_{1}, q_{2}, q_{3}$, as is done on the figure. Also, since $\\frac{1}{2^{i}}+\\cdots+\\frac{1}{2^{i+1}-1}<2^{i} \\cdot \\frac{1}{2^{i}}=1<\\frac{3}{2}$, in each $r_{i}, i \\geq 2$, one can put $q_{2^{i}}, \\ldots, q_{2^{i+1}-1}$. This completes the proof. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-404.jpg?height=252&width=254&top_left_y=1103&top_left_x=664)  Remark. It can be shown that the squares $q_{1}, q_{2}$ cannot fit in any square of side less than $\\frac{3}{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1974", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. I 3 (SWE 3) ${ }^{\\mathrm{IMO}}$ Let $P(x)$ be a polynomial with integer coefficients. If $n(P)$ is the number of (distinct) integers $k$ such that $P^{2}(k)=1$, prove that  $$ n(P)-\\operatorname{deg}(P) \\leq 2 $$  where $\\operatorname{deg}(P)$ denotes the degree of the polynomial $P$.", "solution": "3. For $\\operatorname{deg}(P) \\leq 2$ the statement is obvious, since $n(P) \\leq \\operatorname{deg}\\left(P^{2}\\right)=$ $2 \\operatorname{deg}(P) \\leq \\operatorname{deg}(P)+2$ 。 Suppose now that $\\operatorname{deg}(P) \\geq 3$ and $n(P)>\\operatorname{deg}(P)+2$. Then there is at least one integer $b$ for which $P(b)=-1$, and at least one $x$ with $P(x)=1$. We may assume w.l.o.g. that $b=0$ (if necessary, we consider the polynomial $P(x+b)$ instead). If $k_{1}, \\ldots, k_{m}$ are all integers for which $P\\left(k_{i}\\right)=1$, then $P(x)=Q(x)\\left(x-k_{1}\\right) \\cdots\\left(x-k_{m}\\right)+1$ for some polynomial $Q(x)$ with integer coefficients. Setting $x=0$ we obtain $(-1)^{m} Q(0) k_{1} \\cdots k_{m}=1-P(0)=2$. It follows that $k_{1} \\cdots k_{m} \\mid 2$, and hence $m$ is at most 3 . The same holds for the polynomial $-P(x)$, and thus $P(x)=-1$ also has at most 3 integer solutions. This counts for 6 solutions of $P^{2}(x)=1$ in total, implying the statement for $\\operatorname{deg}(P) \\geq 4$. It remains to verify the statement for $n=3$. If $\\operatorname{deg}(P)=3$ and $n(P)=6$, then it follows from the above consideration that $P(x)$ is either $-\\left(x^{2}-\\right.$ $1)(x-2)+1$ or $\\left(x^{2}-1\\right)(x+2)+1$. It is directly checked that $n(P)$ equals only 4 in both cases.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1974", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. I 4 (USS 4) The sum of the squares of five real numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ equals 1. Prove that the least of the numbers $\\left(a_{i}-a_{j}\\right)^{2}$, where $i, j=$ $1,2,3,4,5$ and $i \\neq j$, does not exceed $1 / 10$.", "solution": "4. Assume w.l.o.g. that $a_{1} \\leq a_{2} \\leq a_{3} \\leq a_{4} \\leq a_{5}$. If $m$ is the least value of $\\left|a_{i}-a_{j}\\right|, i \\neq j$, then $a_{i+1}-a_{i} \\geq m$ for $i=1,2, \\ldots, 5$, and consequently $a_{i}-a_{j} \\geq(i-j) m$ for any $i, j \\in\\{1, \\ldots, 5\\}, i>j$. Then it follows that  $$ \\sum_{i>j}\\left(a_{i}-a_{j}\\right)^{2} \\geq m^{2} \\sum_{i>j}(i-j)^{2}=50 m^{2} $$  On the other hand, by the condition of the problem,  $$ \\sum_{i>j}\\left(a_{i}-a_{j}\\right)^{2}=5 \\sum_{i=1}^{5} a_{i}^{2}-\\left(a_{1}+\\cdots+a_{5}\\right)^{2} \\leq 5 $$  Therefore $50 m^{2} \\leq 5$; i.e., $m^{2} \\leq \\frac{1}{10}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1974", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. I 5 (GBR 3) Let $A_{r}, B_{r}, C_{r}$ be points on the circumference of a given circle $S$. From the triangle $A_{r} B_{r} C_{r}$, called $\\triangle_{r}$, the triangle $\\triangle_{r+1}$ is obtained by constructing the points $A_{r+1}, B_{r+1}, C_{r+1}$ on $S$ such that $A_{r+1} A_{r}$ is parallel to $B_{r} C_{r}, B_{r+1} B_{r}$ is parallel to $C_{r} A_{r}$, and $C_{r+1} C_{r}$ is parallel to $A_{r} B_{r}$. Each angle of $\\triangle_{1}$ is an integer number of degrees and those integers are not multiples of 45 . Prove that at least two of the triangles $\\triangle_{1}, \\triangle_{2}, \\ldots, \\triangle_{15}$ are congruent.", "solution": "5. All the angles are assumed to be oriented and measured modulo $180^{\\circ}$. Denote by $\\alpha_{i}, \\beta_{i}, \\gamma_{i}$ the angles of triangle $\\triangle_{i}$, at $A_{i}, B_{i}, C_{i}$ respectively. Let us determine the angles of $\\triangle_{i+1}$. If $D_{i}$ is the intersection of lines $B_{i} B_{i+1}$ and $C_{i} C_{i+1}$, we have $\\angle B_{i+1} A_{i+1} C_{i+1}=\\angle D_{i} B_{i} C_{i+1}=\\angle B_{i} D_{i} C_{i+1}+$ $\\angle D_{i} C_{i+1} B_{i}=\\angle B_{i} D_{i} C_{i}-\\angle B_{i} C_{i+1} C_{i}=-2 \\angle B_{i} A_{i} C_{i}$. We conclude that  $$ \\alpha_{i+1}=-2 \\alpha_{i}, \\quad \\text { and analogously } \\quad \\beta_{i+1}=-2 \\beta_{i}, \\quad \\gamma_{i+1}=-2 \\gamma_{i} $$  Therefore $\\alpha_{r+t}=(-2)^{t} \\alpha_{r}$. However, since $(-2)^{12} \\equiv 1(\\bmod 45)$ and consequently $(-2)^{14} \\equiv(-2)^{2}(\\bmod 180)$, it follows that $\\alpha_{15}=\\alpha_{3}$, since all values are modulo $180^{\\circ}$. Analogously, $\\beta_{15}=\\beta_{3}$ and $\\gamma_{15}=\\gamma_{3}$, and moreover, $\\triangle_{3}$ and $\\triangle_{15}$ are inscribed in the same circle; hence $\\triangle_{3} \\cong \\triangle_{15}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1974", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. I 6 (ROM 4) ${ }^{\\mathrm{IMO} 3}$ Does there exist a natural number $n$ for which the number  $$ \\sum_{k=0}^{n}\\binom{2 n+1}{2 k+1} 2^{3 k} $$  is divisible by 5 ?", "solution": "6. We set  $$ \\begin{aligned} & x=\\sum_{k=0}^{n}\\binom{2 n+1}{2 k+1} 2^{3 k}=\\frac{1}{\\sqrt{8}} \\sum_{k=0}^{n}\\binom{2 n+1}{2 k+1} \\sqrt{8}^{2 k+1}, \\\\ & y=\\sum_{k=0}^{n}\\binom{2 n+1}{2 k} 2^{3 k}=\\sum_{k=0}^{n}\\binom{2 n+1}{2 k} \\sqrt{8}^{2 k} \\end{aligned} $$  Both $x$ and $y$ are positive integers. Also, from the binomial formula we obtain  $$ y+x \\sqrt{8}=\\sum_{i=0}^{2 n+1}\\binom{2 n+1}{i} \\sqrt{8}^{i}=(1+\\sqrt{8})^{2 n+1} $$  and similarly  $$ y-x \\sqrt{8}=(1-\\sqrt{8})^{2 n+1} . $$  Multiplying these equalities, we get $y^{2}-8 x^{2}=(1+\\sqrt{8})^{2 n+1}(1-\\sqrt{8})^{2 n+1}=$ $-7^{2 n+1}$. Reducing modulo 5 gives us  $$ 3 x^{2}-y^{2} \\equiv 2^{2 n+1} \\equiv 2 \\cdot(-1)^{n} $$  Now we see that if $x$ is divisible by 5 , then $y^{2} \\equiv \\pm 2(\\bmod 5)$, which is impossible. Therefore $x$ is never divisible by 5 . Second solution. Another standard way is considering recurrent formulas. If we set  $$ x_{m}=\\sum_{k}\\binom{m}{2 k+1} 8^{k}, \\quad y_{m}=\\sum_{k}\\binom{m}{2 k} 8^{k} $$  then since $\\binom{a}{b}=\\binom{a-1}{b}+\\binom{a-1}{b-1}$, it follows that $x_{m+1}=x_{m}+y_{m}$ and $y_{m+1}=8 x_{m}+y_{m}$; therefore $x_{m+1}=2 x_{m}+7 x_{m-1}$. We need to show that none of $x_{2 n+1}$ are divisible by 5 . Considering the sequence $\\left\\{x_{m}\\right\\}$ modulo 5 , we get that $x_{m}=0,1,2,1,1,4,0,3,1,3,3,2,0,4,3,4,4,1, \\ldots$ Zeros occur in the initial position of blocks of length 6 , where each subsequent block is obtained by multiplying the previous one by 3 (modulo 5 ). Consequently, $x_{m}$ is divisible by 5 if and only if $m$ is a multiple of 6 , which cannot happen if $m=2 n+1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1974", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. II 1 (POL 2) Let $a_{i}, b_{i}$ be coprime positive integers for $i=1,2, \\ldots, k$, and $m$ the least common multiple of $b_{1}, \\ldots, b_{k}$. Prove that the greatest common divisor of $a_{1} \\frac{m}{b_{1}}, \\ldots, a_{k} \\frac{m}{b_{k}}$ equals the greatest common divisor of $a_{1}, \\ldots, a_{k}$.", "solution": "7. Consider an arbitrary prime number $p$. If $p \\mid m$, then there exists $b_{i}$ that is divisible by the same power of $p$ as $m$. Then $p$ divides neither $a_{i} \\frac{m}{b_{i}}$ nor $a_{i}$, because $\\left(a_{i}, b_{i}\\right)=1$. If otherwise $p \\nmid m$, then $\\frac{m}{b_{i}}$ is not divisible by $p$ for any $i$, hence $p$ divides $a_{i}$ and $a_{i} \\frac{m}{b_{i}}$ to the same power. Therefore $\\left(a_{1}, \\ldots, a_{k}\\right)$ and $\\left(a_{1} \\frac{m}{b_{1}}, \\ldots, a_{k} \\frac{m}{b_{k}}\\right)$ have the same factorization; hence they are equal. Second solution. For $k=2$ we easily verify the formula $\\left(m \\frac{a_{1}}{b_{1}}, m \\frac{a_{2}}{b_{2}}\\right)=$ $\\frac{m}{b_{1} b_{2}}\\left(a_{1} b_{2}, a_{2} b_{1}\\right)=\\frac{1}{b_{1} b_{2}}\\left[b_{1}, b_{2}\\right]\\left(a_{1}, a_{2}\\right)\\left(b_{1}, b_{2}\\right)=\\left(a_{1}, a_{2}\\right)$, since $\\left[b_{1}, b_{2}\\right]$. $\\left(b_{1}, b_{2}\\right)=b_{1} b_{2}$. We proceed by induction:  $$ \\begin{aligned} \\left(a_{1} \\frac{m}{b_{1}}, \\ldots, a_{k} \\frac{m}{b_{k}}, a_{k+1} \\frac{m}{b_{k+1}}\\right) & =\\left(\\frac{m}{\\left[b_{1}, \\ldots, b_{k}\\right]}\\left(a_{1}, \\ldots, a_{k}\\right), a_{k+1} \\frac{m}{b_{k+1}}\\right) \\\\ & =\\left(a_{1}, \\ldots, a_{k}, a_{k+1}\\right) . \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1974", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. II 2 (NET 3) ${ }^{\\mathrm{IMO} 5}$ If $a, b, c, d$ are arbitrary positive real numbers, find all possible values of  $$ S=\\frac{a}{a+b+d}+\\frac{b}{a+b+c}+\\frac{c}{b+c+d}+\\frac{d}{a+c+d} . $$", "solution": "8. It is clear that  $$ \\begin{gathered} \\frac{a}{a+b+c+d}+\\frac{b}{a+b+c+d}+\\frac{c}{a+b+c+d}+\\frac{d}{a+b+c+d}<S \\\\ \\text { and } \\quad S<\\frac{a}{a+b}+\\frac{b}{a+b}+\\frac{c}{c+d}+\\frac{d}{c+d} \\end{gathered} $$  or equivalently, $1<S<2$. On the other hand, all values from $(1,2)$ are attained. Since $S=1$ for $(a, b, c, d)=(0,0,1,1)$ and $S=2$ for $(a, b, c, d)=(0,1,0,1)$, due to continuity all the values from $(1,2)$ are obtained, for example, for $(a, b, c, d)=(x(1-x), x, 1-x, 1)$, where $x$ goes through $(0,1)$. Second solution. Set  $$ S_{1}=\\frac{a}{a+b+d}+\\frac{c}{b+c+d} \\quad \\text { and } \\quad S_{2}=\\frac{b}{a+b+c}+\\frac{d}{a+c+d} $$  We may assume without loss of generality that $a+b+c+d=1$. Putting $a+c=x$ and $b+d=y$ (then $x+y=1$ ), we obtain that the set of values of  $$ S_{1}=\\frac{a}{1-c}+\\frac{c}{1-a}=\\frac{2 a c+x-x^{2}}{a c+1-x} $$  is $\\left(x, \\frac{2 x}{2-x}\\right)$. Having the analogous result for $S_{2}$ in mind, we conclude that the values that $S=S_{1}+S_{2}$ can take are $\\left(x+y, \\frac{2 x}{2-x}+\\frac{2 y}{2-y}\\right]$. Since $x+y=1$ and  $$ \\frac{2 x}{2-x}+\\frac{2 y}{2-y}=\\frac{4-4 x y}{2+x y} \\leq 2 $$  with equality for $x y=0$, the desired set of values for $S$ is $(1,2)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1974", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. II 3 (CUB 3) Let $x, y, z$ be real numbers each of whose absolute value is different from $1 / \\sqrt{3}$ such that $x+y+z=x y z$. Prove that  $$ \\frac{3 x-x^{3}}{1-3 x^{2}}+\\frac{3 y-y^{3}}{1-3 y^{2}}+\\frac{3 z-z^{3}}{1-3 z^{2}}=\\frac{3 x-x^{3}}{1-3 x^{2}} \\cdot \\frac{3 y-y^{3}}{1-3 y^{2}} \\cdot \\frac{3 z-z^{3}}{1-3 z^{2}} $$", "solution": "9. There exist real numbers $a, b, c$ with $\\tan a=x, \\tan b=y, \\tan c=z$. Then using the additive formula for tangents we obtain  $$ \\tan (a+b+c)=\\frac{x+y+z-x y z}{1-x y-x z-y z} $$  We are given that $x y z=x+y+z$. In this case $x y+y z+z x=1$ is impossible; otherwise, $x, y, z$ would be the zeros of a cubic polynomial $t^{3}-\\lambda t^{2}+t-\\lambda=\\left(t^{2}+1\\right)(t-\\lambda)$ (where $\\left.\\lambda=x y z\\right)$, which has only one real root. It follows that  $$ x+y+z=x y z \\Longleftrightarrow \\tan (a+b+c)=0 . $$  Hence $a+b+c=k \\pi$ for some $k \\in \\mathbb{Z}$. We note that $\\frac{3 x-x^{3}}{1-3 x^{2}}$ actually expresses $\\tan 3 a$. Since $3 a+3 b+3 c=3 k \\pi$, the result follows from (1) for the numbers $\\frac{3 x-x^{3}}{1-3 x^{2}}, \\frac{3 y-y^{3}}{1-3 y^{2}}, \\frac{3 z-z^{3}}{1-3 z^{2}}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1975", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (FRA) There are six ports on a lake. Is it possible to organize a series of routes satisfying the following conditions: (i) Every route includes exactly three ports; (ii) No two routes contain the same three ports; (iii) The series offers exactly two routes to each tourist who desires to visit two different arbitrary ports?", "solution": "1. First, we observe that there cannot exist three routes of the form $(A, B, C)$, $(A, B, D),(A, C, D)$, for if $E, F$ are the remaining two ports, there can be only one route covering $A, E$, namely, $(A, E, F)$. Thus if $(A, B, C)$, $(A, B, D)$ are two routes, the one covering $A, C$ must be w.l.o.g. $(A, C, E)$. The other roots are uniquely determined: These are $(A, D, F),(A, E, F)$, $(B, D, E),(B, E, F),(B, C, F),(C, D, E),(C, D, F)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1975", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (GBR) ${ }^{\\mathrm{IMO} 6}$ The function $f(x, y)$ is a homogeneous polynomial of the $n$th degree in $x$ and $y$. If $f(1,0)=1$ and for all $a, b, c$,  $$ f(a+b, c)+f(b+c, a)+f(c+a, b)=0 $$  prove that $f(x, y)=(x-2 y)(x+y)^{n-1}$.", "solution": "10. We shall prove that for all $x, y$ with $x+y=1$ it holds that $f(x, y)=x-2 y$. In this case $f(x, y)=f(x, 1-x)$ can be regarded as a polynomial in $z=x-2 y=3 x-2$, say $f(x, 1-x)=F(z)$. Putting in the given relation $a=b=x / 2, c=1-x$, we obtain $f(x, 1-x)+2 f(1-x / 2, x / 2)=0$; hence $F(z)+2 F(-z / 2)=0$. Now $F(1)=1$, and we get that for all $k$, $F\\left((-2)^{k}\\right)=(-2)^{k}$. Thus $F(z)=z$ for infinitely many values of $z$; hence $F(z) \\equiv z$. Consequently $f(x, y)=x-2 y$ if $x+y=1$. For general $x, y$ with $x+y \\neq 0$, since $f$ is homogeneous, we have $f(x, y)=$ $(x+y)^{n} f\\left(\\frac{x}{x+y}, \\frac{y}{x+y}\\right)=(x+y)^{n}\\left(\\frac{x}{x+y}-2 \\frac{y}{x+y}\\right)=(x+y)^{n-1}(x-2 y)$. The same is true for $x+y=0$, because $f$ is a polynomial.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1975", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. $(\\mathbf{G B R})^{\\mathrm{IMO} 2}$ Let $a_{1}, a_{2}, a_{3}, \\ldots$ be any infinite increasing sequence of positive integers. (For every integer $i>0, a_{i+1}>a_{i}$.) Prove that there are infinitely many $m$ for which positive integers $x, y, h, k$ can be found such that $0<h<k<m$ and $a_{m}=x a_{h}+y a_{k}$.", "solution": "11. Let $\\left(a_{k_{i}}\\right)$ be the subsequence of $\\left(a_{k}\\right)$ consisting of all $a_{k}$ 's that give remainder $r$ upon division by $a_{1}$. For every $i>1, a_{k_{i}} \\equiv a_{k_{1}}\\left(\\bmod a_{1}\\right)$; hence $a_{k_{i}}=a_{k_{1}}+y a_{1}$ for some integer $y>0$. It follows that for every $r=0,1, \\ldots, a_{1}-1$ there is exactly one member of the corresponding $\\left(a_{k_{i}}\\right)_{i \\geq 1}$ that cannot be represented as $x a_{l}+y a_{m}$, and hence at most $a_{1}+1$ members of $\\left(a_{k}\\right)$ in total are not representable in the given form.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1975", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (GRE) Consider on the first quadrant of the trigonometric circle the $\\operatorname{arcs} A M_{1}=x_{1}, A M_{2}=x_{2}, A M_{3}=x_{3}, \\ldots, A M_{\\nu}=x_{\\nu}$, such that $x_{1}<$ $x_{2}<x_{3}<\\cdots<x_{\\nu}$. Prove that  $$ \\sum_{i=0}^{\\nu-1} \\sin 2 x_{i}-\\sum_{i=0}^{\\nu-1} \\sin \\left(x_{i}-x_{i+1}\\right)<\\frac{\\pi}{2}+\\sum_{i=0}^{\\nu-1} \\sin \\left(x_{i}+x_{i+1}\\right) $$", "solution": "12. Since $\\sin 2 x_{i}=2 \\sin x_{i} \\cos x_{i}$ and $\\sin \\left(x_{i}+x_{i+1}\\right)+\\sin \\left(x_{i}-x_{i+1}\\right)=$ $2 \\sin x_{i} \\cos x_{i+1}$, the inequality from the problem is equivalent to  $$ \\begin{gathered} \\left(\\cos x_{1}-\\cos x_{2}\\right) \\sin x_{1}+\\left(\\cos x_{2}-\\cos x_{3}\\right) \\sin x_{2}+\\cdots \\\\ \\cdots+\\left(\\cos x_{\\nu-1}-\\cos x_{\\nu}\\right) \\sin x_{\\nu-1}<\\frac{\\pi}{4} \\end{gathered} $$  Consider the unit circle with center at $O(0,0)$ and points $M_{i}\\left(\\cos x_{i}, \\sin x_{i}\\right)$ on it. Also, choose the points $N_{i}\\left(\\cos x_{i}, 0\\right)$ and $M_{i}^{\\prime}\\left(\\cos x_{i+1}, \\sin x_{i}\\right)$. It is clear that $\\left(\\cos x_{i}-\\cos x_{i+1}\\right) \\sin x_{i}$ is equal to the area of the rectangle $M_{i} N_{i} N_{i+1} M_{i}^{\\prime}$. Since all these rectangles are disjoint and lie inside the quarter circle in the first quadrant whose area is $\\frac{\\pi}{4}$, inequality (1) follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1975", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (ROM) Let $A_{0}, A_{1}, \\ldots, A_{n}$ be points in a plane such that (i) $A_{0} A_{1} \\leq \\frac{1}{2} A_{1} A_{2} \\leq \\cdots \\leq \\frac{1}{2^{n-1}} A_{n-1} A_{n}$ and (ii) $0<\\measuredangle A_{0} A_{1} A_{2}<\\measuredangle A_{1} A_{2} A_{3}<\\cdots<\\measuredangle A_{n-2} A_{n-1} A_{n}<180^{\\circ}$, where all these angles have the same orientation. Prove that the segments $A_{k} A_{k+1}, A_{m} A_{m+1}$ do not intersect for each $k$ and $n$ such that $0 \\leq k \\leq$ $m-2<n-2$.", "solution": "13. Suppose that $A_{k} A_{k+1} \\cap A_{m} A_{m+1} \\neq \\emptyset$ for some $k, m>k+1$. Without loss of generality we may suppose that $k=0, m=n-1$ and that no two segments $A_{k} A_{k+1}$ and $A_{m} A_{m+1}$ intersect for $0 \\leq k<m-1<n-1$ except for $k=0, m=n-1$. Also, shortening $A_{0} A_{1}$, we may suppose that $A_{0} \\in A_{n-1} A_{n}$. Finally, we may reduce the problem to the case that $A_{0} \\ldots A_{n-1}$ is convex: Otherwise, the segment $A_{n-1} A_{n}$ can be prolonged so that it intersects some $A_{k} A_{k+1}, 0<k<n-2$. If $n=3$, then $A_{1} A_{2} \\geq 2 A_{0} A_{1}$ implies $A_{0} A_{2}>A_{0} A_{1}$, hence $\\angle A_{0} A_{1} A_{2}>$ $\\angle A_{1} A_{2} A_{3}$, a contradiction. Let $n=4$. From $A_{3} A_{2}>A_{1} A_{2}$ we conclude that $\\angle A_{3} A_{1} A_{2}>\\angle A_{1} A_{3} A_{2}$. Using the inequality $\\angle A_{0} A_{3} A_{2}>\\angle A_{0} A_{1} A_{2}$ we obtain that $\\angle A_{0} A_{3} A_{1}>$ $\\angle A_{0} A_{1} A_{3}$ implying $A_{0} A_{1}>A_{0} A_{3}$. Now we have $A_{2} A_{3}<A_{3} A_{0}+A_{0} A_{1}+$ $A_{1} A_{2}<2 A_{0} A_{1}+A_{1} A_{2} \\leq 2 A_{1} A_{2} \\leq A_{2} A_{3}$, which is not possible. Now suppose $n \\geq 5$. If $\\alpha_{i}$ is the exterior angle at $A_{i}$, then $\\alpha_{1}>\\cdots>\\alpha_{n-1}$; hence $\\alpha_{n-1}<\\frac{360^{\\circ}}{n-1} \\leq 90^{\\circ}$. Consequently $\\angle A_{n-2} A_{n-1} A_{0} \\geq 90^{\\circ}$ and $A_{0} A_{n-2}>A_{n-1} A_{n-2}$. On the other hand, $A_{0} A_{n-2}<A_{0} A_{1}+A_{1} A_{2}+\\cdots+$ $A_{n-3} A_{n-2}<\\left(\\frac{1}{2^{n-2}}+\\frac{1}{2^{n-3}}+\\cdots+\\frac{1}{2}\\right) A_{n-1} A_{n-2}<A_{n-1} A_{n-2}$, which contradicts the previous relation.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1975", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (YUG) Let $x_{0}=5$ and $x_{n+1}=x_{n}+\\frac{1}{x_{n}}(n=0,1,2, \\ldots)$. Prove that $45<x_{1000}<45,1$.", "solution": "14. We shall prove that for every $n \\in \\mathbb{N}, \\sqrt{2 n+25} \\leq x_{n} \\leq \\sqrt{2 n+25}+0.1$. Note that for $n=1000$ this gives us exactly the desired inequalities.  First, notice that the recurrent relation is equivalent to  $$ 2 x_{k}\\left(x_{k+1}-x_{k}\\right)=2 $$  Since $x_{0}<x_{1}<\\cdots<x_{k}<\\cdots$, from (1) we get $x_{k+1}^{2}-x_{k}^{2}=\\left(x_{k+1}+\\right.$ $\\left.x_{k}\\right)\\left(x_{k+1}-x_{k}\\right)>2$. Adding these up we obtain $x_{n}^{2} \\geq x_{0}^{2}+2 n$, which proves the first inequality. On the other hand, $x_{k+1}=x_{k}+\\frac{1}{x_{k}} \\leq x_{k}+0.2$ (for $x_{k} \\geq 5$ ), and one also deduces from (1) that $x_{k+1}^{2}-x_{k}^{2}-0.2\\left(x_{k+1}-x_{k}\\right)=\\left(x_{k+1}+x_{k}-\\right.$ $0.2)\\left(x_{k+1}-x_{k}\\right) \\leq 2$. Again, adding these inequalities up, $(k=0, \\ldots, n-1)$ yields  $$ x_{n}^{2} \\leq 2 n+x_{0}^{2}+0.2\\left(x_{n}-x_{0}\\right)=2 n+24+0.2 x_{n} $$  Solving the corresponding quadratic equation, we obtain $x_{n}<0.1+$ $\\sqrt{2 n+24.01}<0.1++\\sqrt{2 n+25}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1975", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (USS) ${ }^{\\mathrm{IMO} 5}$ Is it possible to plot 1975 points on a circle with radius 1 so that the distance between any two of them is a rational number (distances have to be measured by chords)?", "solution": "15. Assume that the center of the circle is at the origin $O(0,0)$, and that the points $A_{1}, A_{2}, \\ldots, A_{1975}$ are arranged on the upper half-circle so that $\\angle A_{i} O A_{1}=\\alpha_{i}\\left(\\alpha_{1}=0\\right)$. The distance $A_{i} A_{j}$ equals $2 \\sin \\frac{\\alpha_{j}-\\alpha_{i}}{2}=$ $2 \\sin \\frac{\\alpha_{j}}{2} \\cos \\frac{\\alpha_{i}}{2}-\\cos \\frac{\\alpha_{j}}{2} \\sin \\frac{\\alpha_{i}}{2}$, and it will be rational if all $\\sin \\frac{\\alpha_{k}}{2}, \\cos \\frac{\\alpha_{k}}{2}$ are rational. Finally, observe that there exist infinitely many angles $\\alpha$ such that both $\\sin \\alpha, \\cos \\alpha$ are rational, and that such $\\alpha$ can be arbitrarily small. For example, take $\\alpha$ so that $\\sin \\alpha=\\frac{2 t}{t^{2}+1}$ and $\\cos \\alpha=\\frac{t^{2}-1}{t^{2}+1}$ for any $t \\in \\mathbb{Q}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1975", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (CZS) ${ }^{\\mathrm{IMO} 1}$ Let $x_{1} \\geq x_{2} \\geq \\cdots \\geq x_{n}$ and $y_{1} \\geq y_{2} \\geq \\cdots \\geq y_{n}$ be two $n$-tuples of numbers. Prove that  $$ \\sum_{i=1}^{n}\\left(x_{i}-y_{i}\\right)^{2} \\leq \\sum_{i=1}^{n}\\left(x_{i}-z_{i}\\right)^{2} $$  is true when $z_{1}, z_{2}, \\ldots, z_{n}$ denote $y_{1}, y_{2}, \\ldots, y_{n}$ taken in another order.", "solution": "2. Since there are finitely many arrangements of the $z_{i}$ 's, assume that $z_{1}, \\ldots, z_{n}$ is the one for which $\\sum_{i=1}^{n}\\left(x_{i}-z_{i}\\right)^{2}$ is minimal. We claim that in this case $i<j \\Rightarrow z_{i} \\geq z_{j}$, from which the claim of the problem directly follows. Indeed, otherwise we would have  $$ \\begin{aligned} \\left(x_{i}-z_{j}\\right)^{2}+\\left(x_{j}-z_{i}\\right)^{2}= & \\left(x_{i}-z_{i}\\right)^{2}+\\left(x_{j}-z_{j}\\right)^{2} \\\\ & +2\\left(x_{i} z_{i}+x_{j} z_{j}-x_{i} z_{j}-x_{j} z_{i}\\right) \\\\ = & \\left(x_{i}-z_{i}\\right)^{2}+\\left(x_{j}-z_{j}\\right)^{2}+2\\left(x_{i}-x_{j}\\right)\\left(z_{i}-z_{j}\\right) \\\\ \\leq & \\left(x_{i}-z_{i}\\right)^{2}+\\left(x_{j}-z_{j}\\right)^{2} \\end{aligned} $$  contradicting the assumption.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1975", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (USA) Find the integer represented by $\\left[\\sum_{n=1}^{10^{9}} n^{-2 / 3}\\right]$. Here $[x]$ denotes the greatest integer less than or equal to $x$ (e.g. $[\\sqrt{2}]=1$ ).", "solution": "3. From $\\left((k+1)^{2 / 3}+(k+1)^{1 / 3} k^{1 / 3}+k^{2 / 3}\\right)\\left((k+1)^{1 / 3}-k^{1 / 3}\\right)=1$ and $3 k^{2 / 3}<(k+1)^{2 / 3}+(k+1)^{1 / 3} k^{1 / 3}+k^{2 / 3}<3(k+1)^{2 / 3}$ we obtain  $$ 3\\left((k+1)^{1 / 3}-k^{1 / 3}\\right)<k^{-2 / 3}<3\\left(k^{1 / 3}-(k-1)^{1 / 3}\\right) . $$  Summing from 1 to $n$ we get  $$ 1+3\\left((n+1)^{1 / 3}-2^{1 / 3}\\right)<\\sum_{k=1}^{n} k^{-2 / 3}<1+3\\left(n^{1 / 3}-1\\right) $$  In particular, for $n=10^{9}$ this inequality gives  $$ 2997<1+3\\left(\\left(10^{9}+1\\right)^{1 / 3}-2^{1 / 3}\\right)<\\sum_{k=1}^{10^{9}} k^{-2 / 3}<2998 $$  Therefore $\\left[\\sum_{k=1}^{10^{9}} k^{-2 / 3}\\right]=2997$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1975", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (SWE) Let $a_{1}, a_{2}, \\ldots, a_{n}, \\ldots$ be a sequence of real numbers such that $0 \\leq a_{n} \\leq 1$ and $a_{n}-2 a_{n+1}+a_{n+2} \\geq 0$ for $n=1,2,3, \\ldots$. Prove that  $$ 0 \\leq(n+1)\\left(a_{n}-a_{n+1}\\right) \\leq 2 \\quad \\text { for } n=1,2,3, \\ldots $$", "solution": "4. Put $\\Delta a_{n}=a_{n}-a_{n+1}$. By the imposed condition, $\\Delta a_{n}>\\Delta a_{n+1}$. Suppose that for some $n, \\Delta a_{n}<0$ : Then for each $k \\geq n, \\Delta a_{k}<\\Delta a_{n}$; hence $a_{n}-a_{n+m}=\\Delta a_{n}+\\cdots+\\Delta a_{n+m-1}<m \\Delta a_{n}$. Thus for sufficiently large $m$ it holds that $a_{n}-a_{n+m}<-1$, which is impossible. This proves the first part of the inequality. Next one observes that $n \\geq \\sum_{k=1}^{n} a_{k}=n a_{n+1}+\\sum_{k=1}^{n} k \\Delta a_{k} \\geq(1+2+\\cdots+n) \\Delta a_{n}=\\frac{n(n+1)}{2} \\Delta a_{n}$. Hence $(n+1) \\Delta a_{n} \\leq 2$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1975", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (SWE) Let $M$ be the set of all positive integers that do not contain the digit 9 (base 10). If $x_{1}, \\ldots, x_{n}$ are arbitrary but distinct elements in $M$, prove that  $$ \\sum_{j=1}^{n} \\frac{1}{x_{j}}<80 $$", "solution": "5. There are exactly $8 \\cdot 9^{k-1} k$-digit numbers in $M$ (the first digit can be chosen in 8 ways, while any other position admits 9 possibilities). The least of them is $10^{k}$, and hence  $$ \\begin{aligned} \\sum_{x_{j}<10^{k}} \\frac{1}{x_{j}} & =\\sum_{i=1}^{k} \\sum_{10^{i-1} \\leq x_{j}<10^{i}} \\frac{1}{x_{j}}<\\sum_{i=1}^{k} \\sum_{10^{i-1} \\leq x_{j}<10^{i}} \\frac{1}{10^{i-1}} \\\\ & =\\sum_{i=1}^{k} \\frac{8 \\cdot 9^{i-1}}{10^{i-1}}=80\\left(1-\\frac{9^{k}}{10^{k}}\\right)<80 . \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1975", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (USS) ${ }^{\\mathrm{IMO} 04}$ Let $A$ be the sum of the digits of the number $16^{16}$ and $B$ the sum of the digits of the number $A$. Find the sum of the digits of the number $B$ without calculating $16^{16}$.", "solution": "6. Let us denote by $C$ the sum of digits of $B$. We know that $16^{16} \\equiv A \\equiv$ $B \\equiv C(\\bmod 9)$. Since $16^{16}=2^{64}=2^{6 \\cdot 10+4} \\equiv 2^{4} \\equiv 7$, we get $C \\equiv 7(\\bmod$ 9). Moreover, $16^{16}<100^{16}=10^{32}$, hence $A$ cannot exceed $9 \\cdot 32=288$; consequently, $B$ cannot exceed 19 and $C$ is at most 10 . Therefore $C=7$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1975", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (GDR) Prove that from $x+y=1(x, y \\in \\mathbb{R})$ it follows that  $$ x^{m+1} \\sum_{j=0}^{n}\\binom{m+j}{j} y^{j}+y^{n+1} \\sum_{i=0}^{m}\\binom{n+i}{i} x^{i}=1 \\quad(m, n=0,1,2, \\ldots) . $$", "solution": "7. We use induction on $m$. Denote by $S_{m}$ the left-hand side of the equality to be proved. First $S_{0}=(1-y)\\left(1+y+\\cdots+y^{n}\\right)+y^{n+1}=1$, since $x=1-y$. Furthermore,  $$ \\begin{aligned} & S_{m+1}-S_{m} \\\\ = & \\binom{m+n+1}{m+1} x^{m+1} y^{n+1}+x^{m+1} \\sum_{j=0}^{n}\\left(\\binom{m+1+j}{j} x y^{j}-\\binom{m+j}{j} y^{j}\\right) \\\\ = & \\binom{m+n+1}{m+1} x^{m+1} y^{n+1} \\\\ & +x^{m+1} \\sum_{j=0}^{n}\\left(\\binom{m+1+j}{j} y^{j}-\\binom{m+j}{j} y^{j}-\\binom{m+1+j}{j} y^{j+1}\\right) \\\\ = & x^{m+1}\\left[\\binom{m+n+1}{n} y^{n+1}+\\sum_{j=0}^{n}\\left(\\binom{m+j}{j-1} y^{j}-\\binom{m+j+1}{j} y^{j+1}\\right)\\right] \\\\ = & 0 \\end{aligned} $$  i.e., $S_{m+1}=S_{m}=1$ for every $m$.  Second solution. Let us be given an unfair coin that, when tossed, shows heads with probability $x$ and tails with probability $y$. Note that $x^{m+1}\\binom{m+j}{j} y^{j}$ is the probability that until the moment when the $(m+1)$ th head appears, exactly $j$ tails $(j<n+1)$ have appeared. Similarly, $y^{n+1}\\binom{n+i}{i} x^{i}$ is the probability that exactly $i$ heads will appear before the $(n+1)$ th tail occurs. Therefore, the above sum is the probability that either $m+1$ heads will appear before $n+1$ tails, or vice versa, and this probability is clearly 1.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1975", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (NET) ${ }^{\\mathrm{IMO} 3}$ On the sides of an arbitrary triangle $A B C$, triangles $B P C$, $C Q A$, and $A R B$ are externally erected such that $\\measuredangle P B C=\\measuredangle C A Q=45^{\\circ}$, $\\measuredangle B C P=\\measuredangle Q C A=30^{\\circ}$, $\\measuredangle A B R=\\measuredangle B A R=15^{\\circ}$. Prove that $\\measuredangle Q R P=90^{\\circ}$ and $Q R=R P$.", "solution": "8. Let $K$ and $L$ be the feet of perpendiculars from $P$ and $Q$ to $B C$ and $A C$ respectively.  Let $M, N$ be points on $A B$ (ordered $A-N-M-B$ ) such that $R M N$ is a right isosceles triangle with $\\angle R=90^{\\circ}$. By sine theorem we have $\\frac{B M}{B A}=\\frac{B M}{B R} \\cdot \\frac{B R}{B A}=\\frac{\\sin 15^{\\circ}}{\\sin 45^{\\circ}}$. Since $\\frac{B K}{B C}=\\frac{\\sin 45^{\\circ} \\sin 30^{\\circ}}{\\cos 15^{\\circ}}=\\frac{\\sin 15^{\\circ}}{\\sin 45^{\\circ}}$, we deduce that $M K \\| A C$ and $M K=A L$. Similarly, $N L \\| B C$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-412.jpg?height=308&width=396&top_left_y=250&top_left_x=886) and $N L=B K$. It follows that the vectors $\\overrightarrow{R N}, \\overrightarrow{N L}, \\overrightarrow{L Q}$ are the images of $\\overrightarrow{R M}, \\overrightarrow{K P}, \\overrightarrow{M K}$ respectively under a rotation of $90^{\\circ}$, and consequently the same holds for their sums $\\overrightarrow{R Q}$ and $\\overrightarrow{R P}$. Therefore, $Q R=R P$ and $\\angle Q R P=90^{\\circ}$. Second solution. Let $A B S$ be the equilateral triangle constructed in the exterior of $\\triangle A B C$. Obviously, the triangles $B P C, B R S, A R S, A Q C$ are similar. Let $f$ be the rotational homothety centered at $B$ that maps $P$ onto $C$, and let $g$ be the rotational homothety about $A$ that maps $C$ onto $Q$. The composition $h=g \\circ f$ is also a rotational homothety; its angle is $\\angle P B C+\\angle C A Q=90^{\\circ}$, and the coefficient is $\\frac{B C}{B P} \\cdot \\frac{A Q}{A C}=1$. Moreover, $R$ is a fixed point of $h$ because $f(R)=S$ and $g(S)=R$. Hence $R$ is the center of $h$, and the statement follows from $h(P)=Q$. Remark. There are two more possible approaches: One includes using complex numbers and the other one is mere calculating of $R P, R Q, P Q$ by the cosine theorem. Second remark. The problem allows a generalization: Given that $\\angle C B P=$ $\\angle C A Q=\\alpha, \\angle B C P=\\angle A C Q=\\beta$, and $\\angle R A B=\\angle R B A=90^{\\circ}-\\alpha-\\beta$, show that $R P=R Q$ and $\\angle P R Q=2 \\alpha$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1975", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (NET) Let $f(x)$ be a continuous function defined on the closed interval $0 \\leq x \\leq 1$. Let $G(f)$ denote the graph of $f(x): G(f)=\\left\\{(x, y) \\in \\mathbb{R}^{2} \\mid 0 \\leq\\right.$ $x \\leq 1, y=f(x)\\}$. Let $G_{a}(f)$ denote the graph of the translated function $f(x-a)$ (translated over a distance $a$ ), defined by $G_{a}(f)=\\{(x, y) \\in$ $\\left.\\mathbb{R}^{2} \\mid a \\leq x \\leq a+1, y=f(x-a)\\right\\}$. Is it possible to find for every $a$, $0<a<1$, a continuous function $f(x)$, defined on $0 \\leq x \\leq 1$, such that $f(0)=f(1)=0$ and $G(f)$ and $G_{a}(f)$ are disjoint point sets?", "solution": "9. Suppose $n$ is the natural number with $n a \\leq 1<(n+1) a$. If a function $f$ with the desired properties exists, then $f_{a}(a)=0$ and let w.l.o.g. $f(a)>0$, or equivalently, let the graph of $f_{a}$ lie below the graph of $f$. In this case also $f(2 a)>f(a)$, since otherwise, the graphs of $f$ and $f_{a}$ would intersect between $a$ and $2 a$. Continuing in this way we are led to $0=f(0)<$ $f(a)<f(2 a)<\\cdots<f(n a)$. Thus if $n a=1$, i.e., $a=1 / n$, such an $f$ does not exist. On the other hand, if $a \\neq 1 / n$, then we similarly obtain $f(1)>f(1-a)>f(1-2 a)>\\cdots>f(1-n a)$. Choosing values of $f$ at $i a, 1-i a, i=1, \\ldots, n$, so that they satisfy $f(1-n a)<\\cdots<f(1-a)<$ $0<f(a)<\\cdots<f(n a)$, we can extend $f$ to other values of $[0,1]$ by linear interpolation. A function obtained this way has the desired property.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1976", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (BUL 1) Let $A B C$ be a triangle with bisectors $A A_{1}, B B_{1}, C C_{1}\\left(A_{1} \\in\\right.$ $B C$, etc.) and $M$ their common point. Consider the triangles $M B_{1} A$, $M C_{1} A, M C_{1} B, M A_{1} B, M A_{1} C, M B_{1} C$, and their inscribed circles. Prove that if four of these six inscribed circles have equal radii, then $A B=$ $B C=C A$.", "solution": "1. Let $r$ denote the common inradius. Some two of the four triangles with the inradii $\\rho$ have cross angles at $M$ : Suppose these are $\\triangle A M B_{1}$ and $\\triangle B M A_{1}$. We shall show that $\\triangle A M B_{1} \\cong \\triangle B M A_{1}$. Indeed, the altitudes of these two triangles are both equal to $r$, the inradius of $\\triangle A B C$, and their interior angles at $M$ are equal to some angle $\\varphi$. If $P$ is the point of tangency of the incircle of $\\triangle A_{1} M B$ with $M B$, then $\\frac{r}{\\rho}=\\frac{A_{1} M+B M+A_{1} B}{A_{1} B}$, which also implies $\\frac{r-2 \\rho}{\\rho}=\\frac{A_{1} M+B M-A_{1} B}{A_{1} B}=\\frac{2 M P}{A_{1} B}=\\frac{2 r \\cot (\\varphi / 2)}{A_{1} B}$. Since similarly $\\frac{r-2 \\rho}{\\rho}=\\frac{2 r \\cot (\\varphi / 2)}{B_{1} A}$, we obtain $A_{1} B=B_{1} A$ and consequently $\\triangle A M B_{1} \\cong \\triangle B M A_{1}$. Thus $\\angle B A C=\\angle A B C$ and $C C_{1} \\perp A B$. There are two alternatives for the other two incircles: (i) If the inradii of $A M C_{1}$ and $A M B_{1}$ are equal to $r$, it is easy to obtain that $\\triangle A M C_{1} \\cong \\triangle A M B_{1}$. Hence $\\angle A B_{1} M=\\angle A C_{1} M=90^{\\circ}$, and $\\triangle A B C$ is equilateral. (ii) The inradii of $A M B_{1}$ and $C M B_{1}$ are equal to $r$. Put $x=\\angle M A C_{1}=$ $\\angle M B C_{1}$. In this case $\\varphi=2 x$ and $\\angle B_{1} M C=90^{\\circ}-x$. Now we have $\\frac{A B_{1}}{C B_{1}}=\\frac{S_{A M B_{1}}}{S_{C M B_{1}}}=\\frac{A M+M B_{1}+A B_{1}}{C M+M B_{1}+C B_{1}}=\\frac{A M+M B_{1}-A B_{1}}{C M+M B_{1}-C B_{1}}=\\frac{\\mathrm{cot} x}{\\cot \\left(45^{\\circ}-x / 2\\right)}$. On the other hand, $\\frac{A B_{1}}{C B_{1}}=\\frac{A B}{B C}=2 \\cos 2 x$. Thus we have an equation for $x: \\tan \\left(45^{\\circ}-x / 2\\right)=2 \\cos 2 x \\tan x$, or equivalently  $$ 2 \\tan \\left(45^{\\circ}-\\frac{x}{2}\\right) \\sin \\left(45^{\\circ}-\\frac{x}{2}\\right) \\cos \\left(45^{\\circ}-\\frac{x}{2}\\right)=2 \\cos 2 x \\sin x . $$  Hence $\\sin 3 x-\\sin x=2 \\sin ^{2}\\left(45^{\\circ}-\\frac{x}{2}\\right)=1-\\sin x$, implying $\\sin 3 x=1$, i.e., $x=30^{\\circ}$. Therefore $\\triangle A B C$ is equilateral.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1976", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (USA 6) ${ }^{\\mathrm{IMO}}$ Find the largest number obtainable as the product of positive integers whose sum is 1976.", "solution": "10. Let $a_{1}<a_{2}<\\cdots<a_{n}$ be positive integers whose sum is 1976 . Let $M$ denote the maximal value of $a_{1} a_{2} \\cdots a_{n}$. We make the following observations: (1) $a_{1}=1$ does not yield the maximum, since replacing $1, a_{2}$ by $1+a_{2}$ increases the product. (2) $a_{j}-a_{i} \\geq 2$ does not yield the maximal value, since replacing $a_{i}, a_{j}$ by $a_{i}+1, a_{j}-1$ increases the product. (3) $a_{i} \\geq 5$ does not yield the maximal value, since $2\\left(a_{i}-2\\right)=2 a_{i}-4>a_{i}$. Since $4=2^{2}$, we may assume that all $a_{i}$ are either 2 or 3 , and $M=2^{k} 3^{l}$, where $2 k+3 l=1976$. (4) $k \\geq 3$ does not yield the maximal value, since $2 \\cdot 2 \\cdot 2<3 \\cdot 3$.  Hence $k \\leq 2$ and $2 k \\equiv 1976(\\bmod 3)$ gives us $k=1, l=658$ and $M=2 \\cdot 3^{658}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1976", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (VIE 1) Prove that there exist infinitely many positive integers $n$ such that the decimal representation of $5^{n}$ contains a block of 1976 consecutive zeros.", "solution": "11. We shall show by induction that $5^{2^{k}}-1=2^{k+2} q_{k}$ for each $k=0,1, \\ldots$, where $q_{k} \\in \\mathbb{N}$. Indeed, the statement is true for $k=0$, and if it holds for some $k$ then $5^{2^{k+1}}-1=\\left(5^{2^{k}}+1\\right)\\left(5^{2^{k}}-1\\right)=2^{k+3} d_{k+1}$ where $d_{k+1}=$ $\\left(5^{2^{k}}+1\\right) d_{k} / 2$ is an integer by the inductive hypothesis. Let us now choose $n=2^{k}+k+2$. We have $5^{n}=10^{k+2} q_{k}+5^{k+2}$. It follows from $5^{4}<10^{3}$ that $5^{k+2}$ has at most $[3(k+2) / 4]+2$ nonzero digits, while $10^{k+2} q_{k}$ ends in $k+2$ zeros. Hence the decimal representation of $5^{n}$ contains at least $[(k+2) / 4]-2$ consecutive zeros. Now it suffices to take $k>4 \\cdot 1978$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1976", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (VIE 2) The polynomial $1976\\left(x+x^{2}+\\cdots+x^{n}\\right)$ is decomposed into a sum of polynomials of the form $a_{1} x+a_{2} x^{2}+\\ldots+a_{n} x^{n}$, where $a_{1}, a_{2}, \\cdots, a_{n}$ are distinct positive integers not greater than $n$. Find all values of $n$ for which such a decomposition is possible.", "solution": "12. Suppose the decomposition into $k$ polynomials is possible. The sum of coefficients of each polynomial $a_{1} x+a_{2} x^{2}+\\cdots+a_{n} x^{n}$ equals $1+\\cdots+$ $n=n(n+1) / 2$ while the sum of coefficients of $1976\\left(x+x^{2}+\\cdots+x^{n}\\right)$ is $1976 n$. Hence we must have $1976 n=k n(n+1) / 2$, which reduces to $(n+1) \\mid 3952=2^{4} \\cdot 13 \\cdot 19$. In other words, $n$ is of the form $n=2^{\\alpha} 13^{\\beta} 19^{\\gamma}-1$, with $0 \\leq \\alpha \\leq 4,0 \\leq \\beta \\leq 1,0 \\leq \\gamma \\leq 1$. We can immediately eliminate the values $n=0$ and $n=3951$ that correspond to $\\alpha=\\beta=\\gamma=0$ and $\\alpha=4, \\beta=\\gamma=1$. We claim that all other values $n$ are permitted. There are two cases. $\\alpha \\leq 3$. In this case $k=3952 /(n+1)$ is even. The simple choice of the polynomials $P=x+2 x^{2}+\\cdots+n x^{n}$ and $P^{\\prime}=n x+(n-1) x^{2}+\\cdots+x^{n}$ suffices, since $k\\left(P+P^{\\prime}\\right) / 2=1976\\left(x+x^{2}+\\cdots+x^{n}\\right)$. $\\alpha=4$. Then $k$ is odd. Consider $(k-3) / 2$ pairs $\\left(P, P^{\\prime}\\right)$ of the former case and  $$ \\begin{aligned} P_{1}= & {\\left[n x+(n-1) x^{3}+\\cdots+\\frac{n+1}{2} x^{n}\\right] } \\\\ & +\\left[\\frac{n-1}{2} x^{2}+\\frac{n-3}{2} x^{4}+\\cdots+x^{n-1}\\right] \\\\ P_{2}= & {\\left[\\frac{n+1}{2} x+\\frac{n-1}{2} x^{3}+\\cdots+x^{n}\\right] } \\\\ & +\\left[n x^{2}+(n-1) x^{4}+\\cdots+\\frac{n+3}{2} x^{n-1}\\right] \\end{aligned} $$  Then $P+P_{1}+P_{2}=3(n+1)\\left(x+x^{2}+\\cdots+x^{n}\\right) / 2$ and therefore $(k-3)\\left(P+P^{\\prime}\\right) / 2+\\left(P+P_{1}+P_{2}\\right)=1976\\left(x+x^{2}+\\cdots+x^{n}\\right)$. It follows that the desired decomposition is possible if and only if $1<n<$ 3951 and $n+1 \\mid 2 \\cdot 1976$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1976", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (BUL 3) Let $a_{0}, a_{1}, \\ldots, a_{n}, a_{n+1}$ be a sequence of real numbers satisfying the following conditions:  $$ \\begin{aligned} a_{0} & =a_{n+1}=0 \\\\ \\left|a_{k-1}-2 a_{k}+a_{k+1}\\right| & \\leq 1 \\quad(k=1,2, \\ldots, n) \\end{aligned} $$  Prove that $\\left|a_{k}\\right| \\leq \\frac{k(n+1-k)}{2}(k=0,1, \\ldots, n+1)$.", "solution": "2. Let us put $b_{i}=i(n+1-i) / 2$, and let $c_{i}=a_{i}-b_{i}, i=0,1, \\ldots, n+1$. It is easy to verify that $b_{0}=b_{n+1}=0$ and $b_{i-1}-2 b_{i}+b_{i+1}=-1$. Subtracting this inequality from $a_{i-1}-2 a_{i}+a_{i+1} \\geq-1$, we obtain $c_{i-1}-2 c_{i}+c_{i+1} \\geq 0$, i.e., $2 c_{i} \\leq c_{i-1}+c_{i+1}$. We also have $c_{0}=c_{n+1}=0$.  Suppose that there exists $i \\in\\{1, \\ldots, n\\}$ for which $c_{i}>0$, and let $c_{k}$ be the maximal such $c_{i}$. Assuming w.l.o.g. that $c_{k-1}<c_{k}$, we obtain $c_{k-1}+c_{k+1}<2 c_{k}$, which is a contradiction. Hence $c_{i} \\leq 0$ for all $i$; i.e., $a_{i} \\leq b_{i}$. Similarly, considering the sequence $c_{i}^{\\prime}=a_{i}+b_{i}$ one can show that $c_{i}^{\\prime} \\geq 0$, i.e., $a_{i} \\geq-b_{i}$ for all $i$. This completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1976", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (CZS 3) ${ }^{\\mathrm{IMO} 1}$ In a convex quadrangle with area $32 \\mathrm{~cm}^{2}$, the sum of the lengths of two nonadjacent edges and of the length of one diagonal is equal to 16 cm . (a) What is the length of the other diagonal? (b) What are the lengths of the edges of the quadrangle if the perimeter is a minimum? (c) Is it possible to choose the edges in such a way that the perimeter is a maximum?", "solution": "3. (a) Let $A B C D$ be a quadrangle with $16=d=A B+C D+A C$, and let $S$ be its area. Then $S \\leq(A C \\cdot A B+A C \\cdot C D) / 2=A C(d-A C) / 2 \\leq$ $d^{2} / 8=32$, where equality occurs if and only if $A B \\perp A C \\perp C D$ and $A C=A B+C D=8$. In this case $B D=8 \\sqrt{2}$. (b) Let $A^{\\prime}$ be the point with $\\overrightarrow{D A^{\\prime}}=\\overrightarrow{A C}$. The triangular inequality implies $A D+B C \\geq A A^{\\prime}=8 \\sqrt{5}$. Thus the perimeter attains its minimum for $A B=C D=4$. (c) Let us assume w.l.o.g. that $C D \\leq A B$. Then $C$ lies inside $\\triangle B D A^{\\prime}$ and hence $B C+A D=B C+C A^{\\prime}<B D+D A^{\\prime}$. The maximal value $B D+D A^{\\prime}$ of $B C+A D$ is attained when $C$ approaches $D$, making a degenerate quadrangle.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1976", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (GBR 1a) ${ }^{\\mathrm{IMO} 6}$ For all positive integral $n$, $u_{n+1}=u_{n}\\left(u_{n-1}^{2}-2\\right)-u_{1}$, $u_{0}=2$, and $u_{1}=5 / 2$. Prove that  $$ 3 \\log _{2}\\left[u_{n}\\right]=2^{n}-(-1)^{n} $$  where $[x]$ is the integral part of $x$.", "solution": "4. The first few values are easily verified to be $2^{r_{n}}+2^{-r_{n}}$, where $r_{0}=0$, $r_{1}=r_{2}=1, r_{3}=3, r_{4}=5, r_{5}=11, \\ldots$. Let us put $u_{n}=2^{r_{n}}+2^{-r_{n}}$ (we will show that $r_{n}$ exists and is integer for each $n$ ). A simple calculation gives us $u_{n}\\left(u_{n-1}^{2}-2\\right)=2^{r_{n}+2 r_{n-1}}+2^{-r_{n}-2 r_{n-1}}+2^{r_{n}-2 r_{n-1}}+2^{-r_{n}+2 r_{n-1}}$. If an array $q_{n}$, with $q_{0}=0$ and $q_{1}=1$, is set so as to satisfy the linear recurrence $q_{n+1}=q_{n}+2 q_{n-1}$, then it also satisfies $q_{n}-2 q_{n-1}=-\\left(q_{n-1}-\\right.$ $\\left.2 q_{n-2}\\right)=\\cdots=(-1)^{n-1}\\left(q_{1}-2 q_{0}\\right)=(-1)^{n-1}$. Assuming inductively up to $n r_{i}=q_{i}$, the expression for $u_{n}\\left(u_{n-1}^{2}-2\\right)=u_{n+1}+u_{1}$ reduces to $2^{q_{n+1}}+2^{-q_{n+1}}+u_{1}$. Therefore, $r_{n+1}=q_{n+1}$. The solution to this linear recurrence with $r_{0}=0, r_{1}=1$ is $r_{n}=q_{n}=\\frac{2^{n}-(-1)^{n}}{3}$, and since $\\left[u_{n}\\right]=2^{r_{n}}$ for $n \\geq 0$, the result follows. Remark. One could simply guess that $u_{n}=2^{r_{n}}+2^{-r_{n}}$ for $r_{n}=\\frac{2^{n}-(-1)^{n}}{3}$, and then prove this result by induction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1976", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (NET 3) ${ }^{\\mathrm{IMO} 5}$ Let a set of $p$ equations be given,  $$ \\begin{gathered} a_{11} x_{1}+\\cdots+a_{1 q} x_{q}=0 \\\\ a_{21} x_{1}+\\cdots+a_{2 q} x_{q}=0 \\\\ \\vdots \\\\ a_{p 1} x_{1}+\\cdots+a_{p q} x_{q}=0, \\end{gathered} $$  with coefficients $a_{i j}$ satisfying $a_{i j}=-1,0$, or +1 for all $i=1, \\ldots, p$ and $j=1, \\ldots, q$. Prove that if $q=2 p$, there exists a solution $x_{1}, \\ldots, x_{q}$ of this system such that all $x_{j}(j=1, \\ldots, q)$ are integers satisfying $\\left|x_{j}\\right| \\leq q$ and $x_{j} \\neq 0$ for at least one value of $j$.", "solution": "5. If one substitutes an integer $q$-tuple $\\left(x_{1}, \\ldots, x_{q}\\right)$ satisfying $\\left|x_{i}\\right| \\leq p$ for all $i$ in an equation of the given system, the absolute value of the right-hand member never exceeds $p q$. So for the right-hand member of the system there are $(2 p q+1)^{p}$ possibilities There are $(2 p+1)^{q}$ possible $q$-tuples $\\left(x_{1}, \\ldots, x_{q}\\right)$. Since $(2 p+1)^{q} \\geq(2 p q+1)^{p}$, there are at least two $q$-tuples $\\left(y_{1}, \\ldots, y_{q}\\right)$ and $\\left(z_{1}, \\ldots, z_{q}\\right)$ giving the same right-hand members in the given system. The difference $\\left(x_{1}, \\ldots, x_{q}\\right)=\\left(y_{1}-z_{1}, \\ldots, y_{q}-z_{q}\\right)$ thus satisfies all the requirements of the problem.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1976", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (NET 4) ${ }^{\\mathrm{IMO} 3} \\mathrm{~A}$ rectangular box can be filled completely with unit cubes. If one places cubes with volume 2 in the box such that their edges are parallel to the edges of the box, one can fill exactly $40 \\%$ of the box. Determine all possible (interior) sizes of the box.", "solution": "6. Suppose $a_{1} \\leq a_{2} \\leq a_{3}$ are the dimensions of the box. If we set $b_{i}=$ $\\left[a_{i} / \\sqrt[3]{2}\\right]$, the condition of the problem is equivalent to $\\frac{a_{1}}{b_{1}} \\cdot \\frac{a_{2}}{b_{2}} \\cdot \\frac{a_{3}}{b_{3}}=5$. We list some values of $a, b=[a / \\sqrt[3]{2}]$ and $a / b$ :  | $a$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | $b$ | 1 | 2 | 3 | 3 | 4 | 5 | 6 | 7 | 7 | | $a / b$ | 2 | 1.5 | 1.33 | 1.67 | 1.5 | 1.4 | 1.33 | 1.29 | 1.43 |  We note that if $a>2$, then $a / b \\leq 5 / 3$, and if $a>5$, then $a / b \\leq 3 / 2$. If $a_{1}>2$, then $\\frac{a_{1}}{b_{1}} \\cdot \\frac{a_{2}}{b_{2}} \\cdot \\frac{a_{3}}{b_{3}}<(5 / 3)^{3}<5$, a contradiction. Hence $a_{1}=2$. If also $a_{2}=2$, then $a_{3} / b_{3}=5 / 4 \\leq \\sqrt[3]{2}$, which is impossible. Also, if $a_{2} \\geq 6$, then $\\frac{a_{2}}{b_{2}} \\cdot \\frac{a_{3}}{b_{3}} \\leq(1.5)^{2}<2.5$, again a contradiction. We thus have the following cases: (i) $a_{1}=2, a_{2}=3$, then $a_{3} / b_{3}=5 / 3$, which holds only if $a_{3}=5$; (ii) $a_{1}=2, a_{2}=4$, then $a_{3} / b_{3}=15 / 8$, which is impossible; (iii) $a_{1}=2, a_{2}=5$, then $a_{3} / b_{3}=3 / 2$, which holds only if $a_{3}=6$.  The only possible sizes of the box are therefore $(2,3,5)$ and $(2,5,6)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1976", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (POL 1b) Let $I=(0,1]$ be the unit interval of the real line. For a given number $a \\in(0,1)$ we define a map $T: I \\rightarrow I$ by the formula  $$ T(x, y)= \\begin{cases}x+(1-a) & \\text { if } 0<x \\leq a \\\\ x-a & \\text { if } a<x \\leq 1\\end{cases} $$  Show that for every interval $J \\subset I$ there exists an integer $n>0$ such that $T^{n}(J) \\cap J \\neq \\emptyset$.", "solution": "7. The map $T$ transforms the interval $(0, a]$ onto $(1-a, 1]$ and the interval $(a, 1]$ onto $(0,1-a]$. Clearly $T$ preserves the measure. Since the measure of the interval $[0,1]$ is finite, there exist two positive integers $k, l>k$ such that $T^{k}(J)$ and $T^{l}(J)$ are not disjoint. But the map $T$ is bijective; hence $T^{l-k}(J)$ and $J$ are not disjoint.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1976", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (SWE 3) Let $P$ be a polynomial with real coefficients such that $P(x)>0$ if $x>0$. Prove that there exist polynomials $Q$ and $R$ with nonnegative coefficients such that $P(x)=\\frac{Q(x)}{R(x)}$ if $x>0$.", "solution": "8. Every polynomial with real coefficients can be factored as a product of linear and quadratic polynomials with real coefficients. Thus it suffices to prove the result only for a quadratic polynomial $P(x)=x^{2}-2 a x+b^{2}$, with $a>0$ and $b^{2}>a^{2}$. Using the identity  $$ \\left(x^{2}+b^{2}\\right)^{2 n}-(2 a x)^{2 n}=\\left(x^{2}-2 a x+b^{2}\\right) \\sum_{k=0}^{2 n-1}\\left(x^{2}+b^{2}\\right)^{k}(2 a x)^{2 n-k-1} $$  we have solved the problem if we can choose $n$ such that $b^{2 n}\\binom{2 n}{n}>2^{2 n} a^{2 n}$. However, it is is easy to show that $2 n\\binom{2 n}{n}<2^{2 n}$; hence it is enough to take $n$ such that $(b / a)^{2 n}>2 n$. Since $\\lim _{n \\rightarrow \\infty}(2 n)^{1 /(2 n)}=1<b / a$, such an $n$ always exists.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1976", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (FIN 2) ${ }^{\\mathrm{IMO} 2}$ Let $P_{1}(x)=x^{2}-2, P_{j}(x)=P_{1}\\left(P_{j-1}(x)\\right), j=2,3, \\ldots$. Show that for arbitrary $n$ the roots of the equation $P_{n}(x)=x$ are real and different from one another.", "solution": "9. The equation $P_{n}(x)=x$ is of degree $2^{n}$, and has at most $2^{n}$ distinct roots. If $x>2$, then by simple induction $P_{n}(x)>x$ for all $n$. Similarly, if $x<-1$, then $P_{1}(x)>2$, which implies $P_{n}(x)>2$ for all $n$. It follows that all real roots of the equation $P_{n}(x)=x$ lie in the interval $[-2,2]$, and thus have the form $x=2 \\cos t$. Now we observe that $P_{1}(2 \\cos t)=4 \\cos ^{2} t-2=2 \\cos 2 t$, and in general $P_{n}(2 \\cos t)=2 \\cos 2^{n} t$. Our equation becomes  $$ \\cos 2^{n} t=\\cos t $$  which indeed has $2^{n}$ different solutions $t=\\frac{2 \\pi m}{2^{n}-1}\\left(m=0,1, \\ldots, 2^{n-1}-1\\right)$ and $t=\\frac{2 \\pi m}{2^{n}+1}\\left(m=1,2, \\ldots, 2^{n-1}\\right)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (BUL 1) A pentagon $A B C D E$ inscribed in a circle for which $B C<C D$ and $A B<D E$ is the base of a pyramid with vertex $S$. If $A S$ is the longest edge starting from $S$, prove that $B S>C S$.", "solution": "1. Let $P$ be the projection of $S$ onto the plane $A B C D E$. Obviously $B S>C S$ is equivalent to $B P>C P$. The conditions of the problem imply that $P A>P B$ and $P A>P E$. The locus of such points $P$ is the region of the plane that is determined by the perpendicular bisectors of segments $A B$ and $A E$ and that contains the point diametrically opposite $A$. But since $A B<D E$, the whole of this region lies on one side of the perpendicular bisector of $B C$. The result follows immediately. Remark. The assumption $B C<C D$ is redundant.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (FRG 1) (SL77-3).", "solution": "10. Integers $a, b, q, r$ satisfy  $$ a^{2}+b^{2}=(a+b) q+r, \\quad 0 \\leq r<a+b, \\quad q^{2}+r=1977 . $$  From $q^{2} \\leq 1977$ it follows that $q \\leq 44$, and consequently $a^{2}+b^{2}<$ $45(a+b)$. Having in mind the inequality $(a+b)^{2} \\leq 2\\left(a^{2}+b^{2}\\right)$, we get $(a+b)^{2}<90(a+b)$, i.e., $a+b<90$ and consequently $r<90$. Now from $q^{2}=1977-r>1977-90=1887$ it follows that $q>43$; hence $q=44$ and $r=41$. It remains to find positive integers $a$ and $b$ satisfying $a^{2}+b^{2}=44(a+b)+41$, or equivalently  $$ (a-22)^{2}+(b-22)^{2}=1009 $$  The only solutions to this Diophantine equation are $(|a-22|,|b-22|) \\in$ $\\{(15,28),(28,15)\\}$, which yield $(a, b) \\in\\{(7,50),(37,50),(50,7),(50,37)\\}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (FRG 2) Let $n$ and $z$ be integers greater than 1 and $(n, z)=1$. Prove: (a) At least one of the numbers $z_{i}=1+z+z^{2}+\\cdots+z^{i}, i=0,1, \\ldots, n-1$, is divisible by $n$. (b) If $(z-1, n)=1$, then at least one of the numbers $z_{i}, i=0,1, \\ldots, n-2$, is divisible by $n$.", "solution": "11. (a) Suppose to the contrary that none of the numbers $z_{0}, z_{1}, \\ldots, z_{n-1}$ is divisible by $n$. Then two of these numbers, say $z_{k}$ and $z_{l}(0 \\leq k<l \\leq$ $n-1$ ), are congruent modulo $n$, and thus $n \\mid z_{l}-z_{k}=z^{k+1} z_{l-k-1}$. But since $(n, z)=1$, this implies $n \\mid z_{l-k-1}$, which is a contradiction. (b) Again suppose the contrary, that none of $z_{0}, z_{1}, \\ldots, z_{n-2}$ is divisible by $n$. Since $(z-1, n)=1$, this is equivalent to $n \\nmid(z-1) z_{j}$, i.e., $z^{k} \\not \\equiv 1$ $(\\bmod n)$ for all $k=1,2, \\ldots, n-1$. But since $(z, n)=1$, we also have that $z^{k} \\not \\equiv 0(\\bmod n)$. It follows that there exist $k, l, 1 \\leq k<l \\leq n-1$ such that $z^{k} \\equiv z^{l}$, i.e., $z^{l-k} \\equiv 1(\\bmod n)$, which is a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (FRG 3) Let $z$ be an integer $>1$ and let $M$ be the set of all numbers of the form $z_{k}=1+z+\\cdots+z^{k}, k=0,1, \\ldots$. Determine the set $T$ of divisors of at least one of the numbers $z_{k}$ from $M$.", "solution": "12. According to part (a) of the previous problem we can conclude that $T=$ $\\{n \\in \\mathbb{N} \\mid(n, z)=1\\}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (FRG 4) (SL77-4).", "solution": "13. The figure $\\Phi$ contains two points $A$ and $B$ having maximum distance. Let $h$ be the semicircle with diameter $A B$ that lies in $\\Phi$, and let $k$ be the circle containing $h$. Consider any point $M$ inside $k$. The line passing through $M$ that is orthogonal to $A M$ meets $h$ in some point $P$ (because $\\angle A M B>90^{\\circ}$ ). Let $h^{\\prime}$ and $\\overline{h^{\\prime}}$ be the two semicircles with diameter $A P$, where $M \\in h^{\\prime}$. Since $\\overline{h^{\\prime}}$ contains a point $C$ such that $B C>A B$, it cannot be contained in $\\Phi$, implying that $h^{\\prime} \\subset \\Phi$. Hence $M$ belongs to $\\Phi$. Since $\\Phi$ contains no points outside the circle $k$, it must coincide with the disk determined by $k$. On the other hand, any disk has the required property.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (FRG 5) (SL77-5).", "solution": "14. We prove by induction on $n$ that independently of the word $w_{0}$, the given algorithm generates all words of length $n$. This is clear for $n=1$. Suppose now the statement is true for $n-1$, and that we are given a word $w_{0}=$ $c_{1} c_{2} \\ldots c_{n}$ of length $n$. Obviously, the words $w_{0}, w_{1}, \\ldots, w_{2^{n-1}-1}$ all have the $n$th digit $c_{n}$, and by the inductive hypothesis these are all words whose $n$th digit is $c_{n}$. Similarly, by the inductive hypothesis $w_{2^{n-1}}, \\ldots, w_{2^{n-1}}$ are all words whose $n$th digit is $1-c_{n}$, and the induction is complete.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (GDR 1) Let $n$ be an integer greater than 1 . In the Cartesian coordinate system we consider all squares with integer vertices $(x, y)$ such that $1 \\leq$ $x, y \\leq n$. Denote by $p_{k}(k=0,1,2, \\ldots)$ the number of pairs of points that are vertices of exactly $k$ such squares. Prove that $\\sum_{k}(k-1) p_{k}=0$.", "solution": "15. Each segment is an edge of at most two squares and a diagonal of at most one square. Therefore $p_{k}=0$ for $k>3$, and we have to prove that  $$ p_{0}=p_{2}+2 p_{3} $$  Let us calculate the number $q(n)$ of considered squares. Each of these squares is inscribed in a square with integer vertices and sides parallel to the coordinate axes. There are $(n-s)^{2}$ squares of side $s$ with integer vertices and sides parallel to the coordinate axes, and each of them circumscribes exactly $s$ of the considered squares. It follows that $q(n)=\\sum_{s=1}^{n-1}(n-s)^{2} s=n^{2}\\left(n^{2}-1\\right) / 12$. Computing the number of edges and diagonals of the considered squares in two ways, we obtain that  $$ p_{1}+2 p_{2}+3 p_{3}=6 q(n) $$  On the other hand, the total number of segments with endpoints in the considered integer points is given by  $$ p_{0}+p_{1}+p_{2}+p_{3}=\\binom{n^{2}}{2}=\\frac{n^{2}\\left(n^{2}-1\\right)}{2}=6 q(n) . $$  Now (1) follows immediately from (2) and (3).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (GDR 2) (SL77-6).", "solution": "16. For $i=k$ and $j=l$ the system is reduced to $1 \\leq i, j \\leq n$, and has exactly $n^{2}$ solutions. Let us assume that $i \\neq k$ or $j \\neq l$. The points $A(i, j), B(k, l)$, $C(-j+k+l, i-k+l), D(i-j+l, i+j-k)$ are vertices of a negatively oriented square with integer vertices lying inside the square $[1, n] \\times[1, n]$, and each of these squares corresponds to exactly 4 solutions to the system. By the previous problem there are exactly $q(n)=n^{2}\\left(n^{2}-1\\right) / 12$ such squares. Hence the number of solutions is equal to $n^{2}+4 q(n)=n^{2}\\left(n^{2}+2\\right) / 3$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (GDR 3) A ball $K$ of radius $r$ is touched from the outside by mutually equal balls of radius $R$. Two of these balls are tangent to each other. Moreover, for two balls $K_{1}$ and $K_{2}$ tangent to $K$ and tangent to each other there exist two other balls tangent to $K_{1}, K_{2}$ and also to $K$. How many balls are tangent to $K$ ? For a given $r$ determine $R$.", "solution": "17. Centers of the balls that are tangent to $K$ are vertices of a regular polyhedron with triangular faces, with edge length $2 R$ and radius of circumscribed sphere $r+R$. Therefore the number $n$ of these balls is 4,6 , or 20 . It is straightforward to obtain that: (i) If $n=4$, then $r+R=2 R(\\sqrt{6} / 4)$, whence $R=r(2+\\sqrt{6})$. (ii) If $n=6$, then $r+R=2 R(\\sqrt{2} / 2)$, whence $R=r(1+\\sqrt{2})$. (iii) If $n=20$, then $r+R=2 R \\sqrt{5+\\sqrt{5}} / 8$, whence $R=r[\\sqrt{5-2 \\sqrt{5}}+$  $$ (3-\\sqrt{5}) / 2] $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (GDR 4) Given an isosceles triangle $A B C$ with a right angle at $C$, construct the center $M$ and radius $r$ of a circle cutting on segments $A B, B C, C A$ the segments $D E, F G$, and $H K$, respectively, such that $\\angle D M E+\\angle F M G+\\angle H M K=180^{\\circ}$ and $D E: F G: H K=A B: B C:$ $C A$.", "solution": "18. Let $U$ be the midpoint of the segment $A B$. The point $M$ belongs to $C U$ and $C M=(\\sqrt{5}-1) C U / 2, r=C U \\sqrt{\\sqrt{5}-2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (GBR 1) Given any integer $m>1$ prove that there exist infinitely many positive integers $n$ such that the last $m$ digits of $5^{n}$ are a sequence $a_{m}, a_{m-1}, \\ldots, a_{1}=5\\left(0 \\leq a_{j}<10\\right)$ in which each digit except the last is of opposite parity to its successor (i.e., if $a_{i}$ is even, then $a_{i-1}$ is odd, and if $a_{i}$ is odd, then $a_{i-1}$ is even).", "solution": "19. We shall prove the statement by induction on $m$. For $m=2$ it is trivial, since each power of 5 greater than 5 ends in 25 . Suppose that the statement is true for some $m \\geq 2$, and that the last $m$ digits of $5^{n}$ alternate in parity. It can be shown by induction that the maximum power of 2 that divides $5^{2^{m-2}}-1$ is $2^{m}$, and consequently the difference $5^{n+2^{m-2}}-5^{n}$ is divisible by $10^{m}$ but not by $2 \\cdot 10^{m}$. It follows that the last $m$ digits of the numbers $5^{n+2^{m-2}}$ and $5^{n}$ coincide, but the digits at the position $m+1$ have opposite parity. Hence the last $m+1$ digits of one of these two powers of 5 alternate in parity. The inductive proof is completed.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (BUL 2) (SL77-1).", "solution": "2. We shall prove by induction on $n$ that $f(x)>f(n)$ whenever $x>n$. The case $n=0$ is trivial. Suppose that $n \\geq 1$ and that $x>k$ implies $f(x)>f(k)$ for all $k<n$. It follows that $f(x) \\geq n$ holds for all $x \\geq n$. Let $f(m)=\\min _{x \\geq n} f(x)$. If we suppose that $m>n$, then $m-1 \\geq n$ and consequently $f(m-1) \\geq n$. But in this case the inequality $f(m)>$ $f(f(m-1))$ contradicts the minimality property of $m$. The inductive proof is thus completed. It follows that $f$ is strictly increasing, so $f(n+1)>f(f(n))$ implies that $n+1>f(n)$. But since $f(n) \\geq n$ we must have $f(n)=n$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. (GBR 2) (SL77-7).", "solution": "20. There exist $u, v$ such that $a \\cos x+b \\sin x=r \\cos (x-u)$ and $A \\cos 2 x+$ $B \\sin 2 x=R \\cos 2(x-v)$, where $r=\\sqrt{a^{2}+b^{2}}$ and $R=\\sqrt{A^{2}+B^{2}}$. Then $1-f(x)=r \\cos (x-u)+R \\cos 2(x-v) \\leq 1$ holds for all $x \\in \\mathbb{R}$. There exists $x \\in \\mathbb{R}$ such that $\\cos (x-u) \\geq 0$ and $\\cos 2(x-v)=1$ (indeed, either $x=v$ or $x=v+\\pi$ works). It follows that $R \\leq 1$. Similarly, there exists $x \\in \\mathbb{R}$ such that $\\cos (x-u)=1 / \\sqrt{2}$ and $\\cos 2(x-v) \\geq 0$ (either $x=u-\\pi / 4$ or $x=u+\\pi / 4$ works). It follows that $r \\leq \\sqrt{2}$. Remark. The proposition of this problem contained as an addendum the following, more difficult, inequality:  $$ \\sqrt{a^{2}+b^{2}}+\\sqrt{A^{2}+B^{2}} \\leq 2 . $$  The proof follows from the existence of $x \\in \\mathbb{R}$ such that $\\cos (x-u) \\geq 1 / 2$ and $\\cos 2(x-v) \\geq 1 / 2$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. (GBR 3) Given that $x_{1}+x_{2}+x_{3}=y_{1}+y_{2}+y_{3}=x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}=0$, prove that  $$ \\frac{x_{1}^{2}}{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}+\\frac{y_{1}^{2}}{y_{1}^{2}+y_{2}^{2}+y_{3}^{2}}=\\frac{2}{3} $$", "solution": "21. Let us consider the vectors $v_{1}=\\left(x_{1}, x_{2}, x_{3}\\right), v_{2}=\\left(y_{1}, y_{2}, y_{3}\\right), v_{3}=(1,1,1)$ in space. The given equalities express the condition that these three vectors are mutually perpendicular. Also, $\\frac{x_{1}^{2}}{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}, \\frac{y_{1}^{2}}{y_{1}^{2}+y_{2}^{2}+y_{3}^{2}}$, and $1 / 3$ are the squares of the projections of the vector $(1,0,0)$ onto the directions of $v_{1}, v_{2}, v_{3}$, respectively. The result follows from the fact that the sum of squares of projections of a unit vector on three mutually perpendicular directions is 1.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "22", "problem_type": null, "exam": "IMO", "problem": "22. (GBR 4) (SL77-8).", "solution": "22. Since the quadrilateral $O A_{1} B B_{1}$ is cyclic, $\\angle O A_{1} B_{1}=\\angle O B C$. By using the analogous equalities we obtain $\\angle O A_{4} B_{4}=\\angle O B_{3} C_{3}=\\angle O C_{2} D_{2}=$ $\\angle O D_{1} A_{1}=\\angle O A B$, and similarly $\\angle O B_{4} A_{4}=\\angle O B A$. Hence $\\triangle O A_{4} B_{4} \\sim$ $\\triangle O A B$. Analogously, we have for the other three pairs of triangles $\\triangle O B_{4} C_{4} \\sim \\triangle O B C, \\triangle O C_{4} D_{4} \\sim \\triangle O C D, \\triangle O D_{4} A_{4} \\sim \\triangle O D A$, and consequently $A B C D \\sim A_{4} B_{4} C_{4} D_{4}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "23", "problem_type": null, "exam": "IMO", "problem": "23. (HUN 1) (SL77-9).", "solution": "23. Every polynomial $q\\left(x_{1}, \\ldots, x_{n}\\right)$ with integer coefficients can be expressed in the form $q=r_{1}+x_{1} r_{2}$, where $r_{1}, r_{2}$ are polynomials in $x_{1}, \\ldots, x_{n}$ with integer coefficients in which the variable $x_{1}$ occurs only with even exponents. Thus if $q_{1}=r_{1}-x_{1} r_{2}$, the polynomial $q q_{1}=r_{1}^{2}-x_{1}^{2} r_{2}^{2}$ contains $x_{1}$ only with even exponents. We can continue inductively constructing polynomials $q_{j}, j=2,3, \\ldots, n$, such that $q q_{1} q_{2} \\cdots q_{j}$ contains each of variables $x_{1}, x_{2}, \\ldots, x_{j}$ only with even exponents. Thus the polynomial $q q_{1} \\cdots q_{n}$ is a polynomial in $x_{1}^{2}, \\ldots, x_{n}^{2}$. The polynomials $f$ and $g$ exist for every $n \\in \\mathbb{N}$. In fact, it suffices to construct $q_{1}, \\ldots, q_{n}$ for the polynomial $q=x_{1}+\\cdots+x_{n}$ and take $f=$ $q_{1} q_{2} \\cdots q_{n}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "24", "problem_type": null, "exam": "IMO", "problem": "24. (HUN 2) Determine all real functions $f(x)$ that are defined and continuous on the interval $(-1,1)$ and that satisfy the functional equation  $$ f(x+y)=\\frac{f(x)+f(y)}{1-f(x) f(y)} \\quad(x, y, x+y \\in(-1,1)) . $$", "solution": "24. Setting $x=y=0$ gives us $f(0)=0$. Let us put $g(x)=\\arctan f(x)$. The given functional equation becomes $\\tan g(x+y)=\\tan (g(x)+g(y))$; hence  $$ g(x+y)=g(x)+g(y)+k(x, y) \\pi $$  where $k(x, y)$ is an integer function. But $k(x, y)$ is continuous and $k(0,0)=$ 0 , therefore $k(x, y)=0$. Thus we obtain the classical Cauchy's functional equation $g(x+y)=g(x)+g(y)$ on the interval $(-1,1)$, all of whose continuous solutions are of the form $g(x)=a x$ for some real $a$. Moreover, $g(x) \\in(-\\pi, \\pi)$ implies $|a| \\leq \\pi / 2$. Therefore $f(x)=\\tan a x$ for some $|a| \\leq \\pi / 2$, and this is indeed a solution to the given equation.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "25", "problem_type": null, "exam": "IMO", "problem": "25. (HUN 3) Prove the identity  $$ (z+a)^{n}=z^{n}+a \\sum_{k=1}^{n}\\binom{n}{k}(a-k b)^{k-1}(z+k b)^{n-k} . $$", "solution": "25. Let  $$ f_{n}(z)=z^{n}+a \\sum_{k=1}^{n}\\binom{n}{k}(a-k b)^{k-1}(z+k b)^{n-k} . $$  We shall prove by induction on $n$ that $f_{n}(z)=(z+a)^{n}$. This is trivial for $n=1$. Suppose that the statement is true for some positive integer $n-1$. Then  $$ \\begin{aligned} f_{n}^{\\prime}(z) & =n z^{n-1}+a \\sum_{k=1}^{n-1}\\binom{n}{k}(n-k)(a-k b)^{k-1}(z+k b)^{n-k-1} \\\\ & =n z^{n-1}+n a \\sum_{k=1}^{n-1}\\binom{n-1}{k}(a-k b)^{k-1}(z+k b)^{n-k-1} \\\\ & =n f_{n-1}(z)=n(z+a)^{n-1} \\end{aligned} $$  It remains to prove that $f_{n}(-a)=0$. For $z=-a$ we have by the lemma of (SL81-13),  $$ \\begin{aligned} f_{n}(-a) & =(-a)^{n}+a \\sum_{k=1}^{n}\\binom{n}{k}(-1)^{n-k}(a-k b)^{n-1} \\\\ & =a \\sum_{k=0}^{n}\\binom{n}{k}(-1)^{n-k}(a-k b)^{n-1}=0 . \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "26", "problem_type": null, "exam": "IMO", "problem": "26. (NET 1) Let $p$ be a prime number greater than 5 . Let $V$ be the collection of all positive integers $n$ that can be written in the form $n=k p+1$ or $n=k p-1(k=1,2, \\ldots)$. A number $n \\in V$ is called indecomposable in $V$ if it is impossible to find $k, l \\in V$ such that $n=k l$. Prove that there exists a number $N \\in V$ that can be factorized into indecomposable factors in $V$ in more than one way.", "solution": "26. The result is an immediate consequence (for $G=\\{-1,1\\}$ ) of the following generalization. (1) Let $G$ be a proper subgroup of $\\mathbb{Z}_{n}^{*}$ (the multiplicative group of residue classes modulo $n$ coprime to $n$ ), and let $V$ be the union of elements of $G$. A number $m \\in V$ is called indecomposable in $V$ if there do not exist numbers $p, q \\in V, p, q \\notin\\{-1,1\\}$, such that $p q=m$. There exists a number $r \\in V$ that can be expressed as a product of elements indecomposable in $V$ in more than one way. First proof. We shall start by proving the following lemma. Lemma. There are infinitely many primes not in $V$ that do not divide $n$. Proof. There is at least one such prime: In fact, any number other than $\\pm 1$ not in $V$ must have a prime factor not in $V$, since $V$ is closed under multiplication. If there were a finite number of such primes, say $p_{1}, p_{2}, \\ldots, p_{k}$, then one of the numbers $p_{1} p_{2} \\cdots p_{k}+n, p_{1}^{2} p_{2} \\cdots p_{k}+n$ is not in $V$ and is coprime to $n$ and $p_{1}, \\ldots, p_{k}$, which is a contradiction. [This lemma is actually a direct consequence of Dirichlet's theorem.] Let us consider two such primes $p, q$ that are congruent modulo $n$. Let $p^{k}$ be the least power of $p$ that is in $V$. Then $p^{k}, q^{k}, p^{k-1} q, p q^{k-1}$ belong to $V$ and are indecomposable in $V$. It follows that  $$ r=p^{k} \\cdot q^{k}=p^{k-1} q \\cdot p q^{k-1} $$  has the desired property. Second proof. Let $p$ be any prime not in $V$ that does not divide $n$, and let $p^{k}$ be the least power of $p$ that is in $V$. Obviously $p^{k}$ is indecomposable in $V$. Then the number  $$ r=p^{k} \\cdot\\left(p^{k-1}+n\\right)(p+n)=p\\left(p^{k-1}+n\\right) \\cdot p^{k-1}(p+n) $$  has at least two different factorizations into indecomposable factors.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "27", "problem_type": null, "exam": "IMO", "problem": "27. (NET 2) (SL77-10).", "solution": "27. The result is a consequence of the generalization from the previous problem for $G=\\{1\\}$. Remark. There is an explicit example: $r=(n-1)^{2} \\cdot(2 n-1)^{2}=[(n-$ 1) $(2 n-1)]^{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "28", "problem_type": null, "exam": "IMO", "problem": "28. (NET 3) (SL77-11).", "solution": "28. The recurrent relations give us that  $$ x_{i+1}=\\left[\\frac{x_{i}+\\left[n / x_{i}\\right]}{2}\\right]=\\left[\\frac{x_{i}+n / x_{i}}{2}\\right] \\geq[\\sqrt{n}] $$  On the other hand, if $x_{i}>[\\sqrt{n}]$ for some $i$, then we have $x_{i+1}<x_{i}$. This follows from the fact that $x_{i+1}<x_{i}$ is equivalent to $x_{i}>\\left(x_{i}+n / x_{i}\\right) / 2$, i.e., to $x_{i}^{2}>n$. Therefore $x_{i}=[\\sqrt{n}]$ holds for at least one $i \\leq n-[\\sqrt{n}]+1$. Remark. If $n+1$ is a perfect square, then $x_{i}=[\\sqrt{n}]$ implies $x_{i+1}=$ $[\\sqrt{n}]+1$. Otherwise, $x_{i}=[\\sqrt{n}]$ implies $x_{i+1}=[\\sqrt{n}]$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "29", "problem_type": null, "exam": "IMO", "problem": "29. (NET 4) (SL77-12).", "solution": "29. Let us denote the midpoints of segments $L M, A N, B L, M N, B K, C M$, $N K, C L, D N, K L, D M, A K$ by $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}, P_{6}, P_{7}, P_{8}, P_{9}, P_{10}$, $P_{11}, P_{12}$, respectively.  We shall prove that the dodecagon $P_{1} P_{2} P_{3} \\ldots P_{11} P_{12}$ is regular. From $B L=B A$ and $\\angle A B L=30^{\\circ}$ it follows that $\\angle B A L=75^{\\circ}$. Similarly $\\angle D A M=75^{\\circ}$, and therefore $\\angle L A M=60^{\\circ}$, which together with $A L=A M$ implies that the triangle $A L M$ is equilateral. Now, from the triangles $O L M$ and $A L N$, we get ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-426.jpg?height=383&width=433&top_left_y=233&top_left_x=863) $O P_{1}=L M / 2, O P_{2}=A L / 2$ and $O P_{2} \\| A L$. Hence $O P_{1}=O P_{2}$, $\\angle P_{1} O P_{2}=\\angle P_{1} A L=30^{\\circ}$ and $\\angle P_{2} O M=\\angle L A D=15^{\\circ}$. The desired result follows from symmetry.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (BUL 3) In a company of $n$ persons, each person has no more than $d$ acquaintances, and in that company there exists a group of $k$ persons, $k \\geq d$, who are not acquainted with each other. Prove that the number of acquainted pairs is not greater than $\\left[n^{2} / 4\\right]$.", "solution": "3. Let $v_{1}, v_{2}, \\ldots, v_{k}$ be $k$ persons who are not acquainted with each other. Let us denote by $m$ the number of acquainted couples and by $d_{j}$ the number of acquaintances of person $v_{j}$. Then $m \\leq d_{k+1}+d_{k+2}+\\cdots+d_{n} \\leq d(n-k) \\leq k(n-k) \\leq\\left(\\frac{k+(n-k)}{2}\\right)^{2}=\\frac{n^{2}}{4}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "30", "problem_type": null, "exam": "IMO", "problem": "30. (NET 5) A triangle $A B C$ with $\\angle A=30^{\\circ}$ and $\\angle C=54^{\\circ}$ is given. On $B C$ a point $D$ is chosen such that $\\angle C A D=12^{\\circ}$. On $A B$ a point $E$ is chosen such that $\\angle A C E=6^{\\circ}$. Let $S$ be the point of intersection of $A D$ and $C E$. Prove that $B S=B C$.", "solution": "30. Suppose $\\angle S B A=x$. By the trigonometric form of Ceva's theorem we have  $$ \\frac{\\sin \\left(96^{\\circ}-x\\right)}{\\sin x} \\frac{\\sin 18^{\\circ}}{\\sin 12^{\\circ}} \\frac{\\sin 6^{\\circ}}{\\sin 48^{\\circ}}=1 $$  We claim that $x=12^{\\circ}$ is a solution of this equation. To prove this, it is enough to show that $\\sin 84^{\\circ} \\sin 6^{\\circ} \\sin 18^{\\circ}=\\sin 48^{\\circ} \\sin 12^{\\circ} \\sin 12^{\\circ}$, which is equivalent to $\\sin 18^{\\circ}=2 \\sin 48^{\\circ} \\sin 12^{\\circ}=\\cos 36^{\\circ}-\\cos 60^{\\circ}$. The last equality can be checked directly. Since the equation is equivalent to $\\left(\\sin 96^{\\circ} \\cot x-\\cos 96^{\\circ}\\right) \\sin 6^{\\circ} \\sin 18^{\\circ}=$ $\\sin 48^{\\circ} \\sin 12^{\\circ}$, the solution $x \\in[0, \\pi)$ is unique. Hence $x=12^{\\circ}$. Second solution. We know that if $a, b, c, a^{\\prime}, b^{\\prime}, c^{\\prime}$ are points on the unit circle in the complex plane, the lines $a a^{\\prime}, b b^{\\prime}, c c^{\\prime}$ are concurrent if and only if  $$ \\left(a-b^{\\prime}\\right)\\left(b-c^{\\prime}\\right)\\left(c-a^{\\prime}\\right)=\\left(a-c^{\\prime}\\right)\\left(b-a^{\\prime}\\right)\\left(c-b^{\\prime}\\right) $$  We shall prove that $x=12^{\\circ}$. We may suppose that $A B C$ is the triangle in the complex plane with vertices $a=1, b=\\epsilon^{9}, c=\\epsilon^{14}$, where $\\epsilon=$ $\\cos \\frac{\\pi}{15}+i \\sin \\frac{\\pi}{15}$. If $a^{\\prime}=\\epsilon^{12}, b^{\\prime}=\\epsilon^{28}, c^{\\prime}=\\epsilon$, our task is the same as proving that lines $a a^{\\prime}, b b^{\\prime}, c c^{\\prime}$ are concurrent, or by (1) that  $$ \\left(1-\\epsilon^{28}\\right)\\left(\\epsilon^{9}-\\epsilon\\right)\\left(\\epsilon^{14}-\\epsilon^{12}\\right)-(1-\\epsilon)\\left(\\epsilon^{9}-\\epsilon^{12}\\right)\\left(\\epsilon^{14}-\\epsilon^{28}\\right)=0 . $$  The last equality holds, since the left-hand side is divisible by the minimum polynomial of $\\epsilon: z^{8}+z^{7}-z^{5}-z^{4}-z^{3}+z+1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "31", "problem_type": null, "exam": "IMO", "problem": "31. (POL 1) Let $f$ be a function defined on the set of pairs of nonzero rational numbers whose values are positive real numbers. Suppose that $f$ satisfies the following conditions: (1) $f(a b, c)=f(a, c) f(b, c), f(c, a b)=f(c, a) f(c, b)$; (2) $f(a, 1-a)=1$.  Prove that $f(a, a)=f(a,-a)=1, f(a, b) f(b, a)=1$.", "solution": "31. We obtain from (1) that $f(1, c)=f(1, c) f(1, c)$; hence $f(1, c)=1$ and consequently $f(-1, c) f(-1, c)=f(1, c)=1$, i.e. $f(-1, c)=1$. Analogously, $f(c, 1)=f(c,-1)=1$. Clearly $f(1,1)=f(-1,1)=f(1,-1)=1$. Now let us assume that $a \\neq 1$. Observe that $f\\left(x^{-1}, y\\right)=f\\left(x, y^{-1}\\right)=f(x, y)^{-1}$. Thus by (1) and (2) we get  $$ \\begin{aligned} 1 & =f(a, 1-a) f(1 / a, 1-1 / a) \\\\ & =f(a, 1-a) f\\left(a, \\frac{1}{1-1 / a}\\right)=f\\left(a, \\frac{1-a}{1-1 / a}\\right)=f(a,-a) \\end{aligned} $$  We now have $f(a, a)=f(a,-1) f(a,-a)=1 \\cdot 1=1$ and $1=f(a b, a b)=$ $f(a, a b) f(b, a b)=f(a, a) f(a, b) f(b, a) f(b, b)=f(a, b) f(b, a)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "32", "problem_type": null, "exam": "IMO", "problem": "32. (POL 2) In a room there are nine men. Among every three of them there are two mutually acquainted. Prove that some four of them are mutually acquainted.", "solution": "32. It is a known result that among six persons there are 3 mutually acquainted or 3 mutually unacquainted. By the condition of the problem the last case is excluded. If there is a man in the room who is not acquainted with four of the others, then these four men are mutually acquainted. Otherwise, each man is acquainted with at least five others, and since the sum of numbers of acquaintances of all men in the room is even, one of the men is acquainted with at least six men. Among these six there are three mutually acquainted, and they together with the first one make a group of four mutually acquainted men.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "33", "problem_type": null, "exam": "IMO", "problem": "33. (POL 3) A circle $K$ centered at $(0,0)$ is given. Prove that for every vector $\\left(a_{1}, a_{2}\\right)$ there is a positive integer $n$ such that the circle $K$ translated by the vector $n\\left(a_{1}, a_{2}\\right)$ contains a lattice point (i.e., a point both of whose coordinates are integers).", "solution": "33. Let $r$ be the radius of $K$ and $s>\\sqrt{2} / r$ an integer. Consider the points $A_{k}\\left(k a_{1}-\\left[k a_{1}\\right], k a_{2}-\\left[k a_{2}\\right]\\right)$, where $k=0,1,2, \\ldots, s^{2}$. Since all these points are in the unit square, two of them, say $A_{p}, A_{q}, q>p$, are in a small square with side $1 / s$, and consequently $A_{p} A_{q} \\leq \\sqrt{2} / s<r$. Therefore, for $n=q-p, m_{1}=\\left[q a_{1}\\right]-\\left[p a_{1}\\right]$ and $m_{2}=\\left[q a_{2}\\right]-\\left[p a_{2}\\right]$ the distance between the points $n\\left(a_{1}, a_{2}\\right)$ and $\\left(m_{1}, m_{2}\\right)$ is less then $r$, i.e., the point $\\left(m_{1}, m_{2}\\right)$ is in the circle $K+n\\left(a_{1}, a_{2}\\right)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "34", "problem_type": null, "exam": "IMO", "problem": "34. (POL 4) (SL77-13).", "solution": "34. Let $A$ be the set of the $2^{n}$ sequences of $n$ terms equal to $\\pm 1$. Since there are $k^{2}$ products $a b$ with $a, b \\in B$, by the pigeonhole principle there exists $c \\in A$ such that $a b=c$ holds for at most $k^{2} / 2^{n}$ pairs $(a, b) \\in B \\times B$. Then $c b \\in B$ holds for at most $k^{2} / 2^{n}$ values $b \\in B$, which means that $|B \\cap c B| \\leq k^{2} / 2^{n}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "35", "problem_type": null, "exam": "IMO", "problem": "35. (ROM 1) Find all numbers $N=\\overline{a_{1} a_{2} \\ldots a_{n}}$ for which $9 \\times \\overline{a_{1} a_{2} \\ldots a_{n}}=$ $\\overline{a_{n} \\ldots a_{2} a_{1}}$ such that at most one of the digits $a_{1}, a_{2}, \\ldots, a_{n}$ is zero.", "solution": "35. The solutions are 0 and $N_{k}=10 \\underbrace{99 \\ldots 9}_{k} 89$, where $k=0,1,2, \\ldots$.  Remark. If we omit the condition that at most one of the digits is zero, the solutions are numbers of the form $N_{k_{1}} N_{k_{2}} \\ldots N_{k_{r}}$, where $k_{1}=k_{r}$, $k_{2}=k_{r-1}$ etc. The more general problem $k \\cdot \\overline{a_{1} a_{2} \\ldots a_{n}}=\\overline{a_{n} \\ldots a_{2} a_{1}}$ has solutions only for $k=9$ and for $k=4$ (namely $0,2199 \\ldots 978$ and combinations as above).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "36", "problem_type": null, "exam": "IMO", "problem": "36. (ROM 2) Consider a sequence of numbers $\\left(a_{1}, a_{2}, \\ldots, a_{2^{n}}\\right)$. Define the operation  $$ S\\left(\\left(a_{1}, a_{2}, \\ldots, a_{2^{n}}\\right)\\right)=\\left(a_{1} a_{2}, a_{2} a_{3}, \\ldots, a_{2^{n}-1} a_{2^{n}}, a_{2^{n}} a_{1}\\right) $$  Prove that whatever the sequence $\\left(a_{1}, a_{2}, \\ldots, a_{2^{n}}\\right)$ is, with $a_{i} \\in\\{-1,1\\}$ for $i=1,2, \\ldots, 2^{n}$, after finitely many applications of the operation we get the sequence $(1,1, \\ldots, 1)$.", "solution": "36. It can be shown by simple induction that $S^{m}\\left(a_{1}, \\ldots, a_{2^{n}}\\right)=\\left(b_{1}, \\ldots, b_{2^{n}}\\right)$, where  $$ \\left.b_{k}=\\prod_{i=0}^{m} a_{k+i}^{\\binom{m}{i}} \\text { (assuming that } a_{k+2^{n}}=a_{k}\\right) $$  If we take $m=2^{n}$ all the binomial coefficients $\\binom{m}{i}$ apart from $i=0$ and $i=m$ will be even, and thus $b_{k}=a_{k} a_{k+m}=1$ for all $k$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "37", "problem_type": null, "exam": "IMO", "problem": "37. (ROM 3) Let $A_{1}, A_{2}, \\ldots, A_{n+1}$ be positive integers such that $\\left(A_{i}, A_{n+1}\\right)$ $=1$ for every $i=1,2, \\ldots, n$. Show that the equation  $$ x_{1}^{A_{1}}+x_{2}^{A_{2}}+\\cdots+x_{n}^{A_{n}}=x_{n+1}^{A_{n+1}} $$  has an infinite set of solutions $\\left(x_{1}, x_{2}, \\ldots, x_{n+1}\\right)$ in positive integers.", "solution": "37. We look for a solution with $x_{1}^{A_{1}}=\\cdots=x_{n}^{A_{n}}=n^{A_{1} A_{2} \\cdots A_{n} x}$ and $x_{n+1}=$ $n^{y}$. In order for this to be a solution we must have $A_{1} A_{2} \\cdots A_{n} x+1=$ $A_{n+1} y$. This equation has infinitely many solutions $(x, y)$ in $\\mathbb{N}$, since $A_{1} A_{2} \\cdots A_{n}$ and $A_{n+1}$ are coprime.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "38", "problem_type": null, "exam": "IMO", "problem": "38. (ROM 4) Let $m_{j}>0$ for $j=1,2, \\ldots, n$ and $a_{1} \\leq \\cdots \\leq a_{n}<b_{1} \\leq \\cdots \\leq$ $b_{n}<c_{1} \\leq \\cdots \\leq c_{n}$ be real numbers. Prove:  $$ \\left[\\sum_{j=1}^{n} m_{j}\\left(a_{j}+b_{j}+c_{j}\\right)\\right]^{2}>3\\left(\\sum_{j=1}^{n} m_{j}\\right)\\left[\\sum_{j=1}^{n} m_{j}\\left(a_{j} b_{j}+b_{j} c_{j}+c_{j} a_{j}\\right)\\right] . $$", "solution": "38. The condition says that the quadratic equation $f(x)=0$ has distinct real solutions, where  $$ f(x)=3 x^{2} \\sum_{j=1}^{n} m_{j}-2 x \\sum_{j=1}^{n} m_{j}\\left(a_{j}+b_{j}+c_{j}\\right)+\\sum_{j=1}^{n} m_{j}\\left(a_{j} b_{j}+b_{j} c_{j}+c_{j} a_{j}\\right) $$  It is easy to verify that the function $f$ is the derivative of  $$ F(x)=\\sum_{j=1}^{n} m_{j}\\left(x-a_{j}\\right)\\left(x-b_{j}\\right)\\left(x-c_{j}\\right) $$  Since $F\\left(a_{1}\\right) \\leq 0 \\leq F\\left(a_{n}\\right), F\\left(b_{1}\\right) \\leq 0 \\leq F\\left(b_{n}\\right)$ and $F\\left(c_{1}\\right) \\leq 0 \\leq F\\left(c_{n}\\right)$, $F(x)$ has three distinct real roots, and hence by Rolle's theorem its derivative $f(x)$ has two distinct real roots.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "39", "problem_type": null, "exam": "IMO", "problem": "39. (ROM 5) Consider 37 distinct points in space, all with integer coordinates. Prove that we may find among them three distinct points such that their barycenter has integers coordinates.", "solution": "39. By the pigeonhole principle, we can find 5 distinct points among the given 37 such that their $x$-coordinates are congruent and their $y$-coordinates are congruent modulo 3 . Now among these 5 points either there exist three with $z$-coordinates congruent modulo 3 , or there exist three whose $z$ coordinates are congruent to $0,1,2$ modulo 3 . These three points are the desired ones. Remark. The minimum number $n$ such that among any $n$ integer points in space one can find three points whose barycenter is an integer point is $n=19$. Each proof of this result seems to consist in studying a great number of cases.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (BUL 4) We are given $n$ points in space. Some pairs of these points are connected by line segments so that the number of segments equals $\\left[n^{2} / 4\\right]$, and a connected triangle exists. Prove that any point from which the maximal number of segments starts is a vertex of a connected triangle.", "solution": "4. Consider any vertex $v_{n}$ from which the maximal number $d$ of segments start, and suppose it is not a vertex of a triangle. Let $\\mathcal{A}=$ $\\left\\{v_{1}, v_{2}, \\ldots, v_{d}\\right\\}$ be the set of points that are connected to $v_{n}$, and let $\\mathcal{B}=\\left\\{v_{d+1}, v_{d+2}, \\ldots, v_{n}\\right\\}$ be the set of the other points. Since $v_{n}$ is not a vertex of a triangle, there is no segment both of whose vertices lie in $\\mathcal{A}$; i.e., each segment has an end in $\\mathcal{B}$. Thus, if $d_{j}$ denotes the number of segments at $v_{j}$ and $m$ denotes the total number of segments, we have  $$ m \\leq d_{d+1}+d_{d+2}+\\cdots+d_{n} \\leq d(n-d) \\leq\\left[\\frac{n^{2}}{4}\\right]=m $$  This means that each inequality must be equality, implying that each point in $\\mathcal{B}$ is a vertex of $d$ segments, and each of these segments has the other end in $\\mathcal{A}$. Then there is no triangle at all, which is a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "40", "problem_type": null, "exam": "IMO", "problem": "40. (SWE 1) The numbers $1,2,3, \\ldots, 64$ are placed on a chessboard, one number in each square. Consider all squares on the chessboard of size $2 \\times 2$. Prove that there are at least three such squares for which the sum of the 4 numbers contained exceeds 100.", "solution": "40. Let us divide the chessboard into 16 squares $Q_{1}, Q_{2}, \\ldots, Q_{16}$ of size $2 \\times 2$. Let $s_{k}$ be the sum of numbers in $Q_{k}$, and let us assume that $s_{1} \\geq s_{2} \\geq$ $\\cdots \\geq s_{16}$. Since $s_{4}+s_{5}+\\cdots+s_{16} \\geq 1+2+\\cdots+52=1378$, we must have $s_{4} \\geq 100$ and hence $s_{1}, s_{2}, s_{3} \\geq 100$ as well.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "41", "problem_type": null, "exam": "IMO", "problem": "41. (SWE 2) A wheel consists of a fixed circular disk and a mobile circular ring. On the disk the numbers $1,2,3, \\ldots, N$ are marked, and on the ring $N$ integers $a_{1}, a_{2}, \\ldots, a_{N}$ of sum 1 are marked (see the figure). The ring can be turned into $N$ different positions in which the numbers on the disk and on the ring match each other. Multiply every number on the ring with the corresponding number on the disk and form the sum of $N$ products. In this way a ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-127.jpg?height=349&width=362&top_left_y=824&top_left_x=919) sum is obtained for every position of the ring. Prove that the $N$ sums are different.", "solution": "41. The considered sums are congruent modulo $n$ to $S_{k}=\\sum_{i=1}^{N}(i+k) a_{i}$, $k=0,1, \\ldots, N-1$. Since $S_{k}=S_{0}+k\\left(a_{1}+\\cdots+a_{n}\\right)=S_{0}+k$, all these sums give distinct residues modulo $n$ and therefore are distinct.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "42", "problem_type": null, "exam": "IMO", "problem": "42. (SWE 3) The sequence $a_{n, k}, k=1,2,3, \\ldots, 2^{n}, n=0,1,2, \\ldots$, is defined by the following recurrence formula:  $$ \\begin{aligned} a_{1} & =2, \\quad a_{n, k}=2 a_{n-1, k}^{3}, \\quad a_{n, k+2^{n-1}}=\\frac{1}{2} a_{n-1, k}^{3} \\\\ \\text { for } k & =1,2,3, \\ldots, 2^{n-1}, n=0,1,2, \\ldots . \\end{aligned} $$  Prove that the numbers $a_{n, k}$ are all different.", "solution": "42. It can be proved by induction on $n$ that $\\left\\{a_{n, k} \\mid 1 \\leq k \\leq 2^{n}\\right\\}=\\left\\{2^{m} \\mid m=3^{n}+3^{n-1} s_{1}+\\cdots+3^{1} s_{n-1}+s_{n}\\left(s_{i}= \\pm 1\\right)\\right\\}$. Thus the result is an immediate consequence of the following lemma. Lemma. Each positive integer $s$ can be uniquely represented in the form  $$ s=3^{n}+3^{n-1} s_{1}+\\cdots+3^{1} s_{n-1}+s_{n}, \\quad \\text { where } s_{i} \\in\\{-1,0,1\\} $$  Proof. Both the existence and the uniqueness can be shown by simple induction on $s$. The statement is trivial for $s=1$, while for $s>1$ there exist $q \\in \\mathbb{N}, r \\in\\{-1,0,1\\}$ such that $s=3 q+r$, and $q$ has a unique representation of the form (1).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "43", "problem_type": null, "exam": "IMO", "problem": "43. (FIN 1) Evaluate  $$ S=\\sum_{k=1}^{n} k(k+1) \\cdots(k+p), $$  where $n$ and $p$ are positive integers.", "solution": "43. Since $\\left.k(k+1) \\cdots(k+p)=(p+1)!\\binom{k+p}{p+1}=(p+1)!\\left[\\begin{array}{c}k+p+1 \\\\ p+2\\end{array}\\right)-\\binom{k+p}{p+2}\\right]$, it follows that $\\sum_{k=1}^{n} k(k+1) \\cdots(k+p)=(p+1)!\\binom{n+p+1}{p+2}=\\frac{n(n+1) \\cdots(n+p+1)}{p+2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "44", "problem_type": null, "exam": "IMO", "problem": "44. (FIN 2) Let $E$ be a finite set of points in space such that $E$ is not contained in a plane and no three points of $E$ are collinear. Show that $E$ contains the vertices of a tetrahedron $T=A B C D$ such that $T \\cap E=$ $\\{A, B, C, D\\}$ (including interior points of $T$ ) and such that the projection of $A$ onto the plane $B C D$ is inside a triangle that is similar to the triangle $B C D$ and whose sides have midpoints $B, C, D$.", "solution": "44. Let $d(X, \\sigma)$ denote the distance from a point $X$ to a plane $\\sigma$. Let us consider the pair $(A, \\pi)$ where $A \\in E$ and $\\pi$ is a plane containing some three points $B, C, D \\in E$ such that $d(A, \\pi)$ is the smallest possible. We may suppose that $B, C, D$ are selected such that $\\triangle B C D$ contains no other points of $E$. Let $A^{\\prime}$ be the projection of $A$ on $\\pi$, and let $l_{b}, l_{c}, l_{d}$ be lines through $B, C, D$ parallel to $C D, D B, B C$ respectively. If $A^{\\prime}$ is in the half-plane determined by $l_{d}$ not containing $B C$, then $d(D, A B C) \\leq d\\left(A^{\\prime}, A B C\\right)<d(A, B C D)$, which is impossible. Similarly, $A^{\\prime}$ lies in the half-planes determined by $l_{b}, l_{c}$ that contain $D$, and hence $A^{\\prime}$ is inside the triangle bordered by $l_{b}, l_{c}, l_{d}$. The minimality property of $(A, \\pi)$ and the way in which $B C D$ was selected guarantee that $E \\cap T=\\{A, B, C, D\\}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "45", "problem_type": null, "exam": "IMO", "problem": "45. (FIN 2') (SL77-14).", "solution": "45. As in the previous problem, let us choose the pair $(A, \\pi)$ such that $d(A, \\pi)$ is minimal. If $\\pi$ contains only three points of $E$, we are done. If not, there are four points in $E \\cap P$, say $A_{1}, A_{2}, A_{3}, A_{4}$, such that the quadrilateral $Q=A_{1} A_{2} A_{3} A_{4}$ contains no other points of $E$. Suppose $Q$ is not convex, and that w.l.o.g. $A_{1}$ is inside the triangle $A_{2} A_{3} A_{4}$. If $A_{0}$ is the projection of $A$ on $P$, the point $A_{1}$ belongs to one of the triangles $A_{0} A_{2} A_{3}, A_{0} A_{3} A_{4}$, $A_{0} A_{4} A_{2}$, say $A_{0} A_{2} A_{3}$. Then $d\\left(A_{1}, A A_{2} A_{3}\\right) \\leq d\\left(A_{0}, A A_{2} A_{3}\\right)<A A_{0}$, which is impossible. Hence $Q$ is convex. Also, by the minimality property of $(A, \\pi)$ the pyramid $A A_{1} A_{2} A_{3} A_{4}$ contains no other points of $E$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "46", "problem_type": null, "exam": "IMO", "problem": "46. (FIN 3) Let $f$ be a strictly increasing function defined on the set of real numbers. For $x$ real and $t$ positive, set  $$ g(x, t)=\\frac{f(x+t)-f(x)}{f(x)-f(x-t)} $$  Assume that the inequalities  $$ 2^{-1}<g(x, t)<2 $$  hold for all positive $t$ if $x=0$, and for all $t \\leq|x|$ otherwise. Show that  $$ 14^{-1}<g(x, t)<14 $$  for all real $x$ and positive $t$.", "solution": "46. We need to consider only the case $t>|x|$. There is no loss of generality in assuming $x>0$. To obtain the estimate from below, set  $$ \\begin{aligned} a_{1} & =f\\left(-\\frac{x+t}{2}\\right)-f(-(x+t)), & a_{2}=f(0)-f\\left(-\\frac{x+t}{2}\\right), \\\\ a_{3} & =f\\left(\\frac{x+t}{2}\\right)-f(0), & a_{4}=f(x+t)-f\\left(\\frac{x+t}{2}\\right) . \\end{aligned} $$  Since $-(x+t)<x-t$ and $x<(x+t) / 2$, we have $f(x)-f(x-t) \\leq$ $a_{1}+a_{2}+a_{3}$. Since $2^{-1}<a_{j+1} / a_{j}<2$, it follows that  $$ g(x, t)>\\frac{a_{4}}{a_{1}+a_{2}+a_{3}}>\\frac{a_{3} / 2}{4 a_{3}+2 a_{3}+a_{3}}=14^{-1} . $$  To obtain the estimate from above, set  $$ \\begin{aligned} b_{1} & =f(0)-f\\left(-\\frac{x+t}{3}\\right), & b_{2}=f\\left(\\frac{x+t}{3}\\right)-f(0), \\\\ b_{3} & =f\\left(\\frac{2(x+t)}{3}\\right)-f\\left(\\frac{x+t}{3}\\right), & b_{4}=f(x+t)-f\\left(\\frac{2(x+t)}{3}\\right) . \\end{aligned} $$  If $t<2 x$, then $x-t<-(x+t) / 3$ and therefore $f(x)-f(x-t) \\geq b_{1}$. If $t \\geq 2 x$, then $(x+t) / 3 \\leq x$ and therefore $f(x)-f(x-t) \\geq b_{2}$. Since $2^{-1}<b_{j+1} / b_{j}<2$, we get  $$ g(x, t)<\\frac{b_{2}+b_{3}+b_{4}}{\\min \\left\\{b_{1}, b_{2}\\right\\}}<\\frac{b_{2}+2 b_{2}+4 b_{2}}{b_{2} / 2}=14 $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "47", "problem_type": null, "exam": "IMO", "problem": "47. (USS 1) A square $A B C D$ is given. A line passing through $A$ intersects $C D$ at $Q$. Draw a line parallel to $A Q$ that intersects the boundary of the square at points $M$ and $N$ such that the area of the quadrilateral $A M N Q$ is maximal.", "solution": "47. $M$ lies on $A B$ and $N$ lies on $B C$. If $C Q \\leq 2 C D / 3$, then $B M=C Q / 2$. If $C Q>2 C D / 3$, then $N$ coincides with C .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "48", "problem_type": null, "exam": "IMO", "problem": "48. (USS 2) The intersection of a plane with a regular tetrahedron with edge $a$ is a quadrilateral with perimeter $P$. Prove that $2 a \\leq P \\leq 3 a$.", "solution": "48. Let a plane cut the edges $A B, B C, C D, D A$ at points $K, L, M, N$ respectively. Let $D^{\\prime}, A^{\\prime}, B^{\\prime}$ be distinct points in the plane $A B C$ such that the triangles $B C D^{\\prime}, C D^{\\prime} A^{\\prime}, D^{\\prime} A^{\\prime} B^{\\prime}$ are equilateral, and $M^{\\prime} \\in\\left[C D^{\\prime}\\right], N^{\\prime} \\in\\left[D^{\\prime} A^{\\prime}\\right]$, and $K^{\\prime} \\in\\left[A^{\\prime} B^{\\prime}\\right]$ such that $C M^{\\prime}=C M$, $A^{\\prime} N^{\\prime}=A N$, and $A^{\\prime} K^{\\prime}=A K$. The perimeter $P$ of the quadrilateral $K L M N$ is equal to the length of the polygonal line $K L M^{\\prime} N^{\\prime} K^{\\prime}$, which is not less than $K K^{\\prime}$. It follows that $P \\geq 2 a$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-430.jpg?height=265&width=548&top_left_y=997&top_left_x=810)  Let us consider all quadrilaterals $K L M N$ that are obtained by intersecting the tetrahedron by a plane parallel to a fixed plane $\\alpha$. The lengths of the segments $K L, L M, M N, N K$ are linear functions in $A K$, and so is $P$. Thus $P$ takes its maximum at an endpoint of the interval, i.e., when the plane $K L M N$ passes through one of the vertices $A, B, C, D$, and it is easy to see that in this case $P \\leq 3 a$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "49", "problem_type": null, "exam": "IMO", "problem": "49. (USS 3) Find all pairs of integers $(p, q)$ for which all roots of the trinomials $x^{2}+p x+q$ and $x^{2}+q x+p$ are integers.", "solution": "49. If one of $p, q$, say $p$, is zero, then $-q$ is a perfect square. Conversely, $(p, q)=\\left(0,-t^{2}\\right)$ and $(p, q)=\\left(-t^{2}, 0\\right)$ satisfy the conditions for $t \\in \\mathbb{Z}$. We now assume that $p, q$ are nonzero. If the trinomial $x^{2}+p x+q$ has two integer roots $x_{1}, x_{2}$, then $|q|=\\left|x_{1} x_{2}\\right| \\geq\\left|x_{1}\\right|+\\left|x_{2}\\right|-1 \\geq|p|-1$. Similarly, if $x^{2}+q x+p$ has integer roots, then $|p| \\geq|q|-1$ and $q^{2}-4 p$ is a square. Thus we have two cases to investigate: (i) $|p|=|q|$. Then $p^{2}-4 q=p^{2} \\pm 4 p$ is a square, so $(p, q)=(4,4)$. (ii) $|p|=|q| \\pm 1$. The solutions for $(p, q)$ are $(t,-1-t)$ for $t \\in \\mathbb{Z}$ and $(5,6)$, $(6,5)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (CZS 1) (SL77-2).", "solution": "5. Let us denote by $I$ and $E$ the sets of interior boundary points and exterior boundary points. Let $A B C D$ be the square inscribed in the circle $k$ with sides parallel to the coordinate axes. Lines $A B, B C, C D, D A$ divide the plane into 9 regions: $\\mathcal{R}, \\mathcal{R}_{A}, \\mathcal{R}_{B}$, $\\mathcal{R}_{C}, \\mathcal{R}_{D}, \\mathcal{R}_{A B}, \\mathcal{R}_{B C}, \\mathcal{R}_{C D}, \\mathcal{R}_{D A}$. There is a unique pair of lattice points $A_{I} \\in \\mathcal{R}, A_{E} \\in$ $\\mathcal{R}_{A}$ that are opposite vertices of a unit square. We similarly define $B_{I}, C_{I}, D_{I}, B_{E}, C_{E}, D_{E}$. Let us form a graph $G$ by connecting each point from $E$ lying in $\\mathcal{R}_{A B}$ (respectively $\\mathcal{R}_{B C}, \\mathcal{R}_{C D}, \\mathcal{R}_{D A}$ ) to its up- ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-420.jpg?height=389&width=505&top_left_y=271&top_left_x=825) per (respectively left, lower, right) neighbor point (which clearly belongs to $I$ ). It is easy to see that: (i) All vertices from $I$ other than $A_{I}, B_{I}, C_{I}, D_{I}$ have degree 1. (ii) $A_{E}$ is not in $E$ if and only if $A_{I} \\in I$ and $\\operatorname{deg} A_{I}=2$. (iii) No other lattice points inside $\\mathcal{R}_{A}$ belong to $E$.  Thus if $m$ is the number of edges of the graph $G$ and $s$ is the number of points among $A_{E}, B_{E}, C_{E}$, and $D_{E}$ that are in $E$, using (i)-(iii) we easily obtain $|E|=m+s$ and $|I|=m-(4-s)=|E|+4$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "50", "problem_type": null, "exam": "IMO", "problem": "50. (USS 4) Determine all positive integers $n$ for which there exists a polynomial $P_{n}(x)$ of degree $n$ with integer coefficients that is equal to $n$ at $n$ different integer points and that equals zero at zero.", "solution": "50. Suppose that $P_{n}(x)=n$ for $x \\in\\left\\{x_{1}, x_{2}, \\ldots, x_{n}\\right\\}$. Then  $$ P_{n}(x)=\\left(x-x_{1}\\right)\\left(x-x_{2}\\right) \\cdots\\left(x-x_{n}\\right)+n . $$  From $P_{n}(0)=0$ we obtain $n=\\left|x_{1} x_{2} \\cdots x_{n}\\right| \\geq 2^{n-2}$ (because at least $n-2$ factors are different from $\\pm 1$ ) and therefore $n \\geq 2^{n-2}$. It follows that $n \\leq 4$. For each positive integer $n \\leq 4$ there exists a polynomial $P_{n}$. Here is the list of such polynomials:  $$ \\begin{array}{ll} n=1: \\pm x, & n=2: 2 x^{2}, x^{2} \\pm x,-x^{2} \\pm 3 x \\\\ n=3: \\pm\\left(x^{3}-x\\right)+3 x^{2}, & n=4:-x^{4}+5 x^{2} . \\end{array} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "51", "problem_type": null, "exam": "IMO", "problem": "51. (USS 5) Several segments, which we shall call white, are given, and the sum of their lengths is 1 . Several other segments, which we shall call black, are given, and the sum of their lengths is 1 . Prove that every such system of segments can be distributed on the segment that is 1.51 long in the following way: Segments of the same color are disjoint, and segments of different colors are either disjoint or one is inside the other. Prove that there exists a system that cannot be distributed in that way on the segment that is 1.49 long.", "solution": "51. We shall use the following algorithm:  Choose a segment of maximum length (\"basic\" segment) and put on it unused segments of the opposite color without overlapping, each time of the maximum possible length, as long as it is possible. Repeat the procedure with remaining segments until all the segments are used. Let us suppose that the last basic segment is black. Then the length of the used part of any white basic segment is greater than the free part, and consequently at least one-half of the length of the white segments has been used more than once. Therefore all basic segments have total length at most 1.5 and can be distributed on a segment of length 1.51. On the other hand, if we are given two white segments of lengths 0.5 and two black segments of lengths 0.999 and 0.001 , we cannot distribute them on a segment of length less than 1.499.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "52", "problem_type": null, "exam": "IMO", "problem": "52. (USA 1) Two perpendicular chords are drawn through a given interior point $P$ of a circle with radius $R$. Determine, with proof, the maximum and the minimum of the sum of the lengths of these two chords if the distance from $P$ to the center of the circle is $k R$.", "solution": "52. The maximum and minimum are $2 R \\sqrt{4-2 k^{2}}$ and $2 R\\left(1+\\sqrt{1-k^{2}}\\right)$ respectively.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "53", "problem_type": null, "exam": "IMO", "problem": "53. (USA 2) Find all pairs of integers $a$ and $b$ for which  $$ 7 a+14 b=5 a^{2}+5 a b+5 b^{2} $$", "solution": "53. The discriminant of the given equation considered as a quadratic equation in $b$ is $196-75 a^{2}$. Thus $75 a^{2} \\leq 196$ and hence $-1 \\leq a \\leq 1$. Now the integer solutions of the given equation are easily found: $(-1,3),(0,0),(1,2)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "54", "problem_type": null, "exam": "IMO", "problem": "54. (USA 3) If $0 \\leq a \\leq b \\leq c \\leq d$, prove that  $$ a^{b} b^{c} c^{d} d^{a} \\geq b^{a} c^{b} d^{c} a^{d} $$", "solution": "54. We shall use the following lemma.  Lemma. If a real function $f$ is convex on the interval $I$ and $x, y, z \\in I$, $x \\leq y \\leq z$, then  $$ (y-z) f(x)+(z-x) f(y)+(x-y) f(z) \\leq 0 . $$  Proof. The inequality is obvious for $x=y=z$. If $x<z$, then there exist $p, r$ such that $p+r=1$ and $y=p x+r z$. Then by Jensen's inequality $f(p x+r z) \\leq p f(x)+r f(z)$, which is equivalent to the statement of the lemma. By applying the lemma to the convex function $-\\ln x$ we obtain $x^{y} y^{z} z^{x} \\geq$ $y^{x} z^{y} x^{z}$ for any $0<x \\leq y \\leq z$. Multiplying the inequalities $a^{b} b^{c} c^{a} \\geq b^{a} c^{b} a^{c}$ and $a^{c} c^{d} d^{a} \\geq c^{a} d^{c} a^{d}$ we get the desired inequality. Remark. Similarly, for $0<a_{1} \\leq a_{2} \\leq \\cdots \\leq a_{n}$ it holds that $a_{1}^{a_{2}} a_{2}^{a_{3}} \\cdots a_{n}^{a_{1}} \\geq a_{2}^{a_{1}} a_{3}^{a_{2}} \\cdots a_{1}^{a_{n}}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "55", "problem_type": null, "exam": "IMO", "problem": "55. (USA 4) Through a point $O$ on the diagonal $B D$ of a parallelogram $A B C D$, segments $M N$ parallel to $A B$, and $P Q$ parallel to $A D$, are drawn, with $M$ on $A D$, and $Q$ on $A B$. Prove that diagonals $A O, B P, D N$ (extended if necessary) will be concurrent.", "solution": "55. The statement is true without the assumption that $O \\in B D$. Let $B P \\cap$ $D N=\\{K\\}$. If we denote $\\overrightarrow{A B}=a, \\overrightarrow{A D}=b$ and $\\overrightarrow{A O}=\\alpha a+\\beta b$ for some $\\alpha, \\beta \\in \\mathbb{R}, 1 / \\alpha+1 / \\beta \\neq 1$, by straightforward calculation we obtain that  $$ \\overrightarrow{A K}=\\frac{\\alpha}{\\alpha+\\beta-\\alpha \\beta} a+\\frac{\\beta}{\\alpha+\\beta-\\alpha \\beta} b=\\frac{1}{\\alpha+\\beta-\\alpha \\beta} \\overrightarrow{A O} $$  Hence $A, K, O$ are collinear.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "56", "problem_type": null, "exam": "IMO", "problem": "56. (USA 5) The four circumcircles of the four faces of a tetrahedron have equal radii. Prove that the four faces of the tetrahedron are congruent triangles.", "solution": "56. See the solution to (LL67-38).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "57", "problem_type": null, "exam": "IMO", "problem": "57. (VIE 1) (SL77-15).", "solution": "57. Suppose that there exists a sequence of 17 terms $a_{1}, a_{2}, \\ldots, a_{17}$ satisfying the required conditions. Then the sum of terms in each row of the rectangular array below is positive, while the sum of terms in each column is negative, which is a contradiction.  $$ \\begin{array}{cccc} a_{1} & a_{2} & \\ldots & a_{11} \\\\ a_{2} & a_{3} & \\ldots & a_{12} \\\\ \\vdots & \\vdots & & \\vdots \\\\ a_{7} & a_{8} & \\ldots & a_{17} \\end{array} $$  On the other hand, there exist 16 -term sequences with the required property. An example is $5,5,-13,5,5,5,-13,5,5,-13,5,5,5,-13,5,5$ which can be obtained by solving the system of equations $\\sum_{i=k}^{k+10} a_{i}=1$ $(k=1,2, \\ldots, 6)$ and $\\sum_{i=l}^{l+6} a_{i}=-1(l=1,2, \\ldots, 10)$. Second solution. We shall prove a stronger statement: If 7 and 11 in the question are replaced by any positive integers $m, n$, then the maximum number of terms is $m+n-(m, n)-1$. Let $a_{1}, a_{2}, \\ldots, a_{l}$ be a sequence of real numbers, and let us define $s_{0}=0$ and $s_{k}=a_{1}+\\cdots+a_{k}(k=1, \\ldots, l)$. The given conditions are equivalent to $s_{k}>s_{k+m}$ for $0 \\leq k \\leq l-m$ and $s_{k}<s_{k+n}$ for $0 \\leq k \\leq l-n$. Let $d=(m, n)$ and $m=m^{\\prime} d, n=n^{\\prime} d$. Suppose that there exists a sequence $\\left(a_{k}\\right)$ of length greater than or equal to $l=m+n-d$ satisfying the required conditions. Then the $m^{\\prime}+n^{\\prime}$ numbers $s_{0}, s_{d}, \\ldots, s_{\\left(m^{\\prime}+n^{\\prime}-1\\right) d}$ satisfy $n^{\\prime}$ inequalities $s_{k+m}<s_{k}$ and $m^{\\prime}$ inequalities $s_{k}<s_{k+n}$. Moreover, each term $s_{k d}$ appears twice in these inequalities: once on the left-hand and once on the right-hand side. It follows that there exists a ring of inequalities $s_{i_{1}}<s_{i_{2}}<\\cdots<s_{i_{k}}<s_{i_{1}}$, giving a contradiction. On the other hand, suppose that such a ring of inequalities can be made also for $l=m+n-d-1$, say $s_{i_{1}}<s_{i_{2}}<\\cdots<s_{i_{k}}<s_{i_{1}}$. If there are $p$ inequalities of the form $a_{k+m}<a_{k}$ and $q$ inequalities of the form $a_{k+n}>a_{k}$ in the ring, then $q n=r m$, which implies $m^{\\prime}\\left|q, n^{\\prime}\\right| p$ and thus $k=p+q \\geq m^{\\prime}+n^{\\prime}$. But since all $i_{1}, i_{2}, \\ldots, i_{k}$ are congruent modulo $d$, we have $k \\leq m^{\\prime}+n^{\\prime}-1$, a contradiction. Hence there exists a sequence of length $m+n-d-1$ with the required property.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "58", "problem_type": null, "exam": "IMO", "problem": "58. (VIE 2) Prove that for every triangle the following inequality holds:  $$ \\frac{a b+b c+c a}{4 S} \\geq \\cot \\frac{\\pi}{6} $$  where $a, b, c$ are lengths of the sides and $S$ is the area of the triangle.", "solution": "58. The following inequality (Finsler and Hadwiger, 1938) is sharper than the one we have to prove:  $$ 2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2} \\geq 4 S \\sqrt{3} $$  First proof. Let us set $2 x=b+c-a, 2 y=c+a-b, 2 z=a+b-c$.  Then $x, y, z>0$ and the inequality (1) becomes  $$ y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2} \\geq x y z(x+y+z) $$  which is equivalent to the obvious inequality $(x y-y z)^{2}+(y z-z x)^{2}+$ $(z x-x y)^{2} \\geq 0$. Second proof. Using the known relations for a triangle  $$ \\begin{aligned} a^{2}+b^{2}+c^{2} & =2 s^{2}-2 r^{2}-8 r R, \\\\ a b+b c+c a & =s^{2}+r^{2}+4 r R, \\\\ S & =r s, \\end{aligned} $$  where $r$ and $R$ are the radii of the incircle and the circumcircle, $s$ the semiperimeter and $S$ the area, we can transform (1) into  $$ s \\sqrt{3} \\leq 4 R+r . $$  The last inequality is a consequence of the inequalities $2 r \\leq R$ and $s^{2} \\leq$ $4 R^{2}+4 R r+3 r^{2}$, where the last one follows from the equality $H I^{2}=$ $4 R^{2}+4 R r+3 r^{2}-s^{2}$ ( $H$ and $I$ being the orthocenter and the incenter of the triangle).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "59", "problem_type": null, "exam": "IMO", "problem": "59. (VIE 3) (SL77-16).", "solution": "59. Let us consider the set $R$ of pairs of coordinates of the points from $E$ reduced modulo 3 . If some element of $R$ occurs thrice, then the corresponding points are vertices of a triangle with integer barycenter. Also, no three elements from $E$ can have distinct $x$-coordinates and distinct $y$ coordinates. By an easy discussion we can conclude that the set $R$ contains at most four elements. Hence $|E| \\leq 8$. An example of a set $E$ consisting of 8 points that satisfies the required condition is  $$ E=\\{(0,0),(1,0),(0,1),(1,1),(3,6),(4,6),(3,7),(4,7)\\} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (CZS 2) Let $x_{1}, x_{2}, \\ldots, x_{n}(n \\geq 1)$ be real numbers such that $0 \\leq x_{j} \\leq \\pi$, $j=1,2, \\ldots, n$. Prove that if $\\sum_{j=1}^{n}\\left(\\cos x_{j}+1\\right)$ is an odd integer, then $\\sum_{j=1}^{n} \\sin x_{j} \\geq 1$.", "solution": "6. Let $\\langle y\\rangle$ denote the distance from $y \\in \\mathbb{R}$ to the closest even integer. We claim that  $$ \\langle 1+\\cos x\\rangle \\leq \\sin x \\quad \\text { for all } x \\in[0, \\pi] $$  Indeed, if $\\cos x \\geq 0$, then $\\langle 1+\\cos x\\rangle=1-\\cos x \\leq 1-\\cos ^{2} x=\\sin ^{2} x \\leq$ $\\sin x$; the proof is similar if $\\cos x<0$. We note that $\\langle x+y\\rangle \\leq\\langle x\\rangle+\\langle y\\rangle$ holds for all $x, y \\in \\mathbb{R}$. Therefore  $$ \\sum_{j=1}^{n} \\sin x_{j} \\geq \\sum_{j=1}^{n}\\left\\langle 1+\\cos x_{j}\\right\\rangle \\geq\\left\\langle\\sum_{j=1}^{n}\\left(1+\\cos x_{j}\\right)\\right\\rangle=1 $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "60", "problem_type": null, "exam": "IMO", "problem": "60. (VIE 4) Suppose $x_{0}, x_{1}, \\ldots, x_{n}$ are integers and $x_{0}>x_{1}>\\cdots>x_{n}$. Prove that at least one of the numbers $\\left|F\\left(x_{0}\\right)\\right|,\\left|F\\left(x_{1}\\right)\\right|,\\left|F\\left(x_{2}\\right)\\right|, \\ldots$, $\\left|F\\left(x_{n}\\right)\\right|$, where  $$ F(x)=x^{n}+a_{1} x^{n-1}+\\cdots+a_{n}, \\quad a_{i} \\in \\mathbb{R}, \\quad i=1, \\ldots, n $$  is greater than $\\frac{n!}{2^{n}}$.", "solution": "60. By Lagrange's interpolation formula we have  $$ F(x)=\\sum_{j=0}^{n} F\\left(x_{j}\\right) \\frac{\\prod_{i \\neq j}\\left(x-x_{j}\\right)}{\\prod_{i \\neq j}\\left(x_{i}-x_{j}\\right)} . $$  Since the leading coefficient in $F(x)$ is 1 , it follows that  $$ 1=\\sum_{j=0}^{n} \\frac{F\\left(x_{j}\\right)}{\\prod_{i \\neq j}\\left(x_{i}-x_{j}\\right)} $$  Since  $$ \\left|\\prod_{i \\neq j}\\left(x_{i}-x_{j}\\right)\\right|=\\prod_{i=0}^{j-1}\\left|x_{i}-x_{j}\\right| \\prod_{i=j+1}^{n}\\left|x_{i}-x_{j}\\right| \\geq j!(n-j)! $$  we have  $$ 1 \\leq \\sum_{j=0}^{n} \\frac{\\left|F\\left(x_{j}\\right)\\right|}{\\left|\\prod_{i \\neq j}\\left(x_{i}-x_{j}\\right)\\right|} \\leq \\frac{1}{n!} \\sum_{j=0}^{n}\\binom{n}{j}\\left|F\\left(x_{j}\\right)\\right| \\leq \\frac{2^{n}}{n!} \\max \\left|F\\left(x_{j}\\right)\\right| . $$  Now the required inequality follows immediately.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (CZS 3) Prove the following assertion: If $c_{1}, c_{2}, \\ldots, c_{n}(n \\geq 2)$ are real numbers such that  $$ (n-1)\\left(c_{1}^{2}+c_{2}^{2}+\\cdots+c_{n}^{2}\\right)=\\left(c_{1}+c_{2}+\\cdots+c_{n}\\right)^{2} $$  then either all these numbers are nonnegative or all these numbers are nonpositive.", "solution": "7. Let us suppose that $c_{1} \\leq c_{2} \\leq \\cdots \\leq c_{n}$ and that $c_{1}<0<c_{n}$. There exists $k, 1 \\leq k<n$, such that $c_{k} \\leq 0<c_{k+1}$. Then we have  $$ \\begin{aligned} (n-1)\\left(c_{1}^{2}+c_{2}^{2}+\\cdots+c_{n}^{2}\\right) \\geq & k\\left(c_{1}^{2}+\\cdots+c_{k}^{2}\\right)+(n-k)\\left(c_{k+1}^{2}+\\cdots+c_{n}^{2}\\right) \\\\ \\geq & \\left(c_{1}+\\cdots+c_{k}\\right)^{2}+\\left(c_{k+1}+\\cdots+c_{n}\\right)^{2} \\\\ = & \\left(c_{1}+\\cdots+c_{n}\\right)^{2} \\\\ & -2\\left(c_{1}+\\cdots+c_{k}\\right)\\left(c_{k+1}+\\cdots+c_{n}\\right), \\end{aligned} $$  from which we obtain $\\left(c_{1}+\\cdots+c_{k}\\right)\\left(c_{k+1}+\\cdots+c_{n}\\right) \\geq 0$, a contradiction. Second solution. By the given condition and the inequality between arithmetic and quadratic mean we have  $$ \\begin{aligned} \\left(c_{1}+\\cdots+c_{n}\\right)^{2} & =(n-1)\\left(c_{1}^{2}+\\cdots+c_{n-1}^{2}\\right)+(n-1) c_{n}^{2} \\\\ & \\geq\\left(c_{1}+\\cdots+c_{n-1}\\right)^{2}+(n-1) c_{n}^{2} \\end{aligned} $$  which is equivalent to $2\\left(c_{1}+c_{2}+\\cdots+c_{n}\\right) c_{n} \\geq n c_{n}^{2}$. Similarly, $2\\left(c_{1}+c_{2}+\\right.$ $\\left.\\cdots+c_{n}\\right) c_{i} \\geq n c_{i}^{2}$ for all $i=1, \\ldots, n$. Hence all $c_{i}$ are of the same sign.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (CZS 4) A hexahedron $A B C D E$ is made of two regular congruent tetrahedra $A B C D$ and $A B C E$. Prove that there exists only one isometry $\\mathcal{Z}$ that maps points $A, B, C, D, E$ onto $B, C, A, E, D$, respectively. Find all points $X$ on the surface of hexahedron whose distance from $\\mathcal{Z}(X)$ is minimal.", "solution": "8. There is exactly one point satisfying the given condition on each face of the hexahedron. Namely, on the face $A B D$ it is the point that divides the median from $D$ in the ratio $32: 3$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1977", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (CZS 5) Let $A B C D$ be a regular tetrahedron and $\\mathcal{Z}$ an isometry mapping $A, B, C, D$ into $B, C, D, A$, respectively. Find the set $\\mathcal{M}$ of all points $X$ of the face $A B C$ whose distance from $\\mathcal{Z}(X)$ is equal to a given number $t$. Find necessary and sufficient conditions for the set $\\mathcal{M}$ to be nonempty.", "solution": "9. A necessary and sufficient condition for $\\mathcal{M}$ to be nonempty is that $1 / \\sqrt{10} \\leq t \\leq 1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (BUL 1) The set $M=\\{1,2, \\ldots, 2 n\\}$ is partitioned into $k$ nonintersecting subsets $M_{1}, M_{2}, \\ldots, M_{k}$, where $n \\geq k^{3}+k$. Prove that there exist even numbers $2 j_{1}, 2 j_{2}, \\ldots, 2 j_{k+1}$ in $M$ that are in one and the same subset $M_{i}$ $(1 \\leq i \\leq k)$ such that the numbers $2 j_{1}-1,2 j_{2}-1, \\ldots, 2 j_{k+1}-1$ are also in one and the same subset $M_{j}(1 \\leq j \\leq k)$.", "solution": "1. There exists an $M_{s}$ that contains at least $2 n / k=2\\left(k^{2}+1\\right)$ elements. It follows that $M_{s}$ contains either at least $k^{2}+1$ even numbers or at least $k^{2}+1$ odd numbers. In the former case, consider the predecessors of those $k^{2}+1$ numbers: among them, at least $\\frac{k^{2}+1}{k+1}>k$, i.e., at least $k+1$, belong to the same subset, say $M_{t}$. Then we choose $s, t$. The latter case is similar. Second solution. For all $i, j \\in\\{1,2, \\ldots, k\\}$, consider the set $N_{i j}=\\{r \\mid$ $\\left.2 r \\in M_{i}, 2 r-1 \\in M_{j}\\right\\}$. Then $\\left\\{N_{i j} \\mid i, j\\right\\}$ is a partition of $\\{1,2, \\ldots, n\\}$ into $k^{2}$ subsets. For $n \\geq k^{3}+1$ one of these subsets contains at least $k+1$ elements, and the statement follows. Remark. The statement is not necessarily true when $n=k^{3}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (NET 1) ${ }^{\\text {IMO6 }}$ An international society has its members in 6 different countries. The list of members contains 1978 names, numbered $1,2, \\ldots$, 1978. Prove that there is at least one member whose number is the sum of the numbers of two, not necessarily distinct, of his compatriots.", "solution": "10. Assume the opposite. One of the countries, say $A$, contains at least 330 members $a_{1}, a_{2}, \\ldots, a_{330}$ of the society $(6 \\cdot 329=1974)$. Consider the differences $a_{330}-a_{i},=1,2, \\ldots, 329$ : the members with these numbers are not in $A$, so at least 66 of them, $a_{330}-a_{i_{1}}, \\ldots, a_{330}-a_{i_{66}}$, belong to the same country, say $B$. Then the differences $\\left(a_{i_{66}}-a_{330}\\right)-\\left(a_{i_{j}}-a_{330}\\right)=$ $a_{i_{66}}-a_{i_{j}}, j=1,2, \\ldots, 65$, are neither in $A$ nor in $B$. Continuing this procedure, we find that 17 of these differences are in the same country, say $C$, then 6 among 16 differences of themselves in a country $D$, and 3 among 5 differences of themselves in $E$; finally, one among two differences of these 3 differences belong to country $F$, so that the difference of themselves cannot be in any country. This is a contradiction. Remark. The following stronger $([6!e]=1957)$ statement can be proved in the same way. Schurr's lemma. If $n$ is a natural number and $e$ the logarithm base, then for every partition of the set $\\{1,2, \\ldots,[e n!]\\}$ into $n$ subsets one of these subsets contains some two elements and their difference.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (SWE 2) A function $f: I \\rightarrow \\mathbb{R}$, defined on an interval $I$, is called concave if $f(\\theta x+(1-\\theta) y) \\geq \\theta f(x)+(1-\\theta) f(y)$ for all $x, y \\in I$ and $0 \\leq \\theta \\leq 1$. Assume that the functions $f_{1}, \\ldots, f_{n}$, having all nonnegative values, are concave. Prove that the function $\\left(f_{1} f_{2} \\ldots f_{n}\\right)^{1 / n}$ is concave.", "solution": "11. Set $F(x)=f_{1}(x) f_{2}(x) \\cdots f_{n}(x)$ : we must prove concavity of $F^{1 / n}$. By the assumption,  $$ \\begin{aligned} F(\\theta x+(1-\\theta) y) & \\geq \\prod_{i=1}^{n}\\left[\\theta f_{i}(x)+(1-\\theta) f(y)\\right] \\\\ & =\\sum_{k=0}^{n} \\theta^{k}(1-\\theta)^{n-k} \\sum f_{i_{1}}(x) \\ldots f_{i_{k}}(x) f_{i_{k+1}}(y) f_{i_{n}}(y) \\end{aligned} $$  where the second sum goes through all $\\binom{n}{k} k$-subsets $\\left\\{i_{1}, \\ldots, i_{k}\\right\\}$ of $\\{1, \\ldots, n\\}$. The inequality between the arithmetic and geometric means now gives us  $$ \\sum f_{i_{1}}(x) f_{i_{2}}(x) \\cdots f_{i_{k}}(x) f_{i_{k+1}}(y) f_{i_{n}}(y) \\geq\\binom{ n}{k} F(x)^{k / n} F(y)^{(n-k) / n} $$  Inserting this in the above inequality and using the binomial formula, we finally obtain  $$ \\begin{aligned} F(\\theta x+(1-\\theta) y) & \\geq \\sum_{k=0}^{n} \\theta^{k}(1-\\theta)^{n-k}\\binom{n}{k} F(x)^{k / n} F(y)^{(n-k) / n} \\\\ & =\\left(\\theta F(x)^{1 / n}+(1-\\theta) F(y)^{1 / n}\\right)^{n}, \\end{aligned} $$  which proves the assertion.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (USA 1) ${ }^{\\mathrm{IMO} 4}$ In a triangle $A B C$ we have $A B=A C$. A circle is tangent internally to the circumcircle of $A B C$ and also to the sides $A B, A C$, at $P, Q$ respectively. Prove that the midpoint of $P Q$ is the center of the incircle of $A B C$.", "solution": "12. Let $O$ be the center of the smaller circle, $T$ its contact point with the circumcircle of $A B C$, and $J$ the midpoint of segment $B C$. The figure is symmetric with respect to the line through $A, O, J, T$. A homothety centered at $A$ taking $T$ into $J$ will take the smaller circle into the incircle of $A B C$, hence will take $O$ into the incenter $I$. On the other hand, $\\angle A B T=\\angle A C T=90^{\\circ}$ implies that the quadrilaterals $A B T C$ and $A P O Q$ are similar. Hence the above homothety also maps $O$ to the midpoint of $P Q$. This finishes the proof. Remark. The assertion is true for a nonisosceles triangle $A B C$ as well, and this (more difficult) case is a matter of SL93-3.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (USA 6) ${ }^{\\mathrm{IMO} 2}$ Given any point $P$ in the interior of a sphere with radius $R$, three mutually perpendicular segments $P A, P B, P C$ are drawn terminating on the sphere and having one common vertex in $P$. Consider the rectangular parallelepiped of which $P A, P B, P C$ are coterminal edges. Find the locus of the point $Q$ that is diagonally opposite $P$ in the parallelepiped when $P$ and the sphere are fixed.", "solution": "13. Lemma. If $M N P Q$ is a rectangle and $O$ any point in space, then $O M^{2}+$ $O P^{2}=O N^{2}+O Q^{2}$. Proof. Let $O_{1}$ be the projection of $O$ onto $M N P Q$, and $m, n, p, q$ denote the distances of $O_{1}$ from $M N, N P, P Q, Q M$, respectively. Then $O M^{2}=O O_{1}^{2}+q^{2}+m^{2}, O N^{2}=O O_{1}^{2}+m^{2}+n^{2}, O P^{2}=O O_{1}^{2}+n^{2}+p^{2}$, $O Q^{2}=O O_{1}^{2}+p^{2}+q^{2}$, and the lemma follows immediately. Now we return to the problem. Let $O$ be the center of the given sphere $S$, and $X$ the point opposite $P$ in the face of the parallelepiped through $P, A, B$. By the lemma, we have $O P^{2}+O Q^{2}=O C^{2}+O X^{2}$ and $O P^{2}+$ $O X^{2}=O A^{2}+O B^{2}$. Hence $2 O P^{2}+O Q^{2}=O A^{2}+O B^{2}+O C^{2}=3 R^{2}$, i.e. $O Q=\\sqrt{3 R^{2}-O P^{2}}>R$.  We claim that the locus of $Q$ is the whole sphere $\\left(O, \\sqrt{3 R^{2}-O P^{2}}\\right)$. Choose any point $Q$ on this sphere. Since $O Q>R>O P$, the sphere with diameter $P Q$ intersects $S$ on a circle. Let $C$ be an arbitrary point on this circle, and $X$ the point opposite $C$ in the rectangle $P C Q X$. By the lemma, $O P^{2}+O Q^{2}=O C^{2}+O X^{2}$, hence $O X^{2}=2 R^{2}-O P^{2}>R^{2}$. The plane passing through $P$ and perpendicular to $P C$ intersects $S$ in a circle $\\gamma$; both $P, X$ belong to this plane, $P$ being inside and $X$ outside the circle, so that the circle with diameter $P X$ intersects $\\gamma$ at some point $B$. Finally, we choose $A$ to be the point opposite $B$ in the rectangle $P B X A$ : we deduce that $O A^{2}+O B^{2}=O P^{2}+O X^{2}$, and consequently $A \\in S$. By the construction, there is a rectangular parallelepiped through $P, A, B, C, X, Q$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (VIE 2) Prove that it is possible to place $2 n(2 n+1)$ parallelepipedic (rectangular) pieces of soap of dimensions $1 \\times 2 \\times(n+1)$ in a cubic box with edge $2 n+1$ if and only if $n$ is even or $n=1$. Remark. It is assumed that the edges of the pieces of soap are parallel to the edges of the box.", "solution": "14. We label the cells of the cube by $\\left(a_{1}, a_{2}, a_{3}\\right), a_{i} \\in\\{1,2, \\ldots, 2 n+1\\}$, in a natural way: for example, as Cartesian coordinates of centers of the cells $\\left((1,1,1)\\right.$ is one corner, etc.). Notice that there should be $(2 n+1)^{3}-$ $2 n(2 n+1) \\cdot 2(n+1)=2 n+1$ void cells, i.e., those not covered by any piece of soap. $n=1$. In this case, six pieces of soap $1 \\times 2 \\times 2$ can be placed on the following positions: $[(1,1,1),(2,2,1)],[(3,1,1),(3,2,2)],[(2,3,1),(3,3,2)]$ and the symmetric ones with respect to the center of the box. (Here $[A, B]$ denotes the rectangle with opposite corners at $A, B$.) $n$ is even. Each of the $2 n+1$ planes $P_{k}=\\left\\{\\left(a_{1}, a_{2}, k\\right) \\mid a_{i}=1, \\ldots, 2 n+1\\right\\}$ can receive $2 n$ pieces of soap: In fact, $P_{k}$ can be partitioned into four $n \\times(n+1)$ rectangles at the corners and the central cell, while an $n \\times(n+1)$ rectangle can receive $n / 2$ pieces of soap. $n$ is odd, $n>1$. Let us color a cell $\\left(a_{1}, a_{2}, a_{3}\\right)$ blue, red, or yellow if exactly three, two or one $a_{i}$ respectively is equal to $n+1$. Thus there are 1 blue, $6 n$ red, and $12 n^{2}$ yellow cells. We notice that each piece of soap must contain at least one colored cell (because $2(n+1)>2 n+1)$. Also, every piece of soap contains an even number (actually, $1 \\cdot 2,1(n+1)$, or $2(n+1)$ ) of cells in $P_{k}$. On the other hand, $2 n+1$ cells are void, i.e., one in each plane.  There are several cases for a piece of soap $S$ : (i) $S$ consists of 1 blue, $n+1$ red and $n$ yellow cells; (ii) $S$ consists of 2 red and $2 n$ yellow cells (and no blue cells); (iii) $S$ contains 1 red cell, $n+1$ yellow cells, and the are rest uncolored; (iv) $S$ contains 2 yellow cells and no blue or red ones.  From the descriptions of the last three cases, we can deduce that if $S$ contains $r$ red cells and no blue, then it contains exactly $2+(n-1) r$ red ones. $\\quad(*)$ Now, let $B_{1}, \\ldots, B_{k}$ be all boxes put in the cube, with a possible exception for the one covering the blue cell: thus $k=2 n(2 n+1)$ if the blue cell is void, or $k=2 n(2 n+1)-1$ otherwise. Let $r_{i}$ and $y_{i}$ respectively be the numbers of red and yellow cells inside $B_{i}$. By (*) we have $y_{1}+\\cdots+y_{k}=2 k+(n-1)\\left(r_{1}+\\cdots+r_{k}\\right)$. If the blue cell is void, then $r_{1}+\\cdots+r_{k}=6 n$ and consequently $y_{1}+\\cdots+y_{k}=$ $4 n(2 n+1)+6 n(n-1)=14 n^{2}-2 n$, which is impossible because there are only $12 n^{2}<14 n^{2}-2 n$ yellow cells. Otherwise, $r_{1}+\\cdots+r_{k} \\geq 5 n-2$ (because $n+1$ red cells are covered by the box containing the blue cell, and one can be void) and consequently $y_{1}+\\cdots+y_{k} \\geq 4 n(2 n+$ $1)-2+(n-1)(5 n-2)=13 n^{2}-3 n$; since there are $n$ more yellow cells in the box containing the blue one, this counts for $13 n^{2}-2 n>12 n^{2}$ ( $n \\geq 3$ ), again impossible. Remark. The following solution of the case $n$ odd is simpler, but does not work for $n=3$. For $k=1,2,3$, let $m_{k}$ be the number of pieces whose long sides are perpendicular to the plane $\\pi_{k}\\left(a_{k}=n+1\\right)$. Each of these $m_{k}$ pieces covers exactly 2 cells of $\\pi_{k}$, while any other piece covers $n+1$, $2(n+1)$, or none. It follows that $4 n^{2}+4 n-2 m_{k}$ is divisible by $n+1$, and so is $2 m_{k}$. This further implies that $2 m_{1}+2 m_{2}+2 m_{3}=4 n(2 n+1)$ is a multiple of $n+1$, which is impossible for each odd $n$ except $n=1$ and $n=3$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (YUG 1) Let $p$ be a prime and $A=\\left\\{a_{1}, \\ldots, a_{p-1}\\right\\}$ an arbitrary subset of the set of natural numbers such that none of its elements is divisible by $p$. Let us define a mapping $f$ from $\\mathcal{P}(A)$ (the set of all subsets of $A$ ) to the set $P=\\{0,1, \\ldots, p-1\\}$ in the following way: (i) if $B=\\left\\{a_{i_{1}}, \\ldots, a_{i_{k}}\\right\\} \\subset A$ and $\\sum_{j=1}^{k} a_{i_{j}} \\equiv n(\\bmod p)$, then $f(B)=n$, (ii) $f(\\emptyset)=0, \\emptyset$ being the empty set.  Prove that for each $n \\in P$ there exists $B \\subset A$ such that $f(B)=n$.", "solution": "15. Let $C_{n}=\\left\\{a_{1}, \\ldots, a_{n}\\right\\}\\left(C_{0}=\\emptyset\\right)$ and $P_{n}=\\left\\{f(B) \\mid B \\subseteq C_{n}\\right\\}$. We claim that $P_{n}$ contains at least $n+1$ distinct elements. First note that $P_{0}=\\{0\\}$ contains one element. Suppose that $P_{n+1}=P_{n}$ for some $n$. Since $P_{n+1}=$ $\\left\\{a_{n+1}+r \\mid r \\in P_{n}\\right\\}$, it follows that for each $r \\in P_{n}$, also $r+b_{n} \\in P_{n}$. Then obviously $0 \\in P_{n}$ implies $k b_{n} \\in P_{n}$ for all $k$; therefore $P_{n}=P$ has at least $p \\geq n+1$ elements. Otherwise, if $P_{n+1} \\supset P_{n}$ for all $n$, then $\\left|P_{n+1}\\right| \\geq\\left|P_{n}\\right|+1$ and hence $\\left|P_{n}\\right| \\geq n+1$, as claimed. Consequently, $\\left|P_{p-1}\\right| \\geq p$. (All the operations here are performed modulo $p$.)", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (YUG 2) Determine all the triples $(a, b, c)$ of positive real numbers such that the system  $$ \\begin{array}{r} a x+b y-c z=0 \\\\ a \\sqrt{1-x^{2}}+b \\sqrt{1-y^{2}}-c \\sqrt{1-z^{2}}=0 \\end{array} $$  is compatible in the set of real numbers, and then find all its real solutions.", "solution": "16. Clearly $|x| \\leq 1$. As $x$ runs over $[-1,1]$, the vector $u=\\left(a x, a \\sqrt{1-x^{2}}\\right)$ runs over all vectors of length $a$ in the plane having a nonnegative vertical component. Putting $v=\\left(b y, b \\sqrt{1-y^{2}}\\right), w=\\left(c z, c \\sqrt{1-z^{2}}\\right)$, the system becomes $u+v=w$, with vectors $u, v, w$ of lengths $a, b, c$ respectively in the upper half-plane. Then $a, b, c$ are sides of a (possibly degenerate) triangle; i.e, $|a-b| \\leq c \\leq a+b$ is a necessary condition.  Conversely, if $a, b, c$ satisfy this condition, one constructs a triangle $O M N$ with $O M=a, O N=b, M N=c$. If the vectors $\\overrightarrow{O M}, \\overrightarrow{O N}$ have a positive nonnegative component, then so does their sum. For every such triangle, putting $u=\\overrightarrow{O M}, v=\\overrightarrow{O N}$, and $w=\\overrightarrow{O M}+\\overrightarrow{O N}$ gives a solution, and every solution is given by one such triangle. This triangle is uniquely determined up to congruence: $\\alpha=\\angle M O N=\\angle(u, v)$ and $\\beta=\\angle(u, w)$. Therefore, all solutions of the system are  $$ \\begin{aligned} & x=\\cos t, \\quad y=\\cos (t+\\alpha), \\quad z=y=\\cos (t+\\beta), \\quad t \\in[0, \\pi-\\alpha] \\quad \\text { or } \\\\ & x=\\cos t, \\quad y=\\cos (t-\\alpha), \\quad z=y=\\cos (t-\\beta), \\quad t \\in[\\alpha, \\pi] . \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (FRA 3) Prove that for any positive integers $x, y, z$ with $x y-z^{2}=1$ one can find nonnegative integers $a, b, c, d$ such that $x=a^{2}+b^{2}, y=c^{2}+d^{2}$, $z=a c+b d$. Set $z=(2 q)$ ! to deduce that for any prime number $p=4 q+1, p$ can be represented as the sum of squares of two integers.", "solution": "17. Let $z_{0} \\geq 1$ be a positive integer. Supposing that the statement is true for all triples $(x, y, z)$ with $z<z_{0}$, we shall prove that it is true for $z=z_{0}$ too.  If $z_{0}=1$, verification is trivial, while $x_{0}=y_{0}$ is obviously impossible. So let there be given a triple $\\left(x_{0}, y_{0}, z_{0}\\right)$ with $z_{0}>1$ and $x_{0}<y_{0}$, and define another triple $(x, y, z)$ by  $$ x=z_{0}, \\quad y=x_{0}+y_{0}-2 z_{0}, \\quad \\text { and } \\quad z=z_{0}-x_{0} $$  Then $x, y, z$ are positive integers. This is clear for $x, z$, while $y=x_{0}+y_{0}-$ $2 z_{0} \\geq 2\\left(\\sqrt{x_{0} y_{0}}-z_{0}\\right)>2\\left(z_{0}-z_{0}\\right)=0$. Moreover, $x y-z^{2}=x_{0}\\left(x_{0}+y_{0}-\\right.$ $\\left.2 z_{0}\\right)-\\left(z_{0}-x_{0}\\right)^{2}=x_{0} y_{0}-z_{0}^{2}=1$ and $z<z_{0}$, so that by the assumption, the statement holds for $x, y, z$. Thus for some nonnegative integers $a, b, c, d$ we have  $$ x=a^{2}+b^{2}, \\quad y=c^{2}+d^{2}, \\quad z=a c+b d $$  But then we obtain representations of this sort for $x_{0}, y_{0}, z_{0}$ too:  $$ x_{0}=a^{2}+b^{2}, \\quad y_{0}=(a+c)^{2}+(b+d)^{2}, \\quad z_{0}=a(a+c)+b(b+d) $$  For the second part of the problem, we note that for $z=(2 q)!$,  $$ \\begin{aligned} z^{2} & =(2 q)!(2 q)(2 q-1) \\cdots 1 \\equiv(2 q)!\\cdot(-(2 q+1))(-(2 q+2)) \\cdots(-4 q) \\\\ & =(-1)^{2 q}(4 q)!\\equiv-1(\\bmod p) \\end{aligned} $$  by Wilson's theorem. Hence $p \\mid z^{2}+1=p y$ for some positive integer $y>0$. Now it follows from the first part that there exist integers $a, b$ such that $x=p=a^{2}+b^{2}$. Second solution. Another possibility is using arithmetic of Gaussian integers. Lemma. Suppose $m, n, p, q$ are elements of $\\mathbb{Z}$ or any other unique factorization domain, with $m n=p q$. then there exist elements $a, b, c, d$ such that $m=a b, n=c d, p=a c, q=b d$. Proof is direct, for example using factorization of $a, b, c, d$ into primes. We now apply this lemma to the Gaussian integers in our case (because $\\mathbb{Z}[i]$ has the unique factorization property), having in mind that $x y=$ $z^{2}+1=(z+i)(z-i)$. We obtain (1) $x=a b$, (2) $y=c d$, (3) $z+i=a c$, (4) $z-i=b d$ for some $a, b, c, d \\in \\mathbb{Z}[i]$. Let $a=a_{1}+a_{2} i$, etc. By (3) and (4), $\\operatorname{gcd}\\left(a_{1}, a_{2}\\right)=$ $\\cdots=\\operatorname{gcd}\\left(d_{1}, d_{2}\\right)$. Then (1) and (2) give us $b=\\bar{a}, c=\\bar{d}$. The statement follows at once: $x=a b=a \\bar{a}=a_{1}^{2}+a_{2}^{2}, y=d \\bar{d}=d_{1}^{2}+d_{2}^{2}$ and $z+i=$ $\\left(a_{1} d_{1}+a_{2} d_{2}\\right)+\\imath\\left(a_{2} d_{1}-a_{1} d_{2}\\right) \\Rightarrow z=a_{1} d_{1}+a_{2} d_{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (BUL 4) Two identically oriented equilateral triangles, $A B C$ with center $S$ and $A^{\\prime} B^{\\prime} C$, are given in the plane. We also have $A^{\\prime} \\neq S$ and $B^{\\prime} \\neq S$. If $M$ is the midpoint of $A^{\\prime} B$ and $N$ the midpoint of $A B^{\\prime}$, prove that the triangles $S B^{\\prime} M$ and $S A^{\\prime} N$ are similar.", "solution": "2. Consider the transformation $\\phi$ of the plane defined as the homothety $\\mathcal{H}$ with center $B$ and coefficient 2 followed by the rotation $\\mathcal{R}$ about the center $O$ through an angle of $60^{\\circ}$. Being direct, this mapping must be a rotational homothety. We also see that $\\mathcal{H}$ maps $S$ into the point symmetric to $S$ with respect to $O A$, and $\\mathcal{R}$ takes it back to $S$. Hence $S$ is a fixed point, and is consequently also the center of $\\phi$. Therefore $\\phi$ is the rotational homothety about $S$ with the angle $60^{\\circ}$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-435.jpg?height=319&width=462&top_left_y=920&top_left_x=858) and coefficient 2. (In fact, this could also be seen from the fact that $\\phi$ preserves angles of triangles and maps the segment $S R$ onto $S B$, where $R$ is the midpoint of $A B$.) Since $\\phi(M)=B^{\\prime}$, we conclude that $\\angle M S B^{\\prime}=60^{\\circ}$ and $S B^{\\prime} / S M=2$. Similarly, $\\angle N S A^{\\prime}=60^{\\circ}$ and $S A^{\\prime} / S N=2$, so triangles $M S B^{\\prime}$ and $N S A^{\\prime}$ are indeed similar. Second solution. Probably the simplest way here is using complex numbers. Put the origin at $O$ and complex numbers $a, a^{\\prime}$ at points $A, A^{\\prime}$, and denote the primitive sixth root of 1 by $\\omega$. Then the numbers at $B, B^{\\prime}$, $S$ and $N$ are $\\omega a, \\omega a^{\\prime},(a+\\omega a) / 3$, and $\\left(a+\\omega a^{\\prime}\\right) / 2$ respectively. Now it is easy to verify that $(n-s)=\\omega\\left(a^{\\prime}-s\\right) / 2$, i.e., that $\\angle N S A^{\\prime}=60^{\\circ}$ and $S A^{\\prime} / S N=2$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (CUB 3) ${ }^{\\mathrm{IMO} 1}$ Let $n>m \\geq 1$ be natural numbers such that the groups of the last three digits in the decimal representation of $1978^{m}, 1978^{n}$ coincide. Find the ordered pair $(m, n)$ of such $m, n$ for which $m+n$ is minimal.", "solution": "3. What we need are $m, n$ for which $1978^{m}\\left(1978^{n-m}-1\\right)$ is divisible by $1000=8 \\cdot 125$. Since $1978^{n-m}-1$ is odd, it follows that $1978^{m}$ is divisible by 8 , so $m \\geq 3$. Also, $1978^{n-m}-1$ is divisible by 125 , i.e., $1978^{n-m} \\equiv 1(\\bmod 125)$. Note that $1978 \\equiv-2(\\bmod 5)$, and consequently also $-2^{n-m} \\equiv 1$. Hence $4 \\mid n-m=4 k, k \\geq 1$. It remains to find the least $k$ such that $1978^{4 k} \\equiv 1$ $(\\bmod 125)$. Since $1978^{4} \\equiv(-22)^{4}=484^{2} \\equiv(-16)^{2}=256 \\equiv 6$, we reduce it to $6^{k} \\equiv 1$. Now $6^{k}=(1+5)^{k} \\equiv 1+5 k+25\\binom{k}{2}(\\bmod 125)$, which reduces to $125 \\mid 5 k(5 k-3)$. But $5 k-3$ is not divisible by 5 , and so $25 \\mid k$. Therefore $100 \\mid n-m$, and the desired values are $m=3, n=103$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (CZS 2) Let $T_{1}$ be a triangle having $a, b, c$ as lengths of its sides and let $T_{2}$ be another triangle having $u, v, w$ as lengths of its sides. If $P, Q$ are the areas of the two triangles, prove that  $$ 16 P Q \\leq a^{2}\\left(-u^{2}+v^{2}+w^{2}\\right)+b^{2}\\left(u^{2}-v^{2}+w^{2}\\right)+c^{2}\\left(u^{2}+v^{2}-w^{2}\\right) $$  When does equality hold?", "solution": "4. Let $\\gamma, \\varphi$ be the angles of $T_{1}$ and $T_{2}$ opposite to $c$ and $w$ respectively. By the cosine theorem, the inequality is transformed into  $$ \\begin{aligned} & a^{2}\\left(2 v^{2}-2 u v \\cos \\varphi\\right)+b^{2}\\left(2 u^{2}-2 u v \\cos \\varphi\\right) \\\\ & \\quad+2\\left(a^{2}+b^{2}-2 a b \\cos \\gamma\\right) u v \\cos \\varphi \\geq 4 a b u v \\sin \\gamma \\sin \\varphi \\end{aligned} $$  This is equivalent to $2\\left(a^{2} v^{2}+b^{2} u^{2}\\right)-4 a b u v(\\cos \\gamma \\cos \\varphi+\\sin \\gamma \\sin \\varphi) \\geq 0$, i.e., to  $$ 2(a v-b u)^{2}+4 a b u v(1-\\cos (\\gamma-\\varphi)) \\geq 0 $$  which is clearly satisfied. Equality holds if and only if $\\gamma=\\varphi$ and $a / b=$ $u / v$, i.e., when the triangles are similar, $a$ corresponding to $u$ and $b$ to $v$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (GDR 2) For every integer $d \\geq 1$, let $M_{d}$ be the set of all positive integers that cannot be written as a sum of an arithmetic progression with difference $d$, having at least two terms and consisting of positive integers. Let $A=M_{1}, B=M_{2} \\backslash\\{2\\}, C=M_{3}$. Prove that every $c \\in C$ may be written in a unique way as $c=a b$ with $a \\in A, b \\in B$.", "solution": "5. We first explicitly describe the elements of the sets $M_{1}, M_{2}$. $x \\notin M_{1}$ is equivalent to $x=a+(a+1)+\\cdots+(a+n-1)=n(2 a+n-1) / 2$ for some natural numbers $n, a, n \\geq 2$. Among $n$ and $2 a+n-1$, one is odd and the other even, and both are greater than 1 ; so $x$ has an odd factor $\\geq 3$. On the other hand, for every $x$ with an odd divisor $p>3$ it is easy to see that there exist corresponding $a, n$. Therefore $M_{1}=\\left\\{2^{k} \\mid k=0,1,2, \\ldots\\right\\}$. $x \\notin M_{2}$ is equivalent to $x=a+(a+2)+\\cdots+(a+2(n-1))=n(a+n-1)$, where $n \\geq 2$, i.e. to $x$ being composite. Therefore $M_{2}=\\{1\\} \\cup\\{p \\mid$ $p=$ prime $\\}$. $x \\notin M_{3}$ is equivalent to $x=a+(a+3)+\\cdots+(a+3(n-1))=$ $n(2 a+3(n-1)) / 2$. It remains to show that every $c \\in M_{3}$ can be written as $c=2^{k} p$ with $p$ prime. Suppose the opposite, that $c=2^{k} p q$, where $p, q$ are odd and $q \\geq p \\geq 3$. Then there exist positive integers $a, n(n \\geq 2)$ such that $c=n(2 a+3(n-1)) / 2$ and hence $c \\notin M_{3}$. Indeed, if $k=0$, then $n=2$ and $2 a+3=p q$ work; otherwise, setting $n=p$ one obtains $a=2^{k} q-$ $3(p-1) / 2 \\geq 2 q-3(p-1) / 2 \\geq(p+3) / 2>1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (FRA 2) $)^{\\mathrm{IMO} 5}$ Let $\\varphi:\\{1,2,3, \\ldots\\} \\rightarrow\\{1,2,3, \\ldots\\}$ be injective. Prove that for all $n$,  $$ \\sum_{k=1}^{n} \\frac{\\varphi(k)}{k^{2}} \\geq \\sum_{k=1}^{n} \\frac{1}{k} $$", "solution": "6. For fixed $n$ and the set $\\{\\varphi(1), \\ldots, \\varphi(n)\\}$, there are finitely many possibilities for mapping $\\varphi$ to $\\{1, \\ldots, n\\}$. Suppose $\\varphi$ is the one among these for which $\\sum_{k=1}^{n} \\varphi(k) / k^{2}$ is minimal. If $i<j$ and $\\varphi(i)<\\varphi(j)$ for some $i, j \\in\\{1, \\ldots, n\\}$, define $\\psi$ as $\\psi(i)=\\varphi(j), \\psi(j)=\\varphi(i)$, and $\\psi(k)=\\varphi(k)$ for all other $k$. Then  $$ \\begin{aligned} \\sum \\frac{\\varphi(k)}{k^{2}}-\\sum \\frac{\\psi(k)}{k^{2}} & =\\left(\\frac{\\varphi(i)}{i^{2}}+\\frac{\\varphi(j)}{j^{2}}\\right)-\\left(\\frac{\\varphi(i)}{j^{2}}+\\frac{\\varphi(j)}{i^{2}}\\right) \\\\ & =(i-j)(\\varphi(j)-\\varphi(i)) \\frac{i+j}{i^{2} j^{2}}>0 \\end{aligned} $$  which contradicts the assumption. This shows that $\\varphi(1)<\\cdots<\\varphi(n)$, and consequently $\\varphi(k) \\geq k$ for all $k$. Hence  $$ \\sum_{k=1}^{n} \\frac{\\varphi(k)}{k^{2}} \\geq \\sum_{k=1}^{n} \\frac{k}{k^{2}}=\\sum_{k=1}^{n} \\frac{1}{k} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (FRA 5) We consider three distinct half-lines $O x, O y, O z$ in a plane. Prove the existence and uniqueness of three points $A \\in O x, B \\in O y$, $C \\in O z$ such that the perimeters of the triangles $O A B, O B C, O C A$ are all equal to a given number $2 p>0$.", "solution": "7. Let $x=O A, y=O B, z=O C, \\alpha=\\angle B O C, \\beta=\\angle C O A, \\gamma=\\angle A O B$. The conditions yield the equation $x+y+\\sqrt{x^{2}+y^{2}-2 x y \\cos \\gamma}=2 p$, which transforms to $(2 p-x-y)^{2}=x^{2}+y^{2}-2 x y \\cos \\gamma$, i.e. $(p-x)(p-y)=$ $x y(1-\\cos \\gamma)$. Thus  $$ \\frac{p-x}{x} \\cdot \\frac{p-y}{y}=1-\\cos \\gamma $$  and analogously $\\frac{p-y}{y} \\cdot \\frac{p-z}{z}=1-\\cos \\alpha, \\frac{p-z}{z} \\cdot \\frac{p-x}{x}=1-\\cos \\beta$. Setting $u=\\frac{p-x}{x}, v=\\frac{p-y}{y}, w=\\frac{p-z}{z}$, the above system becomes  $$ u v=1-\\cos \\gamma, \\quad v w=1-\\cos \\alpha, \\quad w u=1-\\cos \\beta $$  This system has a unique solution in positive real numbers $u, v, w$ : $u=\\sqrt{\\frac{(1-\\cos \\beta)(1-\\cos \\gamma)}{1-\\cos \\alpha}}$, etc. Finally, the values of $x, y, z$ are uniquely determined from $u, v, w$. Remark. It is not necessary that the three lines be in the same plane. Also, there could be any odd number of lines instead of three.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (GBR 4) Let $S$ be the set of all the odd positive integers that are not multiples of 5 and that are less than $30 \\mathrm{~m}, \\mathrm{~m}$ being an arbitrary positive integer. What is the smallest integer $k$ such that in any subset of $k$ integers from $S$ there must be two different integers, one of which divides the other?", "solution": "8. Take the subset $\\left\\{a_{i}\\right\\}=\\{1,7,11,13,17,19,23,29, \\ldots, 30 m-1\\}$ of $S$ containing all the elements of $S$ that are not multiples of 3 . There are 8 m such elements. Every element in $S$ can be uniquely expressed as $3^{t} a_{i}$ for some $i$ and $t \\geq 0$. In a subset of $S$ with $8 m+1$ elements, two of them will have the same $a_{i}$, hance one will divide the other. On the other hand, for each $i=1,2, \\ldots, 8 m$ choose $t \\geq 0$ such that $10 \\mathrm{~m}<$ $b_{i}=3^{t} a_{i}<30 \\mathrm{~m}$. Then there are $8 \\mathrm{~m} b_{i}$ 's in the interval $(10 \\mathrm{~m}, 30 \\mathrm{~m})$, and the quotient of any two of them is less than 3 , so none of them can divide any other. Thus the answer is 8 m .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1978", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. $\\mathbf{( G B R} \\mathbf{5})^{\\mathrm{IMO} 3}$ Let $\\{f(n)\\}$ be a strictly increasing sequence of positive integers: $0<f(1)<f(2)<f(3)<\\cdots$. Of the positive integers not belonging to the sequence, the $n$th in order of magnitude is $f(f(n))+1$. Determine $f(240)$.", "solution": "9. Since the $n$th missing number (gap) is $f(f(n))+1$ and $f(f(n))$ is a member of the sequence, there are exactly $n-1$ gaps less than $f(f(n))$. This leads to  $$ f(f(n))=f(n)+n-1 $$  Since 1 is not a gap, we have $f(1)=1$. The first gap is $f(f(1))+1=2$. Two consecutive integers cannot both be gaps (the predecessor of a gap is of the form $f(f(m))$ ). Now we deduce $f(2)=3$; a repeated application of the formula above gives $f(3)=3+1=4, f(4)=4+2=6, f(6)=9$, $f(9)=14, f(14)=22, f(22)=35, f(35)=56, f(56)=90, f(90)=145$, $f(145)=234, f(234)=378$. Also, $f(f(35))+1=91$ is a gap, so $f(57)=92$. Then by $(1), f(92)=148$, $f(148)=239, f(239)=386$. Finally, here $f(f(148))+1=387$ is a gap, so $f(240)=388$.  Second solution. As above, we arrive at formula (1). Then by simple induction it follows that $f\\left(F_{n}+1\\right)=F_{n+1}+1$, where $F_{k}$ is the Fibonacci sequence ( $F_{1}=F_{2}=1$ ). We now prove by induction (on $n$ ) that $f\\left(F_{n}+x\\right)=F_{n+1}+f(x)$ for all $x$ with $1 \\leq x \\leq F_{n-1}$. This is trivially true for $n=0,1$. Supposing that it holds for $n-1$, we shall prove it for $n$ : (i) If $x=f(y)$ for some $y$, then by the inductive assumption and (1)  $$ \\begin{aligned} f\\left(F_{n}+x\\right) & =f\\left(F_{n}+f(y)\\right)=f\\left(f\\left(F_{n-1}+y\\right)\\right) \\\\ & =F_{n}+f(y)+F_{n-1}+y-1=F_{n+1}+f(x) \\end{aligned} $$  (ii) If $x=f(f(y))+1$ is a gap, then $f\\left(F_{n}+x-1\\right)+1=F_{n+1}+f(x-1)+1$ is a gap also:  $$ \\begin{aligned} F_{n+1}+f(x)+1 & =F_{n+1}+f(f(f(y)))+1 \\\\ & =f\\left(F_{n}+f(f(y))\\right)+1=f\\left(f\\left(F_{n-1}+f(y)\\right)\\right)+1 \\end{aligned} $$  It follows that $f\\left(F_{n}+x\\right)=F_{n+1}+f(x-1)+2=F_{n+1}+f(x)$. Now, since we know that each positive integer $x$ is expressible as $x=$ $F_{k_{1}}+F_{k_{2}}+\\cdots+F_{k_{r}}$, where $0<k_{r} \\neq 2, k_{i} \\geq k_{i+1}+2$, we obtain $f(x)=F_{k_{1}+1}+F_{k_{2}+1}+\\cdots+F_{k_{r}+1}$. Particularly, $240=233+5+2$, so $f(240)=377+8+3=388$. Remark. It can be shown that $f(x)=[\\alpha x]$, where $\\alpha=(1+\\sqrt{5}) / 2$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (BEL 1) Prove that in the Euclidean plane every regular polygon having an even number of sides can be dissected into lozenges. (A lozenge is a quadrilateral whose four sides are all of equal length).", "solution": "1. We prove more generally, by induction on $n$, that any $2 n$-gon with equal edges and opposite edges parallel to each other can be dissected. For $n=2$ the only possible such $2 n$-gon is a single lozenge, so our theorem holds in this case. We will now show that it holds for general $n$. Assume by induction that it holds for $n-1$. Let $A_{1} A_{2} \\ldots A_{2 n}$ be an arbitrary $2 n$-gon with equal edges and opposite edges parallel to each other. Then we can construct points $B_{i}$ for $i=3,4, \\ldots, n$ such that $\\overrightarrow{A_{i} B_{i}}=\\overrightarrow{A_{2} A_{1}}=\\overrightarrow{A_{n+1} A_{n+2}}$. We set $B_{2}=A_{2 n+1}=A_{1}$ and $B_{n+1}=A_{n+2}$. It follows that $A_{i} B_{i} B_{i+1} A_{i+1}$ for $i=2,3,4, \\ldots, n$ are all lozenges. It also follows that $B_{i} B_{i+1}$ for $i=2,3,4, \\ldots, n$ are equal to the edges of $A_{1} A_{2} \\ldots A_{2 n}$ and parallel to $A_{i} A_{i+1}$ and hence to $A_{n+i} A_{n+i+1}$. Thus $B_{2} \\ldots B_{n+1} A_{n+3} \\ldots A_{2 n}$ is a $2(n-1)$-gon with equal edges and opposite sides parallel and hence, by the induction hypothesis, can be dissected into lozenges. We have thus provided a dissection for $A_{1} A_{2} \\ldots A_{2 n}$. This completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (FIN 3) Show that for any vectors $a, b$ in Euclidean space,  $$ |a \\times b|^{3} \\leq \\frac{3 \\sqrt{3}}{8}|a|^{2}|b|^{2}|a-b|^{2} $$  Remark. Here $\\times$ denotes the vector product.", "solution": "10. In the cases $a=\\overrightarrow{0}, b=\\overrightarrow{0}$, and $a \\| b$ the inequality is trivial. Otherwise, let us consider a triangle $A B C$ such that $\\overrightarrow{C B}=a$ and $\\overrightarrow{C A}=b$. From this point on we shall refer to $\\alpha, \\beta, \\gamma$ as angles of $A B C$. Since $|a \\times b|=$ $|a||b| \\sin \\gamma$, our inequality reduces to $|a||b| \\sin ^{3} \\gamma \\leq 3 \\sqrt{3}|c|^{2} / 8$, which is further reduced to  $$ \\sin \\alpha \\sin \\beta \\sin \\gamma \\leq \\frac{3 \\sqrt{3}}{8} $$  using the sine law. The last inequality follows immediately from Jensen's inequality applied to the function $f(x)=\\ln \\sin x$, which is concave for $0<x<\\pi$ because $f^{\\prime}(x)=\\cot x$ is strictly decreasing.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (GDR 1) Given real numbers $x_{1}, x_{2}, \\ldots, x_{n}(n \\geq 2)$, with $x_{i} \\geq 1 / n$ $(i=1,2, \\ldots, n)$ and with $x_{1}^{2}+x_{2}^{2}+\\cdots+x_{n}^{2}=1$, find whether the product $P=x_{1} x_{2} x_{3} \\cdots x_{n}$ has a greatest and/or least value and if so, give these values.", "solution": "11. Let us define $y_{i}=x_{i}^{2}$. We thus have $y_{1}+y_{2}+\\cdots+y_{n}=1, y_{i} \\geq 1 / n^{2}$, and $P=\\sqrt{y_{1} y_{2} \\ldots y_{n}}$. The upper bound is obtained immediately from the AM-GM inequality: $P \\leq 1 / n^{n / 2}$, where equality holds when $x_{i}=\\sqrt{y_{i}}=1 / \\sqrt{n}$. For the lower bound, let us assume w.l.o.g. that $y_{1} \\geq y_{2} \\geq \\cdots \\geq y_{n}$. We note that if $a \\geq b \\geq 1 / n^{2}$ and $s=a+b>2 / n^{2}$ is fixed, then $a b=\\left(s^{2}-(a-b)^{2}\\right) / 4$ is minimized when $|a-b|$ is maximized, i.e., when $b=1 / n^{2}$. Hence $y_{1} y_{2} \\cdots y_{n}$ is minimal when $y_{2}=y_{3}=\\cdots=y_{n}=1 / n^{2}$. Then $y_{1}=\\left(n^{2}-n+1\\right) / n^{2}$ and therefore $P_{\\min }=\\sqrt{n^{2}-n+1} / n^{n}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (GDR 3) Let $R$ be a set of exactly 6 elements. A set $F$ of subsets of $R$ is called an $S$-family over $R$ if and only if it satisfies the following three conditions: (i) For no two sets $X, Y$ in $F$ is $X \\subseteq Y$; (ii) For any three sets $X, Y, Z$ in $F, X \\cup Y \\cup Z \\neq R$, (iii) $\\bigcup_{X \\in F} X=R$.  We define $|F|$ to be the number of elements of $F$ (i.e., the number of subsets of $R$ belonging to $F$ ). Determine, if it exists, $h=\\max |F|$, the maximum being taken over all S-families over $R$.", "solution": "12. The first criterion ensures that all sets in an $S$-family are distinct. Since the number of different families of subsets is finite, $h$ has to exist. In fact, we will show that $h=11$. First of all, if there exists $X \\in F$ such that $|X| \\geq 5$, then by (3) there exists $Y \\in F$ such that $X \\cup Y=R$. In this case $|F|$ is at most 2. Similarly, for $|X|=4$, for the remaining two elements either there exists a subset in $F$ that contains both, in which case we obtain the previous case, or there exist different $Y$ and $Z$ containing them, in which case $X \\cup Y \\cup Z=R$, which must not happen. Hence we can assume $|X| \\leq 4$ for all $X \\in F$. Assume $|X|=1$ for some $X$. In that case other sets must not contain that subset and hence must be contained in the remaining 5 -element subset. These elements must not be subsets of each other. From elementary combinatorics, the largest number of subsets of a 5 -element set of which none is subset of another is $\\binom{5}{2}=10$. This occurs when we take all 2-element subsets. These subsets also satisfy (2). Hence $|F|_{\\max }=11$ in this case. Otherwise, let us assume $|X|=3$ for some $X$. Let us define the following families of subsets: $G=\\{Z=Y \\backslash X \\mid Y \\in F\\}$ and $H=\\{Z=Y \\cap X \\mid Y \\in$ $F\\}$. Then no two sets in $G$ must complement each other in $R \\backslash X$, and $G$ must cover this set. Hence $G$ contains exactly the sets of each of the remaining 3 elements. For each element of $G$ no two sets in $H$ of which one is a subset of another may be paired with it. There can be only 3 such subsets selected within a 3 -element set $X$. Hence the number of remaining sets is smaller than $3 \\cdot 3=9$. Hence in this case $|F|_{\\max }=10$.  In the remaining case all subsets have two elements. There are $\\binom{6}{2}=15$ of them. But for every three that complement each other one must be discarded; hence the maximal number for $F$ in this case is $2 \\cdot 15 / 3=10$. It follows that $h=11$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (GRE 1) Show that $\\frac{20}{60}<\\sin 20^{\\circ}<\\frac{21}{60}$.", "solution": "13. From elementary trigonometry we have $\\sin 3 t=3 \\sin t-4 \\sin ^{3} t$. Hence, if we denote $y=\\sin 20^{\\circ}$, we have $\\sqrt{3} / 2=\\sin 60^{\\circ}=3 y-4 y^{3}$. Obviously $0<y<1 / 2=\\sin 30^{\\circ}$. The function $f(x)=3 x-4 x^{3}$ is strictly increasing on $[0,1 / 2)$ because $f^{\\prime}(x)=3-12 x^{2}>0$ for $0 \\leq x<1 / 2$. Now the desired inequality $\\frac{20}{60}=\\frac{1}{3}<\\sin 20^{\\circ}<\\frac{21}{60}=\\frac{7}{20}$ follows from  $$ f\\left(\\frac{1}{3}\\right)<\\frac{\\sqrt{3}}{2}<f\\left(\\frac{7}{20}\\right) $$  which is directly verified.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (GRE 5) Find all bases of logarithms in which a real positive number can be equal to its logarithm or prove that none exist.", "solution": "14. Let us assume that $a \\in \\mathbb{R} \\backslash\\{1\\}$ is such that there exist $a$ and $x$ such that $x=\\log _{a} x$, or equivalently $f(x):=\\ln x / x=\\ln a$. Then $a$ is a value of the function $f(x)$ for $x \\in \\mathbb{R}^{+} \\backslash\\{1\\}$, and the converse also holds. First we observe that $f(x)$ tends to $-\\infty$ as $x \\rightarrow 0$ and $f(x)$ tends to 0 as $x \\rightarrow 1$. Since $f(x)>0$ for $x>1$, the function $f(x)$ takes its maximum at a point $x$ for which $f^{\\prime}(x)=(1-\\ln x) / x^{2}=0$. Hence  $$ \\max f(x)=f(e)=e^{1 / e} . $$  It follows that the set of values of $f(x)$ for $x \\in \\mathbb{R}^{+}$is the interval $\\left(-\\infty, e^{1 / e}\\right)$, and consequently the desired set of bases $a$ of logarithms is $(0,1) \\cup\\left(1, e^{1 / e}\\right]$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (ISR 2) ${ }^{\\mathrm{IMO} 5}$ The nonnegative real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, a$ satisfy the following relations:  $$ \\sum_{i=1}^{5} i x_{i}=a, \\quad \\sum_{i=1}^{5} i^{3} x_{i}=a^{2}, \\quad \\sum_{i=1}^{5} i^{5} x_{i}=a^{3} $$  What are the possible values of $a$ ?", "solution": "15. We note that $\\sum_{i=1}^{5} i\\left(a-i^{2}\\right)^{2} x_{i}=a^{2} \\sum_{i=1}^{5} i x_{i}-2 a \\sum_{i=1}^{5} i^{3} x_{i}+\\sum_{i=1}^{5} i^{5} x_{i}=a^{2} \\cdot a-2 a \\cdot a^{2}+a^{3}=0$. Since the terms in the sum on the left are all nonnegative, it follows that all the terms have to be 0 . Thus, either $x_{i}=0$ for all $i$, in which case $a=0$, or $a=j^{2}$ for some $j$ and $x_{i}=0$ for $i \\neq j$. In this case, $x_{j}=a / j=j$. Hence, the only possible values of $a$ are $\\{0,1,4,9,16,25\\}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (ISR 4) Let $K$ denote the set $\\{a, b, c, d, e\\} . F$ is a collection of 16 different subsets of $K$, and it is known that any three members of $F$ have at least one element in common. Show that all 16 members of $F$ have exactly one element in common.", "solution": "16. Obviously, no two elements of $F$ can be complements of each other. If one of the sets has one element, then the conclusion is trivial. If there exist two different 2-element sets, then they must contain a common element, which in turn must then be contained in all other sets. Thus we can assume that there exists at most one 2 -element subset of $K$ in $F$. Since there can be at most 6 subsets of more than 3 elements of a 5 -element set, it follows that at least 9 out of 10 possible 3 -element subsets of $K$ belong to $F$. Let us assume, without loss of generality, that all sets but $\\{c, d, e\\}$ belong to $F$. Then sets $\\{a, b, c\\},\\{a, d, e\\}$, and $\\{b, c, d\\}$ have no common element, which is a contradiction. Hence it follows that all sets have a common element.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (NET 1) Inside an equilateral triangle $A B C$ one constructs points $P$, $Q$ and $R$ such that  $$ \\begin{aligned} & \\angle Q A B=\\angle P B A=15^{\\circ}, \\\\ & \\angle R B C=\\angle Q C B=20^{\\circ}, \\\\ & \\angle P C A=\\angle R A C=25^{\\circ} . \\end{aligned} $$  Determine the angles of triangle $P Q R$.", "solution": "17. Let $K, L$, and $M$ be intersections of $C Q$ and $B R, A R$ and $C P$, and $A Q$ and $B P$, respectively. Let $\\angle X$ denote the angle of the hexagon $K Q M P L R$ at the vertex $X$, where $X$ is one of the six points. By an elementary calculation of angles we get $\\angle K=140^{\\circ}, \\angle L=130^{\\circ}, \\angle M=150^{\\circ}, \\angle P=100^{\\circ}, \\angle Q=95^{\\circ}, \\angle R=105^{\\circ}$. Since $\\angle K B C=\\angle K C B$, it follows that $K$ is on the symmetry line of $A B C$ through $A$. Analogous statements hold for $L$ and $M$. Let $K_{R}$ and $K_{Q}$ be points symmetric to $K$ with respect to $A R$ and $A Q$, respectively. Since $\\angle A K_{Q} Q=\\angle A K_{Q} K_{R}=70^{\\circ}$ and $\\angle A K_{R} R=\\angle A K_{R} K_{Q}=70^{\\circ}$, it follows that $K_{R}, R, Q$, and $K_{Q}$ are collinear. Hence $\\angle Q R K=$ $2 \\angle R-180^{\\circ}$ and $\\angle R Q K=2 \\angle Q-$ $180^{\\circ}$. We analogously get $\\angle P R L=$ $2 \\angle R-180^{\\circ}, \\angle R P L=2 \\angle P-$ $180^{\\circ}, \\angle Q P M=2 \\angle P-180^{\\circ}$ and $\\angle P Q M=2 \\angle Q-180^{\\circ}$. From these formulas we easily get $\\angle R P Q=$ $60^{\\circ}, \\angle R Q P=75^{\\circ}$, and $\\angle Q R P=$ $45^{\\circ}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (POL 1) Let $m$ positive integers $a_{1}, \\ldots, a_{m}$ be given. Prove that there exist fewer than $2^{m}$ positive integers $b_{1}, \\ldots, b_{n}$ such that all sums of distinct $b_{k}$ 's are distinct and all $a_{i}(i \\leq m)$ occur among them.", "solution": "18. Let us write all $a_{i}$ in binary representation. For $S \\subseteq\\{1,2, \\ldots, m\\}$ let us define $b(S)$ as the number in whose binary representation ones appear in exactly the slots where ones appear in all $a_{i}$ where $i \\subseteq S$ and don't appear in any other $a_{i}$. Some $b(S)$, including $b(\\emptyset)$, will equal 0 , and hence there are fewer than $2^{m}$ different positive $b(S)$. We note that no two positive $b\\left(S_{1}\\right)$ and $b\\left(S_{2}\\right)\\left(S_{1} \\neq S_{2}\\right)$ have ones in the same decimal places. Hence sums of distinct $b(S)$ 's are distinct. Moreover  $$ a_{i}=\\sum_{i \\in S} b(S) $$  and hence the positive $b(S)$ are indeed the numbers $b_{1}, \\ldots, b_{n}$ whose existence we had to prove.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (ROM 1) Consider the sequences $\\left(a_{n}\\right),\\left(b_{n}\\right)$ defined by  $$ a_{1}=3, \\quad b_{1}=100, \\quad a_{n+1}=3^{a_{n}}, \\quad b_{n+1}=100^{b_{n}} . $$  Find the smallest integer $m$ for which $b_{m}>a_{100}$.", "solution": "19. Let us define $i_{j}$ for two positive integers $i$ and $j$ in the following way: $i_{1}=i$ and $i_{j+1}=i^{i_{j}}$ for all positive integers $j$. Thus we must find the smallest $m$ such that $100_{m}>3_{100}$. Since $100_{1}=100>27=3_{2}$, we inductively have $100_{j}=10^{100_{j-1}}>3^{100_{j-1}}>3^{3_{j}}=3_{j+1}$ and hence $m \\leq 99$. We now prove that $m=99$ by proving $100_{98}<3_{100}$. We note that $\\left(100_{1}\\right)^{2}=10^{4}<27^{4}=3^{12}<3^{27}=3_{3}$. We also note for $d>12$ (which trivially holds for all $d=100_{i}$ ) that if $c>d^{2}$, then we have  $$ 3^{c}>3^{d^{2}}>3^{12 d}=\\left(3^{12}\\right)^{d}>10000^{d}=\\left(100^{d}\\right)^{2} $$  Hence from $3_{3}>\\left(100_{1}\\right)^{2}$ it inductively follows that $3_{j}>\\left(100_{j-2}\\right)^{2}>$ $100_{j-2}$ and hence that $100_{99}>3_{100}>100_{98}$. Hence $m=99$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (BEL 4) From a bag containing 5 pairs of socks, each pair a different color, a random sample of 4 single socks is drawn. Any complete pairs in the sample are discarded and replaced by a new pair draw from the bag. The process continues until the bag is empty or there are 4 socks of different colors held outside the bag. What is the probability of the latter alternative?", "solution": "2. The only way to arrive at the latter alternative is to draw four different socks in the first drawing or to draw only one pair in the first drawing and then draw two different socks in the last drawing. We will call these probabilities respectively $p_{1}, p_{2}, p_{3}$. We calculate them as follows:  $$ p_{1}=\\frac{\\binom{5}{4} 2^{4}}{\\binom{10}{4}}=\\frac{8}{21}, \\quad p_{2}=\\frac{5\\binom{4}{2} 2^{2}}{\\binom{10}{4}}=\\frac{4}{7}, \\quad p_{3}=\\frac{4}{\\binom{6}{2}}=\\frac{4}{15} . $$  We finally calculate the desired probability: $P=p_{1}+p_{2} p_{3}=\\frac{8}{15}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. (SWE 2) Given the integer $n>1$ and the real number $a>0$ determine the maximum of $\\sum_{i=1}^{n-1} x_{i} x_{i+1}$ taken over all nonnegative numbers $x_{i}$ with sum $a$.", "solution": "20. Let $x_{k}=\\max \\left\\{x_{1}, x_{2}, \\ldots, x_{n}\\right\\}$. Then $x_{i} x_{i+1} \\leq x_{i} x_{k}$ for $i=1,2, \\ldots, k-1$ and $x_{i} x_{i+1} \\leq x_{k} x_{i+1}$ for $i=k, \\ldots, n-1$. Summing up these inequalities for $i=1,2, \\ldots, n-1$ we obtain  $$ \\sum_{i=1}^{n-1} \\leq x_{k}\\left(x_{1}+\\cdots+x_{k-1}+x_{k+1}+\\cdots+x_{n}\\right)=x_{k}\\left(a-x_{k}\\right) \\leq \\frac{a^{2}}{4} $$  We note that the value $a^{2} / 4$ is attained for $x_{1}=x_{2}=a / 2$ and $x_{3}=\\cdots=$ $x_{n}=0$. Hence $a^{2} / 4$ is the required maximum.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. (USS 1) Let $N$ be the number of integral solutions of the equation  $$ x^{2}-y^{2}=z^{3}-t^{3} $$  satisfying the condition $0 \\leq x, y, z, t \\leq 10^{6}$, and let $M$ be the number of integral solutions of the equation  $$ x^{2}-y^{2}=z^{3}-t^{3}+1 $$  satisfying the condition $0 \\leq x, y, z, t \\leq 10^{6}$. Prove that $N>M$.", "solution": "21. Let $f(n)$ be the number of different ways $n \\in \\mathbb{N}$ can be expressed as $x^{2}+y^{3}$ where $x, y \\in\\left\\{0,1, \\ldots, 10^{6}\\right\\}$. Clearly $f(n)=0$ for $n<0$ or $n>10^{12}+10^{18}$. The first equation can be written as $x^{2}+t^{3}=y^{2}+z^{3}=n$, whereas the second equation can be written as $x^{2}+t^{3}=n+1, y^{2}+z^{3}=n$. Hence we obtain the following formulas for $M$ and $N$ :  $$ M=\\sum_{i=0}^{m} f(i)^{2}, \\quad N=\\sum_{i=0}^{m-1} f(i) f(i+1) . $$  Using the AM-GM inequality we get  $$ \\begin{aligned} N & =\\sum_{i=0}^{m-1} f(i) f(i+1) \\\\ & \\leq \\sum_{i=0}^{m-1} \\frac{f(i)^{2}+f(i+1)^{2}}{2}=\\frac{f(0)^{2}}{2}+\\sum_{i=1}^{m-1} f(i)^{2}+\\frac{f(m)^{2}}{2}<M . \\end{aligned} $$  The last inequality is strong, since $f(0)=1>0$. This completes our proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "22", "problem_type": null, "exam": "IMO", "problem": "22. (USS 3) ${ }^{\\mathrm{IMO} 3}$ There are two circles in the plane. Let a point $A$ be one of the points of intersection of these circles. Two points begin moving simultaneously with constant speeds from the point $A$, each point along its own circle. The two points return to the point $A$ at the same time. Prove that there is a point $P$ in the plane such that at every moment of time the distances from the point $P$ to the moving points are equal.", "solution": "22. Let the centers of the two circles be denoted by $O$ and $O_{1}$ and their respective radii by $r$ and $r_{1}$, and let the positions of the points on the circles at time $t$ be denoted by $M(t)$ and $N(t)$. Let $Q$ be the point such that $O A O_{1} Q$ is a parallelogram. We will show that $Q$ is the point $P$ we are looking for, i.e., that $Q M(t)=$ $Q N(t)$ for all $t$. We note that $O Q=$ $O_{1} A=r_{1}, O_{1} Q=O A=r$ and ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-449.jpg?height=353&width=469&top_left_y=1504&top_left_x=850) $\\angle Q O A=\\angle Q O_{1} A=\\phi$. Since the two points return to $A$ at the same time, it follows that $\\angle M(t) O A=\\angle N(t) O_{1} A=\\omega t$. Therefore $\\angle Q O M(t)=$ $\\angle Q O_{1} N(t)=\\phi+\\omega t$, from which it follows that $\\triangle Q O M(t) \\cong \\triangle Q O_{1} N(t)$. Hence $Q M(t)=Q N(t)$, as we claimed.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "23", "problem_type": null, "exam": "IMO", "problem": "23. (USA 4) Find all natural numbers $n$ for which $2^{8}+2^{11}+2^{n}$ is a perfect square.", "solution": "23. It is easily verified that no solutions exist for $n \\leq 8$. Let us now assume that $n>8$. We note that $2^{8}+2^{11}+2^{n}=2^{8} \\cdot\\left(9+2^{n-8}\\right)$. Hence $9+2^{n-8}$ must also be a square, say $9+2^{n-8}=x^{2}, x \\in \\mathbb{N}$, i.e., $2^{n-8}=x^{2}-9=$ $(x-3)(x+3)$. Thus $x-3$ and $x+3$ are both powers of 2 , which is possible only for $x=5$ and $n=12$. Hence, $n=12$ is the only solution.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "24", "problem_type": null, "exam": "IMO", "problem": "24. (USA 5) A circle O with center $O$ on base $B C$ of an isosceles triangle $A B C$ is tangent to the equal sides $A B, A C$. If point $P$ on $A B$ and point $Q$ on $A C$ are selected such that $P B \\times C Q=(B C / 2)^{2}$, prove that line segment $P Q$ is tangent to circle O , and prove the converse.", "solution": "24. Clearly $O$ is the midpoint of $B C$. Let $M$ and $N$ be the points of tangency of the circle with $A B$ and $A C$, respectively, and let $\\angle B A C=2 \\varphi$. Then $\\angle B O M=\\angle C O N=\\varphi$. Let us assume that $P Q$ touches the circle in $X$. If we set $\\angle P O M=$ $\\angle P O X=x$ and $\\angle Q O N=\\angle Q O X=y$, then $2 x+2 y=\\angle M O N=$ $180^{\\circ}-2 \\varphi$, i.e., $y=90^{\\circ}-\\varphi-x$. It follows that $\\angle O Q C=180^{\\circ}-\\angle Q O C-$ $\\angle O C Q=180^{\\circ}-(\\varphi+y)-\\left(90^{\\circ}-\\varphi\\right)=90^{\\circ}-y=x+\\varphi=\\angle B O P$. Hence the triangles $B O P$ and $C Q O$ are similar, and consequently $B P \\cdot C Q=$ $B O \\cdot C O=(B C / 2)^{2}$. Conversely, let $B P \\cdot C Q=(B C / 2)^{2}$ and let $Q^{\\prime}$ be the point on $(A C)$ such that $P Q^{\\prime}$ is tangent to the circle. Then $B P \\cdot C Q^{\\prime}=(B C / 2)^{2}$, which implies $Q \\equiv Q^{\\prime}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "25", "problem_type": null, "exam": "IMO", "problem": "25. (USA 6) ${ }^{\\mathrm{IMO} 4}$ Given a point $P$ in a given plane $\\pi$ and also a given point $Q$ not in $\\pi$, show how to determine a point $R$ in $\\pi$ such that $\\frac{Q P+P R}{Q R}$ is a maximum.", "solution": "25. Let us first look for such a point $R$ on a line $l$ in $\\pi$ going through $P$. Let $\\angle Q P R=2 \\theta$. Consider a point $Q^{\\prime}$ on $l$ such that $Q^{\\prime} P=Q P$. Then we have  $$ \\frac{Q P+P R}{Q R}=\\frac{R Q^{\\prime}}{Q R}=\\frac{\\sin Q^{\\prime} Q R}{\\sin Q Q^{\\prime} R} $$  Since $Q Q^{\\prime} P$ is fixed, the maximal value of the expression occurs when $\\angle Q Q^{\\prime} R=90^{\\circ}$. In this case $(Q P+P R) / Q R=1 / \\sin \\theta$. Looking at all possible lines $l$, we see that $\\theta$ is minimized when $l$ equals the projection of $P Q$ onto $\\pi$. Hence, the point $R$ is the intersection of the projection of $P Q$ onto $\\pi$ and the plane through $Q$ perpendicular to $P Q$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "26", "problem_type": null, "exam": "IMO", "problem": "26. (YUG 4) Prove that the functional equations  $$ \\begin{aligned} f(x+y) & =f(x)+f(y) \\\\ \\text { and } \\quad f(x+y+x y) & =f(x)+f(y)+f(x y) \\quad(x, y \\in \\mathbb{R}) \\end{aligned} $$  are equivalent.", "solution": "26. Let us assume that $f(x+y)=f(x)+f(y)$ for all reals. In this case we trivially apply the equation to get $f(x+y+x y)=f(x+y)+f(x y)=$ $f(x)+f(y)+f(x y)$. Hence the equivalence is proved in the first direction. Now let us assume that $f(x+y+x y)=f(x)+f(y)+f(x y)$ for all reals. Plugging in $x=y=0$ we get $f(0)=0$. Plugging in $y=-1$ we get $f(x)=-f(-x)$. Plugging in $y=1$ we get $f(2 x+1)=2 f(x)+f(1)$ and hence $f(2(u+v+u v)+1)=2 f(u+v+u v)+f(1)=2 f(u v)+$ $2 f(u)+2 f(v)+f(1)$ for all real $u$ and $v$. On the other hand, plugging in $x=u$ and $y=2 v+1$ we get $f(2(u+v+u v)+1)=f(u+(2 v+$ 1) $+u(2 v+1))=f(u)+2 f(v)+f(1)+f(2 u v+u)$. Hence it follows that $2 f(u v)+2 f(u)+2 f(v)+f(1)=f(u)+2 f(v)+f(1)+f(2 u v+u)$, i.e.,  $$ f(2 u v+u)=2 f(u v)+f(u) $$  Plugging in $v=-1 / 2$ we get $0=2 f(-u / 2)+f(u)=-2 f(u / 2)+f(u)$. Hence, $f(u)=2 f(u / 2)$ and consequently $f(2 x)=2 f(x)$ for all reals. Now (1) reduces to $f(2 u v+u)=f(2 u v)+f(u)$. Plugging in $u=y$ and $x=2 u v$, we obtain $f(x)+f(y)=f(x+y)$ for all nonzero reals $x$ and $y$. Since $f(0)=0$, it trivially holds that $f(x+y)=f(x)+f(y)$ when one of $x$ and $y$ is 0 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (BUL 1) Find all polynomials $f(x)$ with real coefficients for which  $$ f(x) f\\left(2 x^{2}\\right)=f\\left(2 x^{3}+x\\right) $$", "solution": "3. An obvious solution is $f(x)=0$. We now look for nonzero solutions. We note that plugging in $x=0$ we get $f(0)^{2}=f(0)$; hence $f(0)=0$ or $f(0)=1$. If $f(0)=0$, then $f$ is of the form $f(x)=x^{k} g(x)$, where $g(0) \\neq 0$. Plugging this formula into $f(x) f\\left(2 x^{2}\\right)=f\\left(2 x^{3}+x\\right)$ we get $2^{k} x^{2 k} g(x) g\\left(2 x^{2}\\right)=\\left(2 x^{2}+1\\right)^{k} g\\left(2 x^{3}+x\\right)$. Plugging in $x=0$ gives us $g(0)=0$, which is a contradiction. Hence $f(0)=1$. For an arbitrary root $\\alpha$ of the polynomial $f, 2 \\alpha^{3}+\\alpha$ must also be a root. Let $\\alpha$ be a root of the largest modulus. If $|\\alpha|>1$ then $\\left|2 \\alpha^{3}+\\alpha\\right|>$ $2|\\alpha|^{3}-|\\alpha|>|\\alpha|$, which is impossible. It follows that $|\\alpha| \\leq 1$ and hence all roots of $f$ have modules less than or equal to 1 . But the product of all roots of $f$ is $|f(0)|=1$, which implies that all the roots have modulus 1. Consequently, for a root $\\alpha$ it holds that $|\\alpha|=\\left|2 \\alpha^{3}-\\alpha\\right|=1$. This is possible only if $\\alpha= \\pm \\imath$. Since the coefficients of $f$ are real it follows that $f$ must be of the form $f(x)=\\left(x^{2}+1\\right)^{k}$ where $k \\in \\mathbb{N}_{0}$. These polynomials satisfy the original formula. Hence, the solutions for $f$ are $f(x)=0$ and $f(x)=\\left(x^{2}+1\\right)^{k}, k \\in \\mathbb{N}_{0}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (BUL 3) ${ }^{\\mathrm{IMO} 2} \\mathrm{~A}$ pentagonal prism $A_{1} A_{2} \\ldots A_{5} B_{1} B_{2} \\ldots B_{5}$ is given. The edges, the diagonals of the lateral walls and the internal diagonals of the prism are each colored either red or green in such a way that no triangle whose vertices are vertices of the prism has its three edges of the same color. Prove that all edges of the bases are of the same color.", "solution": "4. Let us prove first that the edges $A_{1} A_{2}, A_{2} A_{3}, \\ldots, A_{5} A_{1}$ are of the same color. Assume the contrary, and let w.l.o.g. $A_{1} A_{2}$ be red and $A_{2} A_{3}$ be green. Three of the segments $A_{2} B_{l}(l=1,2,3,4,5)$, say $A_{2} B_{i}, A_{2} B_{j}, A_{2} B_{k}$, have to be of the same color, let it w.l.o.g. be red. Then $A_{1} B_{i}, A_{1} B_{j}, A_{1} B_{k}$ must be green. At least one of the sides of triangle $B_{i} B_{j} B_{k}$, say $B_{i} B_{j}$, must be an edge of the prism. Then looking at the triangles $A_{1} B_{i} B_{j}$ and $A_{2} B_{i} B_{j}$ we deduce that $B_{i} B_{j}$ can be neither green nor red, which is a contradiction. Hence all five edges of the pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$ have the same color. Similarly, all five edges of $B_{1} B_{2} B_{3} B_{4} B_{5}$ have the same color. We now show that the two colors are the same. Assume otherwise, i.e., that w.l.o.g. the $A$ edges are painted red and the $B$ edges green. Let us call segments of the form $A_{i} B_{j}$ diagonal ( $i$ and $j$ may be equal). We now count the diagonal segments by grouping the red segments based on their $A$ point, and the green segments based on their $B$ point. As above, the assumption that three of $A_{i} B_{j}$ for fixed $i$ are red leads to a contradiction. Hence at most two diagonal segments out of each $A_{i}$ may be red, which counts up to at most 10 red segments. Similarly, at most 10 diagonal segments can be green. But then we can paint at most 20 diagonal segments out of 25 , which is a contradiction. Hence all edges in the pentagons $A_{1} A_{2} A_{3} A_{4} A_{5}$ and $B_{1} B_{2} B_{3} B_{4} B_{5}$ have the same color.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (CZS 2) Let $n \\geq 2$ be an integer. Find the maximal cardinality of a set $M$ of pairs $(j, k)$ of integers, $1 \\leq j<k \\leq n$, with the following property: If $(j, k) \\in M$, then $(k, m) \\notin M$ for any $m$.", "solution": "5. Let $A=\\{x \\mid(x, y) \\in M\\}$ and $B=\\{y \\mid(x, y) \\in M$. Then $A$ and $B$ are disjoint and hence  $$ |M| \\leq|A| \\cdot|B| \\leq \\frac{(|A|+|B|)^{2}}{4} \\leq\\left[\\frac{n^{2}}{4}\\right] $$  These cardinalities can be achieved for $M=\\{(a, b) \\mid a=1,2, \\ldots,[n / 2]$, $b=[n / 2]+1, \\ldots, n\\}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (CZS 4) Find the real values of $p$ for which the equation  $$ \\sqrt{2 p+1-x^{2}}+\\sqrt{3 x+p+4}=\\sqrt{x^{2}+9 x+3 p+9} $$  in $x$ has exactly two real distinct roots $(\\sqrt{t}$ means the positive square root of $t$ ).", "solution": "6. Setting $q=x^{2}+x-p$, the given equation becomes  $$ \\sqrt{(x+1)^{2}-2 q}+\\sqrt{(x+2)^{2}-q}=\\sqrt{(2 x+3)^{2}-3 q} . $$  Taking squares of both sides we get $2 \\sqrt{\\left((x+1)^{2}-2 q\\right)\\left((x+2)^{2}-q\\right)}=$ $2(x+1)(x+2)$. Taking squares again we get  $$ q\\left(2 q-2(x+2)^{2}-(x+1)^{2}\\right)=0 $$  If $2 q=2(x+2)^{2}+(x+1)^{2}$, at least one of the expressions under the three square roots in (1) is negative, and in that case the square root is not well-defined. Thus, we must have $q=0$. Now (1) is equivalent to $|x+1|+|x+2|=|2 x+3|$, which holds if and only if $x \\notin(-2,-1)$. The number of real solutions $x$ of $q=x^{2}+x-p=0$ which are not in the interval $(-2,-1)$ is zero if $p<-1 / 4$, one if $p=-1 / 4$ or $0<p<2$, and two otherwise. Hence, the answer is $-1 / 4<p \\leq 0$ or $p \\geq 2$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (FRG 1) ${ }^{\\mathrm{IMO} 1}$ Given that $1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\cdots-\\frac{1}{1318}+\\frac{1}{1319}=\\frac{p}{q}$, where $p$ and $q$ are natural numbers having no common factor, prove that $p$ is divisible by 1979.", "solution": "7. We denote the sum mentioned above by $S$. We have the following equalities:  $$ \\begin{aligned} S & =1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\cdots-\\frac{1}{1318}+\\frac{1}{1319} \\\\ & =1+\\frac{1}{2}+\\cdots+\\frac{1}{1319}-2\\left(\\frac{1}{2}+\\frac{1}{4}+\\cdots+\\frac{1}{1318}\\right) \\\\ & =1+\\frac{1}{2}+\\cdots+\\frac{1}{1319}-\\left(1+\\frac{1}{2}+\\cdots+\\frac{1}{659}\\right) \\\\ & =\\frac{1}{660}+\\frac{1}{661}+\\cdots+\\frac{1}{1319} \\\\ & =\\sum_{i=660}^{989} \\frac{1}{i}+\\frac{1}{1979-i}=\\sum_{i=660}^{989} \\frac{1979}{i \\cdot(1979-i)} \\end{aligned} $$  Since no term in the sum contains a denominator divisible by 1979 (1979 is a prime number), it follows that when $S$ is represented as $p / q$ the numerator $p$ will have to be divisible by 1979.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (FRG 3) For all rational $x$ satisfying $0 \\leq x<1, f$ is defined by  $$ f(x)= \\begin{cases}f(2 x) / 4, & \\text { for } 0 \\leq x<1 / 2 \\\\ 3 / 4+f(2 x-1) / 4, & \\text { for } 1 / 2 \\leq x<1\\end{cases} $$  Given that $x=0 . b_{1} b_{2} b_{3} \\ldots$ is the binary representation of $x$, find $f(x)$.", "solution": "8. By the definition of $f$, it holds that $f\\left(0 . b_{1} b_{2} \\ldots\\right)=3 b_{1} / 4+f\\left(0 . b_{2} b_{3} \\ldots\\right) / 4$ $=0 . b_{1} b_{1}+f\\left(0 . b_{2} b_{3} \\ldots\\right) / 4$. Continuing this argument we obtain  $$ f\\left(0 . b_{1} b_{2} b_{3} \\ldots\\right)=0 . b_{1} b_{1} \\ldots b_{n} b_{n}+\\frac{1}{2^{2 n}} f\\left(0 . b_{n+1} b_{n+2} \\ldots\\right) $$  The binary representation of every rational number is eventually periodic. Let us first determine $f(x)$ for a rational $x$ with the periodic representation $x=0 . \\overline{b_{1} b_{2} \\ldots b_{n}}$. Using (1) we obtain $f(x)=0 . b_{1} b_{1} \\ldots b_{n} b_{n}+f(x) / 2^{2 n}$, and hence $f(x)=\\frac{2^{n}}{2^{n}-1} 0 . b_{1} b_{1} \\ldots b_{n} b_{n}=0 . \\overline{b_{1} b_{1} \\ldots b_{n} b_{n}}$. Now let $x=0 . a_{1} a_{2} \\ldots a_{k} \\overline{b_{1} b_{2} \\ldots b_{n}}$ be an arbitrary rational number. Then it follows from (1) that $f(x)=0 . a_{1} a_{1} \\ldots a_{k} a_{k}+\\frac{1}{2^{2 n}} f\\left(0 . \\overline{b_{1} b_{2} \\ldots b_{n}}\\right)=0 . a_{1} a_{1} \\ldots a_{k} a_{k} \\overline{b_{1} b_{1} \\ldots b_{n} b_{n}}$. Hence $f\\left(0 . b_{1} b_{2} \\ldots\\right)=0 . b_{1} b_{1} b_{2} b_{2} \\ldots$ for every rational number $0 . b_{1} b_{2} \\ldots$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1979", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (FRG 4) ${ }^{\\mathrm{IMO6}}$ Let $S$ and $F$ be two opposite vertices of a regular octagon. A counter starts at $S$ and each second is moved to one of the two neighboring vertices of the octagon. The direction is determined by the toss of a coin. The process ends when the counter reaches $F$. We define $a_{n}$ to be the number of distinct paths of duration $n$ seconds that the counter may take to reach $F$ from $S$. Prove that for $n=1,2,3, \\ldots$, $a_{2 n-1}=0, \\quad a_{2 n}=\\frac{1}{\\sqrt{2}}\\left(x^{n-1}-y^{n-1}\\right), \\quad$ where $x=2+\\sqrt{2}, y=2-\\sqrt{2}$.", "solution": "9. Let us number the vertices, starting from $S$ and moving clockwise. In that case $S=1$ and $F=5$. After an odd number of moves to a neighboring point we can be only on an even point, and hence it follows that $a_{2 n-1}=0$ for all $n \\in \\mathbb{N}$. Let us define respectively $z_{n}$ and $w_{n}$ as the number of paths from $S$ to $S$ in $2 n$ moves and the number of paths from $S$ to points 3 and 7 in $2 n$ moves. We easily derive the following recurrence relations:  $$ a_{2 n+2}=w_{n}, \\quad w_{n+1}=2 w_{n}+2 z_{n}, \\quad z_{n+1}=2 z_{n}+w_{n}, \\quad n=0,1,2, \\ldots $$  By subtracting the second equation from the third we get $z_{n+1}=w_{n+1}-$ $w_{n}$. By plugging this equation into the formula for $w_{n+2}$ we get $w_{n+2}-$ $4 w_{n+1}+2 w_{n}=0$. The roots of the characteristic equation $r^{2}-4 r+2=0$ are $x=2+\\sqrt{2}$ and $y=2-\\sqrt{2}$. From the conditions $w_{0}=0$ and $w_{1}=2$ we easily obtain $a_{2 n}=w_{n-1}=\\left(x^{n-1}-y^{n-1}\\right) / \\sqrt{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (BEL) ${ }^{\\mathrm{IMO} 4}$ (a) For which values of $n>2$ is there a set of $n$ consecutive positive integers such that the largest number in the set is a divisor of the least common multiple of the remaining $n-1$ numbers? (b) For which values of $n>2$ is there a unique set having the stated property?", "solution": "1. Assume that the set $\\{a-n+1, a-n+2, \\ldots, a\\}$ of $n$ consecutive numbers satisfies the condition $a \\mid \\operatorname{lcm}[a-n+1, \\ldots, a-1]$. Let $a=p_{1}^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\ldots p_{r}^{\\alpha_{r}}$ be the canonic representation of $a$, where $p_{1}<p_{2}<\\cdots<p_{r}$ are primes and $\\alpha_{1}, \\cdots, \\alpha_{r}>0$. Then for each $j=1,2, \\ldots, r$, there exists $m, m=$ $1,2, \\ldots, n-1$, such that $p_{j}^{\\alpha_{j}} \\mid a-m$, i.e., such that $p_{j}^{\\alpha_{j}} \\mid m$. Thus $p_{j}^{\\alpha_{j}} \\leq$ $n-1$. If $r=1$, then $a=p_{1}^{\\alpha_{1}} \\leq n-1$, which is impossible. Therefore $r \\geq 2$. But then there must exist two distinct prime numbers less than $n$; hence $n \\geq 4$. For $n=4$, we must have $p_{1}^{\\alpha_{1}}, p_{2}^{\\alpha_{2}} \\leq 3$, which leads to $p_{1}=2, p_{2}=3$, $\\alpha_{1}=\\alpha_{2}=1$. Therefore $a=6$, and $\\{3,4,5,6\\}$ is a unique set satisfying the condition of the problem. For every $n \\geq 5$ there exist at least two such sets. In fact, for $n=5$ we easily find two sets: $\\{2,3,4,5,6\\}$ and $\\{8,9,10,11,12\\}$. Suppose that $n \\geq 6$. Let $r, s, t$ be natural numbers such that $2^{r} \\leq n-1<2^{r+1}$, $3^{s} \\leq n-1<3^{s+1}, 5^{t} \\leq n-1<5^{t+1}$. Taking $a=2^{r} \\cdot 3^{s}$ and $a=2^{r} \\cdot 5^{t}$ we obtain two distinct sets with the required property. Thus the answers are (a) $n \\geq 4$ and (b) $n=4$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (FRA) Determine the smallest natural number $n$ having the following property: For every integer $p, p \\geq n$, it is possible to subdivide (partition) a given square into $p$ squares (not necessarily equal).", "solution": "10. It is easy to see that partitioning into $p=2 k$ squares is possible for $k \\geq 2$ (Fig. 1). Furthermore, whenever it is possible to partition the square into $p$ squares, there is a partition of the square into $p+3$ squares: namely, in the partition into $p$ squares, divide one of them into four new squares. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-455.jpg?height=272&width=240&top_left_y=1678&top_left_x=393)  Fig. 1 ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-455.jpg?height=272&width=308&top_left_y=1678&top_left_x=919)  Fig. 2  This implies that both $p=2 k$ and $p=2 k+3$ are possible if $k \\geq 2$, and therefore all $p \\geq 6$ are possible.  On the other hand, partitioning the square into 5 squares is not possible. Assuming it is possible, one of its sides would be covered by exactly two squares, which cannot be of the same size (Fig. 2). The rest of the big square cannot be partitioned into three squares. Hence, the answer is $n=6$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (NET) On a semicircle with unit radius four consecutive chords $A B, B C$, $C D, D E$ with lengths $a, b, c, d$, respectively, are given. Prove that  $$ a^{2}+b^{2}+c^{2}+d^{2}+a b c+b c d<4 $$", "solution": "11. Let us denote the center of the semicircle by $O$, and $\\angle A O B=2 \\alpha$, $\\angle B O C=2 \\beta, A C=m, C E=n$. We claim that $a^{2}+b^{2}+n^{2}+a b n=4$. Indeed, since $a=2 \\sin \\alpha, b=2 \\sin \\beta$, $n=2 \\cos (\\alpha+\\beta)$, we have  $$ \\begin{aligned} a^{2} & +b^{2}+n^{2}+a b n \\\\ & =4\\left(\\sin ^{2} \\alpha+\\sin ^{2} \\beta+\\cos ^{2}(\\alpha+\\beta)+2 \\sin \\alpha \\sin \\beta \\cos (\\alpha+\\beta)\\right) \\\\ & =4+4\\left(-\\frac{\\cos 2 \\alpha}{2}-\\frac{\\cos 2 \\beta}{2}+\\cos (\\alpha+\\beta) \\cos (\\alpha-\\beta)\\right) \\\\ & =4+4(\\cos (\\alpha+\\beta) \\cos (\\alpha-\\beta)-\\cos (\\alpha+\\beta) \\cos (\\alpha-\\beta))=4 \\end{aligned} $$  Analogously, $c^{2}+d^{2}+m^{2}+c d m=4$. By adding both equalities and subtracting $m^{2}+n^{2}=4$ we obtain  $$ a^{2}+b^{2}+c^{2}+d^{2}+a b n+c d m=4 $$  Since $n>c$ and $m>b$, the desired inequality follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (NET) ${ }^{\\mathrm{IMO} 3}$ Determine the maximum value of $m^{2}+n^{2}$ where $m$ and $n$ are integers satisfying  $$ m, n \\in\\{1,2, \\ldots, 100\\} \\quad \\text { and } \\quad\\left(n^{2}-m n-m^{2}\\right)^{2}=1 $$", "solution": "12. We will solve the contest problem (in which $m, n \\in\\{1,2, \\ldots, 1981\\}$ ). For $m=1, n$ can be either 1 or 2 . If $m>1$, then $n(n-m)=m^{2} \\pm 1>0$; hence $n-m>0$. Set $p=n-m$. Since $m^{2}-m p-p^{2}=m^{2}-p(m+p)=$ $-\\left(n^{2}-n m-m^{2}\\right)$, we see that $(m, n)$ is a solution of the equation if and only if $(p, m)$ is a solution too. Therefore, all the solutions of the equation are given as two consecutive members of the Fibonacci sequence  $$ 1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584, \\ldots $$  So the required maximum is $987^{2}+1597^{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (ROM) Let $P$ be a polynomial of degree $n$ satisfying  $$ P(k)=\\binom{n+1}{k}^{-1} \\quad \\text { for } k=0,1, \\ldots, n $$  Determine $P(n+1)$.", "solution": "13. Lemma. For any polynomial $P$ of degree at most $n$,  $$ \\sum_{i=0}^{n+1}(-1)^{i}\\binom{n+1}{i} P(i)=0 $$  Proof. We shall use induction on $n$. For $n=0$ it is trivial. Assume that it is true for $n=k$ and suppose that $P(x)$ is a polynomial of degree $k+1$. Then $P(x)-P(x+1)$ clearly has degree at most $k$; hence (1) gives  $$ \\begin{aligned} 0 & =\\sum_{i=0}^{k+1}(-1)^{i}\\binom{k+1}{i}(P(i)-P(i+1)) \\\\ & =\\sum_{i=0}^{k+1}(-1)^{i}\\binom{k+1}{i} P(i)+\\sum_{i=1}^{k+2}(-1)^{i}\\binom{k+1}{i-1} P(i) \\\\ & =\\sum_{i=0}^{k+2}(-1)^{i}\\binom{k+2}{i} P(i) \\end{aligned} $$  This completes the proof of the lemma. Now we apply the lemma to obtain the value of $P(n+1)$. Since $P(i)=$ $\\binom{n+1}{i}^{-1}$ for $i=0,1, \\ldots, n$, we have  $$ 0=\\sum_{i=0}^{n+1}(-1)^{i}\\binom{n+1}{i} P(i)=(-1)^{n+1} P(n+1)+ \\begin{cases}1, & 2 \\mid n ; \\\\ 0, & 2 \\nmid n .\\end{cases} $$  It follows that $P(n+1)= \\begin{cases}1, & 2 \\mid n ; \\\\ 0, & 2 \\nmid n .\\end{cases}$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (ROM) Prove that a convex pentagon (a five-sided polygon) $A B C D E$ with equal sides and for which the interior angles satisfy the condition $\\angle A \\geq \\angle B \\geq \\angle C \\geq \\angle D \\geq \\angle E$ is a regular pentagon.", "solution": "14. We need the following lemma.  Lemma. If a convex quadrilateral $P Q R S$ satisfies $P S=Q R$ and $\\angle S P Q \\geq$ $\\angle R Q P$, then $\\angle Q R S \\geq \\angle P S R$. Proof. If the lines $P S$ and $Q R$ are parallel, then this quadrilateral is a parallelogram, and the statement is trivial. Otherwise, let $X$ be the point of intersection of lines $P S$ and $Q R$. Assume that $\\angle S P Q+\\angle R Q P>180^{\\circ}$. Then $\\angle X P Q \\leq \\angle X Q P$ implies that $X P \\geq X Q$, and consequently $X S \\geq X R$. Hence, $\\angle Q R S=$ $\\angle X R S \\geq \\angle X S R=\\angle P S R$. Similarly, if $\\angle S P Q+\\angle R Q P<180^{\\circ}$, then $\\angle X P Q \\geq \\angle X Q P$, from which it follows that $X P \\leq X Q$, and thus $X S \\leq X R$; hence $\\angle Q R S=$ $180^{\\circ}-\\angle X R S \\geq 180^{\\circ}-\\angle X S R=\\angle P S R$. Now we apply the lemma to the quadrilateral $A B C D$. Since $\\angle B \\geq \\angle C$ and $A B=C D$, it follows that $\\angle C D A \\geq \\angle B A D$, which together with $\\angle E D A=\\angle E A D$ gives $\\angle D \\geq \\angle A$. Thus $\\angle A=\\angle B=\\angle C=\\angle D$. Analogously, by applying the lemma to $B C D E$ we obtain $\\angle E \\geq \\angle B$, and hence $\\angle B=\\angle C=\\angle D=\\angle E$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (GBR) ${ }^{\\mathrm{IMO} 1}$ Find the point $P$ inside the triangle $A B C$ for which  $$ \\frac{B C}{P D}+\\frac{C A}{P E}+\\frac{A B}{P F} $$  is minimal, where $P D, P E, P F$ are the perpendiculars from $P$ to $B C, C A$, $A B$ respectively.", "solution": "15. Set $B C=a, C A=b, A B=c$, and denote the area of $\\triangle A B C$ by $P$, and $a / P D+b / P E+c / P F$ by $S$. Since $a \\cdot P D+b \\cdot P E+c \\cdot P F=2 P$, by the Cauchy-Schwarz inequality we have  $$ 2 P S=(a \\cdot P D+b \\cdot P E+c \\cdot P F)\\left(\\frac{a}{P D}+\\frac{b}{P E}+\\frac{c}{P F}\\right) \\geq(a+b+c)^{2} $$  with equality if and only if $P D=P E=P F$, i.e., $P$ is the incenter of $\\triangle A B C$. In that case, $S$ attains its minimum:  $$ S_{\\min }=\\frac{(a+b+c)^{2}}{2 P} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (GBR) A sequence of real numbers $u_{1}, u_{2}, u_{3}, \\ldots$ is determined by $u_{1}$ and the following recurrence relation for $n \\geq 1$ :  $$ 4 u_{n+1}=\\sqrt[3]{64 u_{n}+15} $$  Describe, with proof, the behavior of $u_{n}$ as $n \\rightarrow \\infty$.", "solution": "16. The sequence $\\left\\{u_{n}\\right\\}$ is bounded, whatever $u_{1}$ is. Indeed, assume the opposite, and let $u_{m}$ be the first member of the sequence such that $\\left|u_{m}\\right|>\\max \\left\\{2,\\left|u_{1}\\right|\\right\\}$. Then $\\left|u_{m-1}\\right|=\\left|u_{m}^{3}-15 / 64\\right|>\\left|u_{m}\\right|$, which is impossible. Next, let us see for what values of $u_{m}, u_{m+1}$ is greater, equal, or smaller, respectively. If $u_{m+1}=u_{m}$, then $u_{m}=u_{m+1}^{3}-15 / 64=u_{m}^{3}-15 / 64$; i.e., $u_{m}$ is a root of $x^{3}-x-15 / 64=0$. This equation factors as $(x+1 / 4)\\left(x^{2}-x / 4-\\right.$ $15 / 16)=0$, and hence $u_{m}$ is equal to $x_{1}=(1-\\sqrt{61}) / 8, x_{2}=-1 / 4$, or $x_{3}=(1+\\sqrt{61}) / 8$, and these are the only possible limits of the sequence. Each of $u_{m+1}>u_{m}, u_{m+1}<u_{m}$ is equivalent to $u_{m}^{3}-u_{m}-15 / 64<0$ and $u_{m}^{3}-u_{m}-15 / 64>0$ respectively. Thus the former is satisfied for $u_{m}$ in the interval $I_{1}=\\left(-\\infty, x_{1}\\right)$ or $I_{3}=\\left(x_{2}, x_{3}\\right)$, while the latter is satisfied for $u_{m}$ in $I_{2}=\\left(x_{1}, x_{2}\\right)$ or $I_{4}=\\left(x_{3}, \\infty\\right)$. Moreover, since the function $f(x)=\\sqrt[3]{x+15 / 64}$ is strictly increasing with fixed points $x_{1}, x_{2}, x_{3}$, it follows that $u_{m}$ will never escape from the interval $I_{1}, I_{2}, I_{3}$, or $I_{4}$ to which it belongs initially. Therefore: (1) if $u_{1}$ is one of $x_{1}, x_{2}, x_{3}$, the sequence $\\left\\{u_{m}\\right\\}$ is constant; (2) if $u_{1} \\in I_{1}$, then the sequence is strictly increasing and tends to $x_{1}$; (3) if $u_{1} \\in I_{2}$, then the sequence is strictly decreasing and tends to $x_{1}$; (4) if $u_{1} \\in I_{3}$, then the sequence is strictly increasing and tends to $x_{3}$; (5) if $u_{1} \\in I_{4}$, then the sequence is strictly decreasing and tends to $x_{3}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (USS) ${ }^{\\text {IMO5 }}$ Three equal circles touch the sides of a triangle and have one common point $O$. Show that the center of the circle inscribed in and of the circle circumscribed about the triangle $A B C$ and the point $O$ are collinear.", "solution": "17. Let us denote by $S_{A}, S_{B}, S_{C}$ the centers of the given circles, where $S_{A}$ lies on the bisector of $\\angle A$, etc. Then $S_{A} S_{B}\\left\\|A B, S_{B} S_{C}\\right\\| B C, S_{C} S_{A} \\| C A$, so that the inner bisectors of the angles of triangle $A B C$ are also inner bisectors of the angles of $\\triangle S_{A} S_{B} S_{C}$. These two triangles thus have a common incenter $S$, which is also the center of the homothety $\\chi$ mapping $\\triangle S_{A} S_{B} S_{C}$ onto $\\triangle A B C$. The point $O$ is the circumcenter of triangle $S_{A} S_{B} S_{C}$, and so is mapped by $\\chi$ onto the circumcenter $P$ of $A B C$. This means that $O, P$, and the center $S$ of $\\chi$ are collinear.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (USS) Several equal spherical planets are given in outer space. On the surface of each planet there is a set of points that is invisible from any of the remaining planets. Prove that the sum of the areas of all these sets is equal to the area of the surface of one planet.", "solution": "18. Let $C$ be the convex hull of the set of the planets: its border consists of parts of planes, parts of cylinders, and parts of the surfaces of some planets. These parts of planets consist exactly of all the invisible points; any point on a planet that is inside $C$ is visible. Thus it remains to show that the areas of all the parts of planets lying on the border of $C$ add up to the area of one planet. As we have seen, an invisible part of a planet is bordered by some main spherical arcs, parallel two by two. Now fix any planet $P$, and translate these arcs onto arcs on the surface of $P$. All these arcs partition the surface of $P$ into several parts, each of which corresponds to the invisible part of one of the planets. This correspondence is bijective, and therefore the statement follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (YUG) A finite set of unit circles is given in a plane such that the area of their union $U$ is $S$. Prove that there exists a subset of mutually disjoint circles such that the area of their union is greater that $\\frac{2 S}{9}$.", "solution": "19. Consider the partition of plane $\\pi$ into regular hexagons, each having inradius 2. Fix one of these hexagons, denoted by $\\gamma$. For any other hexagon $x$ in the partition, there exists a unique translation $\\tau_{x}$ taking it onto $\\gamma$. Define the mapping $\\varphi: \\pi \\rightarrow \\gamma$ as follows: If $A$ belongs to the interior of a hexagon $x$, then $\\varphi(A)=\\tau_{x}(A)$ (if $A$ is on the border of some hexagon, it does not actually matter where its image is). The total area of the images of the union of the given circles equals $S$, while the area of the hexagon $\\gamma$ is $8 \\sqrt{3}$. Thus there exists a point $B$ of $\\gamma$ that is covered at least $\\frac{S}{8 \\sqrt{3}}$ times, i.e., such that $\\varphi^{-1}(B)$ consists of at least $\\frac{S}{8 \\sqrt{3}}$ distinct points of the plane that belong to some of the circles. For any of these points, take a circle that contains it. All these circles are disjoint, with total area not less than $\\frac{\\pi}{8 \\sqrt{3}} S \\geq 2 S / 9$. Remark. The statement becomes false if the constant $2 / 9$ is replaced by any number greater than $1 / 4$. In that case a counterexample is, for example, a set of unit circles inside a circle of radius 2 covering a sufficiently large part of its area.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (BUL) A sphere $S$ is tangent to the edges $A B, B C, C D, D A$ of a tetrahedron $A B C D$ at the points $E, F, G, H$ respectively. The points $E, F, G, H$ are the vertices of a square. Prove that if the sphere is tangent to the edge $A C$, then it is also tangent to the edge $B D$.", "solution": "2. Lemma. Let $E, F, G, H, I$, and $K$ be points on edges $A B, B C, C D, D A$, $A C$, and $B D$ of a tetrahedron. Then there is a sphere that touches the edges at these points if and only if  $$ \\begin{aligned} & A E=A H=A I, \\quad B E=B F=B K, \\\\ & C F=C G=C I, \\quad D G=D H=D K . \\end{aligned} $$  Proof. The \"only if\" side of the equivalence is obvious. We now assume (*). Denote by $\\epsilon, \\phi, \\gamma, \\eta, \\iota$, and $\\kappa$ planes through $E, F, G, H, I, K$ perpendicular to $A B, B C, C D, D A, A C$ and $B D$ respectively. Since the three planes $\\epsilon, \\eta$, and $\\iota$ are not mutually parallel, they intersect in a common point $O$. Clearly, $\\triangle A E O \\cong$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-451.jpg?height=324&width=496&top_left_y=1374&top_left_x=845) $\\triangle A H O \\cong \\triangle A I O$; hence $O E=O H=O I=r$, and the sphere $\\sigma(O, r)$ is tangent to $A B, A D, A C$. To prove that $\\sigma$ is also tangent to $B C, C D, B D$ it suffices to show that planes $\\phi, \\gamma$, and $\\kappa$ also pass through $O$. Without loss of generality we can prove this for just $\\phi$. By the conditions for $E, F, I$, these are exactly the points of tangency of the incircle of $\\triangle A B C$ and its sides, and if $S$ is the incenter, then $S E \\perp A B, S F \\perp B C, S I \\perp A C$. Hence $\\epsilon, \\iota$, and $\\phi$ all pass through $S$ and are perpendicular to the plane $A B C$, and consequently all share the line $l$ through $S$ perpendicular to $A B C$.  Since $l=\\epsilon \\cap \\iota$, the point $O$ will be situated on $l$, and hence $\\phi$ will also contain $O$. This completes our proof of the lemma. Let $A H=A E=x, B E=B F=y, C F=C G=z$, and $D G=D H=w$. If the sphere is also tangent to $A C$ at some point $I$, then $A I=x$ and $I C=z$. Using the stated lemma it suffices to prove that if $A C=x+z$, then $B D=y+w$. Let $E F=F G=G H=H I=t, \\angle B A D=\\alpha, \\angle A B C=\\beta, \\angle B C D=\\gamma$, and $\\angle A D C=\\delta$. We get  $$ t^{2}=E H^{2}=A E^{2}+A H^{2}-2 \\cdot A E \\cdot A H \\cos \\alpha=2 x^{2}(1-\\cos \\alpha) $$  We similarly conclude that $t^{2}=2 y^{2}(1-\\cos \\beta)=2 z^{2}(1-\\cos \\gamma)=2 w^{2}(1-$ $\\cos \\delta$ ). Further, using that $A B=x+y, B C=y+z, \\cos \\beta=1-t^{2} / 2 y^{2}$, we obtain  $$ A C^{2}=A B^{2}+B C^{2}-2 A B \\cdot B C \\cos \\beta=(x-z)^{2}+t^{2}\\left(\\frac{x}{y}+1\\right)\\left(\\frac{z}{y}+1\\right) $$  Analogously, from the triangle $A D C$ we get $A C^{2}=(x-z)^{2}+t^{2}(x / w+$ $1)(z / w+1)$, which gives $(x / y+1)(z / y+1)=(x / w+1)(z / w+1)$. Since $f(s)=(x / s+1)(z / s+1)$ is a decreasing function in $s$, it follows that $y=w$; similarly $x=z$. Hence $C F=C G=x$ and $D G=D H=y$. Hence $A C \\| E F$ and $A C: t=$ $A C: E F=A B: E B=(x+y): y$; i.e., $A C=t(x+y) / y$. Similarly, from the triangle $A B D$, we get that $B D=t(x+y) / x$. Hence if $A C=x+z=2 x$, it follows that $2 x=t(x+y) / y \\Rightarrow 2 x y=t(x+y) \\Rightarrow B D=t(x+y) / x=$ $2 y=y+w$. This completes the proof. Second solution. Without loss of generality, assume that $E F=2$. Consider the Cartesian system in which points $O, E, F, G, H$ respectively have coordinates $(0,0,0),(-1,-1, a),(1,-1, a),(1,1, a),(-1,1, a)$. Line $A H$ is perpendicular to $O H$ and $A E$ is perpendicular to $O E$; hence from Pythagoras's theorem $A O^{2}=A H^{2}+H O^{2}=A E^{2}+E O^{2}=A E^{2}+H O^{2}$, which implies $A H=A E$. Therefore the $y$-coordinate of $A$ is zero; analogously the $x$-coordinates of $B$ and $D$ and the $y$-coordinate of $C$ are 0 . Let $A$ have coordinates $\\left(x_{0}, 0, z_{1}\\right)$ : then $\\overrightarrow{E A}\\left(x_{0}+1,1, z_{1}-a\\right) \\perp \\overrightarrow{E O}(1,1,-a)$, i.e., $\\overrightarrow{E A} \\cdot \\overrightarrow{E O}=x_{0}+2+a\\left(a-z_{1}\\right)=0$. Similarly, for $B\\left(0, y_{0}, z_{2}\\right)$ we have $y_{0}+2+a\\left(a-z_{2}\\right)=0$. This gives us  $$ z_{1}=\\frac{x_{0}+a^{2}+2}{a}, \\quad z_{2}=\\frac{y_{0}+a^{2}+2}{a} $$  We haven't used yet that $A\\left(x_{0}, 0, z_{1}\\right), E(-1,-1, a)$ and $B\\left(0, y_{0}, z_{2}\\right)$ are collinear, so let $A^{\\prime}, B^{\\prime}, E^{\\prime}$ be the feet of perpendiculars from $A, B, E$ to the plane $x y$. The line $A^{\\prime} B^{\\prime}$, given by $y_{0} x+x_{0} y=x_{0} y_{0}, z=0$, contains the point $E^{\\prime}(-1,-1,0)$, from which we obtain  $$ \\left(x_{0}+1\\right)\\left(y_{0}+1\\right)=1 $$  In the same way, from the points $B$ and $C$ we get relations similar to (1) and (2) and conclude that $C$ has the coordinates $C\\left(-x_{0}, 0, z_{1}\\right)$. Similarly we get $D\\left(0,-y_{0}, z_{2}\\right)$. The condition that $A C$ is tangent to the sphere $\\sigma(O, O E)$ is equivalent to $z_{1}=\\sqrt{a^{2}+2}$, i.e., to $x_{0}=a \\sqrt{a^{2}+2}-\\left(a^{2}+2\\right)$. But then (2) implies that $y_{0}=-a \\sqrt{a^{2}+2}-\\left(a^{2}+2\\right)$ and $z_{2}=-\\sqrt{a^{2}+2}$, which means that the sphere $\\sigma$ is tangent to $B D$ as well. This finishes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (CAN) Find the minimum value of  $$ \\max (a+b+c, b+c+d, c+d+e, d+e+f, e+f+g) $$  subject to the constraints (i) $a, b, c, d, e, f, g \\geq 0$, (ii) $a+b+c+d+e+f+g=1$.", "solution": "3. Denote $\\max (a+b+c, b+c+d, c+d+e, d+e+f, e+f+g)$ by $p$. We have  $$ (a+b+c)+(c+d+e)+(e+f+g)=1+c+e \\leq 3 p $$  which implies that $p \\geq 1 / 3$. However, $p=1 / 3$ is achieved by taking $(a, b, c, d, e, f, g)=(1 / 3,0,0,1 / 3,0,0,1 / 3)$. Therefore the answer is $1 / 3$. Remark. In fact, one can prove a more general statement in the same way. Given positive integers $n, k, n \\geq k$, if $a_{1}, a_{2}, \\ldots, a_{n}$ are nonnegative real numbers with sum 1 , then the minimum value of $\\max _{i=1, \\ldots, n-k+1}\\left\\{a_{i}+\\right.$ $\\left.a_{i+1}+\\cdots+a_{i+k-1}\\right\\}$ is $1 / r$, where $r$ is the integer with $k(r-1)<n \\leq k r$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (CAN) Let $\\left\\{f_{n}\\right\\}$ be the Fibonacci sequence $\\{1,1,2,3,5, \\ldots\\}$. (a) Find all pairs $(a, b)$ of real numbers such that for each $n, a f_{n}+b f_{n+1}$ is a member of the sequence. (b) Find all pairs $(u, v)$ of positive real numbers such that for each $n$, $u f_{n}^{2}+v f_{n+1}^{2}$ is a member of the sequence.", "solution": "4. We shall use the known formula for the Fibonacci sequence  $$ f_{n}=\\frac{1}{\\sqrt{5}}\\left(\\alpha^{n}-(-1)^{n} \\alpha^{-n}\\right), \\quad \\text { where } \\alpha=\\frac{1+\\sqrt{5}}{2} . $$  (a) Suppose that $a f_{n}+b f_{n+1}=f_{k_{n}}$ for all $n$, where $k_{n}>0$ is an integer depending on $n$. By (1), this is equivalent to $a\\left(\\alpha^{n}-(-1)^{n} \\alpha^{-n}\\right)+$ $b\\left(\\alpha^{n+1}+(-1)^{n} \\alpha^{-n-1}\\right)=\\alpha^{k_{n}}-(-1)^{k_{n}} \\alpha^{-k_{n}}$, i.e.,  $$ \\alpha^{k_{n}-n}=a+b \\alpha-\\alpha^{-2 n}(-1)^{n}\\left(a-b \\alpha^{-1}-(-\\alpha)^{n-k_{n}}\\right) \\rightarrow a+b \\alpha $$  as $n \\rightarrow \\infty$. Hence, since $k_{n}$ is an integer, $k_{n}-n$ must be constant from some point on: $k_{n}=n+k$ and $\\alpha^{k}=a+b \\alpha$. Then it follows from (2) that $\\alpha^{-k}=a-b \\alpha^{-1}$, and from (1) we conclude that $a f_{n}+b f_{n+1}=$ $f_{k+n}$ holds for every $n$. Putting $n=1$ and $n=2$ in the previous relation and solving the obtained system of equations we get $a=f_{k-1}$, $b=f_{k}$. It is easy to verify that such $a$ and $b$ satisfy the conditions. (b) As in (a), suppose that $u f_{n}^{2}+v f_{n+1}^{2}=f_{l_{n}}$ for all $n$. This leads to  $$ \\begin{aligned} u+v \\alpha^{2}-\\sqrt{5} \\alpha^{l_{n}-2 n}= & 2(u-v)(-1)^{n} \\alpha^{-2 n} \\\\ & -\\left(u \\alpha^{-4 n}+v \\alpha^{-4 n-2}+(-1)^{l_{n}} \\sqrt{5} \\alpha^{-l_{n}-2 n}\\right) \\\\ \\rightarrow & 0 \\end{aligned} $$  as $n \\rightarrow \\infty$. Thus $u+v \\alpha^{2}=\\sqrt{5} \\alpha^{l_{n}-2 n}$, and $l_{n}-2 n=k$ is equal to a constant. Putting this into the above equation and multiplying by $\\alpha^{2 n}$ we get $u-v \\rightarrow 0$ as $n \\rightarrow \\infty$, i.e., $u=v$. Finally, substituting $n=1$ and $n=2$ in $u f_{n}^{2}+u f_{n+1}^{2}=f_{l_{n}}$ we easily get that the only possibility is $u=v=1$ and $k=1$. It is easy to verify that such $u$ and $v$ satisfy the conditions.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (COL) A cube is assembled with 27 white cubes. The larger cube is then painted black on the outside and disassembled. A blind man reassembles it. What is the probability that the cube is now completely black on the outside? Give an approximation of the size of your answer.", "solution": "5. There are four types of small cubes upon disassembling: (1) 8 cubes with three faces, painted black, at one corner; (2) 12 cubes with two black faces, both at one edge; (3) 6 cubes with one black face; (4) 1 completely white cube.  All cubes of type (1) must go to corners, and be placed in a correct way (one of three): for this step we have $3^{8} \\cdot 8$ ! possibilities. Further, all cubes of type (2) must go in a correct way (one of two) to edges, admitting $2^{12} \\cdot 12$ ! possibilities; similarly, there are $4^{6} \\cdot 6$ ! ways for cubes of type (3), and 24 ways for the cube of type (4). Thus the total number of good reassemblings is $3^{8} 8!\\cdot 2^{12} 12!\\cdot 4^{6} 6!\\cdot 24$, while the number of all possible reassemblings is $24^{27} \\cdot 27!$. The desired probability is $\\frac{3^{8} 8!\\cdot 2^{12} 12!\\cdot 4^{6} 6!\\cdot 24}{24^{27} \\cdot 27!}$. It is not necessary to calculate these numbers to find out that the blind man practically has no chance to reassemble the cube in a right way: in fact, the probability is of order $1.8 \\cdot 10^{-37}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (CUB) Let $P(z)$ and $Q(z)$ be complex-variable polynomials, with degree not less than 1. Let  $$ P_{k}=\\{z \\in \\mathbb{C} \\mid P(z)=k\\}, \\quad Q_{k}=\\{z \\in \\mathbb{C} \\mid Q(z)=k\\} $$  Let also $P_{0}=Q_{0}$ and $P_{1}=Q_{1}$. Prove that $P(z) \\equiv Q(z)$.", "solution": "6. Assume w.l.o.g. that $n=\\operatorname{deg} P \\geq \\operatorname{deg} Q$, and let $P_{0}=\\left\\{z_{1}, z_{2}, \\ldots, z_{k}\\right\\}$, $P_{1}=\\left\\{z_{k+1}, z_{k+2}, \\ldots z_{k+m}\\right\\}$. The polynomials $P$ and $Q$ match at $k+m$ points $z_{1}, z_{2}, \\ldots, z_{k+m}$; hence if we prove that $k+m>n$, the result will follow. By the assumption, $P(x)=\\left(x-z_{1}\\right)^{\\alpha_{1}} \\cdots\\left(x-z_{k}\\right)^{\\alpha_{k}}=\\left(x-z_{k+1}\\right)^{\\alpha_{k+1}} \\cdots\\left(x-z_{k+m}\\right)^{\\alpha_{k+m}}+1$ for some positive integers $\\alpha_{1}, \\ldots, \\alpha_{k+m}$. Let us consider $P^{\\prime}(x)$. As we know, it is divisible by $\\left(x-z_{i}\\right)^{\\alpha_{i}-1}$ for $i=1,2, \\ldots, k+m$; i.e.,  $$ \\prod_{i=1}^{k+m}\\left(x-z_{i}\\right)^{\\alpha_{i}-1} \\mid P^{\\prime}(x) $$  Therefore $2 n-k-m=\\operatorname{deg} \\prod_{i=1}^{k+m}\\left(x-z_{i}\\right)^{\\alpha_{i}-1} \\leq \\operatorname{deg} P^{\\prime}=n-1$, i.e., $k+m \\geq n+1$, as we claimed.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (FIN) ${ }^{\\mathrm{IMO} 6}$ Assume that $f(x, y)$ is defined for all positive integers $x$ and $y$, and that the following equations are satisfied:  $$ \\begin{aligned} f(0, y) & =y+1 \\\\ f(x+1,0) & =f(x, 1) \\\\ f(x+1, y+1) & =f(x, f(x+1, y)) \\end{aligned} $$  Determine $f(2,2), f(3,3)$ and $f(4,4)$. Alternative version: Determine $f(4,1981)$.", "solution": "7. We immediately find that $f(1,0)=f(0,1)=2$. Then $f(1, y+1)=$ $f(0, f(1, y))=f(1, y)+1$; hence $f(1, y)=y+2$ for $y \\geq 0$. Next we find that $f(2,0)=f(1,1)=3$ and $f(2, y+1)=f(1, f(2, y))=f(2, y)+2$, from which $f(2, y)=2 y+3$. Particularly, $f(2,2)=7$. Further, $f(3,0)=$ $f(2,1)=5$ and $f(3, y+1)=f(2, f(3, y))=2 f(3, y)+3$. This gives by induction $f(3, y)=2^{y+3}-3$. For $y=3, f(3,3)=61$. Finally, from $f(4,0)=f(3,1)=13$ and $f(4, y+1)=f(3, f(4, y))=2^{f(4, y)+3}-3$, we conclude that  $$ f(4, y)=2^{2 y^{2}}-3 \\quad(y+3 \\text { twos }) $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (FRG) ${ }^{\\mathrm{IMO} 2}$ Let $f(n, r)$ be the arithmetic mean of the minima of all $r$ subsets of the set $\\{1,2, \\ldots, n\\}$. Prove that $f(n, r)=\\frac{n+1}{r+1}$.", "solution": "8. Since the number $k, k=1,2, \\ldots, n-r+1$, is the minimum in exactly $\\binom{n-k}{r-1} r$-element subsets of $\\{1,2, \\ldots, n\\}$, it follows that  $$ f(n, r)=\\frac{1}{\\binom{n}{r}} \\sum_{k=1}^{n-r+1} k\\binom{n-k}{r-1} $$  To calculate the sum in the above expression, using the equality $\\binom{r+j}{j}=$ $\\sum_{i=0}^{j}\\binom{r+i-1}{r-1}$, we note that  $$ \\begin{aligned} \\sum_{k=1}^{n-r+1} k\\binom{n-k}{r-1} & =\\sum_{j=0}^{n-r}\\left(\\sum_{i=0}^{j}\\binom{r+i-1}{r-1}\\right) \\\\ & =\\sum_{j=0}^{n-r}\\binom{r+j}{r}=\\binom{n+1}{r+1}=\\frac{n+1}{r+1}\\binom{n}{r} . \\end{aligned} $$  Therefore $f(n, r)=(n+1) /(r+1)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1981", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (FRG) A sequence $\\left(a_{n}\\right)$ is defined by means of the recursion  $$ a_{1}=1, \\quad a_{n+1}=\\frac{1+4 a_{n}+\\sqrt{1+24 a_{n}}}{16} $$  Find an explicit formula for $a_{n}$.", "solution": "9. If we put $1+24 a_{n}=b_{n}^{2}$, the given recurrent relation becomes  $$ \\frac{2}{3} b_{n+1}^{2}=\\frac{3}{2}+\\frac{b_{n}^{2}}{6}+b_{n}=\\frac{2}{3}\\left(\\frac{3}{2}+\\frac{b_{n}}{2}\\right)^{2}, \\quad \\text { i.e., } \\quad b_{n+1}=\\frac{3+b_{n}}{2} $$  where $b_{1}=5$. To solve this recurrent equation, we set $c_{n}=2^{n-1} b_{n}$. From (1) we obtain  $$ \\begin{aligned} c_{n+1} & =c_{n}+3 \\cdot 2^{n-1}=\\cdots=c_{1}+3\\left(1+2+2^{2}+\\cdots+2^{n-1}\\right) \\\\ & =5+3\\left(2^{n}-1\\right)=3 \\cdot 2^{n}+2 \\end{aligned} $$  Therefore $b_{n}=3+2^{-n+2}$ and consequently  $$ a_{n}=\\frac{b_{n}^{2}-1}{24}=\\frac{1}{3}\\left(1+\\frac{3}{2^{n}}+\\frac{1}{2^{2 n-1}}\\right)=\\frac{1}{3}\\left(1+\\frac{1}{2^{n-1}}\\right)\\left(1+\\frac{1}{2^{n}}\\right) . $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "1", "problem_type": "Algebra", "exam": "IMO", "problem": "1. A1 (GBR 3) ${ }^{\\mathrm{IMO}}$ The function $f(n)$ is defined for all positive integers $n$ and takes on nonnegative integer values. Also, for all $m, n$,  $$ \\begin{gathered} f(m+n)-f(m)-f(n)=0 \\text { or } 1 \\\\ f(2)=0, \\quad f(3)>0, \\quad \\text { and } \\quad f(9999)=3333 \\end{gathered} $$  Determine $f(1982)$.", "solution": "1. From $f(1)+f(1) \\leq f(2)=0$ we obtain $f(1)=0$. Since $0<f(3) \\leq$ $f(1)+f(2)+1$, it follows that $f(3)=1$. Note that if $f(3 n) \\geq n$, then $f(3 n+3) \\geq f(3 n)+f(3) \\geq n+1$. Hence by induction $f(3 n) \\geq n$ holds for all $n \\in \\mathbb{N}$. Moreover, if the inequality is strict for some $n$, then it is so for all integers greater than $n$ as well. Since $f(9999)=3333$, we deduce that $f(3 n)=n$ for all $n \\leq 3333$. By the given condition, we have $3 f(n) \\leq f(3 n) \\leq 3 f(n)+2$. Therefore $f(n)=[f(3 n) / 3]=[n / 3]$ for $n \\leq 3333$. In particular, $f(1982)=$ $[1982 / 3]=660$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. B4 (BRA 1) A box contains $p$ white balls and $q$ black balls. Beside the box there is a pile of black balls. Two balls are taken out of the box. If they have the same color, a black ball from the pile is put into the box. If they have different colors, the white ball is put back into the box. This procedure is repeated until the last two balls are removed from the box and one last ball is put in. What is the probability that this last ball is white?", "solution": "10. If the two balls taken from the box are both white, then the number of white balls decreases by two; otherwise, it remains unchanged. Hence the parity of the number of white balls does not change during the procedure. Therefore if $p$ is even, the last ball cannot be white; the probability is 0 . If $p$ is odd, the last ball has to be white; the probability is 1 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. B5 (CAN 3) (a) Find the rearrangement $\\left\\{a_{1}, \\ldots, a_{n}\\right\\}$ of $\\{1,2, \\ldots, n\\}$ that maximizes  $$ a_{1} a_{2}+a_{2} a_{3}+\\cdots+a_{n} a_{1}=Q $$  (b) Find the rearrangement that minimizes $Q$.", "solution": "11. (a) Suppose $\\left\\{a_{1}, a_{2}, \\ldots, a_{n}\\right\\}$ is the arrangement that yields the maximal value $Q_{\\max }$ of $Q$. Note that the value of $Q$ for the rearrangement $\\left\\{a_{1}, \\ldots, a_{i-1}, a_{j}, a_{j-1}, \\ldots, a_{i}, a_{j+1}, \\ldots, a_{n}\\right\\}$ equals $Q_{\\text {max }}-\\left(a_{i}-\\right.$ $\\left.a_{j}\\right)\\left(a_{i-1}-a_{j+1}\\right)$, where $1<i<j<n$. Hence $\\left(a_{i}-a_{j}\\right)\\left(a_{i-1}-a_{j+1}\\right) \\geq 0$ for all $1<i<j<n$.  We may suppose w.l.o.g. that $a_{1}=1$. Let $a_{i}=2$. If $2<i<$ $n$, then $\\left(a_{2}-a_{i}\\right)\\left(a_{1}-a_{i+1}\\right)<0$, which is impossible. Therefore $i$ is either 2 or $n$; let w.l.o.g. $a_{n}=2$. Further, if $a_{j}=3$ for $2<j<n$, then $\\left(a_{1}-a_{j+1}\\right)\\left(a_{2}-a_{j}\\right)<0$, which is impossible; therefore $a_{2}=3$. Continuing this argument we obtain that $A=\\{1,3,5, \\ldots, 2[(n-1) / 2]+1,2[n / 2], \\ldots, 4,2\\}$. (b) A similar argument leads to the minimizing rearrangement $\\{1, n, 2$, $n-1, \\ldots,[n / 2]+1\\}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. B6 (FIN 3) Four distinct circles $C, C_{1}, C_{2}, C_{3}$ and a line $L$ are given in the plane such that $C$ and $L$ are disjoint and each of the circles $C_{1}, C_{2}, C_{3}$ touches the other two, as well as $C$ and $L$. Assuming the radius of $C$ to be 1 , determine the distance between its center and $L$.", "solution": "12. Let $y$ be the line perpendicular to $L$ passing through the center of $C$. It can be shown by a continuity argument that there exists a point $Y \\in y$ such that an inversion $\\Psi$ centered at $Y$ maps $C$ and $L$ onto two concentric circles $\\widehat{C}$ and $\\widehat{L}$. Let $\\widehat{X}$ denote the image of an object $X$ under $\\Psi$. Then the circles $\\widehat{C_{i}}$ touch $\\widehat{C}$ externally and $\\widehat{L}$ internally, and all have the same radius. Let us now rotate the picture around the common center $Z$ of $\\widehat{C}$ and $\\widehat{L}$ so that $\\widehat{C_{3}}$ passes through $Y$. Applying the inversion $\\Psi$ again on the picture thus obtained, $\\widehat{C}$ and $\\widehat{L}$ go back to $C$ and $L$, but $\\widehat{C_{3}}$ goes to a line $C_{3}^{\\prime}$ parallel to $L$, while the images of $\\widehat{C_{1}}$ and $\\widehat{C_{2}}$ go to two equal circles $C_{1}^{\\prime}$ and $C_{2}^{\\prime}$ touching $L, C_{3}^{\\prime}$, and $C$. This way we have achieved that $C_{3}$ becomes a line. Denote by $O_{1}, O_{2}, O$ respectively the centers of the circles $C_{1}^{\\prime}, C_{2}^{\\prime}, C$ and by $T$ the point of tangency of the circles $C_{1}^{\\prime}$ and $C_{2}^{\\prime}$. If $x$ is the common radius of the circles $C_{1}^{\\prime}$ and $C_{2}^{\\prime}$, then from $\\triangle O_{1} T O$ we obtain ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-463.jpg?height=266&width=496&top_left_y=1028&top_left_x=832) that $(x-1)^{2}+x^{2}=(x+1)^{2}$, and thus $x=4$. Hence the distance of $O$ from $L$ equals $2 x-1=7$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "13", "problem_type": "Combinatorics", "exam": "IMO", "problem": "13. C1 (NET 1) ${ }^{\\mathrm{IMO} 2} \\mathrm{~A}$ scalene triangle $A_{1} A_{2} A_{3}$ is given with sides $a_{1}, a_{2}, a_{3}$ ( $a_{i}$ is the side opposite to $A_{i}$ ). For all $i=1,2,3, M_{i}$ is the midpoint of side $a_{i}, T_{i}$ is the point where the incircle touches side $a_{i}$, and the reflection of $T_{i}$ in the interior bisector of $A_{i}$ yields the point $S_{i}$. Prove that the lines $M_{1} S_{1}, M_{2} S_{2}$, and $M_{3} S_{3}$ are concurrent.", "solution": "13. Points $S_{1}, S_{2}, S_{3}$ clearly lie on the inscribed circle. Let $\\widehat{X Y}$ denote the oriented arc $X Y$. The $\\operatorname{arcs} \\widehat{T_{2} S_{1}}$ and $\\widehat{T_{1} T_{3}}$ are equal, since they are symmetric with respect to the bisector of $\\angle A_{1}$. Similarly, $\\widehat{T_{3} T_{2}}=\\widehat{S_{2} T_{1}}$. Therefore $\\widehat{T_{3} S_{1}}=$ $\\widehat{T_{3} T_{2}}+\\widehat{T_{2} S_{1}}=\\widehat{S_{2} T_{1}}+\\widehat{T_{1} T_{3}}=$ $S_{2} T_{3}$. It follows that $S_{1} S_{2}$ is parallel to $A_{1} A_{2}$, and consequently $S_{1} S_{2} \\|$ $M_{1} M_{2}$. Analogously $S_{1} S_{3} \\| M_{1} M_{3}$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-463.jpg?height=358&width=505&top_left_y=1450&top_left_x=827) and $S_{2} S_{3} \\| M_{2} M_{3}$. Since the circumcircles of $\\triangle M_{1} M_{2} M_{3}$ and $\\triangle S_{1} S_{2} S_{3}$ are not equal, these triangles are not congruent and hence they must be homothetic. Then all the lines $M_{i} S_{i}$ pass through the center of homothety. Second solution. Set the complex plane so that the incenter of $\\triangle A_{1} A_{2} A_{3}$ is the unit circle centered at the origin. Let $t_{i}, s_{i}$ respectively denote the complex numbers of modulus 1 corresponding to $T_{i}, S_{i}$. Clearly $t_{1} \\overline{t_{1}}=$ $t_{2} \\overline{t_{2}}=t_{3} \\overline{t_{3}}=1$. Since $T_{2} T_{3}$ and $T_{1} S_{1}$ are parallel, we obtain $t_{2} t_{3}=t_{1} s_{1}$, or $s_{1}=t_{2} t_{3} \\overline{t_{1}}$. Similarly $s_{2}=t_{1} t_{3} \\overline{t_{2}}, s_{3}=t_{1} t_{2} \\overline{t_{3}}$, from which it follows that $s_{2}-s_{3}=t_{1}\\left(t_{3} \\overline{t_{2}}-t_{2} \\overline{t_{3}}\\right)$. Since the number in parentheses is strictly imaginary, we conclude that $O T_{1} \\perp S_{2} S_{3}$ and consequently $S_{2} S_{3} \\| A_{2} A_{3}$. We proceed as in the first solution.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "14", "problem_type": "Combinatorics", "exam": "IMO", "problem": "14. C2 (AUS 4) Let $A B C D$ be a convex plane quadrilateral and let $A_{1}$ denote the circumcenter of $\\triangle B C D$. Define $B_{1}, C_{1}, D_{1}$ in a corresponding way. (a) Prove that either all of $A_{1}, B_{1}, C_{1}, D_{1}$ coincide in one point, or they are all distinct. Assuming the latter case, show that $A_{1}, C_{1}$ are on opposite sides of the line $B_{1} D_{1}$, and similarly, $B_{1}, D_{1}$ are on opposite sides of the line $A_{1} C_{1}$. (This establishes the convexity of the quadrilateral $A_{1} B_{1} C_{1} D_{1}$.) (b) Denote by $A_{2}$ the circumcenter of $B_{1} C_{1} D_{1}$, and define $B_{2}, C_{2}, D_{2}$ in an analogous way. Show that the quadrilateral $A_{2} B_{2} C_{2} D_{2}$ is similar to the quadrilateral $A B C D$.", "solution": "14. (a) If any two of $A_{1}, B_{1}, C_{1}, D_{1}$ coincide, say $A_{1} \\equiv B_{1}$, then $A B C D$ is inscribed in a circle centered at $A_{1}$ and hence all $A_{1}, B_{1}, C_{1}, D_{1}$ coincide. Assume now the opposite, and let w.l.o.g. $\\angle D A B+\\angle D C B<180^{\\circ}$. Then $A$ is outside the circumcircle of $\\triangle B C D$, so $A_{1} A>A_{1} C$. Similarly, $C_{1} C>C_{1} A$. Hence the perpendicular bisector $l_{A C}$ of $A C$ separates points $A_{1}$ and $C_{1}$. Since $B_{1}, D_{1}$ lie on $l_{A C}$, this means that $A_{1}$ and $C_{1}$ are on opposite sides $B_{1} D_{1}$. Similarly one can show that $B_{1}$ and $D_{1}$ are on opposite sides of $A_{1} C_{1}$. (b) Since $A_{2} B_{2} \\perp C_{1} D_{1}$ and $C_{1} D_{1} \\perp A B$, it follows that $A_{2} B_{2} \\| A B$. Similarly $A_{2} C_{2}\\left\\|A C, A_{2} D_{2}\\right\\| A D, B_{2} C_{2}\\left\\|B C, B_{2} D_{2}\\right\\| B D$, and $C_{2} D_{2} \\| C D$. Hence $\\triangle A_{2} B_{2} C_{2} \\sim \\triangle A B C$ and $\\triangle A_{2} D_{2} C_{2} \\sim \\triangle A D C$, and the result follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "15", "problem_type": "Combinatorics", "exam": "IMO", "problem": "15. C3 (CAN 5) Show that  $$ \\frac{1-s^{a}}{1-s} \\leq(1+s)^{a-1} $$  holds for every $1 \\neq s>0$ real and $0<a \\leq 1$ rational.", "solution": "15. Let $a=k / n$, where $n, k \\in \\mathbb{N}, n \\geq k$. Putting $t^{n}=s$, the given inequality becomes $\\frac{1-t^{k}}{1-t^{n}} \\leq\\left(1+t^{n}\\right)^{k / n-1}$, or equivalently  $$ \\left(1+t+\\cdots+t^{k-1}\\right)^{n}\\left(1+t^{n}\\right)^{n-k} \\leq\\left(1+t+\\cdots+t^{n-1}\\right)^{n} $$  This is clearly true for $k=n$. Therefore it is enough to prove that the lefthand side of the above inequality is an increasing function of $k$. We are led to show that $\\left(1+t+\\cdots+t^{k-1}\\right)^{n}\\left(1+t^{n}\\right)^{n-k} \\leq\\left(1+t+\\cdots+t^{k}\\right)^{n}\\left(1+t^{n}\\right)^{n-k-1}$. This is equivalent to $1+t^{n} \\leq A^{n}$, where $A=\\frac{1+t+\\cdots+t^{k}}{1+t+\\cdots+t^{k-1}}$. But this easily follows, since  $$ \\begin{aligned} A^{n}-t^{n} & =(A-t)\\left(A^{n-1}+A^{n-2} t+\\cdots+t^{n-1}\\right) \\\\ & \\geq(A-t)\\left(1+t+\\cdots+t^{n-1}\\right)=\\frac{1+t+\\cdots+t^{n-1}}{1+t+\\cdots+t^{k-1}} \\geq 1 \\end{aligned} $$  Remark. The original problem asked to prove the inequality for real $a$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "16", "problem_type": "Combinatorics", "exam": "IMO", "problem": "16. C4 (GBR 2) ${ }^{\\mathrm{IMO}}$ Prove that if $n$ is a positive integer such that the equation $x^{3}-3 x y^{2}+y^{3}=n$ has a solution in integers $(x, y)$, then it has at least three such solutions. Show that the equation has no solution in integers when $n=2891$.", "solution": "16. It is easy to verify that whenever $(x, y)$ is a solution of the equation $x^{3}-3 x y^{2}+y^{3}=n$, so are the pairs $(y-x,-x)$ and $(-y, x-y)$. No two of these three solutions are equal unless $x=y=n=0$. Observe that $2981 \\equiv 2(\\bmod 9)$. Since $x^{3}, y^{3} \\equiv 0, \\pm 1(\\bmod 9), x^{3}-$ $3 x y^{2}+y^{3}$ cannot give the remainder 2 when divided by 9 . Hence the above equation for $n=2981$ has no integer solutions.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "17", "problem_type": "Combinatorics", "exam": "IMO", "problem": "17. C5 (USS 5) The right triangles $A B C$ and $A B_{1} C_{1}$ are similar and have opposite orientation. The right angles are at $C$ and $C_{1}$, and we also have $\\measuredangle C A B=\\measuredangle C_{1} A B_{1}$. Let $M$ be the point of intersection of the lines $B C_{1}$ and $B_{1} C$. Prove that if the lines $A M$ and $C C_{1}$ exist, they are perpendicular.", "solution": "17. Let $A$ be the origin of the Cartesian plane. Suppose that $B C: A C=k$ and that $(a, b)$ and $\\left(a_{1}, b_{1}\\right)$ are coordinates of the points $C$ and $C_{1}$, respectively. Then the coordinates of the point $B$ are $(a, b)+k(-b, a)=(a-k b, b+k a)$, while the coordinates of $B_{1}$ are $\\left(a_{1}, b_{1}\\right)+k\\left(b_{1},-a_{1}\\right)=\\left(a+k b_{1}, b_{1}-k a_{1}\\right)$. Thus the lines $B C_{1}$ and $C B_{1}$ are given by the equations $\\frac{x-a_{1}}{y-b_{1}}=\\frac{x-(a-k b)}{y-(b+k a)}$ and $\\frac{x-a}{y-b}=\\frac{x-\\left(a_{1}+k b_{1}\\right)}{y-\\left(b_{1}-k a_{1}\\right)}$ respectively. After multiplying, these equations transform into the forms  $$ \\begin{aligned} B C_{1}: & k a x+k b y & =k a a_{1}+k b b_{1}+b a_{1}-a b_{1}-\\left(b-b_{1}\\right) x+\\left(a-a_{1}\\right) y \\\\ C B_{1}: & k a_{1} x+k b_{1} y & =k a a_{1}+k b b_{1}+b a_{1}-a b_{1}-\\left(b-b_{1}\\right) x+\\left(a-a_{1}\\right) y . \\end{aligned} $$  The coordinates $\\left(x_{0}, y_{0}\\right)$ of the point $M$ satisfy these equations, from which we deduce that $k a x_{0}+k b y_{0}=k a_{1} x_{0}+k b_{1} y_{0}$. This yields $\\frac{x_{0}}{y_{0}}=-\\frac{b_{1}-b}{a_{1}-a}$, implying that the lines $C C_{1}$ and $A M$ are perpendicular.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "18", "problem_type": "Combinatorics", "exam": "IMO", "problem": "18. C6 (FRA 2) Let $O$ be a point of three-dimensional space and let $l_{1}, l_{2}, l_{3}$ be mutually perpendicular straight lines passing through $O$. Let $S$ denote the sphere with center $O$ and radius $R$, and for every point $M$ of $S$, let $S_{M}$ denote the sphere with center $M$ and radius $R$. We denote by $P_{1}, P_{2}, P_{3}$ the intersection of $S_{M}$ with the straight lines $l_{1}, l_{2}, l_{3}$, respectively, where we put $P_{i} \\neq O$ if $l_{i}$ meets $S_{M}$ at two distinct points and $P_{i}=O$ otherwise ( $i=$ $1,2,3)$. What is the set of centers of gravity of the (possibly degenerate) triangles $P_{1} P_{2} P_{3}$ as $M$ runs through the points of $S$ ?", "solution": "18. Set the coordinate system with the axes $x, y, z$ along the lines $l_{1}, l_{2}, l_{3}$ respectively. The coordinates $(a, b, c)$ of $M$ satisfy $a^{2}+b^{2}+c^{2}=R^{2}$, and so $S_{M}$ is given by the equation $(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=R^{2}$. Hence the coordinates of $P_{1}$ are $(x, 0,0)$ with $(x-a)^{2}+b^{2}+c^{2}=R^{2}$, implying that either $x=2 a$ or $x=0$. Thus by the definition we obtain $x=2 a$. Similarly, the coordinates of $P_{2}$ and $P_{3}$ are $(0,2 b, 0)$ and $(0,0,2 c)$ respectively. Now, the centroid of $\\triangle P_{1} P_{2} P_{3}$ has the coordinates $(2 a / 3,2 b / 3,2 c / 3)$. Therefore the required locus of points is the sphere with center $O$ and radius $2 R / 3$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "19", "problem_type": "Combinatorics", "exam": "IMO", "problem": "19. C7 (CZS 3) Let $M$ be the set of real numbers of the form $\\frac{m+n}{\\sqrt{m^{2}+n^{2}}}$, where $m$ and $n$ are positive integers. Prove that for every pair $x \\in M$, $y \\in M$ with $x<y$, there exists an element $z \\in M$ such that $x<z<y$.", "solution": "19. Let us set $x=m / n$. Since $f(x)=(m+n) / \\sqrt{m^{2}+n^{2}}=(x+1) / \\sqrt{1+x^{2}}$ is a continuous function of $x, f(x)$ takes all values between any two values of $f$; moreover, the corresponding $x$ can be rational. This completes the proof.  Remark. Since $f$ is increasing for $x \\geq 1,1 \\leq x<z<y$ implies $f(x)<$ $f(z)<f(y)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "2", "problem_type": "Algebra", "exam": "IMO", "problem": "2. A2 (YUG 1) Let $K$ be a convex polygon in the plane and suppose that $K$ is positioned in the coordinate system in such a way that  $$ \\operatorname{area}\\left(K \\cap Q_{i}\\right)=\\frac{1}{4} \\text { area } K(i=1,2,3,4,), $$  where the $Q_{i}$ denote the quadrants of the plane. Prove that if $K$ contains no nonzero lattice point, then the area of $K$ is less than 4.", "solution": "2. Since $K$ does not contain a lattice point other than $O(0,0)$, it is bounded by four lines $u, v, w, x$ that pass through the points $U(1,0), V(0,1)$, $W(-1,0), X(0,-1)$ respectively. Let $P Q R S$ be the quadrilateral formed by these lines, where $U \\in S P, V \\in P Q, W \\in Q R, X \\in R S$. If one of the quadrants, say $Q_{1}$, contains no vertices of $P Q R S$, then $K \\cap Q_{1}$ is contained in $\\triangle O U V$ and hence has area less than $1 / 2$. Consequently the area of $K$ is less than 2 . Let us now suppose that $P, Q, R, S$ lie in different quadrants. One of the angles of $P Q R S$ is at least $90^{\\circ}$ : let it be $\\angle P$. Then $S_{U P V} \\leq P U \\cdot P V / 2 \\leq$ $\\left(P U^{2}+P V^{2}\\right) / 4 \\leq U V^{2} / 4=1 / 2$, which implies that $S_{K \\cap Q_{1}}<S_{O U P V} \\leq$ 1. Hence the area of $K$ is less than 4.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "20", "problem_type": "Combinatorics", "exam": "IMO", "problem": "20. C8 (TUN 3) Let $A B C D$ be a convex quadrilateral and draw regular triangles $A B M, C D P, B C N, A D Q$, the first two outward and the other two inward. Prove that $M N=A C$. What can be said about the quadrilateral $M N P Q$ ?", "solution": "20. Since $M N$ is the image of $A C$ under rotation about $B$ for $60^{\\circ}$, we have $M N=A C$. Similarly, $P Q$ is the image of $A C$ under rotation about $D$ through $60^{\\circ}$, from which it follows that $P Q \\| M N$. Hence either $M, N, P, Q$ are collinear or $M N P Q$ is a parallelogram.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "3", "problem_type": "Algebra", "exam": "IMO", "problem": "3. A3 (USS 4) ${ }^{\\mathrm{IMO} 3}$ Consider the infinite sequences $\\left\\{x_{n}\\right\\}$ of positive real numbers with the following properties:  $$ x_{0}=1 \\quad \\text { and for all } i \\geq 0, x_{i+1} \\leq x_{i} $$  (a) Prove that for every such sequence there is an $n \\geq 1$ such that $\\frac{x_{0}^{2}}{x_{1}}+$ $\\frac{x_{1}^{2}}{x_{2}}+\\cdots+\\frac{x_{n-1}^{2}}{x_{n}} \\geq 3.999$. (b) Find such a sequence for which $\\frac{x_{0}^{2}}{x_{1}}+\\frac{x_{1}^{2}}{x_{2}}+\\cdots+\\frac{x_{n-1}^{2}}{x_{n}}<4$ for all $n$.", "solution": "3. (a) By the Cauchy-Schwarz inequality we have $\\left(x_{0}^{2} / x_{1}+\\cdots+x_{n-1}^{2} / x_{n}\\right)$. $\\left(x_{1}+\\cdots+x_{n}\\right) \\geq\\left(x_{0}+\\cdots+x_{n-1}\\right)^{2}$. Let us set $X_{n-1}=x_{1}+x_{2}+$ $\\cdots+x_{n-1}$. Using $x_{0}=1$, the last inequality can be rewritten as  $$ \\frac{x_{0}^{2}}{x_{1}}+\\cdots+\\frac{x_{n-1}^{2}}{x_{n}} \\geq \\frac{\\left(1+X_{n-1}\\right)^{2}}{X_{n-1}+x_{n}} \\geq \\frac{4 X_{n-1}}{X_{n-1}+x_{n}}=\\frac{4}{1+x_{n} / X_{n-1}} $$  Since $x_{n} \\leq x_{n-1} \\leq \\cdots \\leq x_{1}$, it follows that $X_{n-1} \\geq(n-1) x_{n}$. Now (1) yields $x_{0}^{2} / x_{1}+\\cdots+x_{n-1}^{2} / x_{n} \\geq 4(n-1) / n$, which exceeds 3.999 for $n>4000$. (b) The sequence $x_{n}=1 / 2^{n}$ obviously satisfies the required condition.  Second solution to part (a). For each $n \\in \\mathbb{N}$, let us find a constant $c_{n}$ such that the inequality $x_{0}^{2} / x_{1}+\\cdots+x_{n-1}^{2} / x_{n} \\geq c_{n} x_{0}$ holds for any sequence $x_{0} \\geq x_{1} \\geq \\cdots \\geq x_{n}>0$. For $n=1$ we can take $c_{1}=1$. Assuming that $c_{n}$ exists, we have  $$ \\frac{x_{0}^{2}}{x_{1}}+\\left(\\frac{x_{1}^{2}}{x_{2}}+\\cdots+\\frac{x_{n}^{2}}{x_{n+1}}\\right) \\geq \\frac{x_{0}^{2}}{x_{1}}+c_{n} x_{1} \\geq 2 \\sqrt{x_{0}^{2} c_{n}}=x_{0} \\cdot 2 \\sqrt{c_{n}} $$  Thus we can take $c_{n+1}=2 \\sqrt{c_{n}}$. Then inductively $c_{n}=2^{2-1 / 2^{n-2}}$, and since $c_{n} \\rightarrow 4$ as $n \\rightarrow \\infty$, the result follows. Third solution. Since $\\left\\{x_{n}\\right\\}$ is decreasing, there exists $\\lim _{n \\rightarrow \\infty} x_{n}=x \\geq 0$. If $x>0$, then $x_{n-1}^{2} / x_{n} \\geq x_{n} \\geq x$ holds for each $n$, and the result is trivial.  If otherwise $x=0$, then we note that $x_{n-1}^{2} / x_{n} \\geq 4\\left(x_{n-1}-x_{n}\\right)$ for each $n$, with equality if and only if $x_{n-1}=2 x_{n}$. Hence  $$ \\lim _{n \\rightarrow \\infty} \\sum_{k=1}^{n} \\frac{x_{k-1}^{2}}{x_{k}} \\geq \\lim _{n \\rightarrow \\infty} \\sum_{k=1}^{n} 4\\left(x_{k-1}-x_{k}\\right)=4 x_{0}=4 $$  Equality holds if and only if $x_{n-1}=2 x_{n}$ for all $n$, and consequently $x_{n}=1 / 2^{n}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "4", "problem_type": "Algebra", "exam": "IMO", "problem": "4. A4 (BUL 2) Determine all real values of the parameter $a$ for which the equation  $$ 16 x^{4}-a x^{3}+(2 a+17) x^{2}-a x+16=0 $$  has exactly four distinct real roots that form a geometric progression.", "solution": "4. Suppose that $a$ satisfies the requirements of the problem and that $x, q x$, $q^{2} x, q^{3} x$ are the roots of the given equation. Then $x \\neq 0$ and we may assume that $|q|>1$, so that $|x|<|q x|<\\left|q^{2} x\\right|<\\left|q^{3} x\\right|$. Since the equation is symmetric, $1 / x$ is also a root and therefore $1 / x=q^{3} x$, i.e., $q=x^{-2 / 3}$. It follows that the roots are $x, x^{1 / 3}, x^{-1 / 3}, x^{-1}$. Now by Vieta's formula we have $x+x^{1 / 3}+x^{-1 / 3}+x^{-1}=a / 16$ and $x^{4 / 3}+x^{2 / 3}+2+x^{-2 / 3}+x^{-4 / 3}=$ $(2 a+17) / 16$. On setting $z=x^{1 / 3}+x^{-1 / 3}$ these equations become  $$ \\begin{aligned} z^{3}-2 z & =a / 16 \\\\ \\left(z^{2}-2\\right)^{2}+z^{2}-2 & =(2 a+17) / 16 \\end{aligned} $$  Substituting $a=16\\left(z^{3}-2 z\\right)$ in the second equation leads to $z^{4}-2 z^{3}-$ $3 z^{2}+4 z+15 / 16=0$. We observe that this polynomial factors as $(z+$ $3 / 2)(z-5 / 2)\\left(z^{2}-z-1 / 4\\right)$. Since $|z|=\\left|x^{1 / 3}+x^{-1 / 3}\\right| \\geq 2$, the only viable value is $z=5 / 2$. Consequently $a=170$ and the roots are $1 / 8,1 / 2,2,8$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "5", "problem_type": "Algebra", "exam": "IMO", "problem": "5. A5 (NET 2) ${ }^{\\mathrm{IMO}}$ Let $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ be a regular hexagon. Each of its diagonals $A_{i-1} A_{i+1}$ is divided into the same ratio $\\frac{\\lambda}{1-\\lambda}$, where $0<\\lambda<1$, by a point $B_{i}$ in such a way that $A_{i}, B_{i}$, and $B_{i+2}$ are collinear ( $i \\equiv$ $1, \\ldots, 6(\\bmod 6))$. Compute $\\lambda$.", "solution": "5. We first observe that $\\triangle A_{5} B_{4} A_{4} \\cong$ $\\triangle A_{3} B_{2} A_{2}$. Since $\\angle A_{5} A_{3} A_{2}=90^{\\circ}$, we have $\\angle A_{2} B_{4} A_{4}=\\angle A_{2} B_{4} A_{3}+$ $\\angle A_{3} B_{4} A_{4}=\\left(90^{\\circ}-\\angle B_{2} A_{2} A_{3}\\right)+$ $\\left(\\angle B_{4} A_{5} A_{4}+\\angle A_{5} A_{4} B_{4}\\right)=90^{\\circ}+$ $\\angle B_{4} A_{5} A_{4}=120^{\\circ}$. Hence $B_{4}$ belongs to the circle with center $A_{3}$ and radius $A_{3} A_{4}$, so $A_{3} A_{4}=A_{3} B_{4}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-461.jpg?height=311&width=428&top_left_y=1245&top_left_x=866)  Thus $\\lambda=A_{3} B_{4} / A_{3} A_{5}=A_{3} A_{4} / A_{3} A_{5}=1 / \\sqrt{3}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "6", "problem_type": "Algebra", "exam": "IMO", "problem": "6. A6 (VIE 1) ${ }^{\\text {IMO6 }}$ Let $S$ be a square with sides of length 100 and let $L$ be a path within $S$ that does not meet itself and that is composed of linear segments $A_{0} A_{1}, A_{1} A_{2}, \\ldots, A_{n-1} A_{n}$ with $A_{0} \\neq A_{n}$. Suppose that for every point $P$ of the boundary of $S$ there is a point of $L$ at a distance from $P$ not greater than $\\frac{1}{2}$. Prove that there are two points $X$ and $Y$ in $L$ such that the distance between $X$ and $Y$ is not greater than 1 and the length of that part of $L$ that lies between $X$ and $Y$ is not smaller than 198.", "solution": "6. Denote by $d(U, V)$ the distance between points or sets of points $U$ and $V$. For $P, Q \\in L$ we shall denote by $L_{P Q}$ the part of $L$ between points $P$ and $Q$ and by $l_{P Q}$ the length of this part. Let us denote by $S_{i}(i=1,2,3,4)$ the vertices of $S$ and by $T_{i}$ points of $L$ such that $S_{i} T_{i} \\leq 1 / 2$ in such a way that $l_{A_{0} T_{1}}$ is the least of the $l_{A_{0} T_{i}}$ 's, $S_{2}$ and $S_{4}$ are neighbors of $S_{1}$, and $l_{A_{0} T_{2}}<l_{A_{0} T_{4}}$. Now we shall consider the points of the segment $S_{1} S_{4}$. Let $D$ and $E$ be the sets of points defined as follows: $D=\\left\\{X \\in\\left[S_{1} S_{4}\\right] \\mid d\\left(X, L_{A_{0} T_{2}}\\right) \\leq 1 / 2\\right\\}$ and $E=\\left\\{X \\in\\left[S_{1} S_{4}\\right] \\mid d\\left(X, L_{T_{2} A_{n}}\\right) \\leq 1 / 2\\right\\}$. Clearly $D$ and $E$ are closed, nonempty (indeed, $S_{1} \\in D$ and $S_{4} \\in E$ ) subsets of [ $\\left.S_{1} S_{4}\\right]$. Since their union is a connected set $S_{1} S_{4}$, it follows that they must have a nonempty intersection. Let $P \\in D \\cap E$. Then there exist points $X \\in L_{A_{0} T_{2}}$ and $Y \\in L_{T_{2} A_{n}}$ such that $d(P, X) \\leq 1 / 2, d(P, Y) \\leq 1 / 2$, and consequently $d(X, Y) \\leq 1$. On the other hand, $T_{2}$ lies between $X$ and $Y$ on $L$, and thus $L_{X Y}=L_{X T_{2}}+L_{T_{2} Y} \\geq X T_{2}+T_{2} Y \\geq\\left(P S_{2}-X P-S_{2} T_{2}\\right)+\\left(P S_{2}-Y P-\\right.$ $\\left.S_{2} T_{2}\\right) \\geq 99+99=198$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. B1 (CAN 2) Let $p(x)$ be a cubic polynomial with integer coefficients with leading coefficient 1 and with one of its roots equal to the product of the other two. Show that $2 p(-1)$ is a multiple of $p(1)+p(-1)-2(1+p(0))$.", "solution": "7. Let $a, b, a b$ be the roots of the cubic polynomial $P(x)=(x-a)(x-b)(x-$ $a b)$. Observe that  $$ \\begin{aligned} 2 p(-1) & =-2(1+a)(1+b)(1+a b) \\\\ p(1)+p(-1)-2(1+p(0)) & =-2(1+a)(1+b) \\end{aligned} $$  The statement of the problem is trivial if both the expressions are equal to zero. Otherwise, the quotient $\\frac{2 p(-1)}{p(1)+p(-1)-2(1+p(0))}=1+a b$ is rational and consequently $a b$ is rational. But since $(a b)^{2}=-P(0)$ is an integer, it follows that $a b$ is also an integer. This completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. B2 (POL 4) A convex, closed figure lies inside a given circle. The figure is seen from every point of the circumference at a right angle (that is, the two rays drawn from the point and supporting the convex figure are perpendicular). Prove that the center of the circle is a center of symmetry of the figure.", "solution": "8. Let $\\mathcal{F}$ be the given figure. Consider any chord $A B$ of the circumcircle $\\gamma$ that supports $\\mathcal{F}$. The other supporting lines to $\\mathcal{F}$ from $A$ and $B$ intersect $\\gamma$ again at $D$ and $C$ respectively so that $\\angle D A B=\\angle A B C=90^{\\circ}$. Then $A B C D$ is a rectangle, and hence $C D$ must support $\\mathcal{F}$ as well, from which it follows that $\\mathcal{F}$ is inscribed in the rectangle $A B C D$ touching each of its sides. We easily conclude that $\\mathcal{F}$ is the intersection of all such rectangles. Now, since the center $O$ of $\\gamma$ is the center of symmetry of all these rectangles, it must be so for their intersection $\\mathcal{F}$ as well.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1982", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. B3 (GBR 1) Let $A B C$ be a triangle, and let $P$ be a point inside it such that $\\measuredangle P A C=\\measuredangle P B C$. The perpendiculars from $P$ to $B C$ and $C A$ meet these lines at $L$ and $M$, respectively, and $D$ is the midpoint of $A B$. Prove that $D L=D M$.", "solution": "9. Let $X$ and $Y$ be the midpoints of the segments $A P$ and $B P$. Then $D Y P X$ is a parallelogram. Since $X$ and $Y$ are the circumcenters of $\\triangle A P M$ and $\\triangle B P L$, it follows that $X M=$ $X P=D Y$ and $Y L=Y P=D X$. Furthermore, $\\angle D X M=\\angle D X P+$ $\\angle P X M=\\angle D X P+2 \\angle P A M=$ $\\angle D Y P+2 \\angle P B L=\\angle D Y P+$ $\\angle P Y L=\\angle D Y L$. Therefore, the triangles $D X M$ and $L Y D$ are con- ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-462.jpg?height=379&width=451&top_left_y=1247&top_left_x=863) gruent, implying $D M=D L$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (AUS 1) The localities $P_{1}, P_{2}, \\ldots, P_{1983}$ are served by ten international airlines $A_{1}, A_{2}, \\ldots, A_{10}$. It is noticed that there is direct service (without stops) between any two of these localities and that all airline schedules offer round-trip flights. Prove that at least one of the airlines can offer a round trip with an odd number of landings.", "solution": "1. Suppose that there are $n$ airlines $A_{1}, \\ldots, A_{n}$ and $N>2^{n}$ cities. We shall prove that there is a round trip by at least one $A_{i}$ containing an odd number of stops. For $n=1$ the statement is trivial, since one airline serves at least 3 cities and hence $P_{1} P_{2} P_{3} P_{1}$ is a round trip with 3 landings. We use induction on $n$, and assume that $n>1$. Suppose the contrary, that all round trips by $A_{n}$ consist of an even number of stops. Then we can separate the cities into two nonempty classes $Q=\\left\\{Q_{1}, \\ldots, Q_{r}\\right\\}$ and $R=\\left\\{R_{1}, \\ldots, R_{s}\\right\\}$ (where $r+s=N$ ), so that each flight by $A_{n}$ runs between a $Q$-city and an $R$-city. (Indeed, take any city $Q_{1}$ served by $A_{n}$; include each city linked to $Q_{1}$ by $A_{n}$ in $R$, then include in $Q$ each city linked by $A_{n}$ to any $R$-city, etc. Since all round trips are even, no contradiction can arise.) At least one of $r, s$ is larger than $2^{n-1}$, say $r>2^{n-1}$. But, only $A_{1}, \\ldots, A_{n-1}$ run between cities in $\\left\\{Q_{1}, \\ldots, Q_{r}\\right\\}$; hence by the induction hypothesis at least one of them flies a round trip with an odd number of landings, a contradiction. It only remains to notice that for $n=10,2^{n}=1024<1983$. Remark. If there are $N=2^{n}$ cities, there is a schedule with $n$ airlines that contain no odd round trip by any of the airlines. Let the cities be $P_{k}$, $k=0, \\ldots, 2^{n}-1$, and write $k$ in the binary system as an $n$-digit number $\\overline{a_{1} \\ldots a_{n}}$ (e.g., $1=(0 \\ldots 001)_{2}$ ). Link $P_{k}$ and $P_{l}$ by $A_{i}$ if the $i$ th digits $k$ and $l$ are distinct but the first $i-1$ digits are the same. All round trips under $A_{i}$ are even, since the $i$ th digit alternates.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (FIN 1) Let $p$ and $q$ be integers. Show that there exists an interval $I$ of length $1 / q$ and a polynomial $P$ with integral coefficients such that  $$ \\left|P(x)-\\frac{p}{q}\\right|<\\frac{1}{q^{2}} $$  for all $x \\in I$.", "solution": "10. Choose $P(x)=\\frac{p}{q}\\left((q x-1)^{2 n+1}+1\\right), I=[1 / 2 q, 3 / 2 q]$. Then all the coefficients of $P$ are integers, and  $$ \\left|P(x)-\\frac{p}{q}\\right|=\\left|\\frac{p}{q}(q x-1)^{2 n+1}\\right| \\leq\\left|\\frac{p}{q}\\right| \\frac{1}{2^{2 n+1}}, $$  for $x \\in I$. The desired inequality follows if $n$ is chosen large enough.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (FIN $\\mathbf{2}^{\\prime}$ ) Let $f:[0,1] \\rightarrow \\mathbb{R}$ be continuous and satisfy:  $$ \\begin{aligned} b f(2 x) & =f(x), & & 0 \\leq x \\leq 1 / 2 \\\\ f(x) & =b+(1-b) f(2 x-1), & & 1 / 2 \\leq x \\leq 1 \\end{aligned} $$  where $b=\\frac{1+c}{2+c}, c>0$. Show that $0<f(x)-x<c$ for every $x, 0<x<1$.", "solution": "11. First suppose that the binary representation of $x$ is finite: $x=0, a_{1} a_{2} \\ldots a_{n}$ $=\\sum_{j=1}^{n} a_{j} 2^{-j}, a_{i} \\in\\{0,1\\}$. We shall prove by induction on $n$ that  $$ f(x)=\\sum_{j=1}^{n} b_{0} \\ldots b_{j-1} a_{j}, \\quad \\text { where } b_{k}=\\left\\{\\begin{array}{lr} -b & \\text { if } a_{k}=0 \\\\ 1-b & \\text { if } a_{k}=1 \\end{array}\\right. $$  (Here $a_{0}=0$.) Indeed, by the recursion formula, $a_{1}=0 \\Rightarrow f(x)=b f\\left(\\sum_{j=1}^{n-1} a_{j+1} 2^{-j}\\right)=b \\sum_{j=1}^{n-1} b_{1} \\ldots b_{j} a_{j+1}$ hence $f(x)=$ $\\sum_{j=0}^{n-1} b_{0} \\ldots b_{j} a_{j+1}$ as $b_{0}=b_{1}=b ;$ $a_{1}=1 \\Rightarrow f(x)=b+(1-b) f\\left(\\sum_{j=1}^{n-1} a_{j+1} 2^{-j}\\right)=\\sum_{j=0}^{n-1} b_{0} \\ldots b_{j} a_{j+1}$, as $b_{0}=b, b_{1}=1-b$. Clearly, $f(0)=0, f(1)=1, f(1 / 2)=b>1 / 2$. Assume $x=\\sum_{j=0}^{n} a_{j} 2^{-j}$, and for $k \\geq 2, v=x+2^{-n-k+1}, u=x+2^{-n-k}=(v+x) / 2$. Then $f(v)=$ $f(x)+b_{0} \\ldots b_{n} b^{k-2}$ and $f(u)=f(x)+b_{0} \\ldots b_{n} b^{k-1}>(f(v)+f(x)) / 2$. This means that the point $(u, f(u))$ lies above the line joining $(x, f(x))$ and $(v, f(v))$. By induction, every $(x, f(x))$, where $x$ has a finite binary expansion, lies above the line joining $(0,0)$ and $(1 / 2, b)$ if $0<x<1 / 2$, or above the line joining $(1 / 2, b)$ and $(1,1)$ if $1 / 2<x<1$. It follows immediately that $f(x)>x$. For the second inequality, observe that  $$ \\begin{aligned} f(x)-x & =\\sum_{j=1}^{\\infty}\\left(b_{0} \\ldots b_{j-1}-2^{-j}\\right) a_{j} \\\\ & <\\sum_{j=1}^{\\infty}\\left(b^{j}-2^{-j}\\right) a_{j}<\\sum_{j=1}^{\\infty}\\left(b^{j}-2^{-j}\\right)=\\frac{b}{1-b}-1=c . \\end{aligned} $$  By continuity, these inequalities also hold for $x$ with infinite binary representations.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (GBR 4) $)^{\\mathrm{IMO} 1}$ Find all functions $f$ defined on the positive real numbers and taking positive real values that satisfy the following conditions: (i) $f(x f(y))=y f(x)$ for all positive real $x, y$. (ii) $f(x) \\rightarrow 0$ as $x \\rightarrow+\\infty$.", "solution": "12. Putting $y=x$ in (1) we see that there exist positive real numbers $z$ such that $f(z)=z$ (this is true for every $z=x f(x)$ ). Let $a$ be any of them.  Then $f\\left(a^{2}\\right)=f(a f(a))=a f(a)=a^{2}$, and by induction, $f\\left(a^{n}\\right)=a^{n}$. If $a>1$, then $a^{n} \\rightarrow+\\infty$ as $n \\rightarrow \\infty$, and we have a contradiction with (2). Again, $a=f(a)=f(1 \\cdot a)=a f(1)$, so $f(1)=1$. Then, $a f\\left(a^{-1}\\right)=$ $f\\left(a^{-1} f(a)\\right)=f(1)=1$, and by induction, $f\\left(a^{-n}\\right)=a^{-n}$. This shows that $a \\nless 1$. Hence, $a=1$. It follows that $x f(x)=1$, i.e., $f(x)=1 / x$ for all $x$. This function satisfies (1) and (2), so $f(x)=1 / x$ is the unique solution.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (LUX 2) Let $E$ be the set of $1983^{3}$ points of the space $\\mathbb{R}^{3}$ all three of whose coordinates are integers between 0 and 1982 (including 0 and 1982). A coloring of $E$ is a map from $E$ to the set $\\{$ red, blue $\\}$. How many colorings of $E$ are there satisfying the following property: The number of red vertices among the 8 vertices of any right-angled parallelepiped is a multiple of 4 ?", "solution": "13. Given any coloring of the $3 \\times 1983-2$ points of the axes, we prove that there is a unique coloring of $E$ having the given property and extending this coloring. The first thing to notice is that given any rectangle $R_{1}$ parallel to a coordinate plane and whose edges are parallel to the axes, there is an even number $r_{1}$ of red vertices on $R_{1}$. Indeed, let $R_{2}$ and $R_{3}$ be two other rectangles that are translated from $R_{1}$ orthogonally to $R_{1}$ and let $r_{2}, r_{3}$ be the numbers of red vertices on $R_{2}$ and $R_{3}$ respectively. Then $r_{1}+r_{2}$, $r_{1}+r_{3}$, and $r_{2}+r_{3}$ are multiples of 4 , so $r_{1}=\\left(r_{1}+r_{2}+r_{1}+r_{3}-r_{2}-r_{3}\\right) / 2$ is even. Since any point of a coordinate plane is a vertex of a rectangle whose remaining three vertices lie on the corresponding axes, this determines uniquely the coloring of the coordinate planes. Similarly, the coloring of the inner points of the parallelepiped is completely determined. The solution is hence $2^{3 \\times 1983-2}=2^{5947}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (POL 2) ${ }^{\\mathrm{IMO} 5}$ Prove or disprove: From the interval $[1, \\ldots, 30000]$ one can select a set of 1000 integers containing no arithmetic triple (three consecutive numbers of an arithmetic progression).", "solution": "14. Let $T_{n}$ be the set of all nonnegative integers whose ternary representations consist of at most $n$ digits and do not contain a digit 2 . The cardinality of $T_{n}$ is $2^{n}$, and the greatest integer in $T_{n}$ is $11 \\ldots 1=3^{0}+3^{1}+\\cdots+3^{n-1}=$ $\\left(3^{n}-1\\right) / 2$. We claim that there is no arithmetic triple in $T_{n}$. To see this, suppose $x, y, z \\in T_{n}$ and $2 y=x+z$. Then $2 y$ has only 0 's and 2's in its ternary representation, and a number of this form can be the sum of two integers $x, z \\in T_{n}$ in only one way, namely $x=z=y$. But $\\left|T_{10}\\right|=2^{10}=1024$ and $\\max T_{10}=\\left(3^{10}-1\\right) / 2=29524<30000$. Thus the answer is yes.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (POL 3) Decide whether there exists a set $M$ of natural numbers satisfying the following conditions: (i) For any natural number $m>1$ there are $a, b \\in M$ such that $a+b=m$. (ii) If $a, b, c, d \\in M, a, b, c, d>10$ and $a+b=c+d$, then $a=c$ or $a=d$.", "solution": "15. There is no such set. Suppose $M$ satisfies $(a)$ and $(b)$ and let $q_{n}=$ $|\\{a \\in M: a \\leq n\\}|$. Consider the differences $b-a$, where $a, b \\in M$ and $10<a<b \\leq k$. They are all positive and less than $k$, and (b) implies that they are $\\left(\\frac{q_{k}}{-} \\frac{q_{10}}{2}\\right)$ different integers. Hence $\\binom{q_{k}-q_{10}}{2}<k$, so $q_{k} \\leq \\sqrt{2 k}+10$. It follows from (a) that among the numbers of the form $a+b$, where $a, b \\in M, a \\leq b \\leq n$, or $a \\leq n<b \\leq 2 n$, there are all integers from the interval $[2,2 n+1]$. Thus $\\binom{q_{n}+1}{2}+q_{n}\\left(q_{2 n}-q_{n}\\right) \\geq 2 n$ for every $n \\in \\mathbb{N}$. Set $Q_{k}=\\sqrt{2 k}+10$. We have  $$ \\begin{aligned} \\binom{q_{n}+1}{2}+q_{n}\\left(q_{2 n}-q_{n}\\right) & =\\frac{1}{2} q_{n}+\\frac{1}{2} q_{n}\\left(2 q_{2 n}-q_{n}\\right) \\\\ & \\leq \\frac{1}{2} q_{n}+\\frac{1}{2} q_{n}\\left(2 Q_{2 n}-q_{n}\\right) \\\\ & \\leq \\frac{1}{2} Q_{n}+\\frac{1}{2} Q_{n}\\left(2 Q_{2 n}-Q_{n}\\right) \\\\ & \\leq 2(\\sqrt{2}-1) n+(20+\\sqrt{2} / 2) \\sqrt{n}+55 \\end{aligned} $$  which is less than $n$ for $n$ large enough, a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (GDR 1) Let $F(n)$ be the set of polynomials $P(x)=a_{0}+a_{1} x+\\cdots+a_{n} x^{n}$, with $a_{0}, a_{1}, \\ldots, a_{n} \\in \\mathbb{R}$ and $0 \\leq a_{0}=a_{n} \\leq a_{1}=a_{n-1} \\leq \\cdots \\leq a_{[n / 2]}=$ $a_{[(n+1) / 2]}$. Prove that if $f \\in F(m)$ and $g \\in F(n)$, then $f g \\in F(m+n)$.", "solution": "16. Set $h_{n, i}(x)=x^{i}+\\cdots+x^{n-i}, 2 i \\leq n$. The set $F(n)$ is the set of linear combinations with nonnegative coefficients of the $h_{n, i}$ 's. This is a convex cone. Hence, it suffices to prove that $h_{n, i} h_{m, j} \\in F(m+n)$. Indeed, setting $p=n-2 i$ and $q=m-2 j$ and assuming $p \\leq q$ we obtain  $$ h_{n, i}(x) h_{m, j}(x)=\\left(x^{i}+\\cdots+x^{i+p}\\right)\\left(x^{j}+\\cdots+x^{j+q}\\right)=\\sum_{k=i+j}^{n-i+j} h_{m+n, k} $$  which proves the claim.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (GDR 3) Let $P_{1}, P_{2}, \\ldots, P_{n}$ be distinct points of the plane, $n \\geq 2$. Prove that  $$ \\max _{1 \\leq i<j \\leq n} P_{i} P_{j}>\\frac{\\sqrt{3}}{2}(\\sqrt{n}-1) \\min _{1 \\leq i<j \\leq n} P_{i} P_{j} $$", "solution": "17. Set $a=\\min P_{i} P_{j}, b=\\max P_{i} P_{j}$. We use the following lemma.  Lemma. There exists a disk of radius less than or equal to $b / \\sqrt{3}$ containing all the $P_{i}$ 's. Assuming that this is proved, the disks with center $P_{i}$ and radius $a / 2$ are disjoint and included in a disk of radius $b / \\sqrt{3}+a / 2$; hence comparing areas,  $$ n \\pi \\cdot \\frac{a^{2}}{4}<\\pi \\cdot\\left(\\frac{b}{\\sqrt{3}}+a / 2\\right)^{2} \\quad \\text { and } \\quad b>\\sqrt{3} / 2 \\cdot(\\sqrt{n}-1) a $$  Proof of the lemma. If a nonobtuse triangle with sides $a \\geq b \\geq c$ has a circumscribed circle of radius $R$, we have $R=a /(2 \\sin \\alpha) \\leq a / \\sqrt{3}$. Now we show that there exists a disk $D$ of radius $R$ containing $A=$ $\\left\\{P_{1}, \\ldots, P_{n}\\right\\}$ whose border $C$ is such that $C \\cap A$ is not included in an open semicircle, and hence contains either two diametrically opposite points and $R \\leq b / 2$, or an acute-angled triangle and $R \\leq b / \\sqrt{3}$. Among all disks whose borders pass through three points of $A$ and that contain all of $A$, let $D$ be the one of least radius. Suppose that $C \\cap A$ is contained in an arc of central angle less than $180^{\\circ}$, and that $P_{i}, P_{j}$ are its endpoints. Then there exists a circle through $P_{i}, P_{j}$ of smaller radius that contains $A$, a contradiction. Thus $D$ has the required property, and the assertion follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (FRG 3) ${ }^{\\mathrm{IMO} 3}$ Let $a, b, c$ be positive integers satisfying $(a, b)=(b, c)=$ $(c, a)=1$. Show that $2 a b c-a b-b c-c a$ is the largest integer not representable as  $$ x b c+y c a+z a b $$  with nonnegative integers $x, y, z$.", "solution": "18. Let $\\left(x_{0}, y_{0}, z_{0}\\right)$ be one solution of $b c x+c a y+a b z=n$ (not necessarily nonnegative). By subtracting $b c x_{0}+c a y_{0}+a b z_{0}=n$ we get  $$ b c\\left(x-x_{0}\\right)+c a\\left(y-y_{0}\\right)+a b\\left(z-z_{0}\\right)=0 . $$  Since $(a, b)=(a, c)=1$, we must have $a \\mid x-x_{0}$ or $x-x_{0}=a s$. Substituting this in the last equation gives  $$ b c s+c\\left(y-y_{0}\\right)+b\\left(z-z_{0}\\right)=0 $$  Since $(b, c)=1$, we have $b \\mid y-y_{0}$ or $y-y_{0}=b t$. If we substitute this in the last equation we get $b c s+b c t+b\\left(z-z_{0}\\right)=0$, or $c s+c t+z-z_{0}=0$, or $z-z_{0}=-c(s+t)$. In $x=x_{0}+a s$ and $y=y_{0}+b t$, we can choose $s$ and $t$ such that $0 \\leq x \\leq a-1$ and $0 \\leq y \\leq b-1$. If $n>2 a b c-b c-c a-a b$, then $a b z=n-b c x-a c y>2 a b c-a b-b c-c a-b c(a-1)-c a(b-1)=-a b$ or $z>-1$, i.e., $z \\geq 0$. Hence, it is representable as $b c x+c a y+a b z$ with $x, y, z \\geq 0$. Now we prove that $2 a b c-b c-c a-a b$ is not representable as $b c x+c a y+a b z$ with $x, y, z \\geq 0$. Suppose that $b c x+c a y+a b z=2 a b c-a b-b c-c a$ with $x, y, z \\geq 0$. Then  $$ b c(x+1)+c a(y+1)+a b(z+1)=2 a b c $$  with $x+1, y+1, z+1 \\geq 1$. Since $(a, b)=(a, c)=1$, we have $a \\mid x+1$ and thus $a \\leq x+1$. Similarly $b \\leq y+1$ and $c \\leq z+1$. Thus $b c a+c a b+a b c \\leq 2 a b c$, a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (ROM 1) Let $\\left(F_{n}\\right)_{n \\geq 1}$ be the Fibonacci sequence $F_{1}=F_{2}=1, F_{n+2}=$ $F_{n+1}+F_{n}(n \\geq 1)$, and $P(x)$ the polynomial of degree 990 satisfying  $$ P(k)=F_{k}, \\quad \\text { for } k=992, \\ldots, 1982 $$  Prove that $P(1983)=F_{1983}-1$.", "solution": "19. For all $n$, there exists a unique polynomial $P_{n}$ of degree $n$ such that $P_{n}(k)=F_{k}$ for $n+2 \\leq k \\leq 2 n+2$ and $P_{n}(2 n+3)=F_{2 n+3}-1$. For $n=0$, we have $F_{1}=F_{2}=1, F_{3}=2, P_{0}=1$. Now suppose that $P_{n-1}$ has been constructed and let $P_{n}$ be the polynomial of degree $n$ satisfying $P_{n}(X+2)-P_{n}(X+1)=P_{n-1}(X)$ and $P_{n}(n+2)=F_{n+2}$. (The mapping $\\mathbb{R}_{n}[X] \\rightarrow \\mathbb{R}_{n-1}[X] \\times \\mathbb{R}, P \\mapsto(Q, P(n+2)$ ), where $Q(X)=$ $P(X+2)-P(X+1)$, is bijective, since it is injective and those two spaces have the same dimension; clearly $\\operatorname{deg} Q=\\operatorname{deg} P-1$.) Thus for $n+2 \\leq k \\leq 2 n+2$ we have $P_{n}(k+1)=P_{n}(k)+F_{k-1}$ and $P_{n}(n+2)=F_{n+2}$; hence by induction on $k, P_{n}(k)=F_{k}$ for $n+2 \\leq k \\leq 2 n+2$ and  $$ P_{n}(2 n+3)=F_{2 n+2}+P_{n-1}(2 n+1)=F_{2 n+3}-1 $$  Finally, $P_{990}$ is exactly the polynomial $P$ of the terms of the problem, for $P_{990}-P$ has degree less than or equal to 990 and vanishes at the 991 points $k=992, \\ldots, 1982$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (BEL 1) Let $n$ be a positive integer. Let $\\sigma(n)$ be the sum of the natural divisors $d$ of $n$ (including 1 and $n$ ). We say that an integer $m \\geq 1$ is superabundant (P.Erdös, 1944) if $\\forall k \\in\\{1,2, \\ldots, m-1\\}, \\frac{\\sigma(m)}{m}>\\frac{\\sigma(k)}{k}$. Prove that there exists an infinity of superabundant numbers.", "solution": "2. By definition, $\\sigma(n)=\\sum_{d \\mid n} d=\\sum_{d \\mid n} n / d=n \\sum_{d \\mid n} 1 / d$, hence $\\sigma(n) / n=$ $\\sum_{d \\mid n} 1 / d$. In particular, $\\sigma(n!) / n!=\\sum_{d \\mid n!} 1 / d \\geq \\sum_{k=1}^{n} 1 / k$. It follows that the sequence $\\sigma(n) / n$ is unbounded, and consequently there exist an infinite number of integers $n$ such that $\\sigma(n) / n$ is strictly greater than $\\sigma(k) / k$ for $k<n$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. (ROM 3) Solve the system of equations  $$ \\begin{gathered} x_{1}\\left|x_{1}\\right|=x_{2}\\left|x_{2}\\right|+\\left(x_{1}-a\\right)\\left|x_{1}-a\\right| \\\\ x_{2}\\left|x_{2}\\right|=x_{3}\\left|x_{3}\\right|+\\left(x_{2}-a\\right)\\left|x_{2}-a\\right| \\\\ \\quad \\cdots \\\\ x_{n}\\left|x_{n}\\right|=x_{1}\\left|x_{1}\\right|+\\left(x_{n}-a\\right)\\left|x_{n}-a\\right| \\end{gathered} $$  in the set of real numbers, where $a>0$.", "solution": "20. If $\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)$ satisfies the system with parameter $a$, then $\\left(-x_{1},-x_{2}\\right.$, $\\ldots,-x_{n}$ ) satisfies the system with parameter $-a$. Hence it is sufficient to consider only $a \\geq 0$. Let $\\left(x_{1}, \\ldots, x_{n}\\right)$ be a solution. Suppose $x_{1} \\leq a, x_{2} \\leq a, \\ldots, x_{n} \\leq a$. Summing the equations we get  $$ \\left(x_{1}-a\\right)^{2}+\\cdots+\\left(x_{n}-a\\right)^{2}=0 $$  and see that $(a, a, \\ldots, a)$ is the only such solution. Now suppose that $x_{k} \\geq a$ for some $k$. According to the $k$ th equation,  $$ x_{k+1}\\left|x_{k+1}\\right|=x_{k}^{2}-\\left(x_{k}-a\\right)^{2}=a\\left(2 x_{k}-a\\right) \\geq a^{2} $$  which implies that $x_{k+1} \\geq a$ as well (here $x_{n+1}=x_{1}$ ). Consequently, all $x_{1}, x_{2}, \\ldots, x_{n}$ are greater than or equal to $a$, and as above $(a, a, \\ldots, a)$ is the only solution.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. (SWE 1) Find the greatest integer less than or equal to $\\sum_{k=1}^{2^{1983}} k^{1 / 1983-1}$.", "solution": "21. Using the identity  $$ a^{n}-b^{n}=(a-b) \\sum_{m=0}^{n-1} a^{n-m-1} b^{m} $$  with $a=k^{1 / n}$ and $b=(k-1)^{1 / n}$ one obtains  $$ 1<\\left(k^{1 / n}-(k-1)^{1 / n}\\right) n k^{1-1 / n} \\text { for all integers } n>1 \\text { and } k \\geq 1 $$  This gives us the inequality $k^{1 / n-1}<n\\left(k^{1 / n}-(k-1)^{1 / n}\\right)$ if $n>1$ and $k \\geq 1$. In a similar way one proves that $n\\left((k+1)^{1 / n}-k^{1 / n}\\right)<k^{1 / n-1}$ if $n>1$ and $k \\geq 1$. Hence for $n>1$ and $m>1$ it holds that  $$ \\begin{aligned} n \\sum_{k=1}^{m}\\left((k+1)^{1 / n}-k^{1 / n}\\right) & <\\sum_{k=1}^{m} k^{1 / n-1} \\\\ & <n \\sum_{k=2}^{m}\\left(k^{1 / n}-(k-1)^{1 / n}\\right)+1, \\end{aligned} $$  or equivalently,  $$ n\\left((m+1)^{1 / n}-1\\right)<\\sum_{k=1}^{m} k^{1 / n-1}<n\\left(m^{1 / n}-1\\right)+1 $$  The choice $n=1983$ and $m=2^{1983}$ then gives  $$ 1983<\\sum_{k=1}^{2^{1983}} k^{1 / 1983-1}<1984 $$  Therefore the greatest integer less than or equal to the given sum is 1983.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "22", "problem_type": null, "exam": "IMO", "problem": "22. (SWE 4) Let $n$ be a positive integer having at least two different prime factors. Show that there exists a permutation $a_{1}, a_{2}, \\ldots, a_{n}$ of the integers $1,2, \\ldots, n$ such that  $$ \\sum_{k=1}^{n} k \\cdot \\cos \\frac{2 \\pi a_{k}}{n}=0 $$", "solution": "22. Decompose $n$ into $n=s t$, where the greatest common divisor of $s$ and $t$ is 1 and where $s>1$ and $t>1$. For $1 \\leq k \\leq n$ put $k=v s+u$, where $0 \\leq v \\leq t-1$ and $1 \\leq u \\leq s$, and let $a_{k}=a_{v s+u}$ be the unique integer in the set $\\{1,2,3, \\ldots, n\\}$ such that $v s+u t-a_{v s+u}$ is a multiple of $n$. To prove that this construction gives a permutation, assume that $a_{k_{1}}=a_{k_{2}}$, where $k_{i}=v_{i} s+u_{i}, i=1,2$. Then $\\left(v_{1}-v_{2}\\right) s+\\left(u_{1}-u_{2}\\right) t$ is a multiple of $n=s t$. It follows that $t$ divides $\\left(v_{1}-v_{2}\\right)$, while $\\left|v_{1}-v_{2}\\right| \\leq t-1$, and that $s$ divides $\\left(u_{1}-u_{2}\\right)$, while $\\left|u_{1}-u_{2}\\right| \\leq s-1$. Hence, $v_{1}=v_{2}, u_{1}=u_{2}$, and $k_{1}=k_{2}$. We have proved that $a_{1}, \\ldots, a_{n}$ is a permutation of $\\{1,2, \\ldots, n\\}$ and hence  $$ \\sum_{k=1}^{n} k \\cos \\frac{2 \\pi a_{k}}{n}=\\sum_{v=0}^{t-1}\\left(\\sum_{u=1}^{s}(v s+u) \\cos \\left(\\frac{2 \\pi v}{t}+\\frac{2 \\pi u}{s}\\right)\\right) $$  Using $\\sum_{u=1}^{s} \\cos (2 \\pi u / s)=\\sum_{u=1}^{s} \\sin (2 \\pi u / s)=0$ and the additive formulas for cosine, one finds that  $$ \\begin{aligned} \\sum_{k=1}^{n} k \\cos \\frac{2 \\pi a_{k}}{n}= & \\sum_{v=0}^{t-1}\\left(\\cos \\frac{2 \\pi v}{t} \\sum_{u=1}^{s} u \\cos \\frac{2 \\pi u}{s}-\\sin \\frac{2 \\pi v}{t} \\sum_{u=1}^{s} u \\sin \\frac{2 \\pi u}{s}\\right) \\\\ = & \\left(\\sum_{u=1}^{s} u \\cos \\frac{2 \\pi u}{s}\\right)\\left(\\sum_{v=0}^{t-1} \\cos \\frac{2 \\pi v}{t}\\right) \\\\ & -\\left(\\sum_{u=1}^{s} u \\sin \\frac{2 \\pi u}{s}\\right)\\left(\\sum_{v=0}^{t-1} \\sin \\frac{2 \\pi v}{t}\\right)=0 \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "23", "problem_type": null, "exam": "IMO", "problem": "23. (USS 1) ${ }^{\\mathrm{IMO} 2}$ Let $K$ be one of the two intersection points of the circles $W_{1}$ and $W_{2}$. Let $O_{1}$ and $O_{2}$ be the centers of $W_{1}$ and $W_{2}$. The two common tangents to the circles meet $W_{1}$ and $W_{2}$ respectively in $P_{1}$ and $P_{2}$, the first tangent, and $Q_{1}$ and $Q_{2}$, the second tangent. Let $M_{1}$ and $M_{2}$ be the midpoints of $P_{1} Q_{1}$ and $P_{2} Q_{2}$, respectively. Prove that  $$ \\angle O_{1} K O_{2}=\\angle M_{1} K M_{2} $$", "solution": "23. We note that $\\angle O_{1} K O_{2}=\\angle M_{1} K M_{2}$ is equivalent to $\\angle O_{1} K M_{1}=$ $\\angle O_{2} K M_{2}$. Let $S$ be the intersection point of the common tangents, and let $L$ be the second point of intersection of $S K$ and $W_{1}$. Since $\\triangle S O_{1} P_{1} \\sim \\triangle S P_{1} M_{1}$, we have $S K$. $S L=S P_{1}^{2}=S O_{1} \\cdot S M_{1}$ which implies that points $O_{1}, L, K, M_{1}$ lie on a circle. Hence $\\angle O_{1} K M_{1}=$ $\\angle O_{1} L M_{1}=\\angle O_{2} K M_{2}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-474.jpg?height=295&width=551&top_left_y=959&top_left_x=809)", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "24", "problem_type": null, "exam": "IMO", "problem": "24. (USS 2) Let $d_{n}$ be the last nonzero digit of the decimal representation of $n$ !. Prove that $d_{n}$ is aperiodic; that is, there do not exist $T$ and $n_{0}$ such that for all $n \\geq n_{0}, d_{n+T}=d_{n}$.", "solution": "24. See the solution of (SL91-15).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "25", "problem_type": null, "exam": "IMO", "problem": "25. (USS 3) Prove that every partition of 3-dimensional space into three disjoint subsets has the following property: One of these subsets contains all possible distances; i.e., for every $a \\in \\mathbb{R}_{+}$, there are points $M$ and $N$ inside that subset such that distance between $M$ and $N$ is exactly $a$.", "solution": "25. Suppose the contrary, that $\\mathbb{R}^{3}=P_{1} \\cup P_{2} \\cup P_{3}$ is a partition such that $a_{1} \\in \\mathbb{R}^{+}$is not realized by $P_{1}, a_{2} \\in \\mathbb{R}^{+}$is not realized by $P_{2}$ and $a_{3} \\in \\mathbb{R}^{+}$ not realized by $P_{3}$, where w.l.o.g. $a_{1} \\geq a_{2} \\geq a_{3}$. If $P_{1}=\\emptyset=P_{2}$, then $P_{3}=\\mathbb{R}^{3}$, which is impossible. If $P_{1}=\\emptyset$, and $X \\in P_{2}$, the sphere centered at $X$ with radius $a_{2}$ is included in $P_{3}$ and $a_{3} \\leq a_{2}$ is realized, which is impossible. If $P_{1} \\neq \\emptyset$, let $X_{1} \\in P_{1}$. The sphere $S$ centered in $X_{1}$, of radius $a_{1}$ is included in $P_{2} \\cap P_{3}$. Since $a_{1} \\geq a_{3}, S \\not \\subset P_{3}$. Let $X_{2} \\in P_{2} \\cap S$. The circle $\\left\\{Y \\in S \\mid d\\left(X_{2}, Y\\right)=a_{2}\\right\\}$ is included in $P_{3}$, but $a_{2} \\leq a_{1}$; hence it has radius $r=a_{2} \\sqrt{1-a_{2}^{2} /\\left(4 a_{1}^{2}\\right)} \\geq a_{2} \\sqrt{3} / 2$ and $a_{3} \\leq a_{2} \\leq a_{2} \\sqrt{3}<2 r$; hence $a_{3}$ is realized by $P_{3}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (BEL 3) $)^{\\mathrm{IMO} 4}$ We say that a set $E$ of points of the Euclidian plane is \"Pythagorean\" if for any partition of $E$ into two sets $A$ and $B$, at least one of the sets contains the vertices of a right-angled triangle. Decide whether the following sets are Pythagorean: (a) a circle; (b) an equilateral triangle (that is, the set of three vertices and the points of the three edges).", "solution": "3. (a) A circle is not Pythagorean. Indeed, consider the partition into two semicircles each closed at one and open at the other end. (b) An equilateral triangle, call it $P Q R$, is Pythagorean. Let $P^{\\prime}, Q^{\\prime}$, and $R^{\\prime}$ be the points on $Q R, R P$, and $P Q$ such that $P R^{\\prime}: R^{\\prime} Q=Q P^{\\prime}$ : $P^{\\prime} R=R Q^{\\prime}: Q^{\\prime} P=1: 2$. Then $Q^{\\prime} R^{\\prime} \\perp P Q$, etc. Suppose that $P Q R$ is not Pythagorean, and consider a partition into $A, B$, neither of which contains the vertices of a right-angled triangle. At least two of $P^{\\prime}, Q^{\\prime}$, and $R^{\\prime}$ belong to the same class, say $P^{\\prime}, Q^{\\prime} \\in A$. Then $[P R] \\backslash\\left\\{Q^{\\prime}\\right\\} \\subset B$ and hence $R^{\\prime} \\in A$ (otherwise, if $R^{\\prime \\prime}$ is the foot of the perpendicular from $R^{\\prime}$ to $P R, \\triangle R R^{\\prime} R^{\\prime \\prime}$ is right-angled with all vertices in $B$ ). But this implies again that $[P Q] \\backslash\\left\\{R^{\\prime}\\right\\} \\subset B$, and thus $B$ contains vertices of a rectangular triangle. This is a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (BEL 5) On the sides of the triangle $A B C$, three similar isosceles triangles $A B P(A P=P B), A Q C(A Q=Q C)$, and $B R C(B R=R C)$ are constructed. The first two are constructed externally to the triangle $A B C$, but the third is placed in the same half-plane determined by the line $B C$ as the triangle $A B C$. Prove that $A P R Q$ is a parallelogram.", "solution": "4. The rotational homothety centered at $C$ that sends $B$ to $R$ also sends $A$ to $Q$; hence the triangles $A B C$ and $Q R C$ are similar. For the same reason, $\\triangle A B C$ and $\\triangle P B R$ are similar. Moreover, $B R=C R$; hence $\\triangle C R Q \\cong$ $\\triangle R B P$. Thus $P R=Q C=A Q$ and $Q R=P B=P A$, so $A P Q R$ is a parallelogram.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (BRA 1) Consider the set of all strictly decreasing sequences of $n$ natural numbers having the property that in each sequence no term divides any other term of the sequence. Let $A=\\left(a_{j}\\right)$ and $B=\\left(b_{j}\\right)$ be any two such sequences. We say that $A$ precedes $B$ if for some $k, a_{k}<b_{k}$ and $a_{i}=b_{i}$ for $i<k$. Find the terms of the first sequence of the set under this ordering.", "solution": "5. Each natural number $p$ can be written uniquely in the form $p=2^{q}(2 r-1)$. We call $2 r-1$ the odd part of $p$. Let $A_{n}=\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ be the first sequence. Clearly the terms of $A_{n}$ must have different odd parts, so those parts must be at least $1,3, \\ldots, 2 n-1$. Being the first sequence, $A_{n}$ must have the numbers $2 n-1,2 n-3, \\ldots, 2 k+1$ as terms, where $k=[n+1 / 3]$ (then $3(2 k-1)<2 n-1<3(2 k+1)$ ). Smaller odd numbers $2 s+1$ (with $s<k$ ) obviously cannot be terms of $A_{n}$. In this way we have obtained the $n-k$ odd numbers of $A_{n}$. The other $k$ terms must be even, and by the same reasoning as above they must be precisely the terms of $2 A_{k}$ (twice the terms of $A_{k}$ ). Therefore $A_{n}$ is defined recursively as  $$ \\begin{gathered} A_{0}=\\emptyset, \\quad A_{1}=\\{1\\}, \\quad A_{2}=\\{3,2\\} \\\\ A_{n}=\\{2 n-1,2 n-3, \\ldots, 2 k+1\\} \\cup 2 A_{k} . \\end{gathered} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (CAN 2) Suppose that $\\left\\{x_{1}, x_{2}, \\ldots, x_{n}\\right\\}$ are positive integers for which $x_{1}+x_{2}+\\cdots+x_{n}=2(n+1)$. Show that there exists an integer $r$ with $0 \\leq r \\leq n-1$ for which the following $n-1$ inequalities hold:  $$ \\begin{aligned} x_{r+1}+\\cdots+x_{r+i} & \\leq 2 i+1 & & \\forall i, 1 \\leq i \\leq n-r ; \\\\ x_{r+1}+\\cdots+x_{n}+x_{1}+\\cdots+x_{i} & \\leq 2(n-r+i)+1 & & \\forall i, 1 \\leq i \\leq r-1 . \\end{aligned} $$  Prove that if all the inequalities are strict, then $r$ is unique and that otherwise there are exactly two such $r$.", "solution": "6. The existence of $r$ : Let $S=\\left\\{x_{1}+x_{2}+\\cdots+x_{i}-2 i \\mid i=1,2, \\ldots, n\\right\\}$. Let $\\max S$ be attained for the first time at $r^{\\prime}$. If $r^{\\prime}=n$, then $x_{1}+x_{2}+\\cdots+x_{i}-2 i<2$ for $1 \\leq i \\leq n-1$, so one can take $r=r^{\\prime}$. Suppose that $r^{\\prime}<n$. Then for $l<n-r^{\\prime}$ we have $x_{r^{\\prime}+1}+x_{r^{\\prime}+2}+\\cdots+x_{r^{\\prime}+l}=$ $\\left(x_{1}+\\cdots+x_{r^{\\prime}+l}-2\\left(r^{\\prime}+l\\right)\\right)-\\left(x_{1}+\\cdots+x_{r^{\\prime}}-2 r^{\\prime}\\right)+2 l \\leq 2 l$; also, for $i<r^{\\prime}$ we have $\\left(x_{r^{\\prime}+1}+\\cdots+x_{n}\\right)+\\left(x_{1}+\\cdots+x_{i}-2 i\\right)<\\left(x_{r^{\\prime}+1}+\\cdots+\\right.$ $\\left.x_{n}\\right)+\\left(x_{1}+\\cdots+x_{r^{\\prime}}-2 r^{\\prime}\\right)=\\left(x_{1}+\\cdots+x_{n}\\right)-2 r^{\\prime}=2\\left(n-r^{\\prime}\\right)+2 \\Rightarrow$ $x_{r^{\\prime}+1}+\\cdots+x_{n}+x_{1}+\\cdots+x_{i} \\leq 2\\left(n+i-r^{\\prime}\\right)+1$, so we can again take $r=r^{\\prime}$ 。 For the second part of the problem, we relabel the sequence so that $r=0$ works. Suppose that the inequalities are strict. We have $x_{1}+x_{2}+\\cdots+x_{k} \\leq 2 k$, $k=1, \\ldots, n-1$. Now, $2 n+2=\\left(x_{1}+\\cdots+x_{k}\\right)+\\left(x_{k+1}+\\cdots+x_{n}\\right) \\leq$ $2 k+x_{k+1}+\\cdots+x_{n} \\Rightarrow x_{k+1}+\\cdots+x_{n} \\geq 2(n-k)+2>2(n-k)+1$. So we cannot begin with $x_{k+1}$ for any $k>0$. Now assume that there is an equality for some $k$. There are two cases: (i) Suppose $x_{1}+x_{2}+\\cdots+x_{i} \\leq 2 i(i=1, \\ldots, k)$ and $x_{1}+\\cdots+x_{k}=2 k+1$, $x_{1}+\\cdots+x_{k+l} \\leq 2(k+l)+1(1 \\leq l \\leq n-1-k)$. For $i \\leq k-1$ we have $x_{i+1}+\\cdots+x_{n}=2(n+1)-\\left(x_{1}+\\cdots+x_{i}\\right)>2(n-i)+1$, so we cannot take $r=i$. If there is a $j \\geq 1$ such that $x_{1}+x_{2}+\\cdots+x_{k+j} \\leq 2(k+j)$, then also $x_{k+j+1}+\\cdots+x_{n}>2(n-k-j)+1$. If $(\\forall j \\geq 1) x_{1}+\\cdots+x_{k+j}=$ $2(k+j)+1$, then $x_{n}=3$ and $x_{k+1}=\\cdots=x_{n-1}=2$. In this case we directly verify that we cannot take $r=k+j$. However, we can also take $r=k$ : for $k+l \\leq n-1, x_{k+1}+\\cdots+x_{k+l} \\leq 2(k+l)+1-(2 k+1)=2 l$, also $x_{k+1}+\\cdots+x_{n}=2(n-k)+1$, and moreover $x_{1} \\leq 2, x_{1}+x_{2} \\leq 4, \\ldots$. (ii) Suppose $x_{1}+\\cdots+x_{i} \\leq 2 i(1 \\leq i \\leq n-2)$ and $x_{1}+\\cdots+x_{n-1}=2 n-1$. Then we can obviously take $r=n-1$. On the other hand, for any $1 \\leq i \\leq n-2, x_{i+1}+\\cdots+x_{n-1}+x_{n}=\\left(x_{1}+\\cdots+x_{n-1}\\right)-\\left(x_{1}+\\cdots+\\right.$ $\\left.x_{i}\\right)+3>2(n-i)+1$, so we cannot take another $r \\neq 0$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (CAN 5) Let $a$ be a positive integer and let $\\left\\{a_{n}\\right\\}$ be defined by $a_{0}=0$ and  $$ a_{n+1}=\\left(a_{n}+1\\right) a+(a+1) a_{n}+2 \\sqrt{a(a+1) a_{n}\\left(a_{n}+1\\right)} \\quad(n=1,2 \\ldots) $$  Show that for each positive integer $n, a_{n}$ is a positive integer.", "solution": "7. Clearly, each $a_{n}$ is positive and $\\sqrt{a_{n+1}}=\\sqrt{a_{n}} \\sqrt{a+1}+\\sqrt{a_{n}+1} \\sqrt{a}$. Notice that $\\sqrt{a_{n+1}+1}=\\sqrt{a+1} \\sqrt{a_{n}+1}+\\sqrt{a} \\sqrt{a_{n}}$. Therefore  $$ \\begin{aligned} & (\\sqrt{a+1}-\\sqrt{a})\\left(\\sqrt{a_{n}+1}-\\sqrt{a_{n}}\\right) \\\\ & \\quad=\\left(\\sqrt{a+1} \\sqrt{a_{n}+1}+\\sqrt{a} \\sqrt{a_{n}}\\right)-\\left(\\sqrt{a_{n}} \\sqrt{a+1}+\\sqrt{a_{n}+1} \\sqrt{a}\\right) \\\\ & \\quad=\\sqrt{a_{n+1}+1}-\\sqrt{a_{n+1}} \\end{aligned} $$  By induction, $\\sqrt{a_{n+1}}-\\sqrt{a_{n}}=(\\sqrt{a+1}-\\sqrt{a})^{n}$. Similarly, $\\sqrt{a_{n+1}}+\\sqrt{a_{n}}=$ $(\\sqrt{a+1}+\\sqrt{a})^{n}$. Hence,  $$ \\sqrt{a_{n}}=\\frac{1}{2}\\left[(\\sqrt{a+1}+\\sqrt{a})^{n}-(\\sqrt{a+1}-\\sqrt{a})^{n}\\right] $$  from which the result follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (SPA 2) In a test, $3 n$ students participate, who are located in three rows of $n$ students in each. The students leave the test room one by one. If $N_{1}(t), N_{2}(t), N_{3}(t)$ denote the numbers of students in the first, second, and third row respectively at time $t$, find the probability that for each $t$ during the test,  $$ \\left|N_{i}(t)-N_{j}(t)\\right|<2, \\quad i \\neq j, \\quad i, j=1,2, \\ldots $$", "solution": "8. Situations in which the condition of the statement is fulfilled are the following: $S_{1}: N_{1}(t)=N_{2}(t)=N_{3}(t)$ $S_{2}: N_{i}(t)=N_{j}(t)=h, N_{k}(t)=h+1$, where $(i, j, k)$ is a permutation of the set $\\{1,2,3\\}$. In this case the first student to leave must be from row $k$. This leads to the situation $S_{1}$. $S_{3}: N_{i}(t)=h, N_{j}(t)=N_{k}(t)=h+1,((i, j, k)$ is a permutation of the set $\\{1,2,3\\})$. In this situation the first student leaving the room belongs to row $j$ (or $k$ ) and the second to row $k$ (or $j$ ). After this we arrive at the situation $S_{1}$. Hence, the initial situation is $S_{1}$ and after each triple of students leaving the room the situation $S_{1}$ must recur. We shall compute the probability $P_{h}$ that from a situation $S_{1}$ with $3 h$ students in the room $(h \\leq n)$ one arrives at a situation $S_{1}$ with $3(h-1)$ students in the room:  $$ P_{h}=\\frac{(3 h) \\cdot(2 h) \\cdot h}{(3 h) \\cdot(3 h-1) \\cdot(3 h-2)}=\\frac{3!h^{3}}{3 h(3 h-1)(3 h-2)} $$  Since the room becomes empty after the repetition of $n$ such processes, which are independent, we obtain for the probability sought  $$ P=\\prod_{h=1}^{n} P_{h}=\\frac{(3!)^{n}(n!)^{3}}{(3 n)!} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1983", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (USA 3) ${ }^{\\mathrm{IMO} 6}$ If $a, b$, and $c$ are sides of a triangle, prove that  $$ a^{2} b(a-b)+b^{2} c(b-c)+c^{2} a(c-a) \\geq 0 $$  Determine when there is equality.", "solution": "9. For any triangle of sides $a, b, c$ there exist 3 nonnegative numbers $x, y, z$ such that $a=y+z, b=z+x, c=x+y$ (these numbers correspond to the division of the sides of a triangle by the point of contact of the incircle). The inequality becomes  $$ (y+z)^{2}(z+x)(y-x)+(z+x)^{2}(x+y)(z-y)+(x+y)^{2}(y+z)(x-z) \\geq 0 $$  Expanding, we get $x y^{3}+y z^{3}+z x^{3} \\geq x y z(x+y+z)$. This follows from Cauchy's inequality $\\left(x y^{3}+y z^{3}+z x^{3}\\right)(z+x+y) \\geq(\\sqrt{x y z}(x+y+z))^{2}$ with equality if and only if $x y^{3} / z=y z^{3} / x=z x^{3} / y$, or equivalently $x=y=z$, i.e., $a=b=c$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (FRA 1) Find all solutions of the following system of $n$ equations in $n$ variables:  $$ \\begin{aligned} & x_{1}\\left|x_{1}\\right|-\\left(x_{1}-a\\right)\\left|x_{1}-a\\right|=x_{2}\\left|x_{2}\\right| \\\\ & x_{2}\\left|x_{2}\\right|-\\left(x_{2}-a\\right)\\left|x_{2}-a\\right|=x_{3}\\left|x_{3}\\right| \\\\ & \\cdots \\\\ & x_{n}\\left|x_{n}\\right|-\\left(x_{n}-a\\right)\\left|x_{n}-a\\right|=x_{1}\\left|x_{1}\\right| \\end{aligned} $$  where $a$ is a given number.", "solution": "1. This is the same problem as (SL83-20).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (GBR 1) Prove that the product of five consecutive positive integers cannot be the square of an integer.", "solution": "10. Suppose that the product of some five consecutive numbers is a square. It is easily seen that among them at least one, say $n$, is divisible neither by 2 nor 3 . Since $n$ is coprime to the remaining four numbers, it is itself a square of a number $m$ of the form $6 k \\pm 1$. Thus $n=(6 k \\pm 1)^{2}=24 r+1$, where $r=k(3 k \\pm 1) / 2$. Note that neither of the numbers $24 r-1,24 r+5$ is one of our five consecutive numbers because it is not a square. Hence the five numbers must be $24 r, 24 r+1, \\ldots, 24 r+4$. However, the number $24 r+4=(6 k \\pm 1)^{2}+3$ is divisible by $6 r+1$, which implies that it is a square as well. It follows that these two squares are 1 and 4 , which is impossible.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (CAN 1) Let $n$ be a natural number and $a_{1}, a_{2}, \\ldots, a_{2 n}$ mutually distinct integers. Find all integers $x$ satisfying  $$ \\left(x-a_{1}\\right) \\cdot\\left(x-a_{2}\\right) \\cdots\\left(x-a_{2 n}\\right)=(-1)^{n}(n!)^{2} $$", "solution": "11. Suppose that an integer $x$ satisfies the equation. Then the numbers $x-$ $a_{1}, x-a_{2}, \\ldots, x-a_{2 n}$ are $2 n$ distinct integers whose product is $1 \\cdot(-1)$. $2 \\cdot(-2) \\cdots n \\cdot(-n)$. From here it is obvious that the numbers $x-a_{1}, x-a_{2}, \\ldots, x-a_{2 n}$ are some reordering of the numbers $-n,-n+1, \\ldots,-1,1, \\ldots, n-1, n$. It follows that their sum is 0 , and therefore $x=\\left(a_{1}+a_{2}+\\cdots+a_{2 n}\\right) / 2 n$. This is the only solution if $\\left\\{a_{1}, a_{2}, \\ldots, a_{2 n}\\right\\}=\\{x-n, \\ldots, x-1, x+1, \\ldots, x+n\\}$ for some $x \\in \\mathbb{N}$. Otherwise there is no solution.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (NET 1) ${ }^{\\mathrm{IMO} 2}$ Find two positive integers $a, b$ such that none of the numbers $a, b, a+b$ is divisible by 7 and $(a+b)^{7}-a^{7}-b^{7}$ is divisible by $7^{7}$.", "solution": "12. By the binomial formula we have  $$ \\begin{aligned} (a+b)^{7}-a^{7}-b^{7} & =7 a b\\left[\\left(a^{5}+b^{5}\\right)+3 a b\\left(a^{3}+b^{3}\\right)+5 a^{2} b^{2}(a+b)\\right] \\\\ & =7 a b(a+b)\\left(a^{2}+a b+b^{2}\\right)^{2} \\end{aligned} $$  Thus it will be enough to find $a$ and $b$ such that $7 \\nmid a, b$ and $7^{3} \\mid a^{2}+a b+b^{2}$. Such numbers must satisfy $(a+b)^{2}>a^{2}+a b+b^{2} \\geq 7^{3}=343$, implying $a+b \\geq 19$. Trying $a=1$ we easily find the example $(a, b)=(1,18)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (BUL 5) Prove that the volume of a tetrahedron inscribed in a right circular cylinder of volume 1 does not exceed $\\frac{2}{3 \\pi}$.", "solution": "13. Let $Z$ be the given cylinder of radius $r$, altitude $h$, and volume $\\pi r^{2} h=1, k_{1}$ and $k_{2}$ the circles surrounding its bases, and $V$ the volume of an inscribed tetrahedron $A B C D$. We claim that there is no loss of generality in assuming that $A, B, C, D$ all lie on $k_{1} \\cup k_{2}$. Indeed, if the vertices $A, B, C$ are fixed and $D$ moves along a segment $E F$ parallel to the axis of the cylinder $\\left(E \\in k_{1}, F \\in k_{2}\\right)$, the maximum distance of $D$ from the plane $A B C$ (and consequently the maximum value of $V$ ) is achieved either at $E$ or at $F$. Hence we shall consider only the following two cases: (i) $A, B \\in k_{1}$ and $C, D \\in k_{2}$. Let $P, Q$ be the projections of $A, B$ on the plane of $k_{2}$, and $R, S$ the projections of $C, D$ on the plane of $k_{1}$, respectively. Then $V$ is one-third of the volume $V^{\\prime}$ of the prism $A R B S C P D Q$ with bases $A R B S$ and $C P D Q$. The area of the quadrilateral $A R B S$ inscribed in $k_{1}$ does not exceed the area of the square inscribed therein, which is $2 r^{2}$. Hence $3 V=V^{\\prime} \\leq 2 r^{2} h=2 / \\pi$. (ii) $A, B, C \\in k_{1}$ and $D \\in k_{2}$. The area of the triangle $A B C$ does not exceed the area of an equilateral triangle inscribed in $k_{1}$, which is $3 \\sqrt{3} r^{2} / 4$. Consequently, $V \\leq \\frac{\\sqrt{3}}{4} r^{2} h=\\frac{\\sqrt{3}}{4 \\pi}<\\frac{2}{3 \\pi}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (ROM 5) ${ }^{\\mathrm{IMO} 4}$ Let $A B C D$ be a convex quadrilateral for which the circle with diameter $A B$ is tangent to the line $C D$. Show that the circle with diameter $C D$ is tangent to the line $A B$ if and only if the lines $B C$ and $A D$ are parallel.", "solution": "14. Let $M$ and $N$ be the midpoints of $A B$ and $C D$, and let $M^{\\prime}, N^{\\prime}$ be their projections on $C D$ and $A B$, respectively. We know that $M M^{\\prime}=A B /$, and hence  $$ S_{A B C D}=S_{A M D}+S_{B M C}+S_{C M D}=\\frac{1}{2}\\left(S_{A B D}+S_{A B C}\\right)+\\frac{1}{4} A B \\cdot C D $$  The line $A B$ is tangent to the circle with diameter $C D$ if and only if $N N^{\\prime}=C D / 2$, or equivalently,  $$ S_{A B C D}=S_{A N D}+S_{B N C}+S_{A N B}=\\frac{1}{2}\\left(S_{B C D}+S_{A C D}\\right)+\\frac{1}{4} A B \\cdot C D $$  By (1), this is further equivalent to $S_{A B C}+S_{A B D}=S_{B C D}+S_{A C D}$. But since $S_{A B C}+S_{A C D}=S_{A B D}+S_{B C D}=S_{A B C D}$, this reduces to $S_{A B C}=S_{B C D}$, i.e., to $B C \\| A D$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (LUX 2) Angles of a given triangle $A B C$ are all smaller than $120^{\\circ}$. Equilateral triangles $A F B, B D C$ and $C E A$ are constructed in the exterior of $\\triangle A B C$. (a) Prove that the lines $A D, B E$, and $C F$ pass through one point $S$. (b) Prove that $S D+S E+S F=2(S A+S B+S C)$.", "solution": "15. (a) Since rotation by $60^{\\circ}$ around $A$ transforms the triangle $C A F$ into $\\triangle E A B$, it follows that $\\measuredangle(C F, E B)=60^{\\circ}$. We similarly deduce that $\\measuredangle(E B, A D)=\\measuredangle(A D, F C)=60^{\\circ}$. Let $S$ be the intersection point of $B E$ and $A D$. Since $\\measuredangle C S E=\\measuredangle C A E=60^{\\circ}$, it follows that $E A S C$ is cyclic. Therefore $\\measuredangle(A S, S C)=60^{\\circ}=\\measuredangle(A D, F C)$, which implies that $S$ lies on $C F$ as well. (b) A rotation of $E A S C$ around $E$ by $60^{\\circ}$ transforms $A$ into $C$ and $S$ into a point $T$ for which $S E=S T=S C+C T=S C+S A$. Summing the equality $S E=S C+S A$ and the analogous equalities $S D=S B+S C$ and $S F=S A+S B$ yields the result.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (POL 1) ${ }^{\\mathrm{IMO} 6}$ Let $a, b, c, d$ be odd positive integers such that $a<b<c<$ $d$, $a d=b c$, and $a+d=2^{k}, b+c=2^{m}$ for some integers $k$ and $m$. Prove that $a=1$.", "solution": "16. From the first two conditions we can easily conclude that $a+d>b+c$ (indeed, $(d+a)^{2}-(d-a)^{2}=(c+b)^{2}-(c-b)^{2}=4 a d=4 b c$ and $d-a>c-b>0)$. Thus $k>m$. From $d=2^{k}-a$ and $c=2^{m}-b$ we get $a\\left(2^{k}-a\\right)=b\\left(2^{m}-b\\right)$, or equivalently,  $$ (b+a)(b-a)=2^{m}\\left(b-2^{k-m} a\\right) $$  Since $2^{k-m} a$ is even and $b$ is odd, the highest power of 2 that divides the right-hand side of $(1)$ is $m$. Hence $(b+a)(b-a)$ is divisible by $2^{m}$ but not by $2^{m+1}$, which implies $b+a=2^{m_{1}} p$ and $b-a=2^{m_{2}} q$, where $m_{1}, m_{2} \\geq 1$, $m_{1}+m_{2}=m$, and $p, q$ are odd. Furthermore, $b=\\left(2^{m_{1}} p+2^{m_{2}} q\\right) / 2$ and $a=\\left(2^{m_{1}} p-2^{m_{2}} q\\right) / 2$ are odd, so either $m_{1}=1$ or $m_{2}=1$. Note that $m_{1}=1$ is not possible, since it would imply that $b-a=2^{m-1} q \\geq 2^{m-1}$, although $b+c=2^{m}$ and $b<c$ imply that $b<2^{m-1}$. Hence $m_{2}=1$ and $m_{1}=m-1$. Now since $a+b<b+c=2^{m}$, we obtain $a+b=2^{m-1}$ and $b-a=2 q$, where $q$ is an odd integer. Substituting these into (1) and dividing both sides by $2^{m}$ we get  $$ q=2^{m-2}+q-2^{k-m} a \\quad \\Longrightarrow \\quad 2^{k-m} a=2^{m-2} . $$  Since $a$ is odd and $k>m$, it follows that $a=1$. Remark. Now it is not difficult to prove that all fourtuplets $(a, b, c, d)$ that satisfy the given conditions are of the form $\\left(1,2^{m-1}-1,2^{m-1}+1,2^{2 m-2}-\\right.$ 1 ), where $m \\in \\mathbb{N}, m \\geq 3$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (FRG 3) In a permutation $\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)$ of the set $1,2, \\ldots, n$ we call a pair $\\left(x_{i}, x_{j}\\right)$ discordant if $i<j$ and $x_{i}>x_{j}$. Let $d(n, k)$ be the number of such permutations with exactly $k$ discordant pairs. Find $d(n, 2)$ and $d(n, 3)$.", "solution": "17. For any $m=0,1, \\ldots, n-1$, we shall find the number of permutations $\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)$ with exactly $k$ discordant pairs such that $x_{n}=n-m$. This $x_{n}$ is a member of exactly $m$ discordant pairs, and hence the permutation $\\left(x_{1}, \\ldots, x_{n-1}\\right.$ of the set $\\{1,2, \\ldots, n\\} \\backslash\\{m\\}$ must have exactly $k-m$ discordant pairs: there are $d(n-1, k-m)$ such permutations. Therefore  $$ \\begin{aligned} d(n, k) & =d(n-1, k)+d(n-1, k-1) \\cdots+d(n-1, k-n+1) \\\\ & =d(n-1, k)+d(n, k-1) \\end{aligned} $$  (note that $d(n, k)$ is 0 if $k<0$ or $k>\\binom{n}{2}$ ). We now proceed to calculate $d(n, 2)$ and $d(n, 3)$. Trivially, $d(n, 0)=1$. It follows that $d(n, 1)=d(n-1,1)+d(n, 0)=d(n-1,1)+1$, which yields $d(n, 1)=d(1,1)+n-1=n-1$.  Further, $d(n, 2)=d(n-1,2)+d(n, 1)=d(n-1,2)+n-1=d(2,2)+$ $2+3+\\cdots+n-1=\\left(n^{2}-n-2\\right) / 2$. Finally, using the known formula $1^{2}+2^{2}+\\cdots+k^{2}=k(k+1)(2 k+1) / 6$, we have $d(n, 3)=d(n-1,3)+d(n, 2)=d(n-1,3)+\\left(n^{2}-n-2\\right) / 2=$ $d(2,3)+\\sum_{i=3}^{n}\\left(n^{2}-n-2\\right) / 2=\\left(n^{3}-7 n+6\\right) / 6$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (USA 5) Inside triangle $A B C$ there are three circles $k_{1}, k_{2}, k_{3}$ each of which is tangent to two sides of the triangle and to its incircle $k$. The radii of $k_{1}, k_{2}, k_{3}$ are 1, 4, and 9 . Determine the radius of $k$.", "solution": "18. Suppose that circles $k_{1}\\left(O_{1}, r_{1}\\right), k_{2}\\left(O_{2}, r_{2}\\right)$, and $k_{3}\\left(O_{3}, r_{3}\\right)$ touch the edges of the angles $\\angle B A C, \\angle A B C$, and $\\angle A C B$, respectively. Denote also by $O$ and $r$ the center and radius of the incircle. Let $P$ be the point of tangency of the incircle with $A B$ and let $F$ be the foot of the perpendicular from $O_{1}$ to $O P$. From $\\triangle O_{1} F O$ we obtain $\\cot (\\alpha / 2)=2 \\sqrt{r r_{1}} /\\left(r-r_{1}\\right)$ and analogously $\\cot (\\beta / 2)=2 \\sqrt{r r_{2}} /\\left(r-r_{2}\\right), \\cot (\\gamma / 2)=2 \\sqrt{r r_{3}} /\\left(r-r_{3}\\right)$. We will now use a well-known trigonometric identity for the angles of a triangle:  $$ \\cot \\frac{\\alpha}{2}+\\cot \\frac{\\beta}{2}+\\cot \\frac{\\gamma}{2}=\\cot \\frac{\\alpha}{2} \\cdot \\cot \\frac{\\beta}{2} \\cdot \\cot \\frac{\\gamma}{2} . $$  (This identity follows from $\\tan (\\gamma / 2)=\\cot (\\alpha / 2+\\beta / 2)$ and the formula for the cotangent of a sum.) Plugging in the obtained cotangents, we get  $$ \\begin{aligned} \\frac{2 \\sqrt{r r_{1}}}{r-r_{1}}+\\frac{2 \\sqrt{r r_{2}}}{r-r_{2}}+\\frac{2 \\sqrt{r r_{3}}}{r-r_{3}}= & \\frac{2 \\sqrt{r r_{1}}}{r-r_{1}} \\cdot \\frac{2 \\sqrt{r r_{2}}}{r-r_{2}} \\cdot \\frac{2 \\sqrt{r r_{3}}}{r-r_{3}} \\Rightarrow \\\\ & \\sqrt{r_{1}}\\left(r-r_{2}\\right)\\left(r-r_{3}\\right)+\\sqrt{r_{2}}\\left(r-r_{1}\\right)\\left(r-r_{3}\\right) \\\\ & +\\sqrt{r_{3}}\\left(r-r_{1}\\right)\\left(r-r_{2}\\right)=4 r \\sqrt{r_{1} r_{2} r_{3}} . \\end{aligned} $$  For $r_{1}=1, r_{2}=4$, and $r_{3}=9$ we get $(r-4)(r-9)+2(r-1)(r-9)+3(r-1)(r-4)=24 r \\Rightarrow 6(r-1)(r-11)=0$. Clearly, $r=11$ is the only viable value for $r$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (CAN 5) The triangular array $\\left(a_{n, k}\\right)$ of numbers is given by $a_{n, 1}=1 / n$, for $n=1,2, \\ldots, a_{n, k+1}=a_{n-1, k}-a_{n, k}$, for $1 \\leq k \\leq n-1$. Find the harmonic mean of the 1985th row.", "solution": "19. First, we shall prove that the numbers in the $n$th row are exactly the numbers  $$ \\frac{1}{n\\binom{n-1}{0}}, \\frac{1}{n\\binom{n-1}{1}}, \\frac{1}{n\\binom{n-1}{2}}, \\ldots, \\frac{1}{n\\binom{n-1}{n-1}} $$  The proof of this fact can be done by induction. For small $n$, the statement can be easily verified. Assuming that the statement is true for some $n$, we have that the $k$ th element in the $(n+1)$ st row is, as is directly verified,  $$ \\frac{1}{n\\binom{n-1}{k-1}}-\\frac{1}{(n+1)\\binom{n}{k-1}}=\\frac{1}{(n+1)\\binom{n}{k}} $$  Thus (1) is proved. Now the geometric mean of the elements of the $n$th row becomes:  $$ \\frac{1}{n \\sqrt[n]{\\binom{n-1}{0} \\cdot\\binom{n-1}{1} \\cdots\\binom{n-1}{n-1}}} \\geq \\frac{1}{n\\left(\\frac{\\binom{n-1}{0}+\\binom{n-1}{1}+\\cdots+\\binom{n-1}{n-1}}{n}\\right)}=\\frac{1}{2^{n-1}} $$  The desired result follows directly from substituting $n=1984$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (CAN 2) Prove: (a) There are infinitely many triples of positive integers $m, n, p$ such that $4 m n-m-n=p^{2}-1$. (b) There are no positive integers $m, n, p$ such that $4 m n-m-n=p^{2}$.", "solution": "2. (a) For $m=t(t-1) / 2$ and $n=t(t+1) / 2$ we have $4 m n-m-n=$ $\\left(t^{2}-1\\right)^{2}-1$ (b) Suppose that $4 m n-m-n=p^{2}$, or equivalently, $(4 m-1)(4 n-1)=$ $4 p^{2}+1$. The number $4 m-1$ has at least one prime divisor, say $q$, that is of the form $4 k+3$. Then $4 p^{2} \\equiv-1(\\bmod q)$. However, by Fermat's theorem we have  $$ 1 \\equiv(2 p)^{q-1}=\\left(4 p^{2}\\right)^{\\frac{q-1}{2}} \\equiv(-1)^{\\frac{q-1}{2}}(\\bmod q) $$  which is impossible since $(q-1) / 2=2 k+1$ is odd.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. (USA 2) Determine all pairs $(a, b)$ of positive real numbers with $a \\neq 1$ such that  $$ \\log _{a} b<\\log _{a+1}(b+1) $$  ### 3.26 The Twenty-Sixth IMO", "solution": "20. Define the set $S=\\mathbb{R}^{+} \\backslash\\{1\\}$. The given inequality is equivalent to $\\ln b / \\ln a<\\ln (b+1) / \\ln (a+1)$. If $b=1$, it is obvious that each $a \\in S$ satisfies this inequality. Suppose now that $b$ is also in $S$. Let us define on $S$ a function $f(x)=\\ln (x+1) / \\ln x$. Since $\\ln (x+1)>\\ln x$ and $1 / x>1 / x+1>0$, we have  $$ f^{\\prime}(x)=\\frac{\\frac{\\ln x}{x+1}-\\frac{\\ln (x+1)}{x}}{\\ln ^{2} x}<0 \\quad \\text { for all } x $$  Hence $f$ is always decreasing. We also note that $f(x)<0$ for $x<1$ and that $f(x)>0$ for $x>1$ (at $x=1$ there is a discontinuity). Let us assume $b>1$. From $\\ln b / \\ln a<\\ln (b+1) / \\ln (a+1)$ we get $f(b)>$ $f(a)$. This holds for $b>a$ or for $a<1$. Now let us assume $b<1$. This time we get $f(b)<f(a)$. This holds for $a<b$ or for $a>1$. Hence all the solutions to $\\log _{a} b<\\log _{a+1}(b+1)$ are $\\{b=1, a \\in S\\}$, $\\{a>b>1\\},\\{b>1>a\\},\\{a<b<1\\}$, and $\\{b<1<a\\}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (USS 3) Find all positive integers $n$ such that  $$ n=d_{6}^{2}+d_{7}^{2}-1 $$  where $1=d_{1}<d_{2}<\\cdots<d_{k}=n$ are all positive divisors of the number $n$.", "solution": "3. From the equality $n=d_{6}^{2}+d_{7}^{2}-1$ we see that $d_{6}$ and $d_{7}$ are relatively prime and $d_{7}\\left|d_{6}^{2}-1=\\left(d_{6}-1\\right)\\left(d_{6}+1\\right), d_{6}\\right| d_{7}^{2}-1=\\left(d_{7}-1\\right)\\left(d_{7}+1\\right)$. Suppose that $d_{6}=a b, d_{7}=c d$ with $1<a<b, 1<c<d$. Then $n$ has 7 divisors smaller than $d_{7}$, namely $1, a, b, c, d, a b, a c$, which is impossible. Hence, one of the two numbers $d_{6}$ and $d_{7}$ is either a prime $p$ or the square of a prime $p^{2}$, where $p$ is not 2 . Let it be $d_{i}, i \\in\\{6,7\\}$; then $d_{i} \\mid\\left(d_{j}-1\\right)\\left(d_{j}+1\\right)$ implies that $d_{j} \\equiv \\pm 1\\left(\\bmod d_{i}\\right)$, and consequently $\\left(d_{i}^{2}-1\\right) / d_{j} \\equiv \\pm 1$ as well. But either $d_{j}$ or $\\left(d_{i}^{2}-1\\right) / d_{j}$ is less than $d_{i}$, and therefore equals $d_{i}-1$. We thus conclude that $d_{7}=d_{6}+1$. Setting $d_{6}=x, d_{7}=x+1$ we obtain that $n=x^{2}+(x+1)^{2}-1=2 x(x+1)$ is even. (i) Assume that one of $x, x+1$ is a prime $p$. The other one has at most 6 divisors and hence must be of the form $2^{3}, 2^{4}, 2^{5}, 2 q, 2 q^{2}, 4 q$, where $q$ is an odd prime. The numbers $2^{3}$ and $2^{4}$ are easily eliminated, while $2^{5}$ yields the solution $x=31, x+1=32, n=1984$. Also, $2 q$ is eliminated because $n=4 p q$ then has only 4 divisors less than $x ; 2 q^{2}$ is eliminated because $n=4 p q^{2}$ has at least 6 divisors less than $x ; 4 q$ is also eliminated because $n=8 p q$ has 6 divisors less than $x$. (ii) Assume that one of $x, x+1$ is $p^{2}$. The other one has at most 5 divisors ( $p$ excluded), and hence is of the form $2^{3}, 2^{4}, 2 q$, where $q$ is an odd prime. The number $2^{3}$ yields the solution $x=8, x+1=9, n=144$, while $2^{4}$ is easily eliminated. Also, $2 q$ is eliminated because $n=4 p^{2} q$ has 6 divisors less than $x$. Thus there are two solutions in total: 144 and 1984.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (MON 1) ${ }^{\\mathrm{IMO} 5}$ Let $d$ be the sum of the lengths of all diagonals of a convex polygon of $n(n>3)$ vertices and let $p$ be its perimeter. Prove that  $$ \\frac{n-3}{2}<\\frac{d}{p}<\\frac{1}{2}\\left(\\left[\\frac{n}{2}\\right]\\left[\\frac{n+1}{2}\\right]-2\\right) . $$", "solution": "4. Consider the convex $n$-gon $A_{1} A_{2} \\ldots A_{n}$ (the indices are considered modulo $n)$. For any diagonal $A_{i} A_{j}$ we have $A_{i} A_{j}+A_{i+1} A_{j+1}>A_{i} A_{i+1}+A_{j} A_{j+1}$. Summing all such $n(n-3) / 2$ inequalities, we obtain $2 d>(n-3) p$, proving the first inequality. Let us now prove the second inequality. We notice that for each diagonal $A_{i} A_{i+j}$ (we may assume w.l.o.g. that $j \\leq[n / 2]$ ) the following relation holds:  $$ A_{i} A_{i+j}<A_{i} A_{i+1}+\\cdots+A_{i+j-1} A_{i+j} . $$  If $n=2 k+1$, then summing the inequalities (1) for $j=2,3, \\ldots, k$ and $i=1,2, \\ldots, n$ yields $d<(2+3+\\cdots+k) p=([n / 2][n+1 / 2]-2) p / 2$. If $n=2 k$, then summing the inequalities (1) for $j=2,3, \\ldots, k-1$, $i=1,2, \\ldots, n$ and for $j=k, i=1,2, \\ldots, k$ again yields $d<(2+3+\\cdots+$ $(k-1)+k / 2) p=\\frac{1}{2}([n / 2][n+1 / 2]-2) p$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (FRG 1) ${ }^{\\mathrm{IMO} 1}$ Let $x, y, z$ be nonnegative real numbers with $x+y+z=1$. Show that  $$ 0 \\leq x y+y z+z x-2 x y z \\leq \\frac{7}{27} $$", "solution": "5. Let $f(x, y, z)=x y+y z+z x-2 x y z$. The first inequality follows immediately by adding $x y \\geq x y z, y z \\geq x y z$, and $z x \\geq x y z$ (in fact, a stronger inequality $x y+y z+z x-9 x y z \\geq 0$ holds). Assume w.l.o.g. that $z$ is the smallest of $x, y, z$. Since $x y \\leq(x+y)^{2} / 4=$ $(1-z)^{2} / 4$ and $z \\leq 1 / 2$, we have  $$ \\begin{aligned} x y+y z+z x-2 x y z & =(x+y) z+x y(1-2 z) \\\\ & \\leq(1-z) z+\\frac{(1-z)^{2}(1-2 z)}{4} \\\\ & =\\frac{7}{27}-\\frac{(1-2 z)(1-3 z)^{2}}{108} \\leq \\frac{7}{27} \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (CAN 3) Let $c$ be a positive integer. The sequence $\\left\\{f_{n}\\right\\}$ is defined as follows:  $$ f_{1}=1, \\quad f_{2}=c, \\quad f_{n+1}=2 f_{n}-f_{n-1}+2 \\quad(n \\geq 2) $$  Show that for each $k \\in \\mathbb{N}$ there exists $r \\in \\mathbb{N}$ such that $f_{k} f_{k+1}=f_{r}$.", "solution": "6. From the given recurrence we infer $f_{n+1}-f_{n}=f_{n}-f_{n-1}+2$. Consequently, $f_{n+1}-f_{n}=\\left(f_{2}-f_{1}\\right)+2(n-1)=c-1+2(n-1)$. Summing up for $n=1,2, \\ldots, k-1$ yields the explicit formula  $$ f_{k}=f_{1}+(k-1)(c-1)+(k-1)(k-2)=k^{2}+b k-b, $$  where $b=c-4$. Now we easily obtain $f_{k} f_{k+1}=k^{4}+2(b+1) k^{3}+\\left(b^{2}+b+\\right.$ $1) k^{2}-\\left(b^{2}+b\\right) k-b$. We are looking for an $r$ for which the last expression equals $f_{r}$. Setting $r=k^{2}+p k+q$ we get by a straightforward calculation that $p=b+1, q=-b$, and $r=k^{2}+(b+1) k-b=f_{k}+k$. Hence $f_{k} f_{k+1}=f_{f_{k}+k}$ for all $k$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (FRG 5) (a) Decide whether the fields of the $8 \\times 8$ chessboard can be numbered by the numbers $1,2, \\ldots, 64$ in such a way that the sum of the four numbers in each of its parts of one of the forms ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-186.jpg?height=171&width=587&top_left_y=1708&top_left_x=531) is divisible by four. (b) Solve the analogous problem for ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-186.jpg?height=159&width=575&top_left_y=1974&top_left_x=539)", "solution": "7. It clearly suffices to solve the problem for the remainders modulo 4 (16 of each kind). (a) The remainders can be placed as shown in Figure 1, so that they satisfy the conditions.  | 10321032 |  | | :---: | :---: | | 23012301 |  | | 32103210 |  | | 01230123 | p | | 10321032 | q r s | | 23012301 | t | | 32103210 |  | | 01230123 |  |  Fig. 1 Fig. 2 (b) Suppose that the required numbering exists. Consider a part of the chessboard as in Figure 2. By the stated condition, all the numbers $p+q+r+s, q+r+s+t, p+q+r+t, p+r+s+t$ give the same remainder modulo 4 , and so do $p, q, r, s$. We deduce that all numbers on black cells of the board, except possibly the two corner cells, give the same remainder, which is impossible.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (ROM 2) $)^{\\mathrm{IMO} 3}$ In a plane two different points $O$ and $A$ are given. For each point $X \\neq O$ of the plane denote by $\\alpha(X)$ the angle $A O X$ measured in radians $(0 \\leq \\alpha(X)<2 \\pi)$ and by $C(X)$ the circle with center $O$ and radius $O X+\\frac{\\alpha(X)}{O X}$. Suppose each point of the plane is colored by one of a finite number of colors. Show that there exists a point $X$ with $\\alpha(X)>0$ such that its color appears somewhere on the circle $C(X)$.", "solution": "8. Suppose that the statement of the problem is false. Consider two arbitrary circles $R=(O, r)$ and $S=(O, s)$ with $0<r<s<1$. The point $X \\in R$ with $\\alpha(X)=r(s-r)<2 \\pi$ satisfies that $C(X)=S$. It follows that the color of the point $X$ does not appear on $S$. Consequently, the set of colors that appear on $R$ is not the same as the set of colors that appear on $S$. Hence any two distinct circles with center at $O$ and radii less than 1 have distinct sets of colors. This is a contradiction, since there are infinitely many such circles but only finitely many possible sets of colors.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1984", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (POL 2) Let $a, b, c$ be positive numbers with $\\sqrt{a}+\\sqrt{b}+\\sqrt{c}=\\frac{\\sqrt{3}}{2}$. Prove that the system of equations  $$ \\begin{aligned} & \\sqrt{y-a}+\\sqrt{z-a}=1, \\\\ & \\sqrt{z-b}+\\sqrt{x-b}=1 \\\\ & \\sqrt{x-c}+\\sqrt{y-c}=1 \\end{aligned} $$  has exactly one solution $(x, y, z)$ in real numbers.", "solution": "9. Let us show first that the system has at most one solution. Suppose that $(x, y, z)$ and $\\left(x^{\\prime}, y^{\\prime}, z^{\\prime}\\right)$ are two distinct solutions and that w.l.o.g. $x<x^{\\prime}$. Then the second and third equation imply that $y>y^{\\prime}$ and $z>z^{\\prime}$, but then $\\sqrt{y-a}+\\sqrt{z-a}>\\sqrt{y^{\\prime}-a}+\\sqrt{z^{\\prime}-a}$, which is a contradiction. We shall now prove the existence of at least one solution. Let $P$ be an arbitrary point in the plane and $K, L, M$ points such that $P K=\\sqrt{a}$, $P L=\\sqrt{b}, P M=\\sqrt{c}$, and $\\angle K P L=\\angle L P M=\\angle M P K=120^{\\circ}$. The lines through $K, L, M$ perpendicular respectively to $P K, P L, P M$ form an equilateral triangle $A B C$, where $K \\in B C, L \\in A C$, and $M \\in A B$. Since its area equals $A B^{2} \\sqrt{3} / 4=S_{\\triangle B P C}+S_{\\triangle A P C}+$ $S_{\\triangle A P B}=A B(\\sqrt{a}+\\sqrt{b}+\\sqrt{c}) / 2$, it follows that $A B=1$. Therefore $x=P A^{2}, y=P B^{2}$, and $z=P C^{2}$ is a solution of the system (indeed, $\\sqrt{y-a}+\\sqrt{z-a}=\\sqrt{P B^{2}-P K^{2}}+\\sqrt{P C^{2}-P K^{2}}=B K+C K=1$, etc.).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (MON 1) ${ }^{\\mathrm{IMO} 4}$ Given a set $M$ of 1985 positive integers, none of which has a prime divisor larger than 26 , prove that the set has four distinct elements whose geometric mean is an integer.", "solution": "1. Since there are 9 primes ( $p_{1}=2<p_{2}=3<\\cdots<p_{9}=23$ ) less than 26 , each number $x_{j} \\in M$ is of the form $\\prod_{i=1}^{9} p_{i}^{a_{i j}}$, where $0 \\leq a_{i j}$. Now, $x_{j} x_{k}$ is a square if $a_{i j}+a_{i k} \\equiv 0(\\bmod 2)$ for $i=1, \\ldots, 9$. Since the number of distinct ninetuples modulo 2 is $2^{9}$, any subset of $M$ with at least 513 elements contains two elements with square product. Starting from $M$ and eliminating such pairs, one obtains $(1985-513) / 2=736>513$ distinct two-element subsets of $M$ each having a square as the product of elements. Reasoning as above, we find at least one (in fact many) pair of such squares whose product is a fourth power.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. 2b.(VIE 1) Prove that for every point $M$ on the surface of a regular tetrahedron there exists a point $M^{\\prime}$ such that there are at least three different curves on the surface joining $M$ to $M^{\\prime}$ with the smallest possible length among all curves on the surface joining $M$ to $M^{\\prime}$.", "solution": "10. If $M$ is at a vertex of the regular tetrahedron $A B C D(A B=1)$, then one can take $M^{\\prime}$ at the center of the opposite face of the tetrahedron. Let $M$ be on the face $(A B C)$ of the tetrahedron, excluding the vertices. Consider a continuous mapping $f$ of $\\mathbb{C}$ onto the surface $S$ of $A B C D$ that maps $m+n e^{2 \\pi / 3}$ for $m, n \\in$ $\\mathbb{Z}$ onto $A, B, C, D$ if $(m, n) \\equiv$ $(1,1),(1,0),(0,1),(0,0)(\\bmod 2)$ re- ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-484.jpg?height=464&width=446&top_left_y=1356&top_left_x=857) spectively, and maps each unit equilateral triangle with vertices of the form $m+n e^{2 \\pi / 3}$ isometrically onto the corresponding face of $A B C D$. The point $M$ then has one preimage $M_{j}, j=1,2, \\ldots, 6$, in each of the six preimages of $\\triangle A B C$ having two vertices on the unit circle. The $M_{j}$ 's form a convex centrally symmetric (possibly degenerate) hexagon. Of the triangles formed by two adjacent sides of this hexagon consider the one, say $M_{1} M_{2} M_{3}$, with the smallest radius of circumcircle and denote by $\\widehat{M^{\\prime}}$ its circumcenter. Then we can choose $M^{\\prime}=f\\left(\\widehat{M^{\\prime}}\\right)$. Indeed, the images of the segments $M_{1} \\widehat{M^{\\prime}}, M_{2} \\widehat{M^{\\prime}}, M_{3} \\widehat{M^{\\prime}}$ are three different shortest paths on $S$ from $M$ to $M^{\\prime}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. 3a.(USS 3) Find a method by which one can compute the coefficients of $P(x)=x^{6}+a_{1} x^{5}+\\cdots+a_{6}$ from the roots of $P(x)=0$ by performing not more than 15 additions and 15 multiplications.", "solution": "11. Let $-x_{1}, \\ldots,-x_{6}$ be the roots of the polynomial. Let $s_{k, i}(k \\leq i \\leq 6)$ denote the sum of all products of $k$ of the numbers $x_{1}, \\ldots, x_{i}$. By Vieta's formula we have $a_{k}=s_{k, 6}$ for $k=1, \\ldots, 6$. Since $s_{k, i}=s_{k-1, i-1} x_{i}+$ $s_{k, i-1}$, one can compute the $a_{k}$ by the following scheme (the horizontal and vertical arrows denote multiplications and additions respectively): ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-485.jpg?height=457&width=609&top_left_y=646&top_left_x=486)", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. 3b.(GBR 4) A sequence of polynomials $P_{m}(x, y, z), m=0,1,2, \\ldots$, in $x, y$, and $z$ is defined by $P_{0}(x, y, z)=1$ and by  $$ P_{m}(x, y, z)=(x+z)(y+z) P_{m-1}(x, y, z+1)-z^{2} P_{m-1}(x, y, z) $$  for $m>0$. Prove that each $P_{m}(x, y, z)$ is symmetric, in other words, is unaltered by any permutation of $x, y, z$.", "solution": "12. We shall prove by induction on $m$ that $P_{m}(x, y, z)$ is symmetric and that  $$ (x+y) P_{m}(x, z, y+1)-(x+z) P_{m}(x, y, z+1)=(y-z) P_{m}(x, y, z) $$  holds for all $x, y, z$. This is trivial for $m=0$. Assume now that it holds for $m=n-1$. Since obviously $P_{n}(x, y, z)=P_{n}(y, x, z)$, the symmetry of $P_{n}$ will follow if we prove that $P_{n}(x, y, z)=P_{n}(x, z, y)$. Using (1) we have $P_{n}(x, z, y)-$ $P_{n}(x, y, z)=(y+z)\\left[(x+y) P_{n-1}(x, z, y+1)-(x+z) P_{n-1}(x, y, z+1)\\right]-\\left(y^{2}-\\right.$ $\\left.z^{2}\\right) P_{n-1}(x, y, z)=(y+z)(y-z) P_{n-1}(x, y, z)-\\left(y^{2}-z^{2}\\right) P_{n-1}(x, y, z)=0$. It remains to prove (1) for $m=n$. Using the already established symmetry we have  $$ \\begin{aligned} & (x+y) P_{n}(x, z, y+1)-(x+z) P_{n}(x, y, z+1) \\\\ & =(x+y) P_{n}(y+1, z, x)-(x+z) P_{n}(z+1, y, x) \\\\ & =(x+y)\\left[(y+x+1)(z+x) P_{n-1}(y+1, z, x+1)-x^{2} P_{n-1}(y+1, z, x)\\right] \\\\ & \\quad-(x+z)\\left[(z+x+1)(y+x) P_{n-1}(z+1, y, x+1)-x^{2} P_{n-1}(z+1, y, x)\\right] \\\\ & =(x+y)(x+z)(y-z) P_{n-1}(x+1, y, z)-x^{2}(y-z) P_{n-1}(x, y, z) \\\\ & =(y-z) P_{n}(z, y, x)=(y-z) P_{n}(x, y, z), \\end{aligned} $$  as claimed.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. 4a.(BUL 1) Let $m$ boxes be given, with some balls in each box. Let $n<m$ be a given integer. The following operation is performed: choose $n$ of the boxes and put 1 ball in each of them. Prove: (a) If $m$ and $n$ are relatively prime, then it is possible, by performing the operation a finite number of times, to arrive at the situation that all the boxes contain an equal number of balls. (b) If $m$ and $n$ are not relatively prime, there exist initial distributions of balls in the boxes such that an equal distribution is not possible to achieve.", "solution": "13. If $m$ and $n$ are relatively prime, there exist positive integers $p, q$ such that $p m=q n+1$. Thus by putting $m$ balls in some boxes $p$ times we can achieve that one box receives $q+1$ balls while all others receive $q$ balls. Repeating this process sufficiently many times, we can obtain an equal distribution of the balls. Now assume $\\operatorname{gcd}(m, n)>1$. If initially there is only one ball in the boxes, then after $k$ operations the number of balls will be $1+k m$, which is never divisible by $n$. Hence the task cannot be done.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. 4b.(IRE 4) A set of 1985 points is distributed around the circumference of a circle and each of the points is marked with 1 or -1 . A point is called \"good\" if the partial sums that can be formed by starting at that point and proceeding around the circle for any distance in either direction are all strictly positive. Show that if the number of points marked with -1 is less than 662 , there must be at least one good point.", "solution": "14. It suffices to prove the existence of a good point in the case of exactly 661 -1 's. We prove by induction on $k$ that in any arrangement with $3 k+2$ points $k$ of which are -1 's a good point exists. For $k=1$ this is clear by inspection. Assume that the assertion holds for all arrangements of $3 n+2$ points and consider an arrangement of $3(n+1)+2$ points. Now there exists a sequence of consecutive -1 's surrounded by two +1 's. There is a point $P$ which is good for the arrangement obtained by removing the two +1 's bordering the sequence of -1 's and one of these -1 's. Since $P$ is out of this sequence, clearly the removal either leaves a partial sum as it was or diminishes it by 1 , so $P$ is good for the original arrangement. Second solution. Denote the number on an arbitrary point by $a_{1}$, and the numbers on successive points going in the positive direction by $a_{2}, a_{3}, \\ldots$ (in particular, $a_{k+1985}=a_{k}$ ). We define the partial sums $s_{0}=0, s_{n}=$ $a_{1}+a_{2}+\\cdots+a_{n}$ for all positive integers $n$; then $s_{k+1985}=s_{k}+s_{1985}$ and $s_{1985} \\geq 663$. Since $s_{1985 m} \\geq 663 m$ and $3 \\cdot 663 m>1985(m+2)+1$ for large $m$, not all values $0,1,2, \\ldots 663 m$ can appear thrice among the $1985(m+2)+1$ sums $s_{-1985}, s_{-1984}, \\ldots, s_{1985(m+1)}$ (and none of them appears out of this set). Thus there is an integral value $s>0$ that appears at most twice as a partial sum, say $s_{k}=s_{l}=s, k<l$. Then either $a_{k}$ or $a_{l}$ is a good point. Actually, $s_{i}>s$ must hold for all $i>l$, and $s_{i}<s$ for all $i<k$ (otherwise, the sum $s$ would appear more than twice). Also, for the same reason there cannot exist indices $p, q$ between $k$ and $l$ such that $s_{p}>s$ and $s_{q}<s$; i.e., for $k<p<l, s_{p}$ 's are either all greater than or equal to $s$, or smaller than or equal to $s$. In the former case $a_{k}$ is good, while in the latter $a_{l}$ is good.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. 5a.(FRA 3) Let $K$ and $K^{\\prime}$ be two squares in the same plane, their sides of equal length. Is it possible to decompose $K$ into a finite number of triangles $T_{1}, T_{2}, \\ldots, T_{p}$ with mutually disjoint interiors and find translations $t_{1}, t_{2}, \\ldots, t_{p}$ such that  $$ K^{\\prime}=\\bigcup_{i=1}^{p} t_{i}\\left(T_{i}\\right) ? $$", "solution": "15. There is no loss of generality if we assume $K=A B C D, K^{\\prime}=$ $A B^{\\prime} C^{\\prime} D^{\\prime}$, and that $K^{\\prime}$ is obtained from $K$ bya clockwise rotation around $A$ by $\\phi, 0 \\leq \\phi \\leq \\pi / 4$. Let $C^{\\prime} D^{\\prime}, B^{\\prime} C^{\\prime}$, and the parallel to $A B$ through $D^{\\prime}$ meet the line $B C$ at $E$, $F$, and $G$ respectively. Let us now choose points $E^{\\prime} \\in A B^{\\prime}, G^{\\prime} \\in A B$, $C^{\\prime \\prime} \\in A D^{\\prime}$, and $E^{\\prime \\prime} \\in A D$ such that ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-486.jpg?height=421&width=516&top_left_y=1572&top_left_x=831) the triangles $A E^{\\prime} G^{\\prime}$ and $A C^{\\prime \\prime} E^{\\prime \\prime}$ are translates of the triangles $D^{\\prime} E G$ and $F C^{\\prime} E$ respectively. Since $A E^{\\prime}=D^{\\prime} E$ and $A C^{\\prime \\prime}=F C^{\\prime}$, we have $C^{\\prime \\prime} E^{\\prime \\prime}=C^{\\prime} E=B^{\\prime} E^{\\prime}$ and $C^{\\prime \\prime} D^{\\prime}=B^{\\prime} F$, which imply that $\\triangle E^{\\prime \\prime} C^{\\prime \\prime} D^{\\prime}$ is a translate of $\\triangle E^{\\prime} B^{\\prime} F$, and consequently $E^{\\prime \\prime} D^{\\prime}=E^{\\prime} F$ and $E^{\\prime \\prime} D^{\\prime} \\| E^{\\prime} F$. It follows that there exist points $H \\in C D, H^{\\prime} \\in B F$, and $D^{\\prime \\prime} \\in E^{\\prime} G^{\\prime}$ such that $E^{\\prime \\prime} D^{\\prime} H D$ is a translate of $E^{\\prime} F H^{\\prime} D^{\\prime \\prime}$. The remaining parts of $K$ and $K^{\\prime}$ are the rectangles $D^{\\prime} G C H$ and $D^{\\prime \\prime} H^{\\prime} B G^{\\prime}$ of equal area. We shall now show that two rectangles with parallel sides and equal areas can be decomposed into translation invariant parts. Let the sides of the rectangles $X Y Z T$ and $X^{\\prime} Y^{\\prime} Z^{\\prime} T^{\\prime}\\left(X Y \\| X^{\\prime} Y^{\\prime}\\right)$ satisfy $X^{\\prime} Y^{\\prime}<X Y$, $Y^{\\prime} Z^{\\prime}>Y Z$, and $X^{\\prime} Y^{\\prime} \\cdot Y^{\\prime} Z^{\\prime}=X Y \\cdot Y Z$. Suppose that $2 X^{\\prime} Y^{\\prime}>X Y$ (otherwise, we may cut off congruent rectangles from both the original ones until we reduce them to the case of $2 X^{\\prime} Y^{\\prime}>X Y$ ). Let $U \\in X Y$ and $V \\in Z T$ be points such that $Y U=T V=X^{\\prime} Y^{\\prime}$ and $W \\in X V$ be a point such that $U W \\| X T$. Then translating $\\triangle X U W$ to a triangle $V Z R$ and $\\triangle X V T$ to a triangle $W R S$ results in a rectangle $U Y R S$ congruent to $X^{\\prime} Y^{\\prime} Z^{\\prime} T^{\\prime}$. Thus we have partitioned $K$ and $K^{\\prime}$ into translation-invariant parts. Although not all the parts are triangles, we may simply triangulate them.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. 5b.(BEL 2) If possible, construct an equilateral triangle whose three vertices are on three given circles.", "solution": "16. Let the three circles be $\\alpha(A, a), \\beta(B, b)$, and $\\gamma(C, c)$, and assume $c \\leq a, b$. We denote by $\\mathcal{R}_{X, \\varphi}$ the rotation around $X$ through an angle $\\varphi$. Let $P Q R$ be an equilateral triangle, say of positive orientation (the case of negatively oriented $\\triangle P Q R$ is analogous), with $P \\in \\alpha, Q \\in \\beta$, and $R \\in \\gamma$. Then $Q=\\mathcal{R}_{P,-60^{\\circ}}(R) \\in \\mathcal{R}_{P,-60^{\\circ}}(\\gamma) \\cap \\beta$. Since the center of $\\mathcal{R}_{P,-60^{\\circ}}(\\gamma)$ is $\\mathcal{R}_{P,-60^{\\circ}}(C)=\\mathcal{R}_{C, 60^{\\circ}}(P)$ and it lies on $\\mathcal{R}_{C, 60^{\\circ}}(\\alpha)$, the union of circles $\\mathcal{R}_{P,-60^{\\circ}}(\\gamma)$ as $P$ varies on $\\alpha$ is the annulus $\\mathcal{U}$ with center $A^{\\prime}=\\mathcal{R}_{C, 60^{\\circ}}(A)$ and radii $a-c$ and $a+c$. Hence there is a solution if and only if $\\mathcal{U} \\cap \\beta$ is nonempty.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. 6a. (SWE 3) ${ }^{\\mathrm{IMO} O}$ The sequence $f_{1}, f_{2}, \\ldots, f_{n}, \\ldots$ of functions is defined for $x>0$ recursively by  $$ f_{1}(x)=x, \\quad f_{n+1}(x)=f_{n}(x)\\left(f_{n}(x)+\\frac{1}{n}\\right) . $$  Prove that there exists one and only one positive number $a$ such that $0<f_{n}(a)<f_{n+1}(a)<1$ for all integers $n \\geq 1$.", "solution": "17. The statement of the problem is equivalent to the statement that there is one and only one $a$ such that $1-1 / n<f_{n}(a)<1$ for all $n$. We note that each $f_{n}$ is a polynomial with positive coefficients, and therefore increasing and convex in $\\mathbb{R}^{+}$. Define $x_{n}$ and $y_{n}$ by $f_{n}\\left(x_{n}\\right)=1-1 / n$ and $f_{n}\\left(y_{n}\\right)=1$. Since  $$ f_{n+1}\\left(x_{n}\\right)=\\left(1-\\frac{1}{n}\\right)^{2}+\\left(1-\\frac{1}{n}\\right) \\frac{1}{n}=1-\\frac{1}{n} $$  and $f_{n+1}\\left(y_{n}\\right)=1+1 / n$, it follows that $x_{n}<x_{n+1}<y_{n+1}<y_{n}$. Moreover, the convexity of $f_{n}$ together with the fact that $f_{n}(x)>x$ for all $x>0$ implies that $y_{n}-x_{n}<f_{n}\\left(y_{n}\\right)-f_{n}\\left(x_{n}\\right)=1 / n$. Therefore the sequences have a common limit $a$, which is the only number lying between $x_{n}$ and $y_{n}$ for all $n$. By the definition of $x_{n}$ and $y_{n}$, the statement immediately follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. 6b.(CAN 5) Let $x_{1}, x_{2}, \\ldots, x_{n}$ be positive numbers. Prove that  $$ \\frac{x_{1}^{2}}{x_{1}^{2}+x_{2} x_{3}}+\\frac{x_{2}^{2}}{x_{2}^{2}+x_{3} x_{4}}+\\cdots+\\frac{x_{n-1}^{2}}{x_{n-1}^{2}+x_{n} x_{1}}+\\frac{x_{n}^{2}}{x_{n}^{2}+x_{1} x_{2}} \\leq n-1 . $$  Supplementary Problems", "solution": "18. Set $y_{i}=\\frac{x_{i}^{2}}{x_{i+1} x_{i+2}}$, where $x_{n+i}=x_{i}$. Then $\\prod_{i=1}^{n} y_{i}=1$ and the inequality to be proved becomes $\\sum_{i=1}^{n} \\frac{y_{i}}{1+y_{i}} \\leq n-1$, or equivalently  $$ \\sum_{i=1}^{n} \\frac{1}{1+y_{i}} \\geq 1 $$  We prove this inequality by induction on $n$. Since $\\frac{1}{1+y}+\\frac{1}{1+y^{-1}}=1$, the inequality is true for $n=2$. Assume that it is true for $n-1$, and let there be given $y_{1}, \\ldots, y_{n}>0$ with $\\prod_{i=1}^{n} y_{i}=1$. Then $\\frac{1}{1+y_{n-1}}+\\frac{1}{1+y_{n}}>\\frac{1}{1+y_{n-1} y_{n}}$, which is equivalent to $1+y_{n} y_{n-1}(1+$ $\\left.y_{n}+y_{n-1}\\right)>0$. Hence by the inductive hypothesis  $$ \\sum_{i=1}^{n} \\frac{1}{1+y_{i}} \\geq \\sum_{i=1}^{n-2} \\frac{1}{1+y_{i}}+\\frac{1}{1+y_{n-1} y_{n}} \\geq 1 $$  Remark. The constant $n-1$ is best possible (take for example $x_{i}=a^{i}$ with $a$ arbitrarily large).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (ISR 3) For which integers $n \\geq 3$ does there exist a regular $n$-gon in the plane such that all its vertices have integer coordinates in a rectangular coordinate system?", "solution": "19. Suppose that for some $n>6$ there is a regular $n$-gon with vertices having integer coordinates, and that $A_{1} A_{2} \\ldots A_{n}$ is the smallest such $n$-gon, of side length $a$. If $O$ is the origin and $B_{i}$ the point such that $\\overrightarrow{O B_{i}}=\\overrightarrow{A_{i-1} A_{i}}$, $i=1,2, \\ldots, n$ (where $A_{0}=A_{n}$ ), then $B_{i}$ has integer coordinates and $B_{1} B_{2} \\ldots B_{n}$ is a regular polygon of side length $2 a \\sin (\\pi / n)<a$, which is impossible. It remains to analyze the cases $n \\leq 6$. If $\\mathcal{P}$ is a regular $n$-gon with $n=$ $3,5,6$, then its center $C$ has rational coordinates. We may suppose that $C$ also has integer coordinates and then rotate $\\mathcal{P}$ around $C$ thrice through $90^{\\circ}$, thus obtaining a regular 12 -gon or 20 -gon, which is impossible. Hence we must have $n=4$ which is indeed a solution.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (BRA 3) A polyhedron has 12 faces and is such that: (i) all faces are isosceles triangles, (ii) all edges have length either $x$ or $y$, (iii) at each vertex either 3 or 6 edges meet, and (iv) all dihedral angles are equal.  Find the ratio $x / y$.", "solution": "2. The polyhedron has $3 \\cdot 12 / 2=18$ edges, and by Euler's formula, 8 vertices. Let $v_{1}$ and $v_{2}$ be the numbers of vertices at which respectively 3 and 6 edges meet. Then $v_{1}+v_{2}=8$ and $3 v_{1}+6 v_{2}=2 \\cdot 18$, implying that $v_{1}=4$. Let $A, B, C, D$ be the vertices at which three edges meet. Since the dihedral angles are equal, all the edges meeting at $A$, say $A E, A F, A G$, must have equal length, say $x$. (If $x=A E=A F \\neq A G=y$, and $A E F$, $A F G$, and $A G E$ are isosceles, $\\angle E A F \\neq \\angle F A G$, in contradiction to the equality of the dihedral angles.) It is easy to see that at $E, F$, and $G$ six edges meet. One proceeds to conclude that if $H$ is the fourth vertex of this kind, $E F G H$ must be a regular tetrahedron of edge length $y$, and the other vertices $A, B, C$, and $D$ are tops of isosceles pyramids based on $E F G, E F H, F G H$, and $G E H$. Let the plane through $A, B, C$ meet $E F$, $H F$, and $G F$, at $E^{\\prime}, H^{\\prime}$, and $G^{\\prime}$. Then $A E^{\\prime} B H^{\\prime} C G^{\\prime}$ is a regular hexagon, and since $x=F A=F E^{\\prime}$, we have $E^{\\prime} G^{\\prime}=x$ and $A E^{\\prime}=x / \\sqrt{3}$. From the isosceles triangles $A E F$ and $F A E^{\\prime}$ we obtain finally, with $\\measuredangle E F A=\\alpha$, $\\frac{y}{2 x}=\\cos \\alpha=1-2 \\sin ^{2}(\\alpha / 2), x /(2 x \\sqrt{3})=\\sin (\\alpha / 2)$, and $y / x=5 / 3$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. $(\\mathbf{G B R} 5)^{\\mathrm{IMO}} \\mathrm{A}$ circle whose center is on the side $E D$ of the cyclic quadrilateral $B C D E$ touches the other three sides. Prove that $E B+C D=$ $E D$.", "solution": "20. Let $O$ be the center of the circle touching the three sides of $B C D E$ and let $F \\in(E D)$ be the point such that $E F=E B$. Then $\\angle E F B=90^{\\circ}-$ $\\angle E / 2=\\angle C / 2=\\angle O C B$, which implies that $B, C, F, O$ lie on a circle. It follows that $\\angle D F C=\\angle O B C=\\angle B / 2=90^{\\circ}-\\angle D / 2$ and consequently $\\angle D C F=\\angle D F C$. Hence $E D=E F+F D=E B+C D$. Second solution. Let $r$ be the radius of the small circle and let $M, N$ be the points of tangency of the circle with $B E$ and $C D$ respectively. Then $E M=r \\cot E, D N=r \\cot D, M B=r \\cot (\\angle B / 2)=r \\tan (\\angle D / 2)$, $N C=r \\tan (\\angle E / 2)$, and $E D=E O+O D=r / \\sin D+r / \\sin E$. The statement follows from the identity $\\cot x+\\tan (x / 2)=1 / \\sin x$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. (IRE 1) The tangents at $B$ and $C$ to the circumcircle of the acute-angled triangle $A B C$ meet at $X$. Let $M$ be the midpoint of $B C$. Prove that (a) $\\angle B A M=\\angle C A X$, and (b) $\\frac{A M}{A X}=\\cos \\angle B A C$.", "solution": "21. Let $B_{1}$ and $C_{1}$ be the points on the rays $A C$ and $A B$ respectively such that $X B_{1}=X C=X B=X C_{1}$. Then $\\angle X B_{1} C=\\angle X C B_{1}=\\angle A B C$ and $\\angle X C_{1} B=\\angle X B C_{1}=\\angle A C B$, which imply that $B_{1}, X, C_{1}$ are collinear and $\\triangle A B_{1} C_{1} \\sim \\triangle A B C$. Moreover, $X$ is the midpoint of $B_{1} C_{1}$ because $X B_{1}=X C=X B=X C_{1}$, from which we conclude that $\\triangle A X C_{1} \\sim$ $\\triangle A M C$. Therefore $\\angle B A X=\\angle C A M$ and  $$ \\frac{A M}{A X}=\\frac{C M}{X C_{1}}=\\frac{C M}{X C}=\\cos \\alpha $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "22", "problem_type": null, "exam": "IMO", "problem": "22. (USS 5) ${ }^{\\mathrm{IMO}} \\mathrm{A}$ circle with center $O$ passes through points $A$ and $C$ and intersects the sides $A B$ and $B C$ of the triangle $A B C$ at points $K$ and $N$, respectively. The circumscribed circles of the triangles $A B C$ and $K B N$ intersect at two distinct points $B$ and $M$. Prove that $\\angle O M B=90^{\\circ}$.", "solution": "22. Assume that $\\triangle A B C$ is acute (the case of an obtuse $\\triangle A B C$ is similar). Let $S$ and $R$ be the centers of the circumcircles of $\\triangle A B C$ and $\\triangle K B N$, respectively. Since $\\angle B N K=\\angle B A C$, the triangles $B N K$ and $B A C$ are similar. Now we have $\\angle C B R=\\angle A B S=90^{\\circ}-\\angle A C B$, which gives us $B R \\perp A C$ and consequently $B R \\| O S$. Similarly $B S \\perp K N$ implies that $B S \\| O R$. Hence $B R O S$ is a parallelogram. Let $L$ be the point symmetric to $B$ with respect to $R$. Then $R L O S$ is also a parallelogram, and since $S R \\perp B M$, we obtain $O L \\perp B M$. However, we also have $L M \\perp B M$, from which we conclude that $O, L, M$ are collinear and $O M \\perp B M$. Second solution. The lines $B M, N K$, and $C A$ are the radical axes of pairs of the three circles, and hence they intersect at a single point $P$. Also, the quadrilateral $M N C P$ is cyclic. Let $O A=O C=O K=O N=r$. We then have $B M \\cdot B P=B N \\cdot B C=O B^{2}-r^{2}$, $P M \\cdot P B=P N \\cdot P K=O P^{2}-r^{2}$. It follows that $O B^{2}-O P^{2}=$ $B P(B M-P M)=B M^{2}-P M^{2}$, which implies that $O M \\perp M B$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-489.jpg?height=360&width=532&top_left_y=760&top_left_x=800)", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (NET 3) ${ }^{\\mathrm{IMO} 3}$ The weight $w(p)$ of a polynomial $p, p(x)=\\sum_{i=0}^{n} a_{i} x^{i}$, with integer coefficients $a_{i}$ is defined as the number of its odd coefficients. For $i=0,1,2, \\ldots$, let $q_{i}(x)=(1+x)^{i}$. Prove that for any finite sequence $0 \\leq i_{1}<i_{2}<\\cdots<i_{n}$, the inequality  $$ w\\left(q_{i_{1}}+\\cdots+q_{i_{n}}\\right) \\geq w\\left(q_{i_{1}}\\right) $$  holds.", "solution": "3. We shall write $P \\equiv Q$ for two polynomials $P$ and $Q$ if $P(x)-Q(x)$ has even coefficients. We observe that $(1+x)^{2^{m}} \\equiv 1+x^{2^{m}}$ for every $m \\in \\mathbb{N}$. Consequently, for every polynomial $p$ with degree less than $k=2^{m}, w\\left(p \\cdot q_{k}\\right)=2 w(p)$. Now we prove the inequality from the problem by induction on $i_{n}$. If $i_{n} \\leq 1$, the inequality is trivial. Assume it is true for any sequence with $i_{1}<\\cdots<i_{n}<2^{m}(m \\geq 1)$, and let there be given a sequence with $k=2^{m} \\leq i_{n}<2^{m+1}$. Consider two cases. (i) $i_{1} \\geq k$. Then $w\\left(q_{i_{1}}+\\cdots+q_{i_{n}}\\right)=2 w\\left(q_{i_{1}-k}+\\cdots+q_{i_{n}-k}\\right) \\geq 2 w\\left(q_{i_{1}-k}\\right)=$ $w\\left(q_{i_{1}}\\right)$. (ii) $i_{1}<k$. Then the polynomial $p=q_{i_{1}}+\\cdots+q_{i_{n}}$ has the form  $$ p=\\sum_{i=0}^{k-1} a_{i} x^{i}+(1+x)^{k} \\sum_{i=0}^{k-1} b_{i} x^{i} \\equiv \\sum_{i=0}^{k-1}\\left[\\left(a_{i}+b_{i}\\right) x^{i}+b_{i} x^{i+k}\\right] $$  Whenever some $a_{i}$ is odd, either $a_{i}+b_{i}$ or $b_{i}$ in the above sum will be odd. It follows that $w(p) \\geq w\\left(q_{i_{1}}\\right)$, as claimed.  The proof is complete.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (AUS 1) ${ }^{\\mathrm{IMO} 2}$ Each of the numbers in the set $N=\\{1,2,3, \\ldots, n-1\\}$, where $n \\geq 3$, is colored with one of two colors, say red or black, so that: (i) $i$ and $n-i$ always receive the same color, and (ii) for some $j \\in N$, relatively prime to $n, i$ and $|j-i|$ receive the same color for all $i \\in N, i \\neq j$. Prove that all numbers in $N$ must receive the same color.", "solution": "4. Let $\\langle x\\rangle$ denote the residue of an integer $x$ modulo $n$. Also, we write $a \\sim b$ if $a$ and $b$ receive the same color. We claim that all the numbers $\\langle i j\\rangle$, $i=1,2, \\ldots, n-1$, are of the same color. Since $j$ and $n$ are coprime, this will imply the desired result. We use induction on $i$. For $i=1$ the statement is trivial. Assume now that the statement is true for $i=1, \\ldots, k-1$. For $1<k<n$ we have $\\langle k j\\rangle \\neq j$. If $\\langle k j\\rangle>j$, then by (ii), $\\langle k j\\rangle \\sim\\langle k j\\rangle-j=\\langle(k-1) j\\rangle$. If otherwise $\\langle k j\\rangle<j$, then by (ii) and (i), $\\langle k j\\rangle \\sim j-\\langle k j\\rangle \\sim n-j+\\langle k j\\rangle=\\langle(k-1) j\\rangle$. This completes the induction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (ROM 1) Let $D$ be the interior of the circle $C$ and let $A \\in C$. Show that the function $f: D \\rightarrow \\mathbb{R}, f(M)=\\frac{|M A|}{\\left|M M^{\\prime}\\right|}$, where $M^{\\prime}=(A M \\cap C$, is strictly convex; i.e., $f(P)<\\frac{f\\left(M_{1}\\right)+f\\left(M_{2}\\right)}{2}, \\forall M_{1}, M_{2} \\in D, M_{1} \\neq M_{2}$, where $P$ is the midpoint of the segment $M_{1} M_{2}$.", "solution": "5. Let w.l.o.g. circle $C$ have unit radius. For each $m \\in \\mathbb{R}$, the locus of points $M$ such that $f(M)=m$ is the circle $C_{m}$ with radius $r_{m}=m /(m+1)$, that is tangent to $C$ at $A$. Let $O_{m}$ be the center of $C_{m}$. We have to show that if $M \\in C_{m}$ and $N \\in C_{n}$, where $m, n>0$, then the midpoint $P$ of $M N$ lies inside the circle $C_{(m+n) / 2}$. This is trivial if $m=n$, so let $m \\neq n$. For fixed $M, P$ is in the image $C_{n}^{\\prime}$ of $C_{n}$ under the homothety with center $M$ and coefficient $1 / 2$. The center of the circle $C_{n}^{\\prime}$ is at the midpoint of $O_{n} M$. If we let both $M$ and $N$ vary, $P$ will be on the union of circles with radius $r_{n} / 2$ and centers in the image of $C_{m}$ under the homothety with center $O_{n}$ and coefficient $1 / 2$. Hence $P$ is not outside the circle centered at the midpoint $O_{m} O_{n}$ and with radius $\\left(r_{m}+r_{n}\\right) / 2$. It remains to show that $r_{(m+n) / 2}>\\left(r_{m}+r_{n}\\right) / 2$. But this inequality is easily reduced to $(m-n)^{2}>0$, which is true.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (POL 1) Let $x_{n}=\\sqrt[2]{2+\\sqrt[3]{3+\\ldots+\\sqrt[n]{n}}}$. Prove that  $$ x_{n+1}-x_{n}<\\frac{1}{n!}, \\quad n=2,3, \\ldots $$  ## Alternatives", "solution": "6. Let us set  $$ \\begin{aligned} & x_{n, i}=\\sqrt[i]{i+\\sqrt[i+1]{i+1+\\cdots+\\sqrt[n]{n}}} \\\\ & y_{n, i}=x_{n+1, i}^{i-1}+x_{n+1, i}^{i-2} x_{n, i}+\\cdots+x_{n, i}^{i-1} \\end{aligned} $$  In particular, $x_{n, 2}=x_{n}$ and $x_{n, i}=0$ for $i>n$. We observe that for $n>i>2$,  $$ x_{n+1, i}-x_{n, i}=\\frac{x_{n+1, i}^{i}-x_{n, i}^{i}}{y_{n, i}}=\\frac{x_{n+1, i+1}-x_{n, i+1}}{y_{n, i}} . $$  Since $y_{n, i}>i x_{n, i}^{i-1} \\geq i^{1+(i-1) / i} \\geq i^{3 / 2}$ and $x_{n+1, n+1}-x_{n, n+1}=\\sqrt[n+1]{n+1}$, simple induction gives  $$ x_{n+1}-x_{n} \\leq \\frac{\\sqrt[n+1]{n+1}}{(n!)^{3 / 2}}<\\frac{1}{n!} \\quad \\text { for } n>2 $$  The inequality for $n=2$ is directly verified.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. 1a.(CZS 3) The positive integers $x_{1}, \\ldots, x_{n}, n \\geq 3$, satisfy $x_{1}<x_{2}<$ $\\cdots<x_{n}<2 x_{1}$. Set $P=x_{1} x_{2} \\cdots x_{n}$. Prove that if $p$ is a prime number, $k$ a positive integer, and $P$ is divisible by $p^{k}$, then $\\frac{P}{p^{k}} \\geq n!$.", "solution": "7. Let $k_{i} \\geq 0$ be the largest integer such that $p^{k_{i}} \\mid x_{i}, i=1, \\ldots, n$, and $y_{i}=x_{i} / p^{k_{i}}$. We may assume that $k=k_{1}+\\cdots+k_{n}$. All the $y_{i}$ must be distinct. Indeed, if $y_{i}=y_{j}$ and $k_{i}>k_{j}$, then $x_{i} \\geq p x_{j} \\geq 2 x_{i} \\geq 2 x_{1}$, which is impossible. Thus $y_{1} y_{2} \\ldots y_{n}=P / p^{k} \\geq n!$. If equality holds, we must have $y_{i}=1, y_{j}=2$ and $y_{k}=3$ for some $i, j, k$. Thus $p \\geq 5$, which implies that either $y_{i} / y_{j} \\leq 1 / 2$ or $y_{i} / y_{j} \\geq 5 / 2$, which is impossible. Hence the inequality is strict.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. 1b.(TUR 5) Find the smallest positive integer $n$ such that (i) $n$ has exactly 144 distinct positive divisors, and (ii) there are ten consecutive integers among the positive divisors of $n$.", "solution": "8. Among ten consecutive integers that divide $n$, there must exist numbers divisible by $2^{3}, 3^{2}, 5$, and 7 . Thus the desired number has the form $n=$ $2^{\\alpha_{1}} 3^{\\alpha_{2}} 5^{\\alpha_{3}} 7^{\\alpha_{4}} 11^{\\alpha_{5}} \\cdots$, where $\\alpha_{1} \\geq 3, \\alpha_{2} \\geq 2, \\alpha_{3} \\geq 1, \\alpha_{4} \\geq 1$. Since $n$ has $\\left(\\alpha_{1}+1\\right)\\left(\\alpha_{2}+1\\right)\\left(\\alpha_{3}+1\\right) \\cdots$ distinct factors, and $\\left(\\alpha_{1}+1\\right)\\left(\\alpha_{2}+1\\right)\\left(\\alpha_{3}+\\right.$ $1)\\left(\\alpha_{4}+1\\right) \\geq 48$, we must have $\\left(\\alpha_{5}+1\\right) \\cdots \\leq 3$. Hence at most one $\\alpha_{j}$, $j>4$, is positive, and in the minimal $n$ this must be $\\alpha_{5}$. Checking through the possible combinations satisfying $\\left(\\alpha_{1}+1\\right)\\left(\\alpha_{2}+1\\right) \\cdots\\left(\\alpha_{5}+1\\right)=144$ one finds that the minimal $n$ is $2^{5} \\cdot 3^{2} \\cdot 5 \\cdot 7 \\cdot 11=110880$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1985", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. 2a.(USA 3) Determine the radius of a sphere $S$ that passes through the centroids of each face of a given tetrahedron $T$ inscribed in a unit sphere with center $O$. Also, determine the distance from $O$ to the center of $S$ as a function of the edges of $T$.", "solution": "9. Let $\\vec{a}, \\vec{b}, \\vec{c}, \\vec{d}$ denote the vectors $\\overrightarrow{O A}, \\overrightarrow{O B}, \\overrightarrow{O C}, \\overrightarrow{O D}$ respectively. Then $|\\vec{a}|=|\\vec{b}|=|\\vec{c}|=|\\vec{d}|=1$. The centroids of the faces are $(\\vec{b}+\\vec{c}+\\vec{d}) / 3$, $(\\vec{a}+\\vec{c}+\\vec{d}) / 3$, etc., and each of these is at distance $1 / 3$ from $P=$ $(\\vec{a}+\\vec{b}+\\vec{c}+\\vec{d}) / 3$; hence the required radius is $1 / 3$. To compute $|P|$ as a function of the edges of $A B C D$, observe that $A B^{2}=(\\vec{b}-\\vec{a})^{2}=$ $2-2 \\vec{a} \\cdot \\vec{b}$ etc. Now  $$ \\begin{aligned} P^{2} & =\\frac{|\\vec{a}+\\vec{b}+\\vec{c}+\\vec{d}|^{2}}{9} \\\\ & =\\frac{16-2\\left(A B^{2}+B C^{2}+A C^{2}+A D^{2}+B D^{2}+C D^{2}\\right)}{9} \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (GBR 3) ${ }^{\\text {IMO5 }}$ Find, with proof, all functions $f$ defined on the nonnegative real numbers and taking nonnegative real values such that (i) $f[x f(y)] f(y)=f(x+y)$, (ii) $f(2)=0$ but $f(x) \\neq 0$ for $0 \\leq x<2$.", "solution": "1. If $w>2$, then setting in (i) $x=w-2, y=2$, we get $f(w)=f((w-$ 2) $f(w)) f(2)=0$. Thus  $$ f(x)=0 \\quad \\text { if and only if } \\quad x \\geq 2 $$  Now let $0 \\leq y<2$ and $x \\geq 0$. The LHS in (i) is zero if and only if $x f(y) \\geq 2$, while the RHS is zero if and only if $x+y \\geq 2$. It follows that $x \\geq 2 / f(y)$ if and only if $x \\geq 2-y$. Therefore  $$ f(y)=\\left\\{\\begin{array}{cl} \\frac{2}{2-y} & \\text { for } 0 \\leq y<2 \\\\ 0 & \\text { for } y \\geq 2 \\end{array}\\right. $$  The confirmation that $f$ satisfies the given conditions is straightforward.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (HUN 2) Three persons $A, B, C$, are playing the following game: A $k$ element subset of the set $\\{1, \\ldots, 1986\\}$ is randomly chosen, with an equal probability of each choice, where $k$ is a fixed positive integer less than or equal to 1986. The winner is $A, B$ or $C$, respectively, if the sum of the chosen numbers leaves a remainder of 0,1 , or 2 when divided by 3 . For what values of $k$ is this game a fair one? (A game is fair if the three outcomes are equally probable.)", "solution": "10. The set $X=\\{1, \\ldots, 1986\\}$ splits into triads $T_{1}, \\ldots, T_{662}$, where $T_{j}=$ $\\{3 j-2,3 j-1,3 j\\}$. Let $\\mathcal{F}$ be the family of all $k$-element subsets $P$ such that $\\left|P \\cap T_{j}\\right|=1$ or 2 for some index $j$. If $j_{0}$ is the smallest such $j_{0}$, we define $P^{\\prime}$ to be the $k$-element set obtained from $P$ by replacing the elements of $P \\cap T_{j_{0}}$ by the ones following cyclically inside $T_{j_{0}}$. Let $s(P)$ denote the remainder modulo 3 of the sum of elements of $P$. Then $s(P), s\\left(P^{\\prime}\\right), s\\left(P^{\\prime \\prime}\\right)$ are distinct, and $P^{\\prime \\prime \\prime}=P$. Thus the operator ${ }^{\\prime}$ gives us a bijective correspondence between the sets $X \\in \\mathcal{F}$ with $s(P)=0$, those with $s(P)=1$, and those with $s(P)=2$. If $3 \\nmid k$ is not divisible by 3 , then each $k$-element subset of $X$ belongs to $\\mathcal{F}$, and the game is fair. If $3 \\mid k$, then $k$-element subsets not belonging to $\\mathcal{F}$ are those that are unions of several triads. Since every such subset has the sum of elements divisible by 3 , it follows that player $A$ has the advantage.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (BUL 1) Let $f(n)$ be the least number of distinct points in the plane such that for each $k=1,2, \\ldots, n$ there exists a straight line containing exactly $k$ of these points. Find an explicit expression for $f(n)$. Simplified version. Show that $f(n)=\\left[\\frac{n+1}{2}\\right]\\left[\\frac{n+2}{2}\\right]$ ( $[x]$ denoting the greatest integer not exceeding $x$ ).", "solution": "11. Let $X$ be a finite set in the plane and $l_{k}$ a line containing exactly $k$ points of $X(k=1, \\ldots, n)$. Then $l_{n}$ contains $n$ points, $l_{n-1}$ contains at least $n-2$ points not lying on $l_{n}, l_{n-2}$ contains at least $n-4$ points not lying on $l_{n}$ or $l_{n-1}$, etc. It follows that  $$ |X| \\geq g(n)=n+(n-2)+(n-4)+\\cdots+\\left(n-2\\left[\\frac{n}{2}\\right]\\right) . $$  Hence $f(n) \\geq g(n)=\\left[\\frac{n+1}{2}\\right]\\left[\\frac{n+2}{2}\\right]$, where the last equality is easily proved by induction. We claim that $f(n)=g(n)$. To prove this, we shall inductively construct a set $X_{n}$ of cardinality $g(n)$ with the required property. For $n \\leq 2$ a one-point and two-point set satisfy the requirements. Assume that $X_{n}$ is a set of $g(n)$ points and that $l_{k}$ is a line containing exactly $k$ points of $X_{n}, k=1, \\ldots, n$. Consider any line $l$ not parallel to any of the $l_{k}$ 's and not containing any point of $X_{n}$ or any intersection point of the $l_{k}$. Let $l$ intersect $l_{k}$ in a point $P_{k}, k=1, \\ldots, n$, and let $P_{n+1}, P_{n+2}$ be two points on $l$ other than $P_{1}, \\ldots, P_{n}$. We define $X_{n+2}=X_{n} \\cup\\left\\{P_{1}, \\ldots, P_{n+2}\\right\\}$. The set $X_{n+2}$ consists of $g(n)+(n+2)=g(n+2)$ points. Since the lines $l, l_{n}, \\ldots, l_{2}, l_{1}$ meet $X_{n}$ in $n+2, n+1, \\ldots, 3,2$ points respectively (and there clearly exists a line containing only one point of $X_{n+2}$ ), this set also meets the demands.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (GDR 3) ${ }^{\\mathrm{IMO} 3}$ To each vertex $P_{i}(i=1, \\ldots, 5)$ of a pentagon an integer $x_{i}$ is assigned, the sum $s=\\sum x_{i}$ being positive. The following operation is allowed, provided at least one of the $x_{i}$ 's is negative: Choose a negative $x_{i}$, replace it by $-x_{i}$, and add the former value of $x_{i}$ to the integers assigned to the two neighboring vertices of $P_{i}$ (the remaining two integers are left unchanged). This operation is to be performed repeatedly until all negative integers disappear. Decide whether this procedure must eventually terminate.", "solution": "12. We define $f\\left(x_{1}, \\ldots, x_{5}\\right)=\\sum_{i=1}^{5}\\left(x_{i+1}-x_{i-1}\\right)^{2}\\left(x_{0}=x_{5}, x_{6}=x_{1}\\right)$. Assuming that $x_{3}<0$, according to the rules the lattice vector $X=$ $\\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\\right)$ changes into $Y=\\left(x_{1}, x_{2}+x_{3},-x_{3}, x_{4}+x_{3}, x_{5}\\right)$. Then  $$ \\begin{aligned} f(Y)-f(X)= & \\left(x_{2}+x_{3}-x_{5}\\right)^{2}+\\left(x_{1}+x_{3}\\right)^{2}+\\left(x_{2}-x_{4}\\right)^{2} \\\\ & +\\left(x_{3}+x_{5}\\right)^{2}+\\left(x_{1}-x_{3}-x_{4}\\right)^{2}-\\left(x_{2}-x_{5}\\right)^{2} \\\\ & -\\left(x_{3}-x_{1}\\right)^{2}-\\left(x_{4}-x_{2}\\right)^{2}-\\left(x_{5}-x_{3}\\right)^{2}-\\left(x_{1}-x_{4}\\right)^{2} \\\\ = & 2 x_{3}\\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\\right)=2 x_{3} S<0 \\end{aligned} $$  Thus $f$ strictly decreases after each step, and since it takes only positive integer values, the number of steps must be finite. Remark. One could inspect the behavior of $g(x)=\\sum_{i=1}^{5} \\sum_{j=1}^{5} \\mid x_{i}+x_{i+1}+$ $\\cdots+x_{j-1} \\mid$ instead. Then $g(Y)-g(X)=\\left|S+x_{3}\\right|-\\left|S-x_{3}\\right|>0$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (FRG 3) A particle moves from $(0,0)$ to $(n, n)$ directed by a fair coin. For each head it moves one step east and for each tail it moves one step north. At $(n, y), y<n$, it stays there if a head comes up and at $(x, n)$, $x<n$, it stays there if a tail comes up. Let $k$ be a fixed positive integer. Find the probability that the particle needs exactly $2 n+k$ tosses to reach $(n, n)$.", "solution": "13. Let us consider the infinite integer lattice and assume that having reached a point $(x, n)$ or $(n, y)$, the particle continues moving east and north following the rules of the game. The required probability $p_{k}$ is equal to the probability of getting to one of the points $E_{1}(n, n+k), E_{2}(n+k, n)$, but without passing through $(n, n+k-1)$ or $(n+k-1, n)$. Thus $p$ is equal to the probability $p_{1}$ of getting to $E_{1}(n, n+k)$ via $D_{1}(n-1, n+k)$ plus the probability $p_{2}$ of getting to $E_{2}(n+k, n)$ via $D_{2}(n+k, n-1)$. Both $p_{1}$ and $p_{2}$ are easily seen to be equal to $\\binom{2 n+k-1}{n-1} 2^{-2 n-k}$, and therefore $p=\\binom{2 n+k-1}{n-1} 2^{-2 n-k+1}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (IRE 1) The circle inscribed in a triangle $A B C$ touches the sides $B C, C A, A B$ in $D, E, F$, respectively, and $X, Y, Z$ are the midpoints of $E F, F D, D E$, respectively. Prove that the centers of the inscribed circle and of the circles around $X Y Z$ and $A B C$ are collinear.", "solution": "14. We shall use the following simple fact.  Lemma. If $\\widehat{k}$ is the image of a circle $k$ under an inversion centered at a point $Z$, and $O_{1}, O_{2}$ are centers of $k$ and $\\widehat{k}$, then $O_{1}, O_{2}$, and $Z$ are collinear. Proof. The result follows immediately from the symmetry with respect to the line $Z O_{1}$. Let $I$ be the center of the inscribed circle $i$. Since $I X \\cdot I A=I E^{2}$, the inversion with respect to $i$ takes points $A$ into $X$, and analogously $B, C$ into $Y, Z$ respectively. It follows from the lemma that the center of circle $A B C$, the center of circle $X Y Z$, and point $I$ are collinear.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (NET 2) Let $A B C D$ be a convex quadrilateral whose vertices do not lie on a circle. Let $A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime}$ be a quadrangle such that $A^{\\prime}, B^{\\prime}, C^{\\prime}, D^{\\prime}$ are the centers of the circumcircles of triangles $B C D, A C D, A B D$, and $A B C$. We write $T(A B C D)=A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime}$. Let us define $A^{\\prime \\prime} B^{\\prime \\prime} C^{\\prime \\prime} D^{\\prime \\prime}=$ $T\\left(A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime}\\right)=T(T(A B C D))$. (a) Prove that $A B C D$ and $A^{\\prime \\prime} B^{\\prime \\prime} C^{\\prime \\prime} D^{\\prime \\prime}$ are similar. (b) The ratio of similitude depends on the size of the angles of $A B C D$. Determine this ratio.", "solution": "15. (a) This is the same problem as $S L 82$-14. (b) If $S$ is the midpoint of $A C$, we have $B^{\\prime} S=A C \\frac{\\cos \\angle D}{2 \\sin \\angle D}, D^{\\prime} S=$ $A C \\frac{\\cos \\angle B}{2 \\sin \\angle B}, B^{\\prime} D^{\\prime}=A C\\left|\\frac{\\sin (\\angle B+\\angle D)}{2 \\sin \\angle B \\sin \\angle D}\\right|$. These formulas are true also if $\\angle B>90^{\\circ}$ or $\\angle D>90^{\\circ}$. We similarly obtain that $A^{\\prime \\prime} C^{\\prime \\prime}=$ $B^{\\prime} D^{\\prime}\\left|\\frac{\\sin \\left(\\angle A^{\\prime}+\\angle C^{\\prime}\\right)}{2 \\sin \\angle A^{\\prime} \\sin \\angle C^{\\prime}}\\right|$. Therefore  $$ A^{\\prime \\prime} C^{\\prime \\prime}=A C \\frac{\\sin ^{2}(\\angle A+\\angle C)}{4 \\sin \\angle A \\sin \\angle B \\sin \\angle C \\sin \\angle D} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (ISR 1) ${ }^{\\mathrm{IMO}}$ Let $A, B$ be adjacent vertices of a regular $n$-gon in the plane and let $O$ be its center. Now let the triangle $A B O$ glide around the polygon in such a way that the points $A$ and $B$ move along the whole circumference of the polygon. Describe the figure traced by the vertex $O$.", "solution": "16. Let $Z$ be the center of the polygon.  Suppose that at some moment we have $A \\in P_{i-1} P_{i}$ and $B \\in$ $P_{i} P_{i+1}$, where $P_{i-1}, P_{i}, P_{i+1}$ are adjacent vertices of the polygon. Since $\\angle A O B=180^{\\circ}-\\angle P_{i-1} P_{i} P_{i+1}$, the quadrilateral $A P_{i} B O$ is cyclic. Hence $\\angle A P_{i} O=\\angle A B O=\\angle A P_{i} Z$, which means that $O \\in P_{i} Z$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-495.jpg?height=329&width=390&top_left_y=251&top_left_x=903)  Moreover, from $O P_{i}=2 r \\sin \\angle P_{i} A O$, where $r$ is the radius of circle $A P_{i} B O$, we obtain that $Z P_{i} \\leq O P_{i} \\leq Z P_{i} / \\cos (\\pi / n)$. Thus $O$ traces a segment $Z Z_{i}$ as $A$ and $B$ move along $P_{i-1} P_{i}$ and $P_{i} P_{i+1}$ respectively, where $Z_{i}$ is a point on the ray $P_{i} Z$ with $P_{i} Z_{i} \\cos (\\pi / n)=P_{i} Z$. When $A, B$ move along the whole circumference of the polygon, $O$ traces an asterisk consisting of $n$ segments of equal length emanating from $Z$ and pointing away from the vertices.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. ( $\\mathbf{C H N} 3)^{\\mathrm{IMO} 2}$ Let $A, B, C$ be fixed points in the plane. A man starts from a certain point $P_{0}$ and walks directly to $A$. At $A$ he turns his direction by $60^{\\circ}$ to the left and walks to $P_{1}$ such that $P_{0} A=A P_{1}$. After he does the same action 1986 times successively around the points $A, B, C, A, B, C, \\ldots$, he returns to the starting point. Prove that $\\triangle A B C$ is equilateral and that the vertices $A, B, C$ are arranged counterclockwise.", "solution": "17. We use complex numbers to represent the position of a point in the plane. For convenience, let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, \\ldots$ be $A, B, C, A, B, \\ldots$ respectively, and let $P_{0}$ be the origin. After the $k$ th step, the position of $P_{k}$ will be $P_{k}=A_{k}+\\left(P_{k-1}-A_{k}\\right) u, k=1,2,3, \\ldots$, where $u=e^{4 \\pi \\tau / 3}$. We easily obtain  $$ P_{k}=(1-u)\\left(A_{k}+u A_{k-1}+u^{2} A_{k-2}+\\cdots+u^{k-1} A_{1}\\right) $$  The condition $P_{0} \\equiv P_{1986}$ is equivalent to $A_{1986}+u A_{1985}+\\cdots+u^{1984} A_{2}+$ $u^{1985} A_{1}=0$, which, having in mind that $A_{1}=A_{4}=A_{7}=\\cdots, A_{2}=A_{5}=$ $A_{8}=\\cdots, A_{3}=A_{6}=A_{9}=\\cdots$, reduces to  $$ 662\\left(A_{3}+u A_{2}+u^{2} A_{1}\\right)=\\left(1+u^{3}+\\cdots+u^{1983}\\right)\\left(A_{3}+u A_{2}+u^{2} A_{1}\\right)=0 $$  It follows that $A_{3}-A_{1}=u\\left(A_{1}-A_{2}\\right)$, and the assertion follows. Second solution. Let $f_{P}$ denote the rotation with center $P$ through $120^{\\circ}$ clockwise. Let $f_{1}=f_{A}$. Then $f_{1}\\left(P_{0}\\right)=P_{1}$. Let $B^{\\prime}=f_{1}(B), C^{\\prime}=f_{1}(C)$, and $f_{2}=f_{B^{\\prime}}$. Then $f_{2}\\left(P_{1}\\right)=P_{2}$ and $f_{2}\\left(A B^{\\prime} C^{\\prime}\\right)=A^{\\prime} B^{\\prime} C^{\\prime \\prime}$. Finally, let $f_{3}=f_{C^{\\prime \\prime}}$ and $f_{3}\\left(A^{\\prime} B^{\\prime} C^{\\prime \\prime}\\right)=A^{\\prime \\prime} B^{\\prime \\prime} C^{\\prime \\prime}$. Then $g=f_{3} f_{2} f_{1}$ is a translation sending $P_{0}$ to $P_{3}$ and $C$ to $C^{\\prime \\prime}$. Now $P_{1986}=P_{0}$ implies that $g^{662}$ is the identity, and thus $C=C^{\\prime \\prime}$. Let $K$ be such that $A B K$ is equilateral and positively oriented. We observe that $f_{2} f_{1}(K)=K$; therefore the rotation $f_{2} f_{1}$ satisfies $f_{2} f_{1}(P) \\neq P$ for $P \\neq K$. Hence $f_{2} f_{1}(C)=C^{\\prime \\prime}=C$ implies $K=C$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (TUR 1) Let $A X, B Y, C Z$ be three cevians concurrent at an interior point $D$ of a triangle $A B C$. Prove that if two of the quadrangles $D Y A Z, D Z B X, D X C Y$ are circumscribable, so is the third.", "solution": "18. We shall use the following criterion for a quadrangle to be circumscribable. Lemma. The quadrangle $A Y D Z$ is circumscribable if and only if $D B-$ $D C=A B-A C$. Proof. Suppose that $A Y D Z$ is circumscribable and that the incircle is tangent to $A Z, Z D, D Y, Y A$ at $M, N, P, Q$ respectively. Then $D B-D C=P B-N C=M B-Q C=A B-A C$. Conversely, assume that $D B-D C=A B-A C$ and let a tangent from $D$ to the incircle of the triangle $A C Z$ meet $C Z$ and $C A$ at $D^{\\prime} \\neq Z$ and $Y^{\\prime} \\neq A$ respectively. According to the first part we have $D^{\\prime} B-D^{\\prime} C=A B-A C$. It follows that $\\left|D^{\\prime} B-D B\\right|=$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-496.jpg?height=315&width=387&top_left_y=236&top_left_x=900) $\\left|D^{\\prime} C-D C\\right|=D D^{\\prime}$, implying that $D^{\\prime} \\equiv D$. Let us assume that $D Z B X$ and $D X C Y$ are circumscribable. Using the lemma we obtain $D C-D A=B C-B A$ and $D A-D B=C A-C B$. Adding these two inequalities yields $D C-D B=A C-A B$, and the statement follows from the lemma.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (BUL 2) A tetrahedron $A B C D$ is given such that $A D=B C=a$; $A C=B D=b ; A B \\cdot C D=c^{2}$. Let $f(P)=A P+B P+C P+D P$, where $P$ is an arbitrary point in space. Compute the least value of $f(P)$.", "solution": "19. Let $M$ and $N$ be the midpoints of segments $A B$ and $C D$, respectively. The given conditions imply that $\\triangle A B D \\cong \\triangle B A C$ and $\\triangle C D A \\cong \\triangle D C B$; hence $M C=M D$ and $N A=N B$. It follows that $M$ and $N$ both lie on the perpendicular bisectors of $A B$ and $C D$, and consequently $M N$ is the common perpendicular bisector of $A B$ and $C D$. Points $B$ and $C$ are symmetric to $A$ and $D$ with respect to $M N$. Now if $P$ is a point in space and $P^{\\prime}$ the point symmetric to $P$ with respect to $M N$, we have $B P=A P^{\\prime}, C P=D P^{\\prime}$, and thus $f(P)=A P+A P^{\\prime}+D P+D P^{\\prime}$. Let $P P^{\\prime}$ intersect $M N$ in $Q$. Then $A P+A P^{\\prime} \\geq 2 A Q$ and $D P+D P^{\\prime} \\geq 2 D Q$, from which it follows that $f(P) \\geq 2(A Q+D Q)=f(Q)$. It remains to minimize $f(Q)$ with $Q$ moving along the line $M N$. Let us rotate point $D$ around $M N$ to a point $D^{\\prime}$ that belongs to the plane $A M N$, on the side of $M N$ opposite to $A$. Then $f(Q)=2\\left(A Q+D^{\\prime} Q\\right) \\geq$ $A D^{\\prime}$, and equality occurs when $Q$ is the intersection of $A D^{\\prime}$ and $M N$. Thus $\\min f(Q)=A D^{\\prime}$. We note that $4 M D^{2}=2 A D^{2}+2 B D^{2}-A B^{2}=$ $2 a^{2}+2 b^{2}-A B^{2}$ and $4 M N^{2}=4 M D^{2}-C D^{2}=2 a^{2}+2 b^{2}-A B^{2}-C D^{2}$. Now, $A D^{\\prime 2}=\\left(A M+D^{\\prime} N\\right)^{2}+M N^{2}$, which together with $A M+D^{\\prime} N=$ $(a+b) / 2$ gives us  $$ A D^{\\prime 2}=\\frac{a^{2}+b^{2}+A B \\cdot C D}{2}=\\frac{a^{2}+b^{2}+c^{2}}{2} $$  We conclude that $\\min f(Q)=\\sqrt{\\left(a^{2}+b^{2}+c^{2}\\right) / 2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (SWE 2) Let $f(x)=x^{n}$ where $n$ is a fixed positive integer and $x=$ $1,2, \\ldots$. Is the decimal expansion $a=0 . f(1) f(2) f(3) \\ldots$ rational for any value of $n$ ? The decimal expansion of $a$ is defined as follows: If $f(x)=d_{1}(x) d_{2}(x) \\ldots$ $\\ldots d_{r(x)}(x)$ is the decimal expansion of $f(x)$, then $a=0.1 d_{1}(2) d_{2}(2) \\ldots$ $\\ldots d_{r(2)}(2) d_{1}(3) \\ldots d_{r(3)}(3) d_{1}(4) \\ldots$.", "solution": "2. No. If $a$ were rational, its decimal expansion would be periodic from some point. Let $p$ be the number of decimals in the period. Since $f\\left(10^{2 p}\\right)$ has $2 n p$ zeros, it contains a full periodic part; hence the period would consist only of zeros, which is impossible.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. (CAN 3) Prove that the sum of the face angles at each vertex of a tetrahedron is a straight angle if and only if the faces are congruent triangles.", "solution": "20. If the faces of the tetrahedron $A B C D$ are congruent triangles, we must have $A B=C D, A C=B D$, and $A D=B C$. Then the sum of angles at $A$ is $\\angle B A C+\\angle C A D+\\angle D A B=\\angle B D C+\\angle C B D+\\angle D C B=180^{\\circ}$. We now assume that the sum of angles at each vertex is $180^{\\circ}$. Let us construct triangles $B C D^{\\prime}, C A D^{\\prime \\prime}, A B D^{\\prime \\prime \\prime}$ in the plane $A B C$, exterior to $\\triangle A B C$, such that $\\triangle B C D^{\\prime} \\cong \\triangle B C D, \\triangle C A D^{\\prime \\prime} \\cong \\triangle C A D$, and $\\triangle A B D^{\\prime \\prime \\prime} \\cong \\triangle A B D$. Then by the assumption, $A \\in D^{\\prime \\prime} D^{\\prime \\prime \\prime}, B \\in D^{\\prime \\prime \\prime} D^{\\prime}$, and $C \\in D^{\\prime} D^{\\prime \\prime}$. Since also $D^{\\prime \\prime} A=D^{\\prime \\prime \\prime} A=D A$, etc., $A, B, C$ are the mid- points of segments $D^{\\prime \\prime} D^{\\prime \\prime \\prime}, D^{\\prime \\prime \\prime} D^{\\prime}, D^{\\prime} D^{\\prime \\prime}$ respectively. Thus the triangles $A B C, B C D^{\\prime}, C A D^{\\prime \\prime}, A B D^{\\prime \\prime \\prime}$ are congruent, and the statement follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. (TUR 2) Let $A B C D$ be a tetrahedron having each sum of opposite sides equal to 1. Prove that  $$ r_{A}+r_{B}+r_{C}+r_{D} \\leq \\frac{\\sqrt{3}}{3} $$  where $r_{A}, r_{B}, r_{C}, r_{D}$ are the inradii of the faces, equality holding only if $A B C D$ is regular.", "solution": "21. Since the sum of all edges of $A B C D$ is 3 , the statement of the problem is an immediate consequence of the following statement: Lemma. Let $r$ be the inradius of a triangle with sides $a, b, c$. Then $a+$ $b+c \\geq 6 \\sqrt{3} \\cdot r$, with equality if and only if the triangle is equilateral. Proof. If $S$ and $p$ denotes the area and semiperimeter of the triangle, by Heron's formula and the AM-GM inequality we have  $$ \\begin{aligned} p r & =S=\\sqrt{p(p-a)(p-b)(p-c)} \\\\ & \\leq \\sqrt{p\\left(\\frac{(p-a)+(p-b)+(p-c)}{3}\\right)^{3}}=\\sqrt{\\frac{p^{4}}{27}}=\\frac{p^{2}}{3 \\sqrt{3}}, \\end{aligned} $$  i.e., $p \\geq 3 \\sqrt{3} \\cdot r$, which is equivalent to the claim.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (USA 3) Let $A, B$, and $C$ be three points on the edge of a circular chord such that $B$ is due west of $C$ and $A B C$ is an equilateral triangle whose side is 86 meters long. A boy swam from $A$ directly toward $B$. After covering a distance of $x$ meters, he turned and swam westward, reaching the shore after covering a distance of $y$ meters. If $x$ and $y$ are both positive integers, determine $y$.", "solution": "3. Let $E$ be the point where the boy turned westward, reaching the shore at $D$. Let the ray $D E$ cut $A C$ at $F$ and the shore again at $G$. Then $E F=$ $A E=x$ (because $A E F$ is an equilateral triangle) and $F G=D E=y$. From $A E \\cdot E B=D E \\cdot E G$ we obtain $x(86-x)=y(x+y)$. If $x$ is odd, then $x(86-x)$ is odd, while $y(x+y)$ is even. Hence $x$ is even, and so $y$ must also be even. Let $y=2 y_{1}$. The above equation can be rewritten as  $$ \\left(x+y_{1}-43\\right)^{2}+\\left(2 y_{1}\\right)^{2}=\\left(43-y_{1}\\right)^{2} \\text {. } $$  Since $y_{1}<43$, we have $\\left(2 y_{1}, 43-y_{1}\\right)=1$, and thus $\\left(\\left|x+y_{1}-43\\right|, 2 y_{1}, 43-\\right.$ $\\left.y_{1}\\right)$ is a primitive Pythagorean triple. Consequently there exist integers $a>b>0$ such that $y_{1}=a b$ and $43-y_{1}=a^{2}+b^{2}$. We obtain that $a^{2}+b^{2}+a b=43$, which has the unique solution $a=6, b=1$. Hence $y=12$ and $x=2$ or $x=72$. Remark. The Diophantine equation $x(86-x)=y(x+y)$ can be also solved directly. Namely, we have that $x(344-3 x)=(2 y+x)^{2}$ is a square, and since $x$ is even, we have $(x, 344-3 x)=2$ or 4 . Consequently $x, 344-3 x$ are either both squares or both two times squares. The rest is easy.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (CZS 3) Let $n$ be a positive integer and let $p$ be a prime number, $p>3$. Find at least $3(n+1)$ [easier version: $2(n+1)]$ sequences of positive integers $x, y, z$ satisfying  $$ x y z=p^{n}(x+y+z) $$  that do not differ only by permutation.", "solution": "4. Let $x=p^{\\alpha} x^{\\prime}, y=p^{\\beta} y^{\\prime}, z=p^{\\gamma} z^{\\prime}$ with $p \\nmid x^{\\prime} y^{\\prime} z^{\\prime}$ and $\\alpha \\geq \\beta \\geq \\gamma$. From the given equation it follows that $p^{n}(x+y)=z\\left(x y-p^{n}\\right)$ and consequently $z^{\\prime} \\mid x+y$. Since also $p^{\\gamma} \\mid x+y$, we have $z \\mid x+y$, i.e., $x+y=q z$. The given equation together with the last condition gives us  $$ x y=p^{n}(q+1) \\quad \\text { and } \\quad x+y=q z . $$  Conversely, every solution of (1) gives a solution of the given equation.  For $q=1$ and $q=2$ we obtain the following classes of $n+1$ solutions each:  $$ \\begin{array}{ll} q=1:(x, y, z)=\\left(2 p^{i}, p^{n-i}, 2 p^{i}+p^{n-i}\\right) & \\text { for } i=0,1,2, \\ldots, n \\\\ q=2:(x, y, z)=\\left(3 p^{j}, p^{n-j}, \\frac{3 p^{j}+p^{n-j}}{2}\\right) & \\text { for } j=0,1,2, \\ldots, n \\end{array} $$  For $n=2 k$ these two classes have a common solution $\\left(2 p^{k}, p^{k}, 3 p^{k}\\right)$; otherwise, all these solutions are distinct. One further solution is given by $(x, y, z)=\\left(1, p^{n}\\left(p^{n}+3\\right) / 2, p^{2}+2\\right)$, not included in the above classes for $p>3$. Thus we have found $2(n+1)$ solutions. Another type of solution is obtained if we put $q=p^{k}+p^{n-k}$. This yields the solutions  $$ (x, y, z)=\\left(p^{k}, p^{n}+p^{n-k}+p^{2 n-2 k}, p^{n-k}+1\\right) \\quad \\text { for } k=0,1, \\ldots, n $$  For $k<n$ these are indeed new solutions. So far, we have found $3(n+1)-1$ or $3(n+1)$ solutions. One more solution is given by $(x, y, z)=\\left(p, p^{n}+\\right.$ $\\left.p^{n-1}, p^{n-1}+p^{n-2}+1\\right)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (FRG 1) ${ }^{\\mathrm{IMO1}}$ The set $S=\\{2,5,13\\}$ has the property that for every $a, b \\in S, a \\neq b$, the number $a b-1$ is a perfect square. Show that for every positive integer $d$ not in $S$, the set $S \\cup\\{d\\}$ does not have the above property.", "solution": "5. Suppose that for every $a, b \\in\\{2,5,13, d\\}, a \\neq b$, the number $a b-1$ is a perfect square. In particular, for some integers $x, y, z$ we have  $$ 2 d-1=x^{2}, \\quad 5 d-1=y^{2}, \\quad 13 d-1=z^{2} $$  Since $x$ is clearly odd, $d=\\left(x^{2}+1\\right) / 2$ is also odd because $4 \\nmid x^{2}+1$. It follows that $y$ and $z$ are even, say $y=2 y_{1}$ and $z=2 z_{1}$. Hence $\\left(z_{1}-\\right.$ $\\left.y_{1}\\right)\\left(z_{1}+y_{1}\\right)=\\left(z^{2}-y^{2}\\right) / 4=2 d$. But in this case one of the factors $z_{1}-y_{1}$, $z_{1}+y_{1}$ is odd and the other one is even, which is impossible.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (NET 1) Find four positive integers each not exceeding 70000 and each having more than 100 divisors.", "solution": "6 . There are five such numbers:  $$ \\begin{aligned} 69300 & =2^{2} \\cdot 3^{2} \\cdot 5^{2} \\cdot 7 \\cdot 11: & 3 \\cdot 3 \\cdot 3 \\cdot 2 \\cdot 2 & =108 \\text { divisors; } \\\\ 50400 & =2^{5} \\cdot 3^{2} \\cdot 5^{2} \\cdot 7: & 6 \\cdot 3 \\cdot 3 \\cdot 2 & =108 \\text { divisors; } \\\\ 60480 & =2^{6} \\cdot 3^{3} \\cdot 5 \\cdot 7: & 7 \\cdot 4 \\cdot 2 \\cdot 2 & =112 \\text { divisors; } \\\\ 55440 & =2^{4} \\cdot 3^{2} \\cdot 5 \\cdot 7 \\cdot 11: & 5 \\cdot 3 \\cdot 2 \\cdot 2 \\cdot 2 & =120 \\text { divisors; } \\\\ 65520 & =2^{4} \\cdot 3^{2} \\cdot 5 \\cdot 7 \\cdot 13: & 5 \\cdot 3 \\cdot 2 \\cdot 2 \\cdot 2 & =120 \\text { divisors. } \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (FRA 5) Let real numbers $x_{1}, x_{2}, \\ldots, x_{n}$ satisfy $0<x_{1}<x_{2}<\\cdots<$ $x_{n}<1$ and set $x_{0}=0, x_{n+1}=1$. Suppose that these numbers satisfy the following system of equations:  $$ \\sum_{j=0, j \\neq i}^{n+1} \\frac{1}{x_{i}-x_{j}}=0 \\quad \\text { where } i=1,2, \\ldots, n $$  Prove that $x_{n+1-i}=1-x_{i}$ for $i=1,2, \\ldots, n$.", "solution": "7. Let $P(x)=\\left(x-x_{0}\\right)\\left(x-x_{1}\\right) \\cdots\\left(x-x_{n}\\right)\\left(x-x_{n+1}\\right)$. Then  $$ P^{\\prime}(x)=\\sum_{j=0}^{n+1} \\frac{P(x)}{x-x_{j}} \\quad \\text { and } \\quad P^{\\prime \\prime}(x)=\\sum_{j=0}^{n+1} \\sum_{k \\neq j} \\frac{P(x)}{\\left(x-x_{j}\\right)\\left(x-x_{k}\\right)} . $$  Therefore  $$ P^{\\prime \\prime}\\left(x_{i}\\right)=2 P^{\\prime}\\left(x_{i}\\right) \\sum_{j \\neq i} \\frac{1}{\\left(x_{i}-x_{j}\\right)} $$  for $i=0,1, \\ldots, n+1$, and the given condition implies $P^{\\prime \\prime}\\left(x_{i}\\right)=0$ for $i=1,2, \\ldots, n$. Consequently,  $$ x(x-1) P^{\\prime \\prime}(x)=(n+2)(n+1) P(x) . $$  It is easy to observe that there is a unique monic polynomial of degree $n+2$ satisfying differential equation (1). On the other hand, the polynomial $Q(x)=(-1)^{n} P(1-x)$ also satisfies this equation, is monic, and $\\operatorname{deg} Q=$ $n+2$. Therefore $(-1)^{n} P(1-x)=P(x)$, and the result follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (USA 1) From a collection of $n$ persons $q$ distinct two-member teams are selected and ranked $1, \\ldots, q$ (no ties). Let $m$ be the least integer larger than or equal to $2 q / n$. Show that there are $m$ distinct teams that may be listed so that (i) each pair of consecutive teams on the list have one member in common and (ii) the chain of teams on the list are in rank order. Alternative formulation. Given a graph with $n$ vertices and $q$ edges numbered $1, \\ldots, q$, show that there exists a chain of $m$ edges, $m \\geq \\frac{2 q}{n}$, each two consecutive edges having a common vertex, arranged monotonically with respect to the numbering.", "solution": "8. We shall solve the problem in the alternative formulation. Let $L_{G}(v)$ denote the length of the longest directed chain of edges in the given graph $G$ that begins in a vertex $v$ and is arranged decreasingly relative to the numbering. By the pigeonhole principle it suffices to show that $\\sum_{v} L(v) \\geq 2 q$ in every such graph. We do this by induction on $q$. For $q=1$ the claim is obvious. We assume that it is true for $q-1$ and consider a graph $G$ with $q$ edges numbered $1, \\ldots, q$. Let the edge number $q$ connect vertices $u$ and $w$. Removing this edge, we get a graph $G^{\\prime}$ with $q-1$ edges. We then have  $$ L_{G}(u) \\geq L_{G^{\\prime}}(w)+1, L_{G}(w) \\geq L_{G^{\\prime}}(u)+1, L_{G}(v) \\geq L_{G^{\\prime}}(v) \\text { for other } v $$  Since $\\sum L_{G^{\\prime}}(v) \\geq 2(q-1)$ by inductive assumption, it follows that $\\sum L_{G}(v) \\geq 2(q-1)+2=2 q$ as desired. Second solution. Let us place a spider at each vertex of the graph. Let us now interchange the positions of the two spiders at the endpoints of each edge, listing the edges increasingly with respect to the numbering. This way we will move spiders exactly $2 q$ times (two for each edge). Hence there is a spider that will be moved at least $2 q / n$ times. All that remains is to notice that the path of each spider consists of edges numbered in increasing order. Remark. A chain of the stated length having all vertices distinct does not necessarily exist. An example is $n=4, q=6$ with the numbering following the order $a b, c d, a c, b d, a d, b c$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1986", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (GDR 1) ${ }^{\\mathrm{IMO} 6}$ Prove or disprove: Given a finite set of points with integer coordinates in the plane, it is possible to color some of these points red and the remaining ones white in such a way that for any straight line $L$ parallel to one of the coordinate axes, the number of red colored points and the number of white colored points on $L$ differ by at most 1.", "solution": "9. We shall use induction on the number $n$ of points. The case $n=1$ is trivial. Let us suppose that the statement is true for all $1,2, \\ldots, n-1$, and that we are given a set $T$ of $n$ points. If there exists a point $P \\in T$ and a line $l$ that is parallel to an axis and contains $P$ and no other points of $T$, then by the inductive hypothesis we can color the set $T \\backslash\\{P\\}$ and then use a suitable color for $P$. Let us now suppose that whenever a line parallel to an axis contains a point of $T$, it contains another point of $T$. It follows that for an arbitrary point $P_{0} \\in T$ we can choose points $P_{1}, P_{2}, \\ldots$ such that $P_{k} P_{k+1}$ is parallel to the $x$-axis for $k$ even, and to the $y$-axis for $k$ odd. We eventually come to a pair of integers $(r, s)$ of the same parity, $0 \\leq r<s$, such that lines $P_{r} P_{r+1}$ and $P_{s} P_{s+1}$ coincide. Hence the closed polygonal line $P_{r+1} P_{r+2} \\ldots P_{s} P_{r+1}$ is of even length. Thus we may color the points of this polygonal line alternately and then apply the inductive assumption for the rest of the set $T$. The induction is complete.  Second solution. Let $P_{1}, P_{2}, \\ldots, P_{k}$ be the points lying on a line $l$ parallel to an axis, going from left to right or from up to down. We draw segments joining $P_{1}$ with $P_{2}, P_{3}$ with $P_{4}$, and generally $P_{2 i-1}$ with $P_{2 i}$. Having this done for every such line $l$, we obtain a set of segments forming certain polygonal lines. If one of these polygonal lines is closed, then it must have an even number of vertices. Thus, we can color the vertices on each of the polygonal lines alternately (a point not lying on any of the polygonal lines may be colored arbitrarily). The obtained coloring satisfies the conditions.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (AUS 6) Let $f$ be a function that satisfies the following conditions: (i) If $x>y$ and $f(y)-y \\geq v \\geq f(x)-x$, then $f(z)=v+z$, for some number $z$ between $x$ and $y$. (ii) The equation $f(x)=0$ has at least one solution, and among the solutions of this equation, there is one that is not smaller than all the other solutions; (iii) $f(0)=1$. (iv) $f(1987) \\leq 1988$. (v) $f(x) f(y)=f(x f(y)+y f(x)-x y)$.  Find $f(1987)$.", "solution": "1. By (ii), $f(x)=0$ has at least one solution, and there is the greatest among them, say $x_{0}$. Then by (v), for any $x$,  $$ 0=f(x) f\\left(x_{0}\\right)=f\\left(x f\\left(x_{0}\\right)+x_{0} f(x)-x_{0} x\\right)=f\\left(x_{0}(f(x)-x)\\right) $$  It follows that $x_{0} \\geq x_{0}(f(x)-x)$. Suppose $x_{0}>0$. By (i) and (iii), since $f\\left(x_{0}\\right)-x_{0}<0<f(0)-0$, there is a number $z$ between 0 and $x_{0}$ such that $f(z)=z$. By (1), $0=f\\left(x_{0}(f(z)-\\right.$ $z))=f(0)=1$, a contradiction. Hence, $x_{0}<0$. Now the inequality $x_{0} \\geq x_{0}(f(x)-x)$ gives $f(x)-x \\geq 1$ for all $x$; so, $f(1987) \\geq 1988$. Therefore $f(1987)=1988$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (ICE 3) Let $S_{1}$ and $S_{2}$ be two spheres with distinct radii that touch externally. The spheres lie inside a cone $C$, and each sphere touches the cone in a full circle. Inside the cone there are $n$ additional solid spheres arranged in a ring in such a way that each solid sphere touches the cone $C$, both of the spheres $S_{1}$ and $S_{2}$ externally, as well as the two neighboring solid spheres. What are the possible values of $n$ ?", "solution": "10. Denote by $r, R$ (take w.l.o.g. $r<R$ ) the radii and by $A, B$ the centers of the spheres $S_{1}, S_{2}$ respectively. Let $s$ be the common radius of the spheres in the ring, $C$ the center of one of them, say $S$, and $D$ the foot of the perpendicular from $C$ to $A B$. The centers of the spheres in the ring form a regular $n$-gon with center $D$, and thus $\\sin (\\pi / n)=s / C D$. Using Heron's formula on the triangle $A B C$, we obtain $(r+R)^{2} C D^{2}=$ $4 r R s(r+R+s)$, and hence ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-502.jpg?height=565&width=438&top_left_y=1118&top_left_x=865)  $$ \\sin ^{2} \\frac{\\pi}{n}=\\frac{s^{2}}{C D^{2}}=\\frac{(r+R)^{2} s}{4(r+R+s) r R} $$  Choosing the unit of length so that $r+R=2$, for simplicity of writing, we write (1) as $1 / \\sin ^{2}(\\pi / n)=r R(1+2 / s)$. Let now $v$ be half the angle at the top of the cone. Then clearly $R-r=(R+r) \\sin v=2 \\sin v$, giving us $R=1+\\sin v, r=1-\\sin v$. It follows that  $$ \\frac{1}{\\sin ^{2} \\frac{\\pi}{n}}=\\left(1+\\frac{2}{s}\\right) \\cos ^{2} v $$  We need to express $s$ as a function of $R$ and $r$. Let $E_{1}, E_{2}, E$ be collinear points of tangency of $S_{1}, S_{2}$, and $S$ with the cone. Obviously, $E_{1} E_{2}=$ $E_{1} E+E_{2} E$, i.e., $2 \\sqrt{r s}+2 \\sqrt{R s}=2 \\sqrt{R r}=(R+r) \\cos v=2 \\cos v$. Hence,  $$ \\cos ^{2} v=s(\\sqrt{R}+\\sqrt{r})^{2}=s(R+r+2 \\sqrt{R r})=s(2+2 \\cos v) . $$  Substituting this into (2), we obtain $2+\\cos v=1 / \\sin (\\pi / n)$. Therefore $1 / 3<\\sin (\\pi / n)<1 / 2$, and we conclude that the possible values for $n$ are 7,8 , and 9 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (POL 1) Find the number of partitions of the set $\\{1,2, \\ldots, n\\}$ into three subsets $A_{1}, A_{2}, A_{3}$, some of which may be empty, such that the following conditions are satisfied: (i) After the elements of every subset have been put in ascending order, every two consecutive elements of any subset have different parity. (ii) If $A_{1}, A_{2}, A_{3}$ are all nonempty, then in exactly one of them the minimal number is even.", "solution": "11. Let $A_{1}$ be the set that contains 1 , and let the minimal element of $A_{2}$ be less than that of $A_{3}$. We shall construct the partitions with required properties by allocating successively numbers to the subsets that always obey the rules. The number 1 must go to $A_{1}$; we show that for every subsequent number we have exactly two possibilities. Actually, while $A_{2}$ and $A_{3}$ are both empty, every successive number can enter either $A_{1}$ or $A_{2}$. Further, when $A_{2}$ is no longer empty, we use induction on the number to be placed, denote it by $m$ : if $m$ can enter $A_{i}$ or $A_{j}$ but not $A_{k}$, and it enters $A_{i}$, then $m+1$ can be placed in $A_{i}$ or $A_{k}$, but not in $A_{j}$. The induction step is finished. This immediately gives us that the final answer is $2^{n-1}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (POL 5) Given a nonequilateral triangle $A B C$, the vertices listed counterclockwise, find the locus of the centroids of the equilateral triangles $A^{\\prime} B^{\\prime} C^{\\prime}$ (the vertices listed counterclockwise) for which the triples of points $A, B^{\\prime}, C^{\\prime} ; A^{\\prime}, B, C^{\\prime}$; and $A^{\\prime}, B^{\\prime}, C$ are collinear.", "solution": "12. Here all angles will be oriented and measured counterclockwise.  Note that $\\measuredangle C A^{\\prime} B=\\measuredangle A B^{\\prime} C=$ $\\measuredangle B C^{\\prime} A=\\pi / 3$. Let $a^{\\prime}, b^{\\prime}, c^{\\prime}$ denote respectively the inner bisectors of angles $A^{\\prime}, B^{\\prime}, C^{\\prime}$ in triangle $A^{\\prime} B^{\\prime} C^{\\prime}$. The lines $a^{\\prime}, b^{\\prime}, c^{\\prime}$ meet at the centroid $X$ of $A^{\\prime} B^{\\prime} C^{\\prime}$, and $\\measuredangle\\left(a^{\\prime}, b^{\\prime}\\right)=$ $\\measuredangle\\left(b^{\\prime}, c^{\\prime}\\right)=\\measuredangle\\left(c^{\\prime}, a^{\\prime}\\right)=2 \\pi / 3$. Now let $K, L, M$ be the points such that $K B=K C, L C=L A, M A=M B$, and $\\measuredangle B K C=\\measuredangle C L A=\\measuredangle A M B=$ $2 \\pi / 3$, and let $C_{1}, C_{2}, C_{3}$ be the circles circumscribed about triangles ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-503.jpg?height=511&width=491&top_left_y=1136&top_left_x=834) $B K C, C L A$, and $A M B$ respectively. These circles are characterized by $C_{1}=\\{Z \\mid \\measuredangle B Z C=2 \\pi / 3\\}$, etc.; hence we deduce that they meet at a point $P$ such that $\\measuredangle B P C=\\measuredangle C P A=\\measuredangle A P B=2 \\pi / 3$ (Torricelli's point). Points $A^{\\prime}, B^{\\prime}, C^{\\prime}$ run over $C_{1} \\backslash\\{P\\}, C_{2} \\backslash\\{P\\}, C_{3} \\backslash\\{P\\}$ respectively. As for $a^{\\prime}, b^{\\prime}, c^{\\prime}$, we see that $K \\in a^{\\prime}, L \\in b^{\\prime}, M \\in c^{\\prime}$, and also that they can take all possible directions except $K P, L P, M P$ respectively (if $K=P, K P$ is assumed to be the corresponding tangent at $K$ ). Then, since $\\measuredangle K X L=$ $2 \\pi / 3, X$ runs over the circle defined by $\\{Z \\mid \\measuredangle K Z L=2 \\pi / 3\\}$, without $P$. But analogously, $X$ runs over the circle $\\{Z \\mid \\measuredangle L Z M=2 \\pi / 3\\}$, from which we can conclude that these two circles are the same, both equal to the circumcircle of $K L M$, and consequently also that triangle $K L M$ is equilateral (which is, anyway, a well-known fact). Therefore, the locus of the points $X$ is the circumcircle of $K L M$ minus point $P$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (GDR 2) ${ }^{\\mathrm{IMO} 5}$ Is it possible to put 1987 points in the Euclidean plane such that the distance between each pair of points is irrational and each three points determine a nondegenerate triangle with rational area?", "solution": "13. We claim that the points $P_{i}\\left(i, i^{2}\\right), i=1,2, \\ldots, 1987$, satisfy the conditions. In fact: (i) $\\overline{P_{i} P_{j}}=\\sqrt{(i-j)^{2}+\\left(i^{2}-j^{2}\\right)^{2}}=|i-j| \\sqrt{1+(i+j)^{2}}$.  It is known that for each positive integer $n, \\sqrt{n}$ is either an integer or an irrational number. Since $i+j<\\sqrt{1+(i+j)^{2}}<i+j+1$, $\\sqrt{1+(i+j)^{2}}$ is not an integer, it is irrational, and so is $\\overline{P_{i} P_{j}}$. (ii) The area $A$ of the triangle $P_{i} P_{j} P_{k}$, for distinct $i, j, k$, is given by  $$ \\begin{aligned} A & =\\left|\\frac{i^{2}+j^{2}}{2}(i-j)+\\frac{j^{2}+k^{2}}{2}(j-k)+\\frac{k^{2}+i^{2}}{2}(k-i)\\right| \\\\ & =\\left|\\frac{(i-j)(j-k)(k-i)}{2}\\right| \\in \\mathbb{Q} \\backslash\\{0\\}, \\end{aligned} $$  also showing that this triangle is nondegenerate.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (FRG 1) How many words with $n$ digits can be formed from the alphabet $\\{0,1,2,3,4\\}$, if neighboring digits must differ by exactly one?", "solution": "14. Let $x_{n}$ be the total number of counted words of length $n$, and $y_{n}, z_{n}, u_{n}$, $z_{n}, y_{n}$ the numbers of counted words of length $n$ starting with $0,1,2,3,4$, respectively (indeed, by symmetry, words starting with 0 are equally numbered as those starting with 4 , etc.). We have the clear relations  $$ \\begin{array}{ll} \\text { (1) } y_{n}=z_{n-1} ; & \\text { (2) } z_{n}=y_{n-1}+u_{n-1} \\\\ \\text { (3) } u_{n}=2 z_{n-1} ; & \\text { (4) } x_{n}=2 y_{n}+2 z_{n}+u_{n} \\end{array} $$  From (1), (2), and (3) we get $z_{n}=z_{n-2}+2 z_{n-2}=3 z_{n-2}$, with $z_{1}=1$, $z_{2}=2$, which gives  $$ z_{2 n}=2 \\cdot 3^{n-1}, \\quad z_{2 n+1}=3^{n} $$  Then (1), (3), and (4) obviously imply  $$ \\begin{array}{ll} y_{2 n}=3^{n-1}, & y_{2 n+1}=2 \\cdot 3^{n-1} \\\\ u_{2 n}=2 \\cdot 3^{n-1}, & u_{2 n+1}=4 \\cdot 3^{n-1} \\\\ x_{2 n}=8 \\cdot 3^{n-1}, & x_{2 n+1}=14 \\cdot 3^{n-1} \\end{array} $$  with the initial number $x_{1}=5$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (FRG 2) $)^{\\mathrm{IMO} 3}$ Suppose $x_{1}, x_{2}, \\ldots, x_{n}$ are real numbers with $x_{1}^{2}+x_{2}^{2}+$ $\\cdots+x_{n}^{2}=1$. Prove that for any integer $k>1$ there are integers $e_{i}$ not all 0 and with $\\left|e_{i}\\right|<k$ such that  $$ \\left|e_{1} x_{1}+e_{2} x_{2}+\\cdots+e_{n} x_{n}\\right| \\leq \\frac{(k-1) \\sqrt{n}}{k^{n}-1} $$", "solution": "15. Since $x_{1}^{2}+x_{2}^{2}+\\cdots+x_{n}^{2}=1$, we get by the Cauchy-Schwarz inequality  $$ \\left|x_{1}\\right|+\\left|x_{2}\\right|+\\cdots+\\left|x_{n}\\right| \\leq \\sqrt{n\\left(x_{1}^{2}+x_{2}^{2}+\\cdots+x_{n}^{2}\\right)}=\\sqrt{n} $$  Hence all $k^{n}$ sums of the form $e_{1} x_{1}+e_{2} x_{2}+\\cdots+e_{n} x_{n}$, with $e_{i} \\in$ $\\{0,1,2, \\ldots, k-1\\}$, must lie in some closed interval $\\Im$ of length $(k-1) \\sqrt{n}$. This interval can be covered with $k^{n}-1$ closed subintervals of length $\\frac{k-1}{k^{n}-1} \\sqrt{n}$. By the pigeonhole principle there must be two of these sums lying in the same subinterval. Their difference, which is of the form $e_{1} x_{1}+e_{2} x_{2}+\\cdots+e_{n} x_{n}$ where $e_{i} \\in\\{0, \\pm 1, \\ldots, \\pm(k-1)\\}$, satisfies  $$ \\left|e_{1} x_{1}+e_{2} x_{2}+\\cdots+e_{n} x_{n}\\right| \\leq \\frac{(k-1) \\sqrt{n}}{k^{n}-1} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (FRG 3) ${ }^{\\mathrm{IMO1}}$ Let $S$ be a set of $n$ elements. We denote the number of all permutations of $S$ that have exactly $k$ fixed points by $p_{n}(k)$. Prove: (a) $\\sum_{k=0}^{n} k p_{n}(k)=n$ !; (b) $\\sum_{k=0}^{n}(k-1)^{2} p_{n}(k)=n$ !.", "solution": "16. We assume that $S=\\{1,2, \\ldots, n\\}$, and use the obvious fact  $$ \\sum_{k=0}^{n} p_{n}(k)=n! $$  (a) To each permutation $\\pi$ of $S$ we assign an $n$-vector $\\left(e_{1}, e_{2}, \\ldots, e_{n}\\right)$, where $e_{i}$ is 1 if $i$ is a fixed point of $\\pi$, and 0 otherwise. Since exactly $p_{n}(k)$ of the assigned vectors contain exactly $k$ \" 1 \"s, the considered sum $\\sum_{k=0}^{n} k p_{n}(k)$ counts all the \" 1 \"s occurring in all the $n$ ! assigned vectors. But for each $i, 1 \\leq i \\leq n$, there are exactly $(n-1)$ ! permutations that fix $i$; i.e., exactly $(n-1)$ ! of the vectors have $e_{i}=1$. Therefore the total number of \" 1 \"s is $n \\cdot(n-1)$ ! = n!, implying  $$ \\sum_{k=0}^{n} k p_{n}(k)=n! $$  (b) In this case, to each permutation $\\pi$ of $S$ we assign a vector $\\left(d_{1}, \\ldots, d_{n}\\right)$ instead, with $d_{i}=k$ if $i$ is a fixed point of $\\pi$, and $d_{i}=0$ otherwise, where $k$ is the number of fixed points of $\\pi$. Let us count the sum $Z$ of all components $d_{i}$ for all the $n$ ! permutations. There are $p_{n}(k)$ such vectors with exactly $k$ components equal to $k$, and sums of components equal to $k^{2}$. Thus, $Z=\\sum_{k=0}^{n} k^{2} p_{n}(k)$. On the other hand, we may first calculate the sum of all components $d_{i}$ for fixed $i$. In fact, the value $d_{i}=k>0$ will occur exactly $p_{n-1}(k-1)$ times, so that the sum of the $d_{i}$ 's is $\\sum_{k=1}^{n} k p_{n-1}(k-1)=\\sum_{k=0}^{n-1}(k+$ 1) $p_{n-1}(k)=2(n-1)$ !. Summation over $i$ yields  $$ Z=\\sum_{k=0}^{n} k^{2} p_{n}(k)=2 n!. $$  From (0), (1), and (2), we conclude that  $$ \\sum_{k=0}^{n}(k-1)^{2} p_{n}(k)=\\sum_{k=0}^{n} k^{2} p_{n}(k)-2 \\sum_{k=0}^{n} k p_{n}(k)+\\sum_{k=0}^{n} p_{n}(k)=n! $$  Remark. Only the first part of this problem was given on the IMO.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (ROM 1) Prove that there exists a four-coloring of the set $M=$ $\\{1,2, \\ldots, 1987\\}$ such that any arithmetic progression with 10 terms in the set $M$ is not monochromatic. Alternative formulation. Let $M=\\{1,2, \\ldots, 1987\\}$. Prove that there is a function $f: M \\rightarrow\\{1,2,3,4\\}$ that is not constant on every set of 10 terms from $M$ that form an arithmetic progression.", "solution": "17. The number of 4 -colorings of the set $M$ is equal to $4^{1987}$. Let $A$ be the number of arithmetic progressions in $M$ with 10 terms. The number of colorings containing a monochromatic arithmetic progression with 10 terms is less than $4 A \\cdot 4^{1977}$. So, if $A<4^{9}$, then there exist 4 -colorings with the required property.  Now we estimate the value of $A$. If the first term of a 10-term progression is $k$ and the difference is $d$, then $1 \\leq k \\leq 1978$ and $d \\leq\\left[\\frac{1987-k}{9}\\right]$; hence  $$ A=\\sum_{k=1}^{1978}\\left[\\frac{1987-k}{9}\\right]<\\frac{1986+1985+\\cdots+9}{9}=\\frac{1995 \\cdot 1978}{18}<4^{9} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (ROM 4) For any integer $r \\geq 1$, determine the smallest integer $h(r) \\geq 1$ such that for any partition of the set $\\{1,2, \\ldots, h(r)\\}$ into $r$ classes, there are integers $a \\geq 0,1 \\leq x \\leq y$, such that $a+x, a+y, a+x+y$ belong to the same class.", "solution": "18. Note first that the statement that some $a+x, a+y, a+x+y$ belong to a class $C$ is equivalent to the following statement: (1) There are positive integers $p, q \\in C$ such that $p<q \\leq 2 p$.  Indeed, given $p, q$, take simply $x=y=q-p, a=2 p-q$; conversely, if $a, x, y(x \\leq y)$ exist such that $a+x, a+y, a+x+y \\in C$, take $p=a+y$, $q=a+x+y$ : clearly, $p<q \\leq 2 p$. We will show that $h(r)=2 r$. Let $\\{1,2, \\ldots, 2 r\\}=C_{1} \\cup C_{2} \\cup \\cdots \\cup C_{r}$ be an arbitrary partition into $r$ classes. By the pigeonhole principle, two among the $r+1$ numbers $r, r+1, \\ldots, 2 r$ belong to the same class, say $i, j \\in C_{k}$. If w.l.o.g. $i<j$, then obviously $i<j \\leq 2 i$, and so by (1) this $C_{k}$ has the required property. On the other hand, we consider the partition  $$ \\{1,2, \\ldots, 2 r-t\\}=\\bigcup_{k=1}^{r-t}\\{k, k+r\\} \\cup\\{r-t+1\\} \\cup \\cdots \\cup\\{r\\} $$  and prove that (1), and thus also the required property, does not hold. In fact, none of the classes in the partition contains $p$ and $q$ with $p<q \\leq 2 p$, because $k+r>2 k$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (USS 2) Let $\\alpha, \\beta, \\gamma$ be positive real numbers such that $\\alpha+\\beta+\\gamma<\\pi$, $\\alpha+\\beta>\\gamma, \\beta+\\gamma>\\alpha, \\gamma+\\alpha>\\beta$. Prove that with the segments of lengths $\\sin \\alpha, \\sin \\beta, \\sin \\gamma$ we can construct a triangle and that its area is not greater than  $$ \\frac{1}{8}(\\sin 2 \\alpha+\\sin 2 \\beta+\\sin 2 \\gamma) $$", "solution": "19. The facts given in the problem allow us to draw a triangular pyramid with angles $2 \\alpha, 2 \\beta, 2 \\gamma$ at the top and lateral edges of length $1 / 2$. At the base there is a triangle whose side lengths are exactly $\\sin \\alpha, \\sin \\beta, \\sin \\gamma$. The area of this triangle does not exceed the sum of areas of the lateral sides, which equals $(\\sin 2 \\alpha+\\sin 2 \\beta+\\sin 2 \\gamma) / 8$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (USA 3) At a party attended by $n$ married couples, each person talks to everyone else at the party except his or her spouse. The conversations involve sets of persons or cliques $C_{1}, C_{2}, \\ldots, C_{k}$ with the following property: no couple are members of the same clique, but for every other pair of persons there is exactly one clique to which both members belong. Prove that if $n \\geq 4$, then $k \\geq 2 n$.", "solution": "2. Let $d_{i}$ denote the number of cliques of which person $i$ is a member. Clearly $d_{i} \\geq 2$. We now distinguish two cases: (i) For some $i, d_{i}=2$. Suppose that $i$ is a member of two cliques, $C_{p}$ and $C_{q}$. Then $\\left|C_{p}\\right|=\\left|C_{q}\\right|=n$, since for each couple other than $i$ and his/her spouse, one member is in $C_{p}$ and one in $C_{q}$. There are thus $(n-1)(n-2)$ pairs $(r, s)$ of nonspouse persons distinct from $i$, where $r \\in C_{p}, s \\in C_{q}$. We observe that each such pair accounts for a different clique. Otherwise, we find two members of $C_{p}$ or $C_{q}$ who belong to one other clique. It follows that $k \\geq 2+(n-1)(n-2) \\geq 2 n$ for $n \\geq 4$. (ii) For every $i, d_{i} \\geq 3$. Suppose that $k<2 n$. For $i=1,2, \\ldots, 2 n$ assign to person $i$ an indeterminant $x_{i}$, and for $j=1,2, \\ldots, k$ set $y=\\sum_{i \\in C_{j}} x_{i}$. From linear algebra, we know that if $k<2 n$, then there exist $x_{1}, x_{2}, \\ldots, x_{2 n}$, not all zero, such that $y_{1}=y_{2}=\\cdots=y_{k}=0$. On the other hand, suppose that $y_{1}=y_{2}=\\cdots=y_{k}=0$. Let $M$ be the set of the couples and $M^{\\prime}$ the set of all other pairs of persons. Then  $$ \\begin{aligned} 0 & =\\sum_{j=1}^{k} y_{j}^{2}=\\sum_{i=1}^{2 n} d_{i} x_{i}^{2}+2 \\sum_{(i, j) \\in M^{\\prime}} x_{i} x_{j} \\\\ & =\\sum_{i=1}^{2 n}\\left(d_{i}-2\\right) x_{i}^{2}+\\left(x_{1}+x_{2}+\\cdots+x_{2 n}\\right)^{2}+\\sum_{(i, j) \\in M}\\left(x_{i}-x_{j}\\right)^{2} \\\\ & \\geq \\sum_{i=1}^{2 n} x_{i}^{2}>0 \\end{aligned} $$  if not all $x_{1}, x_{2}, \\ldots, x_{2 n}$ are zero, which is a contradiction. Hence $k \\geq$ $2 n$. Remark. The condition $n \\geq 4$ is essential. For a party attended by 3 couples $\\{(1,4),(2,5),(3,6)\\}$, there is a collection of 4 cliques satisfying the conditions: $\\{(1,2,3),(3,4,5),(5,6,1),(2,4,6)\\}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. (USS 3) ${ }^{\\text {IMO6 }}$ Let $f(x)=x^{2}+x+p, p \\in \\mathbb{N}$. Prove that if the numbers $f(0), f(1), \\ldots, f([\\sqrt{p / 3}])$ are primes, then all the numbers $f(0), f(1), \\ldots$, $f(p-2)$ are primes.", "solution": "20. Let $y$ be the smallest nonnegative integer with $y \\leq p-2$ for which $f(y)$ is a composite number. Denote by $q$ the smallest prime divisor of $f(y)$. We claim that $y<q$. Suppose the contrary, that $y \\geq q$. Let $r$ be a positive integer such that $y \\equiv r(\\bmod q)$. Then $f(y) \\equiv f(r) \\equiv 0(\\bmod q)$, and since $q \\leq y \\leq p-2 \\leq$ $f(r)$, we conclude that $q \\mid f(r)$, which is a contradiction to the minimality of $y$. Now, we will prove that $q>2 y$. Suppose the contrary, that $q \\leq 2 y$. Since  $$ f(y)-f(x)=(y-x)(y+x+1) $$  we observe that $f(y)-f(q-1-y)=(2 y-q+1) q$, from which it follows that $f(q-1-y)$ is divisible by $q$. But by the assumptions, $q-1-y<y$, implying that $f(q-1-y)$ is prime and therefore equal to $q$. This is impossible, because  $$ f(q-1-y)=(q-1-y)^{2}+(q-1-y)+p>q+p-y-1 \\geq q . $$  Therefore $q \\geq 2 y+1$. Now, since $f(y)$, being composite, cannot be equal to $q$, and $q$ is its smallest prime divisor, we obtain that $f(y) \\geq q^{2}$. Consequently,  $$ y^{2}+y+p \\geq q^{2} \\geq(2 y+1)^{2}=4 y^{2}+4 y+1 \\Rightarrow 3\\left(y^{2}+y\\right) \\leq p-1 $$  and from this we easily conclude that $y<\\sqrt{p / 3}$, which contradicts the condition of the problem. In this way, all the numbers  $$ f(0), f(1), \\ldots, f(p-2) $$  must be prime.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. (USS 4) ${ }^{\\mathrm{IMO} 2}$ The prolongation of the bisector $A L(L \\in B C)$ in the acuteangled triangle $A B C$ intersects the circumscribed circle at point $N$. From point $L$ to the sides $A B$ and $A C$ are drawn the perpendiculars $L K$ and $L M$ respectively. Prove that the area of the triangle $A B C$ is equal to the area of the quadrilateral $A K N M$.", "solution": "21. Let $P$ be the second point of intersection of segment $B C$ and the circle circumscribed about quadrilateral $A K L M$. Denote by $E$ the intersection point of the lines $K N$ and $B C$ and by $F$ the intersection point of the lines $M N$ and $B C$. Then $\\angle B C N=\\angle B A N$ and $\\angle M A L=$ $\\angle M P L$, as angles on the same arc. Since $A L$ is a bisector, $\\angle B C N=$ $\\angle B A L=\\angle M A L=\\angle M P L$, and ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-507.jpg?height=452&width=414&top_left_y=813&top_left_x=866) consequently $P M \\| N C$. Similarly we prove $K P \\| B N$. Then the quadrilaterals $B K P N$ and $N P M C$ are trapezoids; hence  $$ S_{B K E}=S_{N P E} \\quad \\text { and } \\quad S_{N P F}=S_{C M F} $$  Therefore $S_{A B C}=S_{A K N M}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "22", "problem_type": null, "exam": "IMO", "problem": "22. (VIE 3) ${ }^{\\mathrm{IMO} 4}$ Does there exist a function $f: \\mathbb{N} \\rightarrow \\mathbb{N}$, such that $f(f(n))=$ $n+1987$ for every natural number $n$ ?", "solution": "22. Suppose that there exists such function $f$. Then we obtain  $$ f(n+1987)=f(f(f(n)))=f(n)+1987 \\quad \\text { for all } n \\in \\mathbb{N} $$  and from here, by induction, $f(n+1987 t)=f(n)+1987 t$ for all $n, t \\in \\mathbb{N}$. Further, for any $r \\in\\{0,1, \\ldots, 1986\\}$, let $f(r)=1987 k+l, k, l \\in \\mathbb{N}$, $l \\leq 1986$. We have  $$ r+1987=f(f(r))=f(l+1987 k)=f(l)+1987 k, $$  and consequently there are two possibilities: (i) $k=1 \\Rightarrow f(r)=l+1987$ and $f(l)=r$; (ii) $k=0 \\Rightarrow f(r)=l$ and $f(l)=r+1987$; in both cases, $r \\neq l$. In this way, the set $\\{0,1, \\ldots, 1986\\}$ decomposes to pairs $\\{a, b\\}$ such that  $$ f(a)=b \\text { and } f(b)=a+1987, \\quad \\text { or } \\quad f(b)=a \\text { and } f(a)=b+1987 . $$  But the set $\\{0,1, \\ldots, 1986\\}$ has an odd number of elements, and cannot be decomposed into pairs. Contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "23", "problem_type": null, "exam": "IMO", "problem": "23. (YUG 2) Prove that for every natural number $k(k \\geq 2)$ there exists an irrational number $r$ such that for every natural number $m$,  $$ \\left[r^{m}\\right] \\equiv-1 \\quad(\\bmod k) $$  Remark. An easier variant: Find $r$ as a root of a polynomial of second degree with integer coefficients.", "solution": "23. If we prove the existence of $p, q \\in \\mathbb{N}$ such that the roots $r, s$ of  $$ f(x)=x^{2}-k p \\cdot x+k q=0 $$  are irrational real numbers with $0<s<1$ (and consequently $r>1$ ), then we are done, because from $r+s, r s \\equiv 0(\\bmod k)$ we get $r^{m}+s^{m} \\equiv 0$ $(\\bmod k)$, and $0<s^{m}<1$ yields the assertion. To prove the existence of such natural numbers $p$ and $q$, we can take them such that $f(0)>0>f(1)$, i.e.,  $$ k q>0>k(q-p)+1 \\quad \\Rightarrow \\quad p>q>0 $$  The irrationality of $r$ can be obtained by taking $q=p-1$, because the discriminant $D=(k p)^{2}-4 k p+4 k$, for $(k p-2)^{2}<D<(k p-1)^{2}$, is not a perfect square for $p \\geq 2$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (FIN 3) Does there exist a second-degree polynomial $p(x, y)$ in two variables such that every nonnegative integer $n$ equals $p(k, m)$ for one and only one ordered pair $(k, m)$ of nonnegative integers?", "solution": "3. The answer: yes. Set  $$ p(k, m)=k+[1+2+\\cdots+(k+m)]=\\frac{(k+m)^{2}+3 k+m}{2} . $$  It is obviously of the desired type.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (FRA 5) Let $A B C D E F G H$ be a parallelepiped with $A E\\|B F\\| C G \\| D H$. Prove the inequality  $$ A F+A H+A C \\leq A B+A D+A E+A G $$  In what cases does equality hold?", "solution": "4. Setting $x_{1}=\\overrightarrow{A B}, x_{2}=\\overrightarrow{A D}, x_{3}=\\overrightarrow{A E}$, we have to prove that  $$ \\left\\|x_{1}+x_{2}\\right\\|+\\left\\|x_{2}+x_{3}\\right\\|+\\left\\|x_{3}+x_{1}\\right\\| \\leq\\left\\|x_{1}\\right\\|+\\left\\|x_{2}\\right\\|+\\left\\|x_{3}\\right\\|+\\left\\|x_{1}+x_{2}+x_{3}\\right\\| . $$  We have  $$ \\begin{aligned} & \\left(\\left\\|x_{1}\\right\\|+\\left\\|x_{2}\\right\\|+\\left\\|x_{3}\\right\\|\\right)^{2}-\\left\\|x_{1}+x_{2}+x_{3}\\right\\|^{2} \\\\ & \\quad=2 \\sum_{1 \\leq i<j \\leq 3}\\left(\\left\\|x_{i}\\right\\|\\left\\|x_{j}\\right\\|-\\left\\langle x_{i}, x_{j}\\right\\rangle\\right)=\\sum_{1 \\leq i<j \\leq 3}\\left[\\left(\\left\\|x_{i}\\right\\|+\\left\\|x_{j}\\right\\|\\right)^{2}-\\left\\|x_{i}+x_{j}\\right\\|^{2}\\right] \\\\ & \\quad=\\sum_{1 \\leq i<j \\leq 3}\\left(\\left\\|x_{i}\\right\\|+\\left\\|x_{j}\\right\\|+\\left\\|x_{i}+x_{j}\\right\\|\\right)\\left(\\left\\|x_{i}\\right\\|+\\left\\|x_{j}\\right\\|-\\left\\|x_{i}+x_{j}\\right\\|\\right) . \\end{aligned} $$  The following two inequalities are obvious:  $$ \\begin{array}{r} \\left\\|x_{i}\\right\\|+\\left\\|x_{j}\\right\\|-\\left\\|x_{i}+x_{j}\\right\\| \\geq 0 \\\\ \\left\\|x_{i}\\right\\|+\\left\\|x_{j}\\right\\|+\\left\\|x_{i}+x_{j}\\right\\| \\leq\\left\\|x_{1}\\right\\|+\\left\\|x_{2}\\right\\|+\\left\\|x_{3}\\right\\|+\\left\\|x_{1}+x_{2}+x_{3}\\right\\| \\end{array} $$  It follows that  $$ \\begin{aligned} & \\left(\\left\\|x_{1}\\right\\|+\\left\\|x_{2}\\right\\|+\\left\\|x_{3}\\right\\|\\right)^{2}-\\left\\|x_{1}+x_{2}+x_{3}\\right\\|^{2} \\\\ & \\quad \\leq\\left(\\sum_{i=1}^{3}\\left\\|x_{i}\\right\\|+\\left\\|\\sum_{i=1}^{3} x_{i}\\right\\|\\right)\\left(2 \\sum_{i=1}^{3}\\left\\|x_{i}\\right\\|-\\sum_{1 \\leq i<j \\leq 3}\\left\\|x_{i}+x_{j}\\right\\|\\right) \\end{aligned} $$  and dividing by the positive number $\\sum_{i=1}^{3}\\left\\|x_{i}\\right\\|+\\left\\|\\sum_{i=1}^{3} x_{i}\\right\\|$ we obtain  $$ \\sum_{i=1}^{3}\\left\\|x_{i}\\right\\|-\\left\\|\\sum_{i=1}^{3} x_{i}\\right\\| \\leq 2 \\sum_{i=1}^{3}\\left\\|x_{i}\\right\\|-\\sum_{1 \\leq i<j \\leq 3}\\left\\|x_{i}+x_{j}\\right\\| $$  The inequality is proven. Let us analyze the cases of equality. If one of the vectors is null, then equality obviously holds. Suppose that $x_{i} \\neq 0$, $i=1,2,3$. For every $i, j$, at least one of (1) and (2) is equality. Equality in (1) holds if and only if $x_{i}$ and $x_{j}$ are collinear with the same direction, while in (2) it holds if and only if $-x_{k}$ and $x_{1}+x_{2}+x_{3}$ are collinear with the same direction. If not all the vectors are collinear, then there are at least two distinct pairs $x_{i}, x_{j}, i<j$, for which (2) is an equality, so at least two of $x_{i}$ are collinear with $x_{1}+x_{2}+x_{3}$, but then so is the third; hence, the sum $x_{1}+x_{2}+x_{3}$ must be 0 . Thus the cases of equality are (a) the vectors are collinear with the same direction; (b) the vectors are collinear, two of them have the same direction, say $x_{i}, x_{j}$, and $\\left\\|x_{k}\\right\\| \\geq\\left\\|x_{i}\\right\\|+\\left\\|x_{j}\\right\\|$; (c) one of the vectors is 0 ; (d) their sum is 0 .  Second solution. The following technique, although not quite elementary, is often used to effectively reduce geometric inequalities of first degree, like this one, to the one-dimensional case. Let $\\sigma$ be a fixed sphere with center $O$. For an arbitrary segment $d$ in space, and any line $l$, we denote by $\\pi_{l}(d)$ the length of the projection of $d$ onto $l$. Consider the integral of lengths of these projections on all possible directions of $O P$, with $P$ moving on the sphere: $\\int_{\\sigma} \\pi_{O P}(d) d \\sigma$. It is clear that this value depends only on the length of $d$ (because of symmetry); hence  $$ \\int_{\\sigma} \\pi_{O P} d \\sigma=c \\cdot|d| \\quad \\text { for some constant } c \\neq 0 $$  Notice that by the one-dimensional case, for any point $P \\in \\sigma$,  $$ \\begin{gathered} \\pi_{O P}\\left(x_{1}\\right)+\\pi_{O P}\\left(x_{2}\\right)+\\pi_{O P}\\left(x_{3}\\right)+\\pi_{O P}\\left(x_{1}+x_{2}+x_{3}\\right) \\\\ \\geq \\pi_{O P}\\left(x_{1}+x_{2}\\right)+\\pi_{O P}\\left(x_{1}+x_{3}\\right)+\\pi_{O P}\\left(x_{2}+x_{3}\\right) . \\end{gathered} $$  By integration on $\\sigma$, using (1), we obtain $c\\left(\\left\\|x_{1}\\right\\|+\\left\\|x_{2}\\right\\|+\\left\\|x_{3}\\right\\|+\\left\\|x_{1}+x_{2}+x_{3}\\right\\|\\right) \\geq c\\left(\\left\\|x_{1}+x_{2}\\right\\|+\\left\\|x_{1}+x_{3}\\right\\|+\\left\\|x_{2}+x_{3}\\right\\|\\right)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (GBR 1) Find, with proof, the point $P$ in the interior of an acute-angled triangle $A B C$ for which $B L^{2}+C M^{2}+A N^{2}$ is a minimum, where $L, M, N$ are the feet of the perpendiculars from $P$ to $B C, C A, A B$ respectively.", "solution": "5. Assuming the notation $a=\\overline{B C}, b=\\overline{A C}, c=\\overline{A B} ; x=\\overline{B L}, y=\\overline{C M}$, $z=\\overline{A N}$, from the Pythagorean theorem we obtain  $$ \\begin{aligned} (a-x)^{2}+(b-y)^{2} & +(c-z)^{2}=x^{2}+y^{2}+z^{2} \\\\ & =\\frac{x^{2}+(a-x)^{2}+y^{2}+(b-y)^{2}+z^{2}+(c-z)^{2}}{2} \\end{aligned} $$  Since $x^{2}+(a-x)^{2}=a^{2} / 2+(a-2 x)^{2} / 2 \\geq a^{2} / 2$ and similarly $y^{2}+(b-y)^{2} \\geq$ $b^{2} / 2$ and $z^{2}+(c-z)^{2} \\geq c^{2} / 2$, we get  $$ x^{2}+y^{2}+z^{2} \\geq \\frac{a^{2}+b^{2}+c^{2}}{4} $$  Equality holds if and only if $P$ is the circumcenter of the triangle $A B C$, i.e., when $x=a / 2, y=b / 2, z=c / 2$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (GRE 4) Show that if $a, b, c$ are the lengths of the sides of a triangle and if $2 S=a+b+c$, then  $$ \\frac{a^{n}}{b+c}+\\frac{b^{n}}{c+a}+\\frac{c^{n}}{a+b} \\geq\\left(\\frac{2}{3}\\right)^{n-2} S^{n-1}, \\quad n \\geq 1 $$", "solution": "6. Suppose w.l.o.g. that $a \\geq b \\geq c$. Then $1 /(b+c) \\geq 1 /(a+c) \\geq 1 /(a+b)$. Chebyshev's inequality yields  $$ \\frac{a^{n}}{b+c}+\\frac{b^{n}}{a+c}+\\frac{c^{n}}{a+b} \\geq \\frac{1}{3}\\left(a^{n}+b^{n}+c^{n}\\right)\\left(\\frac{1}{b+c}+\\frac{1}{a+c}+\\frac{1}{a+b}\\right) . $$  By the Cauchy-Schwarz inequality we have  $$ 2(a+b+c)\\left(\\frac{1}{b+c}+\\frac{1}{a+c}+\\frac{1}{a+b}\\right) \\geq 9 $$  and the mean inequality yields $\\left(a^{n}+b^{n}+c^{n}\\right) / 3 \\geq[(a+b+c) / 3]^{n}$. We obtain from (1) that  $$ \\begin{aligned} \\frac{a^{n}}{b+c}+\\frac{b^{n}}{a+c}+\\frac{c^{n}}{a+b} & \\geq\\left(\\frac{a+b+c}{3}\\right)^{n}\\left(\\frac{1}{b+c}+\\frac{1}{a+c}+\\frac{1}{a+b}\\right) \\\\ & \\geq \\frac{3}{2}\\left(\\frac{a+b+c}{3}\\right)^{n-1}=\\left(\\frac{2}{3}\\right)^{n-2} S^{n-1} . \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (NET 1) Given five real numbers $u_{0}, u_{1}, u_{2}, u_{3}, u_{4}$, prove that it is always possible to find five real numbers $v_{0}, v_{1}, v_{2}, v_{3}, v_{4}$ that satisfy the following conditions: (i) $u_{i}-v_{i} \\in \\mathbb{N}$. (ii) $\\sum_{0 \\leq i<j \\leq 4}\\left(v_{i}-v_{j}\\right)^{2}<4$.", "solution": "7. For all real numbers $v$ the following inequality holds:  $$ \\sum_{0 \\leq i<j \\leq 4}\\left(v_{i}-v_{j}\\right)^{2} \\leq 5 \\sum_{i=0}^{4}\\left(v_{i}-v\\right)^{2} $$  Indeed,  $$ \\begin{aligned} \\sum_{0 \\leq i<j \\leq 4}\\left(v_{i}-v_{j}\\right)^{2} & =\\sum_{0 \\leq i<j \\leq 4}\\left[\\left(v_{i}-v\\right)-\\left(v_{j}-v\\right)\\right]^{2} \\\\ & =5 \\sum_{i=0}^{4}\\left(v_{i}-v\\right)^{2}-\\left(\\sum_{i=0}^{4}\\left(v_{i}-v\\right)\\right)^{2} \\leq 5 \\sum_{i=0}^{4}\\left(v_{i}-v\\right)^{2} . \\end{aligned} $$  Let us first take $v_{i}$ 's, satisfying condition (1), so that w.l.o.g. $v_{0} \\leq v_{1} \\leq$ $v_{2} \\leq v_{3} \\leq v_{4} \\leq 1+v_{0}$. Defining $v_{5}=1+v_{0}$, we see that one of the differences $v_{j+1}-v_{j}, j=0, \\ldots, 4$, is at most $1 / 5$. Take $v=\\left(v_{j+1}+v_{j}\\right) / 2$, and then place the other three $v_{j}$ 's in the segment $[v-1 / 2, v+1 / 2]$. Now we have $\\left|v-v_{j}\\right| \\leq 1 / 10,\\left|v-v_{j+1}\\right| \\leq 1 / 10$, and $\\left|v-v_{k}\\right| \\leq 1 / 2$, for any $k$ different from $j, j+1$. The $v_{i}$ 's thus obtained have the required property. In fact, using the inequality (1), we obtain  $$ \\sum_{0 \\leq i<j \\leq 4}\\left(v_{i}-v_{j}\\right)^{2} \\leq 5\\left(2\\left(\\frac{1}{10}\\right)^{2}+3\\left(\\frac{1}{2}\\right)^{2}\\right)=3.85<4 $$  Remark. The best possible estimate for the right-hand side is 2 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (HUN 1) (a) Let $(m, k)=1$. Prove that there exist integers $a_{1}, a_{2}, \\ldots, a_{m}$ and $b_{1}, b_{2}, \\ldots, b_{k}$ such that each product $a_{i} b_{j}(i=1,2, \\ldots, m ; j=$ $1,2, \\ldots, k)$ gives a different residue when divided by $m k$. (b) Let $(m, k)>1$. Prove that for any integers $a_{1}, a_{2}, \\ldots, a_{m}$ and $b_{1}, b_{2}$, $\\ldots, b_{k}$ there must be two products $a_{i} b_{j}$ and $a_{s} b_{t}((i, j) \\neq(s, t))$ that give the same residue when divided by $m k$.", "solution": "8. (a) Consider  $$ a_{i}=i k+1, \\quad i=1,2, \\ldots, m ; \\quad b_{j}=j m+1, \\quad j=1,2, \\ldots, k $$  Assume that $m k \\mid a_{i} b_{j}-a_{s} b_{t}=(i k+1)(j m+1)-(s k+1)(t m+1)=$ $k m(i j-s t)+m(j-t)+k(i-s)$. Since $m$ divides this sum, we get that $m \\mid k(i-s)$, or, together with $\\operatorname{gcd}(k, m)=1$, that $i=s$. Similarly $j=t$, which proves part (a). (b) Suppose the opposite, i.e., that all the residues are distinct. Then the residue 0 must also occur, say at $a_{1} b_{1}: m k \\mid a_{1} b_{1}$; so, for some $a^{\\prime}$ and $b^{\\prime}, a^{\\prime}\\left|a_{1}, b^{\\prime}\\right| b_{1}$, and $a^{\\prime} b^{\\prime}=m k$. Assuming that for some $i, s \\neq i$, $a^{\\prime} \\mid a_{i}-a_{s}$, we obtain $m k=a^{\\prime} b^{\\prime} \\mid a_{i} b_{1}-a_{s} b_{1}$, a contradiction. This shows that $a^{\\prime} \\geq m$ and similarly $b^{\\prime} \\geq k$, and thus from $a^{\\prime} b^{\\prime}=m k$ we have $a^{\\prime}=m, b^{\\prime}=k$. We also get (1): all $a_{i}$ 's give distinct residues modulo $m=a^{\\prime}$, and all $b_{j}$ 's give distinct residues modulo $k=b^{\\prime}$. Now let $p$ be a common prime divisor of $m$ and $k$. By $(*)$, exactly $\\frac{p-1}{p} m$ of $a_{i}$ 's and exactly $\\frac{p-1}{p} k$ of $b_{j}$ 's are not divisible by $p$. Therefore there are precisely $\\frac{(p-1)^{2}}{p^{2}} m k$ products $a_{i} b_{j}$ that are not divisible by $p$, although from the assumption that they all give distinct residues it follows that the number of such products is $\\frac{p-1}{p} m k \\neq \\frac{(p-1)^{2}}{p^{2}} m k$. We have arrived at a contradiction, thus proving (b).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1987", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (HUN 2) Does there exist a set $M$ in usual Euclidean space such that for every plane $\\lambda$ the intersection $M \\cap \\lambda$ is finite and nonempty?", "solution": "9. The answer is yes. Consider the curve  $$ C=\\left\\{(x, y, z) \\mid x=t, y=t^{3}, z=t^{5}, \\quad t \\in \\mathbb{R}\\right\\} $$  Any plane defined by an equation of the form $a x+b y+c z+d=0$ intersects the curve $C$ at points $\\left(t, t^{3}, t^{5}\\right)$ with $t$ satisfying $c t^{5}+b t^{3}+a t+d=0$. This last equation has at least one but only finitely many solutions.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (BUL 1) An integer sequence is defined by  $$ a_{n}=2 a_{n-1}+a_{n-2} \\quad(n>1), \\quad a_{0}=0, \\quad a_{1}=1 $$  Prove that $2^{k}$ divides $a_{n}$ if and only if $2^{k}$ divides $n$.", "solution": "1. Assume that $p$ and $q$ are real and $b_{0}, b_{1}, b_{2}, \\ldots$ is a sequence such that $b_{n}=p b_{n-1}+q b_{n-2}$ for all $n>1$. From the equalities $b_{n}=p b_{n-1}+q b_{n-2}$, $b_{n+1}=p b_{n}+q b_{n-1}, b_{n+2}=p b_{n+1}+q b_{n}$, eliminating $b_{n+1}$ and $b_{n-1}$ we obtain that $b_{n+2}=\\left(p^{2}+2 q\\right) b_{n}-q^{2} b_{n-2}$. So the sequence $b_{0}, b_{2}, b_{4}, \\ldots$ has the property  $$ b_{2 n}=P b_{2 n-2}+Q b_{2 n-4}, \\quad P=p^{2}+2 q, \\quad Q=-q^{2} . $$  We shall solve the problem by induction. The sequence $a_{n}$ has $p=2$, $q=1$, and hence $P=6, Q=-1$. Let $k=1$. Then $a_{0}=0, a_{1}=1$, and $a_{n}$ is of the same parity as $a_{n-2}$; i.e., it is even if and only if $n$ is even. Let $k \\geq 1$. We assume that for $n=2^{k} m$, the numbers $a_{n}$ are divisible by $2^{k}$, but divisible by $2^{k+1}$ if and only if $m$ is even. We assume also that the sequence $c_{0}, c_{1}, \\ldots$, with $c_{m}=a_{m \\cdot 2^{k}}$, satisfies the condition $c_{n}=$ $p c_{n-1}-c_{n-2}$, where $p \\equiv 2(\\bmod 4)($ for $k=1$ it is true). We shall prove the same statement for $k+1$. According to (1), $c_{2 n}=P c_{2 n-2}-c_{2 n-4}$, where $P=p^{2}-2$. Obviously $P \\equiv 2(\\bmod 4)$. Since $P=4 s+2$ for some integer $s$, and $c_{2 n}=2^{k+1} d_{2 n}, c_{0}=0, c_{1} \\equiv 2^{k}\\left(\\bmod 2^{k+1}\\right)$, and $c_{2}=p c_{1} \\equiv 2^{k+1}$ $\\left(\\bmod 2^{k+2}\\right)$, we have  $$ c_{2 n}=(4 s+2) 2^{k+1} d_{2 n-2}-c_{2 n-4} \\equiv c_{2 n-4}\\left(\\bmod 2^{k+2}\\right) $$  i.e., $0 \\equiv c_{0} \\equiv c_{4} \\equiv c_{8} \\equiv \\cdots$ and $2^{k+1} \\equiv c_{2} \\equiv c_{6} \\equiv \\cdots\\left(\\bmod 2^{k+2}\\right)$, which proves the statement. Second solution. The recursion is solved by  $$ a_{n}=\\frac{1}{2 \\sqrt{2}}\\left((1+\\sqrt{2})^{n}-(1-\\sqrt{2})^{n}\\right)=\\binom{n}{1}+2\\binom{n}{3}+2^{2}\\binom{n}{5}+\\cdots . $$  Let $n=2^{k} m$ with $m$ odd; then for $p>0$ the summand  $$ 2^{p}\\binom{n}{2 p+1}=2^{k+p} m \\frac{(n-1) \\ldots(n-2 p)}{(2 p+1)!}=2^{k+p} \\frac{m}{2 p+1}\\binom{n-1}{2 p} $$  is divisible by $2^{k+p}$, because the denominator $2 p+1$ is odd. Hence  $$ a_{n}=n+\\sum_{p>0} 2^{p}\\binom{n}{2 p+1}=2^{k} m+2^{k+1} N $$  for some integer $N$, so that $a_{n}$ is exactly divisible by $2^{k}$. Third solution. It can be proven by induction that $a_{2 n}=2 a_{n}\\left(a_{n}+a_{n+1}\\right)$. The required result follows easily, again by induction on $k$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (GDR 1) Let $N=\\{1,2, \\ldots, n\\}, n \\geq 2$. A collection $F=\\left\\{A_{1}, \\ldots, A_{t}\\right\\}$ of subsets $A_{i} \\subseteq N, i=1, \\ldots, t$, is said to be separating if for every pair $\\{x, y\\} \\subseteq N$, there is a set $A_{i} \\in F$ such that $A_{i} \\cap\\{x, y\\}$ contains just one element. A collection $F$ is said to be covering if every element of $N$ is contained in at least one set $A_{i} \\in F$. What is the smallest value $f(n)$ of $t$ such that there is a set $F=\\left\\{A_{1}, \\ldots, A_{t}\\right\\}$ that is simultaneously separating and covering?", "solution": "10. We claim that if the family $\\left\\{A_{1}, \\ldots, A_{t}\\right\\}$ separates the $n$-set $N$, then $2^{t} \\geq n$. The proof goes by induction. The case $t=1$ is clear, so suppose that the claim holds for $t-1$. Since $A_{t}$ does not separate elements of its own or its complement, it follows that $\\left\\{A_{1}, \\ldots, A_{t-1}\\right\\}$ is separating for both $A_{t}$ and $N \\backslash A_{t}$, so that $\\left|A_{t}\\right|,\\left|N \\backslash A_{t}\\right| \\leq 2^{t-1}$. Then $|N| \\leq 2 \\cdot 2^{t-1}=2^{t}$, as claimed. Also, if the set $N$ with $N=2^{t}$ is separated by $\\left\\{A_{1}, \\ldots, A_{t}\\right\\}$, then (precisely) one element of $N$ is not covered. To show this, we again use induction. This is trivial for $t=1$, so let $t \\geq 1$. Since $A_{1}, \\ldots, A_{t-1}$ separate both $A_{t}$ and $N \\backslash A_{t}, N \\backslash A_{t}$ must have exactly $2^{t-1}$ elements, and thus one of its elements is not covered by $A_{1}, \\ldots, A_{t-1}$, and neither is covered by $A_{t}$. We conclude that a separating and covering family of $t$ subsets can exist only if $n \\leq 2^{t}-1$. We now construct such subsets for the set $N$ if $2^{t-1} \\leq n \\leq 2^{t}-1, t \\geq 1$. For $t=1$, put $A_{1}=\\{1\\}$. In the step from $t$ to $t+1$, let $N=N^{\\prime} \\cup N^{\\prime \\prime} \\cup\\{y\\}$, where $\\left|N^{\\prime}\\right|,\\left|N^{\\prime \\prime}\\right| \\leq 2^{t-1}$; let $A_{1}^{\\prime}, \\ldots, A_{t}^{\\prime}$ be subsets covering and separating $N^{\\prime}$ and $A_{1}^{\\prime \\prime}, \\ldots, A_{t}^{\\prime \\prime}$ such subsets for $N^{\\prime \\prime}$. Then the subsets $A_{i}=A_{i}^{\\prime} \\cup A_{i}^{\\prime \\prime}$ $(i=1, \\ldots, t)$ and $A_{t+1}=N^{\\prime \\prime} \\cup\\{y\\}$ obviously separate and cover $N$. The answer: $t=\\left[\\log _{2} n\\right]+1$. Second solution. Suppose that the sets $A_{1}, \\ldots, A_{t}$ cover and separate $N$. Label each element $x \\in N$ with a string of $\\left(x_{1} x_{2} \\ldots x_{t}\\right)$ of 0 's and 1's, where $x_{i}$ is 1 when $x \\in A_{i}, 0$ otherwise. Since the $A_{i}$ 's separate, these strings are distinct; since they cover, the string ( $00 \\ldots 0$ ) does not occur.  Hence $n \\leq 2^{t}-1$. Conversely, for $2^{t-1} \\leq n<2^{t}$, represent the elements of $N$ in base 2 as strings of 0's and 1's of length $t$. For $1 \\leq i \\leq t$, take $A_{i}$ to be the set of numbers in $N$ whose binary string has a 1 in the $i$ th place. These sets clearly cover and separate.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (GDR 3) The lock on a safe consists of three wheels, each of which may be set in eight different positions. Due to a defect in the safe mechanism the door will open if any two of the three wheels are in the correct position. What is the smallest number of combinations that must be tried if one is to guarantee being able to open the safe (assuming that the \"right combination\" is not known)?", "solution": "11. The answer is 32 . Write the combinations as triples $k=(x, y, z), 0 \\leq$ $x, y, z \\leq 7$. Define the sets $K_{1}=\\{(1,0,0),(0,1,0),(0,0,1),(1,1,1)\\}$, $K_{2}=\\{(2,0,0),(0,2,0),(0,0,2),(2,2,2)\\}, K_{3}=\\{(0,0,0),(4,4,4)\\}$, and $K=\\left\\{k=k_{1}+k_{2}+k_{3} \\mid k_{i} \\in K_{i}, i=1,2,3\\right\\}$. There are 32 combinations in $K$. We shall prove that these combinations will open the safe in every case. Let $t=(a, b, c)$ be the right combination. Set $k_{3}=(0,0,0)$ if at least two of $a, b, c$ are less than 4 , and $k_{3}=(4,4,4)$ otherwise. In either case, the difference $t-k_{3}$ contains two nonnegative elements not greater than 3 . Choosing a suitable $k_{2}$ we can achieve that $t-k_{3}-k_{2}$ contains two elements that are 0,1 . So, there exists $k_{1}$ such that $t-k_{3}-k_{2}-k_{1}=t-k$ contains two zeros, for $k \\in K$. This proves that 32 is sufficient. Suppose that $K$ is a set of at most 31 combinations. We say that $k \\in K$ covers the combination $k_{1}$ if $k$ and $k_{1}$ differ in at most one position. One of the eight sets $M_{i}=\\{(i, y, z) \\mid 0 \\leq y, z \\leq 7\\}, i=0,1, \\ldots, 7$, contains at most three elements of $K$. Suppose w.l.o.g. that this is $M_{0}$. Further, among the eight sets $N_{j}=\\{(0, j, z) \\mid 0 \\leq z \\leq 7\\}, j=0, \\ldots, 7$, there are at least five, say w.l.o.g. $N_{0}, \\ldots, N_{4}$, not containing any of the combinations from $K$. Of the 40 elements of the set $N=\\{(0, y, z) \\mid 0 \\leq y \\leq 4,0 \\leq z \\leq 7\\}$, at most $5 \\cdot 3=15$ are covered by $K \\cap M_{0}$, and at least 25 aren't. Consequently, the intersection of $K$ with $L=\\{(x, y, z) \\mid 1 \\leq x \\leq 7,0 \\leq y \\leq 4,0 \\leq z \\leq 7\\}$ contains at least 25 elements. So $K$ has at most $31-25=6$ elements in the set $P=\\{(x, y, z) \\mid 0 \\leq x \\leq 7,5 \\leq y \\leq 7,0 \\leq z \\leq 7\\}$. This implies that for some $j \\in\\{5,6,7\\}$, say w.l.o.g. $j=7, K$ contains at most two elements in $Q_{j}=\\{(x, y, z) \\mid 0 \\leq x, z \\leq 7, y=j\\}$; denote them by $l_{1}, l_{2}$. Of the 64 elements of $Q_{7}$, at most 30 are covered by $l_{1}$ and $l_{2}$. But then there remain 34 uncovered elements, which must be covered by different elements of $K \\backslash Q_{7}$, having itself less at most 29 elements. Contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (GRE 2) In a triangle $A B C$, choose any points $K \\in B C, L \\in A C$, $M \\in A B, N \\in L M, R \\in M K$, and $F \\in K L$. If $E_{1}, E_{2}, E_{3}, E_{4}, E_{5}$, $E_{6}$, and $E$ denote the areas of the triangles $A M R, C K R, B K F, A L F$, $B N M, C L N$, and $A B C$ respectively, show that  $$ E \\geq 8 \\sqrt[6]{E_{1} E_{2} E_{3} E_{4} E_{5} E_{6}} $$  Remark. Points $K, L, M, N, R, F$ lie on segments $B C, A C, A B, L M$, $M K, K L$ respectively.", "solution": "12. Let $E(X Y Z)$ stand for the area of a triangle $X Y Z$. We have  $$ \\begin{gathered} \\frac{E_{1}}{E}=\\frac{E(A M R)}{E(A M K)} \\cdot \\frac{E(A M K)}{E(A B K)} \\cdot \\frac{E(A B K)}{E(A B C)}=\\frac{M R}{M K} \\cdot \\frac{A M}{A B} \\cdot \\frac{B K}{B C} \\Rightarrow \\\\ \\left(\\frac{E_{1}}{E}\\right)^{1 / 3} \\leq \\frac{1}{3}\\left(\\frac{M R}{M K}+\\frac{A M}{A B}+\\frac{B K}{B C}\\right) \\end{gathered} $$  We similarly obtain  $$ \\left(\\frac{E_{2}}{E}\\right)^{1 / 3} \\leq \\frac{1}{3}\\left(\\frac{K R}{M K}+\\frac{B M}{A B}+\\frac{C K}{B C}\\right) $$  Therefore $\\left(E_{1} / E\\right)^{1 / 3}+\\left(E_{2} / E\\right)^{1 / 3} \\leq 1$, i.e., $\\sqrt[3]{E_{1}}+\\sqrt[3]{E_{2}} \\leq \\sqrt[3]{E}$. Analogously, $\\sqrt[3]{E_{3}}+\\sqrt[3]{E_{4}} \\leq \\sqrt[3]{E}$ and $\\sqrt[3]{E_{5}}+\\sqrt[3]{E_{6}} \\leq \\sqrt[3]{E}$; hence  $$ \\begin{aligned} & 8 \\sqrt[6]{E_{1} E_{2} E_{3} E_{4} E_{5} E_{6}} \\\\ & \\quad=2\\left(\\sqrt[3]{E_{1}} \\sqrt[3]{E_{2}}\\right)^{1 / 2} \\cdot 2\\left(\\sqrt[3]{E_{3}} \\sqrt[3]{E_{4}}\\right)^{1 / 2} \\cdot 2\\left(\\sqrt[3]{E_{5}} \\sqrt[3]{E_{6}}\\right)^{1 / 2} \\\\ & \\quad \\leq\\left(\\sqrt[3]{E_{1}}+\\sqrt[3]{E_{2}}\\right) \\cdot\\left(\\sqrt[3]{E_{3}}+\\sqrt[3]{E_{4}}\\right) \\cdot\\left(\\sqrt[3]{E_{5}}+\\sqrt[3]{E_{6}}\\right) \\leq E \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (GRE 3) ${ }^{\\mathrm{IMO5}}$ In a right-angled triangle $A B C$, let $A D$ be the altitude drawn to the hypotenuse and let the straight line joining the incenters of the triangles $A B D, A C D$ intersect the sides $A B, A C$ at the points $K, L$ respectively. If $E$ and $E_{1}$ denote the areas of the triangles $A B C$ and $A K L$ respectively, show that $\\frac{E}{E_{1}} \\geq 2$.", "solution": "13. Let $A B=c, A C=b, \\angle C B A=\\beta, B C=a$, and $A D=h$.  Let $r_{1}$ and $r_{2}$ be the inradii of $A B D$ and $A D C$ respectively and $O_{1}$ and $O_{2}$ the centers of the respective incircles. We obviously have $r_{1} / r_{2}=$ $c / b$. We also have $D O_{1}=\\sqrt{2} r_{1}$, $D O_{2}=\\sqrt{2} r_{2}$, and $\\angle O_{1} D A=$ $\\angle O_{2} D A=45^{\\circ}$. Hence $\\angle O_{1} D O_{2}=$ $90^{\\circ}$ and $D O_{1} / D O_{2}=c / b$ from which it follows that $\\triangle O_{1} D O_{2} \\sim$ $\\triangle B A C$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-516.jpg?height=308&width=520&top_left_y=648&top_left_x=831)  We now define $P$ as the intersection of the circumcircle of $\\triangle O_{1} D O_{2}$ with $D A$. From the above similarity we have $\\angle D P O_{2}=\\angle D O_{1} O_{2}=\\beta=$ $\\angle D A C$. It follows that $P O_{2} \\| A C$ and from $\\angle O_{1} P O_{2}=90^{\\circ}$ it also follows that $P O_{1} \\| A B$. We also have $\\angle P O_{1} O_{2}=\\angle P O_{2} O_{1}=45^{\\circ}$; hence $\\angle L K A=\\angle K L A=45^{\\circ}$, and thus $A K=A L$. From $\\angle O_{1} K A=\\angle O_{1} D A=$ $45^{\\circ}, O_{1} A=O_{1} A$, and $\\angle O_{1} K A=\\angle O_{1} D A$ we have $\\triangle O_{1} K A \\cong \\triangle O_{1} D A$ and hence $A L=A K=A D=h$. Thus  $$ \\frac{E}{E_{1}}=\\frac{a h / 2}{h^{2} / 2}=\\frac{a}{h}=\\frac{a^{2}}{a h}=\\frac{b^{2}+c^{2}}{b c} \\geq 2 . $$  Remark. It holds that for an arbitrary triangle $A B C, A K=A L$ if and only if $A B=A C$ or $\\measuredangle B A C=90^{\\circ}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (HUN 1) For what values of $n$ does there exist an $n \\times n$ array of entries $-1,0$, or 1 such that the $2 n$ sums obtained by summing the elements of the rows and the columns are all different?", "solution": "14. Consider an array $\\left[a_{i j}\\right]$ of the given property and denote the sums of the rows and the columns by $r_{i}$ and $c_{j}$ respectively. Among the $r_{i}$ 's and $c_{j}$ 's, one element of $[-n, n]$ is missing, so that there are at least $n$ nonnegative and $n$ nonpositive sums. By permuting rows and columns we can obtain an array in which $r_{1}, \\ldots, r_{k}$ and $c_{1}, \\ldots, c_{n-k}$ are nonnegative. Clearly  $$ \\sum_{i=1}^{n}\\left|r_{i}\\right|+\\sum_{j=1}^{n}\\left|c_{j}\\right| \\geq \\sum_{r=-n}^{n}|r|-n=n^{2} $$  But on the other hand,  $$ \\begin{aligned} \\sum_{i=1}^{n}\\left|r_{i}\\right|+\\sum_{j=1}^{n}\\left|c_{j}\\right| & =\\sum_{i=1}^{k} r_{i}-\\sum_{i=k+1}^{n} r_{i}+\\sum_{j=1}^{n-k} c_{j}-\\sum_{j=n-k+1}^{n} c_{j}= \\\\ & =\\sum_{i \\leq k} a_{i j}-\\sum_{i>k} a_{i j}+\\sum_{j \\leq n-k} a_{i j}-\\sum_{j>n-k} a_{i j}= \\\\ & =2 \\sum_{i=1}^{k} \\sum_{j=1}^{n-k} a_{i j}-2 \\sum_{i=k+1}^{n} \\sum_{j=n-k+1}^{n} a_{i j} \\leq 4 k(n-k) . \\end{aligned} $$  This yields $n^{2} \\leq 4 k(n-k)$, i.e., $(n-2 k)^{2} \\leq 0$, and thus $n$ must be even. We proceed to show by induction that for all even $n$ an array of the given type exists. For $n=2$ the array in Fig. 1 is good. Let such an $n \\times n$ array be given for some even $n \\geq 2$, with $c_{1}=n, c_{2}=-n+1, c_{3}=$ $n-2, \\ldots, c_{n-1}=2, c_{n}=-1$ and $r_{1}=n-1, r_{2}=-n+2, \\ldots, r_{n-1}=1$, $r_{n}=0$. Upon enlarging this array as indicated in Fig. 2, the positive sums are increased by 2 , the nonpositive sums are decreased by 2 , and the missing sums $-1,0,1,2$ occur in the new rows and columns, so that the obtained array $(n+2) \\times(n+2)$ is of the same type. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-517.jpg?height=89&width=87&top_left_y=1080&top_left_x=487)  Fig. 1  | $n \\times n$ |  |  | 1 | 1 | | :---: | :---: | :---: | :---: | :---: | |  |  |  |  |  | |  |  |  | 1 | 1 | |  |  |  | - |  | -1 | | 1)-1 | 1 | - | 1 | 0 | | 1)-1 |  | 1. | 1 | -1 |  Fig. 2", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (ICE 1) Let $A B C$ be an acute-angled triangle. Three lines $L_{A}, L_{B}$, and $L_{C}$ are constructed through the vertices $A, B$, and $C$ respectively according to the following prescription: Let $H$ be the foot of the altitude drawn from the vertex $A$ to the side $B C$; let $S_{A}$ be the circle with diameter $A H$; let $S_{A}$ meet the sides $A B$ and $A C$ at $M$ and $N$ respectively, where $M$ and $N$ are distinct from $A$; then $L_{A}$ is the line through $A$ perpendicular to $M N$. The lines $L_{B}$ and $L_{C}$ are constructed similarly. Prove that $L_{A}$, $L_{B}$, and $L_{C}$ are concurrent.", "solution": "15. Referring to the description of $L_{A}$, we have $\\angle A M N=\\angle A H N=90^{\\circ}-$ $\\angle H A C=\\angle C$, and similarly $\\angle A N M=\\angle B$. Since the triangle $A B C$ is acute-angled, the line $L_{A}$ lies inside the angle $A$. Hence if $P=L_{A} \\cap B C$ and $Q=L_{B} \\cap A C$, we get $\\angle B A P=90^{\\circ}-\\angle C$; hence $A P$ passes through the circumcenter $O$ of $\\triangle A B C$. Similarly we prove that $L_{B}$ and $L_{C}$ contains the circumcenter $O$ also. It follows that $L_{A}, L_{B}$ and $L_{C}$ intersect at the point $O$. Remark. Without identifying the point of intersection, one can prove the concurrence of the three lines using Ceva's theorem, in usual or trigonometric form.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (IRE 1) $)^{\\mathrm{IMO} 4}$ Show that the solution set of the inequality  $$ \\sum_{k=1}^{70} \\frac{k}{x-k} \\geq \\frac{5}{4} $$  is a union of disjoint intervals the sum of whose lengths is 1988.", "solution": "16. Let $f(x)=\\sum_{k=1}^{70} \\frac{k}{x-k}$. For all integers $i=1, \\ldots, 70$ we have that $f(x)$ tends to plus infinity as $x$ tends downward to $i$, and $f(x)$ tends to minus infinity as $x$ tends upward to $i$. As $x$ tends to infinity, $f(x)$ tends to 0 . Hence it follows that there exist $x_{1}, x_{2}, \\ldots, x_{70}$ such that $1<x_{1}<2<$ $x_{2}<3<\\cdots<x_{69}<70<x_{70}$ and $f\\left(x_{i}\\right)=\\frac{5}{4}$ for all $i=1, \\ldots, 70$. Then the solution to the inequality is given by $S=\\bigcup_{i=1}^{70}\\left(i, x_{i}\\right]$. For numbers $x$ for which $f(x)$ is well-defined, the equality $f(x)=\\frac{5}{4}$ is equivalent to  $$ p(x)=\\prod_{j=1}^{70}(x-j)-\\frac{4}{5} \\sum_{k=1}^{70} k \\prod_{\\substack{j=1 \\\\ j \\neq k}}^{70}(x-j)=0 $$  The numbers $x_{1}, x_{2}, \\ldots, x_{70}$ are then the zeros of this polynomial. The sum $\\sum_{i=1}^{70} x_{i}$ is then equal to minus the coefficient of $x^{69}$ in $p$, which equals $\\sum_{i=1}^{70}\\left(i+\\frac{4}{5} i\\right)$. Finally,  $$ |S|=\\sum_{i=1}^{70}\\left(x_{i}-i\\right)=\\frac{4}{5} \\cdot \\sum_{i=1}^{70} i=\\frac{4}{5} \\cdot \\frac{70 \\cdot 71}{2}=1988 $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (ISR 2) In the convex pentagon $A B C D E$, the sides $B C, C D, D E$ have the same length. Moreover, each diagonal of the pentagon is parallel to a side ( $A C$ is parallel to $D E, B D$ is parallel to $A E$, etc.). Prove that $A B C D E$ is a regular pentagon.", "solution": "17. Let $A C$ and $A D$ meet $B E$ in $R, S$, respectively. Then by the conditions of the problem,  $$ \\begin{aligned} & \\angle A E B=\\angle E B D=\\angle B D C=\\angle D B C=\\angle A D B=\\angle E A D=\\alpha, \\\\ & \\angle A B E=\\angle B E C=\\angle E C D=\\angle C E D=\\angle A C E=\\angle B A C=\\beta \\\\ & \\angle B C A=\\angle C A D=\\angle A D E=\\gamma \\end{aligned} $$  Since $\\angle S A E=\\angle S E A$, it follows that $A S=S E$, and analogously $B R=$ $R A$. But $B S D C$ and $R E D C$ are parallelograms; hence $B S=C D=R E$, giving us $B R=S E$ and $A R=A S$. Then also $A C=A D$, because $R S \\|$ $C D$. We deduce that $2 \\beta=\\angle A C D=\\angle A D C=2 \\alpha$, i.e., $\\alpha=\\beta$. It will be sufficient to show that $\\alpha=\\gamma$, since that will imply $\\alpha=\\beta=\\gamma=$ $36^{\\circ}$. We have that the sum of the interior angles of $A C D$ is $4 \\alpha+\\gamma=180^{\\circ}$. We have  $$ \\frac{\\sin \\gamma}{\\sin \\alpha}=\\frac{A E}{D E}=\\frac{A E}{C D}=\\frac{A E}{R E}=\\frac{\\sin (2 \\alpha+\\gamma)}{\\sin (\\alpha+\\gamma)} $$  i.e., $\\cos \\alpha-\\cos (\\alpha+2 \\gamma)=2 \\sin \\gamma \\sin (\\alpha+\\gamma)=2 \\sin \\alpha \\sin (2 \\alpha+\\gamma)=\\cos (\\alpha+$ $\\gamma)-\\cos (3 \\alpha+\\gamma)$. From $4 \\alpha+\\gamma=180^{\\circ}$ we obtain $-\\cos (3 \\alpha+\\gamma)=\\cos \\alpha$. Hence  $$ \\cos (\\alpha+\\gamma)+\\cos (\\alpha+2 \\gamma)=2 \\cos \\frac{\\gamma}{2} \\cos \\frac{2 \\alpha+3 \\gamma}{2}=0 $$  so that $2 \\alpha+3 \\gamma=180^{\\circ}$. It follows that $\\alpha=\\gamma$. Second solution. We have $\\angle B E C=\\angle E C D=\\angle D E C=\\angle E C A=$ $\\angle C A B$, and hence the trapezoid $B A E C$ is cyclic; consequently, $A E=$ $B C$. Similarly $A B=E D$, and $A B C D$ is cyclic as well. Thus $A B C D E$ is cyclic and has all sides equal; i.e., it is regular.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (LUX 1) ${ }^{\\mathrm{IMO}}$ Consider two concentric circles of radii $R$ and $r(R>r)$ with center $O$. Fix $P$ on the small circle and consider the variable chord $P A$ of the small circle. Points $B$ and $C$ lie on the large circle; $B, P, C$ are collinear and $B C$ is perpendicular to $A P$. (a) For what value(s) of $\\angle O P A$ is the sum $B C^{2}+C A^{2}+A B^{2}$ extremal? (b) What are the possible positions of the midpoints $U$ of $B A$ and $V$ of $A C$ as $\\angle O P A$ varies?", "solution": "18. (i) Define $\\angle A P O=\\phi$ and $S=A B^{2}+A C^{2}+B C^{2}$. We calculate $P A=$ $2 r \\cos \\phi$ and $P B, P C=\\sqrt{R^{2}-r^{2} \\cos ^{2} \\phi} \\pm r \\sin \\phi$. We also have $A B^{2}=$ $P A^{2}+P B^{2}, A C^{2}=P A^{2}+P C^{2}$ and $B C=B P+P C$. Combining all these we obtain  $$ \\begin{aligned} S & =A B^{2}+A C^{2}+B C^{2}=2\\left(P A^{2}+P B^{2}+P C^{2}+P B \\cdot P C\\right) \\\\ & =2\\left(4 r^{2} \\cos ^{2} \\phi+2\\left(R^{2}-r^{2} \\cos ^{2} \\phi+r^{2} \\sin ^{2} \\phi\\right)+R^{2}-r^{2}\\right) \\\\ & =6 R^{2}+2 r^{2} . \\end{aligned} $$  Hence it follows that $S$ is constant; i.e., it does not depend on $\\phi$. (ii) Let $B_{1}$ and $C_{1}$ respectively be points such that $A P B B_{1}$ and $A P C C_{1}$ are rectangles. It is evident that $B_{1}$ and $C_{1}$ lie on the larger circle and that $\\overrightarrow{P U}=\\frac{1}{2} \\overrightarrow{P B_{1}}$ and $\\overrightarrow{P V}=\\frac{1}{2} \\overrightarrow{P C_{1}}$. It is evident that we can arrange for an arbitrary point on the larger circle to be $B_{1}$ or $C_{1}$. Hence, the locus of $U$ and $V$ is equal to the circle obtained when the larger circle is shrunk by a factor of $1 / 2$ with respect to point $P$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (MEX 1) Let $f(n)$ be a function defined on the set of all positive integers and having its values in the same set. Suppose that $f(f(m)+f(n))=m+n$ for all positive integers $n, m$. Find all possible values for $f(1988)$.", "solution": "19. We will show that $f(n)=n$ for every $n$ (thus also $f(1988)=1988$ ).  Let $f(1)=r$ and $f(2)=s$. We obtain respectively the following equalities: $f(2 r)=f(r+r)=2 ; f(2 s)=f(s+s)=4 ; f(4)=f(2+2)=4 r ; f(8)=$ $f(4+4)=4 s ; f(5 r)=f(4 r+r)=5 ; f(r+s)=3 ; f(8)=f(5+3)=6 r+s$. Then $4 s=6 r+s$, which means that $s=2 r$. Now we prove by induction that $f(n r)=n$ and $f(n)=n r$ for every $n \\geq 4$. First we have that $f(5)=f(2+3)=3 r+s=5 r$, so that the statement is true for $n=4$ and $n=5$. Suppose that it holds for $n-1$ and $n$. Then $f(n+1)=f(n-1+2)=(n-1) r+2 r=(n+1) r$, and $f((n+1) r)=f((n-1) r+2 r)=(n-1)+2=n+1$. This completes the induction. Since $4 r \\geq 4$, we have that $f(4 r)=4 r^{2}$, and also $f(4 r)=4$. Then $r=1$, and consequently $f(n)=n$ for every natural number $n$. Second solution. $f(f(1)+n+m)=f(f(1)+f(f(n)+f(m)))=1+f(n)+$ $f(m)$, so $f(n)+f(m)$ is a function of $n+m$. Hence $f(n+1)+f(1)=$ $f(n)+f(2)$ and $f(n+1)-f(n)=f(2)-f(1)$, implying that $f(n)=A n+B$ for some constants $A, B$. It is easy to check that $A=1, B=0$ is the only possibility.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (BUL 3) Let $n$ be a positive integer. Find the number of odd coefficients of the polynomial  $$ u_{n}(x)=\\left(x^{2}+x+1\\right)^{n} $$", "solution": "2. For polynomials $f(x), g(x)$ with integer coefficients, we use the notation $f(x) \\sim g(x)$ if all the coefficients of $f-g$ are even. Let $n=2^{s}$. It is immediately shown by induction that $\\left(x^{2}+x+1\\right)^{2^{s}} \\sim x^{2^{s+1}}+x^{2^{s}}+1$, and the required number for $n=2^{s}$ is 3 .  Let $n=2^{s}-1$. If $s$ is odd, then $n \\equiv 1(\\bmod 3)$, while for $s$ even, $n \\equiv 0$ $(\\bmod 3)$. Consider the polynomial  $$ R_{s}(x)= \\begin{cases}(x+1)\\left(x^{2 n-1}+x^{2 n-4}+\\cdots+x^{n+3}\\right)+x^{n+1} & \\\\ +x^{n}+x^{n-1}+(x+1)\\left(x^{n-4}+x^{n-7}+\\cdots+1\\right), & 2 \\nmid s \\\\ (x+1)\\left(x^{2 n-1}+x^{2 n-4}+\\cdots+x^{n+2}\\right)+x^{n} & \\\\ +(x+1)\\left(x^{n-3}+x^{n-6}+\\cdots+1\\right), & 2 \\mid s\\end{cases} $$  It is easily checked that $\\left(x^{2}+x+1\\right) R_{s}(x) \\sim x^{2^{s+1}}+x^{2^{s}}+1 \\sim\\left(x^{2}+x+1\\right)^{2^{s}}$, so that $R_{s}(x) \\sim\\left(x^{2}+x+1\\right)^{2^{s}-1}$. In this case, the number of odd coefficients is $\\left(2^{s+2}-(-1)^{s}\\right) / 3$. Now we pass to the general case. Let the number $n$ be represented in the binary system as  $$ n=\\underbrace{11 \\ldots 1}_{a_{k}} \\underbrace{00 \\ldots 0}_{b_{k}} \\underbrace{11 \\ldots 1}_{a_{k-1}} \\underbrace{00 \\ldots 0}_{b_{k-1}} \\cdots \\underbrace{11 \\ldots 1}_{a_{1}} \\underbrace{00 \\ldots 0}_{b_{1}} $$  $b_{i}>0(i>1), b_{1} \\geq 0$, and $a_{i}>0$. Then $n=\\sum_{i=1}^{k} 2^{s_{i}}\\left(2^{a_{i}}-1\\right)$, where $s_{i}=b_{1}+a_{1}+b_{2}+a_{2}+\\cdots+b_{i}$, and hence  $$ u_{n}(x)=\\left(x^{2}+x+1\\right)^{n}=\\prod_{i=1}^{k}\\left(x^{2}+x+1\\right)^{2^{s_{i}}\\left(2^{a_{i}}-1\\right)} \\sim \\prod_{i=1}^{k} R_{a_{i}}\\left(x^{2^{s_{i}}}\\right) $$  Let $R_{a_{i}}\\left(x^{2^{s_{i}}}\\right) \\sim x^{r_{i, 1}}+\\cdots+x^{r_{i, d_{i}}}$; clearly $r_{i, j}$ is divisible by $2^{s_{i}}$ and $r_{i, j} \\leq 2^{s_{i}+1}\\left(2^{a_{i}}-1\\right)<2^{s_{i+1}}$, so that for any $j, r_{i, j}$ can have nonzero binary digits only in some position $t, s_{i} \\leq t \\leq s_{i+1}-1$. Therefore, in  $$ \\prod_{i=1}^{k} R_{a_{i}}\\left(x^{2^{s_{i}}}\\right) \\sim \\prod_{i=1}^{k}\\left(x^{r_{i, 1}}+\\cdots+x^{r_{i, d_{i}}}\\right)=\\sum_{i=1}^{k} \\sum_{p_{i}=1}^{d_{i}} x^{r_{1, p_{1}}+r_{2, p_{2}}+\\cdots+r_{k, p_{k}}} $$  all the exponents $r_{1, p_{1}}+r_{2, p_{2}}+\\cdots+r_{k, p_{k}}$ are different, so that the number of odd coefficients in $u_{n}(x)$ is  $$ \\prod_{i=1}^{k} d_{i}=\\prod_{i=1}^{k} \\frac{2^{a_{i}+2}-(-1)^{a_{i}}}{3} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. (MON 4) Find the least natural number $n$ such that if the set $\\{1,2, \\ldots, n\\}$ is arbitrarily divided into two nonintersecting subsets, then one of the subsets contains three distinct numbers such that the product of two of them equals the third.", "solution": "20. Suppose that $A_{n}=\\{1,2, \\ldots, n\\}$ is partitioned into $B_{n}$ and $C_{n}$, and that neither $B_{n}$ nor $C_{n}$ contains 3 distinct numbers one of which is equal to the product of the other two. If $n \\geq 96$, then the divisors of 96 must be split up. Let w.l.o.g. $2 \\in B_{n}$. There are four cases. (i) $3 \\in B_{n}, 4 \\in B_{n}$. Then $6,8,12 \\in C_{n} \\Rightarrow 48,96 \\in B_{n}$. A contradiction for $96=2 \\cdot 48$. (ii) $3 \\in B_{n}, 4 \\in C_{n}$. Then $6 \\in C_{n}, 24 \\in B_{n}, 8,12,48 \\in C_{n}$. A contradiction for $48=6 \\cdot 8$. (iii) $3 \\in C_{n}, 4 \\in B_{n}$. Then $8 \\in C_{n}, 24 \\in B_{n}, 6,48 \\in C_{n}$. A contradiction for $48=6 \\cdot 8$. (iv) $3 \\in C_{n}, 4 \\in C_{n}$. Then $12 \\in B_{n}, 6,24 \\in C_{n}$. A contradiction for $24=4 \\cdot 6$. If $n=95$, there is a very large number of ways of partitioning $A_{n}$. For example, $B_{n}=\\left\\{1, p, p^{2}, p^{3} q^{2}, p^{4} q, p^{2} q r \\mid p, q, r=\\operatorname{distinct}\\right.$ primes $\\}$, $C_{n}=\\left\\{p^{3}, p^{4}, p^{5}, p^{6}, p q, p^{2} q, p^{3} q, p^{2} q^{2}, p q r \\mid p, q, r=\\right.$ distinct primes $\\}$. Then $B_{95}=\\{1,2,3,4,5,7,9,11,13,17,19,23,25,29,31,37,41$, $43,47,48,49,53,59,60,61,67,71,72,73,79,80,83,84,89,90\\}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. (POL 4) Forty-nine students solve a set of three problems. The score for each problem is a whole number of points from 0 to 7 . Prove that there exist two students $A$ and $B$ such that for each problem, $A$ will score at least as many points as $B$.", "solution": "21. Let $X$ be the set of all ordered triples $a=\\left(a_{1}, a_{2}, a_{3}\\right)$ for $a_{i} \\in\\{0,1, \\ldots, 7\\}$. Write $a \\prec b$ if $a_{i} \\leq b_{i}$ for $i=1,2,3$ and $a \\neq b$. Call a subset $Y \\subset X$ independent if there are no $a, b \\in Y$ with $a \\prec b$. We shall prove that an independent set contains at most 48 elements. For $j=0,1, \\ldots, 21$ let $X_{j}=\\left\\{\\left(a_{1}, a_{2}, a_{3}\\right) \\in X \\mid a_{1}+a_{2}+a_{3}=j\\right\\}$. If $x \\prec y$ and $x \\in X_{j}, y \\in X_{j+1}$ for some $j$, then we say that $y$ is a successor of $x$, and $x$ a predecessor of $y$. Lemma. If $A$ is an $m$-element subset of $X_{j}$ and $j \\leq 10$, then there are at least $m$ distinct successors of the elements of $A$. Proof. For $k=0,1,2,3$ let $X_{j, k}=\\left\\{\\left(a_{1}, a_{2}, a_{3}\\right) \\in X_{j} \\mid \\min \\left(a_{1}, a_{2}, a_{3}, 7-\\right.\\right.$ $\\left.\\left.a_{1}, 7-a_{2}, 7-a_{3}\\right)=k\\right\\}$. It is easy to verify that every element of $X_{j, k}$ has at least two successors in $X_{j+1, k}$ and every element of $X_{j+1, k}$ has at most two predecessors in $X_{j, k}$. Therefore the number of elements of $A \\cap X_{j, k}$ is not greater than the number of their successors. Since $X_{j}$ is a disjoint union of $X_{j, k}, k=0,1,2,3$, the lemma follows. Similarly, elements of an $m$-element subset of $X_{j}, j \\geq 11$, have at least $m$ predecessors. Let $Y$ be an independent set, and let $p, q$ be integers such that $p<10<q$. We can transform $Y$ by replacing all the elements of $Y \\cap X_{p}$ with their successors, and all the elements of $Y \\cap X_{q}$ with their predecessors. After this transformation $Y$ will still be independent, and by the lemma its size will not be reduced. Every independent set can be eventually transformed in this way into a subset of $X_{10}$, and $X_{10}$ has exactly 48 elements.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "22", "problem_type": null, "exam": "IMO", "problem": "22. (KOR 2) Let $p$ be the product of two consecutive integers greater than 2. Show that there are no integers $x_{1}, x_{2}, \\ldots, x_{p}$ satisfying the equation  $$ \\sum_{i=1}^{p} x_{i}^{2}-\\frac{4}{4 p+1}\\left(\\sum_{i=1}^{p} x_{i}\\right)^{2}=1 $$  Alternative formulation. Show that there are only two values of $p$ for which there are integers $x_{1}, x_{2}, \\ldots, x_{p}$ satisfying the above inequality.", "solution": "22. Set $X=\\sum_{i=1}^{p} x_{i}$ and w.l.o.g. assume that $X \\geq 0$ (if $\\left(x_{1}, \\ldots, x_{p}\\right)$ is a solution, then $\\left(-x_{1}, \\ldots,-x_{p}\\right)$ is a solution too). Since $x^{2} \\geq x$ for all integers $x$, it follows that $\\sum_{i=1}^{p} x_{i}^{2} \\geq X$. If the last inequality is an equality, then all $x_{i}$ 's are 0 or 1 ; then, taking that there are $a 1$ 's, the equation becomes $4 p+1=4(a+1)+\\frac{4}{a-1}$, which forces $p=6$ and $a=5$. Otherwise, we have $X+1 \\leq \\sum_{i=1}^{p} x_{i}^{2}=\\frac{4}{4 p+1} X^{2}+1$, so $X \\geq p+1$. Also, by the Cauchy-Schwarz inequality, $X^{2} \\leq p \\sum_{i=1}^{p} x_{i}^{2}=\\frac{4 p}{4 p+1} X^{2}+p$, so $X^{2} \\leq 4 p^{2}+p$ and $X \\leq 2 p$. Thus $1 \\leq X / p \\leq 2$. However,  $$ \\begin{aligned} \\sum_{i=1}^{p}\\left(x_{i}-\\frac{X}{p}\\right)^{2} & =\\sum x_{i}^{2}-\\frac{2 X}{p} \\sum x_{i}+\\frac{X^{2}}{p} \\\\ & =\\sum x_{i}^{2}-p \\frac{X^{2}}{p^{2}}=1-\\frac{X^{2}}{p(4 p+1)}<1 \\end{aligned} $$  and we deduce that $-1<x_{i}-X / p<1$ for all $i$. This finally gives $x_{i} \\in\\{1,2\\}$. Suppose there are $b 2$ 's. Then $3 b+p=4(b+p)^{2} /(4 p+1)+1$, so $p=b+1 /(4 b-3)$, which leads to $p=2, b=1$. Thus there are no solutions for any $p \\notin\\{2,6\\}$. Remark. The condition $p=n(n+1), n \\geq 3$, was unnecessary in the official solution, too (its only role was to simplify showing that $X \\neq p-1$ ).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "23", "problem_type": null, "exam": "IMO", "problem": "23. (SIN 2) Let $Q$ be the center of the inscribed circle of a triangle $A B C$. Prove that for any point $P$, $a(P A)^{2}+b(P B)^{2}+c(P C)^{2}=a(Q A)^{2}+b(Q B)^{2}+c(Q C)^{2}+(a+b+c)(Q P)^{2}$, where $a=B C, b=C A$, and $c=A B$.", "solution": "23. Denote by $R$ the intersection point of lines $A Q$ and $B C$. We know that $B R: R C=c: b$ and $A Q: Q R=(b+c): a$. By applying Stewart's theorem to $\\triangle P B C$ and $\\triangle P A R$ we obtain  $$ \\begin{aligned} a \\cdot A P^{2} & +b \\cdot B P^{2}+c \\cdot C P^{2}=a P A^{2}+(b+c) P R^{2}+(b+c) R B \\cdot R C \\\\ & =(a+b+c) Q P^{2}+(b+c) R B \\cdot R C+(a+b+c) Q A \\cdot Q R \\end{aligned} $$  On the other hand, putting $P=Q$ into (1), we get that  $$ a \\cdot A Q^{2}+b \\cdot B Q^{2}+c \\cdot C Q^{2}=(b+c) R B \\cdot R C+(a+b+c) Q A \\cdot Q R, $$  and the required statement follows. Second solution. At vertices $A, B, C$ place weights equal to $a, b, c$ in some units respectively, so that $Q$ is the center of gravity of the system. The left side of the equality to be proved is in fact the moment of inertia of the system about the axis through $P$ and perpendicular to the plane $A B C$. On the other side, the right side expresses the same, due to the parallel axes theorem. Alternative approach. Analytical geometry. The fact that all the variable segments appear squared usually implies that this is a good approach. Assign coordinates $A\\left(x_{a}, y_{a}\\right), B\\left(x_{b}, y_{b}\\right), C\\left(x_{c}, y_{c}\\right)$, and $P(x, y)$, use that $(a+b+c) \\mathbf{Q}=a \\mathbf{A}+b \\mathbf{B}+c \\mathbf{C}$, and calculate. Alternatively, differentiate $f(x, y)=a \\cdot A P^{2}+b \\cdot B P^{2}+c \\cdot C P^{2}-(a+b+c) Q P^{2}$ and show that it is constant.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "24", "problem_type": null, "exam": "IMO", "problem": "24. (SWE 2) Let $\\left\\{a_{k}\\right\\}_{1}^{\\infty}$ be a sequence of nonnegative real numbers such that $a_{k}-2 a_{k+1}+a_{k+2} \\geq 0$ and $\\sum_{j=1}^{k} a_{j} \\leq 1$ for all $k=1,2, \\ldots$. Prove that $0 \\leq\\left(a_{k}-a_{k+1}\\right)<\\frac{2}{k^{2}}$ for all $k=1,2, \\ldots$.", "solution": "24. The first condition means in fact that $a_{k}-a_{k+1}$ is decreasing. In particular, if $a_{k}-a_{k+1}=-\\delta<0$, then $a_{k}-a_{k+m}=\\left(a_{k}-a_{k+1}\\right)+\\cdots+\\left(a_{k+m-1}-\\right.$ $\\left.a_{k+m}\\right)<-m \\delta$, which implies that $a_{k+m}>a_{k}+m \\delta$, and consequently $a_{k+m}>1$ for large enough $m$, a contradiction. Thus $a_{k}-a_{k+1} \\geq 0$ for all $k$. Suppose that $a_{k}-a_{k+1}>2 / k^{2}$. Then for all $i<k, a_{i}-a_{i+1}>2 / k^{2}$, so that $a_{i}-a_{k+1}>2(k+1-i) / k^{2}$, i.e., $a_{i}>2(k+1-i) / k^{2}, i=1,2, \\ldots, k$. But this implies $a_{1}+a_{2}+\\cdots+a_{k}>2 / k^{2}+4 / k^{2}+\\cdots+2 k / k^{2}=k(k+1) / k^{2}$, which is impossible. Therefore $a_{k}-a_{k+1} \\leq 2 / k^{2}$ for all $k$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "25", "problem_type": null, "exam": "IMO", "problem": "25. (GBR 1) A positive integer is called a double number if its decimal representation consists of a block of digits, not commencing with 0 , followed immediately by an identical block. For instance, 360360 is a double number, but 36036 is not. Show that there are infinitely many double numbers that are perfect squares.", "solution": "25. Observe that $1001=7 \\cdot 143$, i.e., $10^{3}=-1+7 a, a=143$. Then by the binomial theorem, $10^{21}=(-1+7 a)^{7}=-1+7^{2} b$ for some integer $b$, so that we also have $10^{21 n} \\equiv-1(\\bmod 49)$ for any odd integer $n>0$. Hence $N=\\frac{9}{49}\\left(10^{21 n}+1\\right)$ is an integer of $21 n$ digits, and $N\\left(10^{21 n}+1\\right)=$ $\\left(\\frac{3}{7}\\left(10^{21 n}+1\\right)\\right)^{2}$ is a double number that is a perfect square.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "26", "problem_type": null, "exam": "IMO", "problem": "26. (GBR 2) ${ }^{\\mathrm{IMO} 3} \\mathrm{~A}$ function $f$ defined on the positive integers (and taking positive integer values) is given by  $$ \\begin{aligned} f(1) & =1, \\quad f(3)=3 \\\\ f(2 n) & =f(n) \\\\ f(4 n+1) & =2 f(2 n+1)-f(n) \\\\ f(4 n+3) & =3 f(2 n+1)-2 f(n) \\end{aligned} $$  for all positive integers $n$. Determine with proof the number of positive integers less than or equal to 1988 for which $f(n)=n$.", "solution": "26. The overline in this problem will exclusively denote binary representation. We will show by induction that if $n=\\overline{c_{k} c_{k-1} \\ldots c_{0}}=\\sum_{i=0}^{k} c_{i} 2^{i}$ is the binary representation of $n\\left(c_{i} \\in\\{0,1\\}\\right)$, then $f(n)=\\overline{c_{0} c_{1} \\ldots c_{k}}=$ $\\sum_{i=0}^{k} c_{i} 2^{k-i}$ is the number whose binary representation is the palindrome of the binary representation of $n$. This evidently holds for $n \\in\\{1,2,3\\}$.  Let us assume that the claim holds for all numbers up to $n-1$ and show it holds for $n=\\overline{c_{k} c_{k-1} \\ldots c_{0}}$. We observe three cases: (i) $c_{0}=0 \\Rightarrow n=2 m \\Rightarrow f(n)=f(m)=\\overline{0 c_{1} \\ldots c_{k}}=\\overline{c_{0} c_{1} \\ldots c_{k}}$. (ii) $c_{0}=1, c_{1}=0 \\Rightarrow n=4 m+1 \\Rightarrow f(n)=2 f(2 m+1)-f(m)=$ $2 \\cdot \\overline{1 c_{2} \\ldots c_{k}}-\\overline{c_{2} \\ldots c_{k}}=2^{k}+2 \\cdot \\overline{c_{2} \\ldots c_{k}}-\\overline{c_{2} \\ldots c_{k}}=\\overline{10 c_{2} \\ldots c_{k}}=$ $\\overline{c_{0} c_{1} \\ldots c_{k}}$. (iii) $c_{0}=1, c_{1}=1 \\Rightarrow n=4 m+3 \\Rightarrow f(n)=3 f(2 m+1)-2 f(m)=$ $3 \\cdot \\overline{1 c_{2} \\ldots c_{k}}-2 \\cdot \\overline{c_{2} \\ldots c_{k}}=2^{k}+2^{k-1}+3 \\cdot \\overline{c_{2} \\ldots c_{k}}-2 \\cdot \\overline{c_{2} \\ldots c_{k}}=$ $\\overline{11 c_{2} \\ldots c_{k}}=\\overline{c_{0} c_{1} \\ldots c_{k}}$. We thus have to find the number of palindromes in binary representation smaller than $1998=\\overline{11111000100}$. We note that for all $m \\in \\mathbb{N}$ the numbers of $2 m$ - and $(2 m-1)$-digit binary palindromes are both equal to $2^{m-1}$. We also note that $\\overline{11111011111}$ and $\\overline{11111111111}$ are the only 11-digit palindromes larger than 1998. Hence we count all palindromes of up to 11 digits and exclude the largest two. The number of $n \\leq 1998$ such that $f(n)=n$ is thus equal to $1+1+2+2+4+4+8+8+16+16+32-2=92$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "27", "problem_type": null, "exam": "IMO", "problem": "27. (GBR 4) The triangle $A B C$ is acute-angled. Let $L$ be any line in the plane of the triangle and let $u, v, w$ be the lengths of the perpendiculars from $A, B, C$ respectively to $L$. Prove that  $$ u^{2} \\tan A+v^{2} \\tan B+w^{2} \\tan C \\geq 2 \\Delta $$  where $\\Delta$ is the area of the triangle, and determine the lines $L$ for which equality holds.", "solution": "27. Consider a Cartesian system with the $x$-axis on the line $B C$ and origin at the foot of the perpendicular from $A$ to $B C$, so that $A$ lies on the $y$-axis. Let $A$ be $(0, \\alpha), B(-\\beta, 0), C(\\gamma, 0)$, where $\\alpha, \\beta, \\gamma>0$ (because $A B C$ is acute-angled). Then $\\tan B=\\frac{\\alpha}{\\beta}, \\quad \\tan C=\\frac{\\alpha}{\\gamma} \\quad$ and $\\quad \\tan A=-\\tan (B+C)=\\frac{\\alpha(\\beta+\\gamma)}{\\alpha^{2}-\\beta \\gamma} ;$ here $\\tan A>0$, so $\\alpha^{2}>\\beta \\gamma$. Let $L$ have equation $x \\cos \\theta+y \\sin \\theta+p=0$. Then  $$ \\begin{aligned} & u^{2} \\tan A+v^{2} \\tan B+w^{2} \\tan C \\\\ & =\\frac{\\alpha(\\beta+\\gamma)}{\\alpha^{2}-\\beta \\gamma}(\\alpha \\sin \\theta+p)^{2}+\\frac{\\alpha}{\\beta}(-\\beta \\cos \\theta+p)^{2}+\\frac{\\alpha}{\\gamma}(\\gamma \\cos \\theta+p)^{2} \\\\ & =\\left(\\alpha^{2} \\sin ^{2} \\theta+2 \\alpha p \\sin \\theta+p^{2}\\right) \\frac{\\alpha(\\beta+\\gamma)}{\\alpha^{2}-\\beta \\gamma}+\\alpha(\\beta+\\gamma) \\cos ^{2} \\theta+\\frac{\\alpha(\\beta+\\gamma)}{\\beta \\gamma} p^{2} \\\\ & =\\frac{\\alpha(\\beta+\\gamma)}{\\beta \\gamma\\left(\\alpha^{2}-\\beta \\gamma\\right)}\\left(\\alpha^{2} p^{2}+2 \\alpha p \\beta \\gamma \\sin \\theta+\\alpha^{2} \\beta \\gamma \\sin ^{2} \\theta+\\beta \\gamma\\left(\\alpha^{2}-\\beta \\gamma\\right) \\cos ^{2} \\theta\\right) \\\\ & =\\frac{\\alpha(\\beta+\\gamma)}{\\beta \\gamma\\left(\\alpha^{2}-\\beta \\gamma\\right)}\\left[(\\alpha p+\\beta \\gamma \\sin \\theta)^{2}+\\beta \\gamma\\left(\\alpha^{2}-\\beta \\gamma\\right)\\right] \\geq \\alpha(\\beta+\\gamma)=2 \\Delta \\end{aligned} $$  with equality when $\\alpha p+\\beta \\gamma \\sin \\theta=0$, i.e., if and only if $L$ passes through $(0, \\beta \\gamma / \\alpha)$, which is the orthocenter of the triangle.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "28", "problem_type": null, "exam": "IMO", "problem": "28. (GBR 5) The sequence $\\left\\{a_{n}\\right\\}$ of integers is defined by $a_{1}=2, a_{2}=7$, and  $$ -\\frac{1}{2}<a_{n+1}-\\frac{a_{n}^{2}}{a_{n-1}} \\leq \\frac{1}{2}, \\quad \\text { for } n \\geq 2 $$  Prove that $a_{n}$ is odd for all $n>1$.", "solution": "28. The sequence is uniquely determined by the conditions, and $a_{1}=2, a_{2}=$ $7, a_{3}=25, a_{4}=89, a_{5}=317, \\ldots$; it satisfies $a_{n}=3 a_{n-1}+2 a_{n-2}$ for $n=3,4,5$. We show that the sequence $b_{n}$ given by $b_{1}=2, b_{2}=7$, $b_{n}=3 b_{n-1}+2 b_{n-2}$ has the same inequality property, i.e., that $b_{n}=a_{n}$ : $b_{n+1} b_{n-1}-b_{n}^{2}=\\left(3 b_{n}+2 b_{n-1}\\right) b_{n-1}-b_{n}\\left(3 b_{n-1}+2 b_{n-2}\\right)=-2\\left(b_{n} b_{n-2}-b_{n-1}^{2}\\right)$ for $n>2$ gives that $b_{n+1} b_{n-1}-b_{n}^{2}=(-2)^{n-2}$ for all $n \\geq 2$. But then  $$ \\left|b_{n+1}-\\frac{b_{n}^{2}}{b_{n-1}}\\right|=\\frac{2^{n-2}}{b_{n-1}}<\\frac{1}{2}, $$  since it is easily shown that $b_{n-1}>2^{n-1}$ for all $n$. It is obvious that $a_{n}=b_{n}$ are odd for $n>1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "29", "problem_type": null, "exam": "IMO", "problem": "29. (USA 3) A number of signal lights are equally spaced along a one-way railroad track, labeled in order $1,2, \\ldots, N(N \\geq 2)$. As a safety rule, a train is not allowed to pass a signal if any other train is in motion on the length of track between it and the following signal. However, there is no limit to the number of trains that can be parked motionless at a signal, one behind the other. (Assume that the trains have zero length.) A series of $K$ freight trains must be driven from Signal 1 to Signal $N$. Each train travels at a distinct but constant speed (i.e., the speed is fixed and different from that of each of the other trains) at all times when it is not blocked by the safety rule. Show that regardless of the order in which the trains are arranged, the same time will elapse between the first train's departure from Signal 1 and the last train's arrival at Signal $N$.", "solution": "29. Let the first train start from Signal 1 at time 0, and let $t_{j}$ be the time it takes for the $j$ th train in the series to travel from one signal to the next. By induction on $k$, we show that Train $k$ arrives at signal $n$ at time $s_{k}+(n-2) m_{k}$, where $s_{k}=t_{1}+\\cdots+t_{k}$ and $m_{k}=\\max _{j=1, \\ldots, k} t_{j}$. For $k=1$ the statement is clear. We now suppose that it is true for $k$ trains and for every $n$, and add a $(k+1)$ th train behind the others at Signal 1. There are two cases to consider: (i) $t_{k+1} \\geq m_{k}$, i.e., $m_{k+1}=t_{k+1}$. Then Train $k+1$ leaves Signal 1 when all the others reach Signal 2, which by the induction happens at time $s_{k}$. Since by the induction hypothesis Train $k$ arrives at Signal $i+1$ at time $s_{k}+(i-1) m_{k} \\leq s_{k}+(i-1) t_{k+1}$, Train $k+1$ is never forced to stop. The journey finishes at time $s_{k}+(n-1) t_{k+1}=s_{k+1}+(n-2) m_{k+1}$. (ii) $t_{k+1}<m_{k}$, i.e., $m_{k+1}=m_{k}$. Train $k+1$ leaves Signal 1 at time $s_{k}$, and reaches Signal 2 at time $s_{k}+t_{k+1}$, but must wait there until all the other trains get to Signal 3, i.e., until time $s_{k}+m_{k}$ (by the induction hypothesis). So it reaches Signal 3 only at time $s_{k}+m_{k}+$ $t_{k+1}$. Similarly, it gets to Signal 4 at time $s_{k}+2 m_{k}+t_{k+1}$, etc. Thus the entire schedule finishes at time $s_{k}+(n-2) m_{k}+t_{k+1}=s_{k+1}+$ $(n-2) m_{k+1}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (CAN 1) The triangle $A B C$ is inscribed in a circle. The interior bisectors of the angles $A, B$, and $C$ meet the circle again at $A^{\\prime}, B^{\\prime}$, and $C^{\\prime}$ respectively. Prove that the area of triangle $A^{\\prime} B^{\\prime} C^{\\prime}$ is greater than or equal to the area of triangle $A B C$.", "solution": "3. Let $R$ be the circumradius, $r$ the inradius, $s$ the semiperimeter, $\\Delta$ the area of $A B C$ and $\\Delta^{\\prime}$ the area of $A^{\\prime} B^{\\prime} C^{\\prime}$. The angles of triangle $A^{\\prime} B^{\\prime} C^{\\prime}$ are $A^{\\prime}=90^{\\circ}-A / 2, B^{\\prime}=90^{\\circ}-B / 2$, and $C^{\\prime}=90^{\\circ}-C / 2$, and hence  $$ \\Delta=2 R^{2} \\sin A \\sin B \\sin C $$  and $\\Delta^{\\prime}=2 R^{2} \\sin A^{\\prime} \\sin B^{\\prime} \\sin C^{\\prime}=2 R^{2} \\cos \\frac{A}{2} \\cos \\frac{B}{2} \\cos \\frac{C}{2}$. Hence,  $$ \\frac{\\Delta}{\\Delta^{\\prime}}=\\frac{\\sin A \\sin B \\sin C}{\\cos \\frac{A}{2} \\cos \\frac{B}{2} \\cos \\frac{C}{2}}=8 \\sin \\frac{A}{2} \\sin \\frac{B}{2} \\sin \\frac{C}{2}=\\frac{2 r}{R} $$  where we have used that $r=A I \\sin (A / 2)=\\cdots=4 R \\sin (A / 2) \\cdot \\sin (B / 2)$. $\\sin (C / 2)$. Euler's inequality $2 r \\leq R$ shows that $\\Delta \\leq \\Delta^{\\prime}$. Second solution. Let $H$ be orthocenter of triangle $A B C$, and $H_{a}, H_{b}, H_{c}$ points symmetric to $H$ with respect to $B C, C A, A B$, respectively. Since $\\angle B H_{a} C=\\angle B H C=180^{\\circ}-\\angle A$, points $H_{a}, H_{b}, H_{c}$ lie on the circumcircle of $A B C$, and the area of the hexagon $A H_{c} B H_{a} C H_{b}$ is double the area of $A B C$. (1) Let us apply the analogous result for the triangle $A^{\\prime} B^{\\prime} C^{\\prime}$. Since its orthocenter is the incenter $I$ of $A B C$, and the point symmetric to $I$ with respect to $B^{\\prime} C^{\\prime}$ is the point $A$, we find by (1) that the area of the hexagon $A C^{\\prime} B A^{\\prime} C B^{\\prime}$ is double the area of $A^{\\prime} B^{\\prime} C^{\\prime}$. But it is clear that the area of $\\Delta C H_{a} B$ is less than or equal to the area of $\\Delta C A^{\\prime} B$ etc.; hence, the area of $A H_{c} B H_{a} C H_{b}$ does not exceed the area of $A C^{\\prime} B A^{\\prime} C B^{\\prime}$. The statement follows immediately.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "30", "problem_type": null, "exam": "IMO", "problem": "30. (USS 1) A point $M$ is chosen on the side $A C$ of the triangle $A B C$ in such a way that the radii of the circles inscribed in the triangles $A B M$ and $B M C$ are equal. Prove that  $$ B M^{2}=\\Delta \\cot \\frac{B}{2} $$  where $\\Delta$ is the area of the triangle $A B C$.", "solution": "30. Let $\\Delta_{1}, s_{1}, r^{\\prime}$ denote the area, semiperimeter, and inradius of triangle $A B M, \\Delta_{2}, s_{2}, r^{\\prime}$ the same quantities for triangle $M B C$, and $\\Delta, s, r$ those for $\\triangle A B C$. Also, let $P^{\\prime}$ and $Q^{\\prime}$ be the points of tangency of the incircle of $\\triangle A B M$ with the side $A B$ and of the incircle of $\\triangle M B C$ with the side $B C$, respectively, and let $P, Q$ be the points of tangency of the incircle of $\\triangle A B C$ with the sides $A B, B C$. We have $\\Delta_{1}=s_{1} r^{\\prime}, \\Delta_{2}=s_{2} r^{\\prime}, \\Delta=s r$, so that $s r=\\left(s_{1}+s_{2}\\right) r^{\\prime}$. Then  $$ s_{1}+s_{2}=s+B M \\quad \\Rightarrow \\quad \\frac{r^{\\prime}}{r}=\\frac{s}{s+B M} $$  On the other hand, from similarity of triangles it follows that $A P^{\\prime} / A P=$ $C Q^{\\prime} / C Q=r^{\\prime} / r$. By a well-known formula we find that $A P=s-B C$, $C Q=s-A B, A P^{\\prime}=s_{1}-B M, C Q^{\\prime}=s_{2}-B M$, and therefore deduce that  $$ \\frac{r^{\\prime}}{r}=\\frac{s_{1}-B M}{s-B C}=\\frac{s_{2}-B M}{s-A B} \\Rightarrow \\frac{r^{\\prime}}{r}=\\frac{s_{1}+s_{2}-2 B M}{2 s-A B-B C}=\\frac{s-B M}{A C} . $$  It follows from (1) and (2) that $(s-B M) / A C=s /(s+B M)$, giving us $s^{2}-B M^{2}=s \\cdot A C$. Finally,  $$ B M^{2}=s(s-A C)=s \\cdot B P=s \\cdot r \\cot \\frac{B}{2}=\\Delta \\cot \\frac{B}{2} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "31", "problem_type": null, "exam": "IMO", "problem": "31. (USS 2) Around a circular table an even number of persons have a discussion. After a break they sit again around the circular table in a different order. Prove that there are at least two people such that the number of participants sitting between them before and after the break is the same.", "solution": "31. Denote the number of participants by $2 n$, and assign to each seat one of the numbers $1,2, \\ldots, 2 n$. Let the participant who was sitting at the seat $k$ before the break move to seat $\\pi(k)$. It suffices to prove that for every permutation $\\pi$ of the set $\\{1,2, \\ldots, 2 n\\}$, there exist distinct $i, j$ such that $\\pi(i)-\\pi(j)= \\pm(i-j)$, the differences being calculated modulo $2 n$. If there are distinct $i$ and $j$ such that $\\pi(i)-i=\\pi(j)-j$ modulo $2 n$, then we are done. Suppose that all the differences $\\pi(i)-i$ are distinct modulo $2 n$. Then they take values $0,1, \\ldots, 2 n-1$ in some order, and consequently  $$ \\sum_{i=1}^{2 n}(\\pi(i)-i)=0+1+\\cdots+(2 n-1) \\equiv n(2 n-1)(\\bmod 2 n) $$  On the other hand, $\\sum_{i=1}^{2 n}(\\pi(i)-i)=\\sum \\pi(i)-\\sum i=0$, which is a contradiction because $n(2 n-1)$ is not divisible by $2 n$. Remark. For an odd number of participants, the statement is false. For example, the permutation $(a, 2 a, \\ldots,(2 n+1) a)$ of $(1,2, \\ldots, 2 n+1)$ modulo $2 n+1$ does not satisfy the statement when $\\operatorname{gcd}\\left(a^{2}-1,2 n+1\\right)=1$. Check that such an $a$ always exists.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (CZS 1) An $n \\times n$ chessboard $(n \\geq 2)$ is numbered by the numbers $1,2, \\ldots, n^{2}$ (every number occurs once). Prove that there exist two neighboring (which share a common edge) squares such that their numbers differ by at least $n$.", "solution": "4. Suppose that the numbers of any two neighboring squares differ by at most $n-1$. For $k=1,2, \\ldots, n^{2}-n$, let $A_{k}, B_{k}$, and $C_{k}$ denote, respectively, the sets of squares numbered by $1,2, \\ldots, k$; of squares numbered by $k+$ $n, k+n+1, \\ldots, n^{2}$; and of squares numbered by $k+1, \\ldots, k+n-1$. By the assumption, the squares from $A_{k}$ and $B_{k}$ have no edge in common; $C_{k}$ has $n-1$ elements only. Consequently, for each $k$ there exists a row and a column all belonging either to $A_{k}$, or to $B_{k}$. For $k=1$, it must belong to $B_{k}$, while for $k=n^{2}-n$ it belongs to $A_{k}$. Let $k$ be the smallest index such that $A_{k}$ contains a whole row and a whole column. Since $B_{k-1}$ has that property too, it must have at least two squares in common with $A_{k}$, which is impossible.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (CZS 2) ${ }^{\\mathrm{IMO} 2}$ Let $n$ be an even positive integer. Let $A_{1}, A_{2}, \\ldots, A_{n+1}$ be sets having $n$ elements each such that any two of them have exactly one element in common while every element of their union belongs to at least two of the given sets. For which $n$ can one assign to every element of the union one of the numbers 0 and 1 in such a manner that each of the sets has exactly $n / 2$ zeros?", "solution": "5. Let $n=2 k$ and let $A=\\left\\{A_{1}, \\ldots, A_{2 k+1}\\right\\}$ denote the family of sets with the desired properties. Since every element of their union $B$ belongs to at least two sets of $A$, it follows that $A_{j}=\\bigcup_{i \\neq j} A_{i} \\cap A_{j}$ holds for every $1 \\leq j \\leq 2 k+1$. Since each intersection in the sum has at most one element and $A_{j}$ has $2 k$ elements, it follows that every element of $A_{j}$, i.e., in general of $B$, is a member of exactly two sets. We now prove that $k$ is even, assuming that the marking described in the problem exists. We have already shown that for every two indices $1 \\leq j \\leq 2 k+1$ and $i \\neq j$ there exists a unique element contained in both $A_{i}$ and $A_{j}$. On a $2 k \\times 2 k$ matrix let us mark in the $i$ th column and $j$ th row for $i \\neq j$ the number that was joined to the element of $B$ in $A_{i} \\cap A_{j}$. In the $i$ th row and column let us mark the number of the element of $B$ in $A_{i} \\cap A_{2 k+1}$. In each row from the conditions of the marking there must be an even number of zeros. Hence, the total number of zeros in the matrix is even. The matrix is symmetric with respect to its main diagonal; hence it has an even number of zeros outside its main diagonal. Hence, the number of zeros on the main diagonal must also be even and this number equals the number of elements in $A_{2 k+1}$ that are marked with 0 , which is $k$. Hence $k$ must be even.  For even $k$ we note that the dimensions of a $2 k \\times 2 k$ matrix are divisible by 4 . Tiling the entire matrix with the $4 \\times 4$ submatrix  $$ Q=\\left[\\begin{array}{llll} 0 & 1 & 0 & 1 \\\\ 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 1 & 0 & 0 & 1 \\end{array}\\right] $$  we obtain a marking that indeed satisfies all the conditions of the problem; hence we have shown that the marking is possible if and only if $k$ is even.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (CZS 3) In a given tetrahedron $A B C D$ let $K$ and $L$ be the centers of edges $A B$ and $C D$ respectively. Prove that every plane that contains the line $K L$ divides the tetrahedron into two parts of equal volume.", "solution": "6. Let $\\omega$ be the plane through $A B$, parallel to $C D$. Define the point transformation $f: X \\mapsto X^{\\prime}$ in space as follows. If $X \\in K L$, then $X^{\\prime}=X$; otherwise, let $\\omega_{X}$ be the plane through $X$ parallel to $\\omega$ : then $X^{\\prime}$ is the point symmetric to $X$ with respect to the intersection point of $K L$ with $\\omega_{X}$. Clearly, $f(A)=B, f(B)=A, f(C)=D, f(D)=C$; hence $f$ maps the tetrahedron onto itself. We shall show that $f$ preserves volumes. Let $s: X \\mapsto X^{\\prime \\prime}$ denote the symmetry with respect to $K L$, and $g$ the transformation mapping $X^{\\prime \\prime}$ into $X^{\\prime}$; then $f=g \\circ s$. If points $X_{1}^{\\prime \\prime}=s\\left(X_{1}\\right)$ and $X_{2}^{\\prime \\prime}=s\\left(X_{2}\\right)$ have the property that $X_{1}^{\\prime \\prime} X_{2}^{\\prime \\prime}$ is parallel to $K L$, then the segments $X_{1}^{\\prime \\prime} X_{2}^{\\prime \\prime}$ and $X_{1}^{\\prime} X_{2}^{\\prime}$ have the same length and lie on the same line. Then by Cavalieri's principle $g$ preserves volume, and so does $f$. Now, if $\\alpha$ is any plane containing the line $K L$, the two parts of the tetrahedron on which it is partitioned by $\\alpha$ are transformed into each other by $f$, and therefore have the same volumes. Second solution. Suppose w.l.o.g. that the plane $\\alpha$ through $K L$ meets the interiors of edges $A C$ and $B D$ at $X$ and $Y$. Let $\\overrightarrow{A X}=\\lambda \\overrightarrow{A C}$ and $\\overrightarrow{B Y}=\\mu \\overrightarrow{B D}$, for $0 \\leq \\lambda, \\mu \\leq 1$. Then the vectors $\\overrightarrow{K X}=\\lambda \\overrightarrow{A C}-\\overrightarrow{A B} / 2$, $\\overrightarrow{K Y}=\\mu \\overrightarrow{B D}+\\overrightarrow{A B} / 2, \\overrightarrow{K L}=\\overrightarrow{A C} / 2+$ $\\overrightarrow{B D} / 2$ are coplanar; i.e., there exist real numbers $a, b, c$, not all zero, such that ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-512.jpg?height=426&width=529&top_left_y=1294&top_left_x=829)  $$ \\overrightarrow{0}=a \\overrightarrow{K X}+b \\overrightarrow{K Y}+c \\overrightarrow{K L}=(\\lambda a+c / 2) \\overrightarrow{A C}+(\\mu b+c / 2) \\overrightarrow{B D}+\\frac{b-a}{2} \\overrightarrow{A B} $$  Since $\\overrightarrow{A C}, \\overrightarrow{B D}, \\overrightarrow{A B}$ are linearly independent, we must have $a=b$ and $\\lambda=\\mu$. We need to prove that the volume of the polyhedron $K X L Y B C$, which is one of the parts of the tetrahedron $A B C D$ partitioned by $\\alpha$, equals half of the volume $V$ of $A B C D$. Indeed, we obtain  $$ V_{K X L Y B C}=V_{K X L C}+V_{K B Y L C}=\\frac{1}{4}(1-\\lambda) V+\\frac{1}{4}(1+\\mu) V=\\frac{1}{2} V . $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (FRA 2) Let $a$ be the greatest positive root of the equation $x^{3}-3 x^{2}+1=$ 0 . Show that $\\left[a^{1788}\\right]$ and $\\left[a^{1988}\\right]$ are both divisible by 17 . ( $[x]$ denotes the integer part of $x$.)", "solution": "7. The algebraic equation $x^{3}-3 x^{2}+1=0$ admits three real roots $\\beta, \\gamma, a$, with  $$ -0.6<\\beta<-0.5, \\quad 0.6<\\gamma<0.7, \\quad \\sqrt{8}<a<3 . $$  Define, for all integers $n$,  $$ u_{n}=\\beta^{n}+\\gamma^{n}+a^{n} . $$  It holds that $u_{n+3}=3 u_{n+2}-u_{n}$. Obviously, $0<\\beta^{n}+\\gamma^{n}<1$ for all $n \\geq 2$, and we see that $u_{n}-1=\\left[a^{n}\\right]$ for $n \\geq 2$. It is now a question whether $u_{1788}-1$ and $u_{1988}-1$ are divisible by 17 . Working modulo 17 , we get $u_{0} \\equiv 3, u_{1} \\equiv 3, u_{2} \\equiv 9, u_{3} \\equiv 7, u_{4} \\equiv$ $1, \\ldots, u_{16}=3, u_{17}=3, u_{18}=9$. Thus, $u_{n}$ is periodic modulo 17 , with period 16. Since $1788=16 \\cdot 111+12,1988=16 \\cdot 124+4$, it follows that $u_{1788} \\equiv u_{12} \\equiv 1$ and $u_{1988} \\equiv u_{4}=1$. So, $\\left[a^{1788}\\right]$ and $\\left[a^{1988}\\right]$ are divisible by 17 . Second solution. The polynomial $x^{3}-3 x^{2}+1$ allows the factorization modulo 17 as $(x-4)(x-5)(x+6)$. Hence it is easily seen that $u_{n} \\equiv$ $4^{n}+5^{n}+(-6)^{n}$. Fermat's theorem gives us $4^{n} \\equiv 5^{n} \\equiv(-6)^{n} \\equiv 1$ for $16 \\mid n$, and the rest follows easily. Remark. In fact, the roots of $x^{3}-3 x^{2}+1=0$ are $\\frac{1}{2 \\sin 10^{\\circ}}, \\frac{1}{2 \\sin 50^{\\circ}}$, and $-\\frac{1}{2 \\sin 70^{\\circ}}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (FRA 3) Let $u_{1}, u_{2}, \\ldots, u_{m}$ be $m$ vectors in the plane, each of length less than or equal to 1 , which add up to zero. Show that one can rearrange $u_{1}, u_{2}, \\ldots, u_{m}$ as a sequence $v_{1}, v_{2}, \\ldots, v_{m}$ such that each partial sum $v_{1}, v_{1}+v_{2}, v_{1}+v_{2}+v_{3}, \\ldots, v_{1}+v_{2}+\\cdots+v_{m}$ has length less than or equal to $\\sqrt{5}$.", "solution": "8. Consider first the case that the vectors are on the same line. Then if $e$ is a unit vector, we can write $u_{1}=x_{1} e, \\ldots, u_{n}=x_{n} e$ for scalars $x_{i},\\left|x_{i}\\right| \\leq 1$, with zero sum. It is now easy to permute $x_{1}, x_{2}, \\ldots, x_{n}$ into $z_{1}, z_{2}, \\ldots z_{n}$ so that $\\left|z_{1}\\right| \\leq 1,\\left|z_{1}+z_{2}\\right| \\leq 1, \\ldots,\\left|z_{1}+z_{2}+\\cdots+z_{n-1}\\right| \\leq 1$. Indeed, suppose w.l.o.g. that $z_{1}=x_{1} \\geq 0$; then we choose $z_{2}, \\ldots, z_{r}$ from the $x_{i}$ 's to be negative, until we get to the first $r$ with $x_{1}+x_{2}+\\cdots+x_{r} \\leq 0$; we continue successively choosing positive $z_{j}$ 's from the remaining $x_{i}$ 's until we get the first partial sum that is positive, and so on. It is easy to verify that $\\left|z_{1}+z_{2}+\\cdots+z_{j}\\right| \\leq 1$ for all $j=1,2, \\ldots, n$. Now we pass to the general case. Let $s$ be the longest vector that can be obtained by summing a subset of $u_{1}, \\ldots, u_{m}$, and assume w.l.o.g. that $s=u_{1}+\\cdots+u_{p}$. Further, let $\\delta$ and $\\delta^{\\prime}$ respectively be the lines through the origin $O$ in the direction of $s$ and perpendicular to $s$, and $e, e^{\\prime}$ respectively the unit vectors on $\\delta$ and $\\delta^{\\prime}$. Put $u_{i}=x_{i} e+y_{i} e^{\\prime}$, $i=1,2, \\ldots, m$. By the definition of $\\delta$ and $\\delta^{\\prime}$, we have $\\left|x_{i}\\right|,\\left|y_{i}\\right| \\leq 1$; $x_{1}+\\cdots+x_{m}=y_{1}+\\cdots+y_{m}=0 ; y_{1}+\\cdots+y_{p}=y_{p+1}+\\cdots+y_{m}=0$; we also have $x_{p+1}, \\ldots, x_{m} \\leq 0$ (otherwise, if $x_{i}>0$ for some $i$, then $\\left|s+v_{i}\\right|>|s|$ ), and similarly $x_{1}, \\ldots, x_{p} \\geq 0$. Finally, suppose by the one-dimensional case that $y_{1}, \\ldots, y_{p}$ and $y_{p+1}, \\ldots, y_{m}$ are permuted in such a way that all the sums $y_{1}+\\cdots+y_{i}$ and $y_{p+1}+\\cdots+y_{p+i}$ are $\\leq 1$ in absolute value. We apply the construction of the one-dimensional case to $x_{1}, \\ldots, x_{m}$ taking, as described above, positive $z_{i}$ 's from $x_{1}, x_{2}, \\ldots, x_{p}$ and negative ones from $x_{p+1}, \\ldots, x_{m}$, but so that the order is preserved; this way we get a permutation $x_{\\sigma_{1}}, x_{\\sigma_{2}}, \\ldots, x_{\\sigma_{m}}$. It is then clear that each sum $y_{\\sigma_{1}}+y_{\\sigma_{2}}+$ $\\cdots+y_{\\sigma_{k}}$ decomposes into the sum $\\left(y_{1}+y_{2}+\\cdots+y_{l}\\right)+\\left(y_{p+1}+\\cdots+y_{p+n}\\right)$ (because of the preservation of order), and that each of these sums is less than or equal to 1 in absolute value. Thus each sum $u_{\\sigma_{1}}+\\cdots+u_{\\sigma_{k}}$ is composed of a vector of length at most 2 and an orthogonal vector of length at most 1 , and so is itself of length at most $\\sqrt{5}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1988", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (FRG 1) ${ }^{\\text {IMO6 }}$ Let $a$ and $b$ be two positive integers such that $a b+1$ divides $a^{2}+b^{2}$. Show that $\\frac{a^{2}+b^{2}}{a b+1}$ is a perfect square.", "solution": "9. Let us assume $\\frac{a^{2}+b^{2}}{a b+1}=k \\in \\mathbb{N}$. We then have $a^{2}-k a b+b^{2}=k$. Let us assume that $k$ is not an integer square, which implies $k \\geq 2$. Now we observe the minimal pair $(a, b)$ such that $a^{2}-k a b+b^{2}=k$ holds. We may assume w.l.o.g. that $a \\geq b$. For $a=b$ we get $k=(2-k) a^{2} \\leq 0$; hence we must have $a>b$. Let us observe the quadratic equation $x^{2}-k b x+b^{2}-k=0$, which has solutions $a$ and $a_{1}$. Since $a+a_{1}=k b$, it follows that $a_{1} \\in \\mathbb{Z}$. Since $a>k b$ implies $k>a+b^{2}>k b$ and $a=k b$ implies $k=b^{2}$, it follows that $a<k b$ and thus $b^{2}>k$. Since $a a_{1}=b^{2}-k>0$ and $a>0$, it follows that $a_{1} \\in \\mathbb{N}$ and $a_{1}=\\frac{b^{2}-k}{a}<\\frac{a^{2}-1}{a}<a$. We have thus found an integer pair $\\left(a_{1}, b\\right)$ with $0<a_{1}<a$ that satisfies the original equation. This is a contradiction of the initial assumption that $(a, b)$ is minimal. Hence $k$ must be an integer square.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (AUS 2) ${ }^{\\mathrm{IMO} 2}$ Let $A B C$ be a triangle. The bisector of angle $A$ meets the circumcircle of triangle $A B C$ in $A_{1}$. Points $B_{1}$ and $C_{1}$ are defined similarly. Let $A A_{1}$ meet the lines that bisect the two external angles at $B$ and $C$ in point $A^{0}$. Define $B^{0}$ and $C^{0}$ similarly. If $S_{X_{1} X_{2} \\ldots X_{n}}$ denotes the area of the polygon $X_{1} X_{2} \\ldots X_{n}$, prove that  $$ S_{A^{0} B^{0} C^{0}}=2 S_{A C_{1} B A_{1} C B_{1}} \\geq 4 S_{A B C} . $$", "solution": "1. Let $I$ denote the intersection of the three internal bisectors. Then $I A_{1}=A_{1} A^{0}$. One way proving this is to realize that the circumcircle of $A B C$ is the nine-point circle of $A^{0} B^{0} C^{0}$, hence it bisects $I A^{0}$, since $I$ is the orthocenter of $A^{0} B^{0} C^{0}$. Another way is through noting that $I A_{1}=A_{1} B$, which follows from $\\angle A_{1} I B=\\angle I B A_{1}=(\\angle A+\\angle B) / 2$, and $A_{1} B=A_{1} A^{0}$ which follows from $\\angle A_{1} A^{0} B=\\angle A_{1} B A^{0}=90^{\\circ}-$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-525.jpg?height=443&width=566&top_left_y=350&top_left_x=792) $\\angle I B A_{1}$. Hence, we obtain $S_{I A_{1} B}=S_{A^{0} A_{1} B}$. Repeating this argument for the six triangles that have a vertex at $I$ and adding them up gives us $S_{A^{0} B^{0} C^{0}}=2 S_{A C_{1} B A_{1} C B_{1}}$. To prove $S_{A C_{1} B A_{1} C B_{1}} \\geq 2 S_{A B C}$, draw the three altitudes in triangle $A B C$ intersecting in $H$. Let $X, Y$, and $Z$ be the symmetric points of $H$ with respect the sides $B C, C A$, and $A B$, respectively. Then $X, Y, Z$ are points on the circumcircle of $\\triangle A B C$ (because $\\left.\\angle B X C=\\angle B H C=180^{\\circ}-\\angle A\\right)$. Since $A_{1}$ is the midpoint of the arc $B C$, we have $S_{B A_{1} C} \\geq S_{B X C}$. Hence  $$ S_{A C_{1} B A_{1} C B_{1}} \\geq S_{A Z B X C Y}=2\\left(S_{B H C}+S_{C H A}+S_{A H B}\\right)=2 S_{A B C} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (GRE 3) Let $g: \\mathbb{C} \\rightarrow \\mathbb{C}, w \\in \\mathbb{C}, a \\in \\mathbb{C}, w^{3}=1(w \\neq 1)$. Show that there is one and only one function $f: \\mathbb{C} \\rightarrow \\mathbb{C}$ such that  $$ f(z)+f(w z+a)=g(z), \\quad z \\in \\mathbb{C} $$  Find the function $f$.", "solution": "10. Plugging in $w z+a$ instead of $z$ into the functional equation, we obtain  $$ f(w z+a)+f\\left(w^{2} z+w a+a\\right)=g(w z+a) . $$  By repeating this process, this time in (1), we get  $$ f\\left(w^{2} z+w a+a\\right)+f(z)=g\\left(w^{2} z+w a+a\\right) $$  Solving the system of linear equations (1), (2) and the original functional equation, we easily get  $$ f(z)=\\frac{g(z)+g\\left(w^{2} z+w a+a\\right)-g(w z+a)}{2} . $$  This function thus uniquely satisfies the original functional equation.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (HUN 1) Define sequence $a_{n}$ by $\\sum_{d \\mid n} a_{d}=2^{n}$. Show that $n \\mid a_{n}$.", "solution": "11. Call a binary sequence $S$ of length $n$ repeating if for some $d \\mid n, d>1, S$ can be split into $d$ identical blocks. Let $x_{n}$ be the number of nonrepeating binary sequences of length $n$. The total number of binary sequences of length $n$ is obviously $2^{n}$. Any sequence of length $n$ can be produced by repeating its unique longest nonrepeating initial block according to need. Hence, we obtain the recursion relation $\\sum_{d \\mid n} x_{d}=2^{n}$. This, along with $x_{1}=2$, gives us $a_{n}=x_{n}$ for all $n$. We now have that the sequences counted by $x_{n}$ can be grouped into groups of $n$, the sequences in the same group being cyclic shifts of each other. Hence, $n \\mid x_{n}=a_{n}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (HUN 3) At $n$ distinct points of a circular race course there are $n$ cars ready to start. Each car moves at a constant speed and covers the circle in an hour. On hearing the initial signal, each of them selects a direction and starts moving immediately. If two cars meet, both of them change directions and go on without loss of speed. Show that at a certain moment each car will be at its starting point.", "solution": "12. Assume that each car starts with a unique ranking number. Suppose that while turning back at a meeting point two cars always exchanged their ranking numbers. We can observe that ranking numbers move at a constant speed and direction. One hour later, after several exchanges, each starting point will be occupied by a car of the same ranking number and proceeding in the same direction as the one that started from there one hour ago. We now give the cars back their original ranking numbers. Since the sequence of the cars along the track cannot be changed, the only possibility is that the original situation has been rotated, maybe onto itself. Hence for some $d \\mid n$, after $d$ hours each car will be at its starting position and orientation.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (ICE 3) ${ }^{\\mathrm{IMO}}$ The quadrilateral $A B C D$ has the following properties: (i) $A B=A D+B C$; (ii) there is a point $P$ inside it at a distance $x$ from the side $C D$ such that $A P=x+A D$ and $B P=x+B C$. Show that  $$ \\frac{1}{\\sqrt{x}} \\geq \\frac{1}{\\sqrt{A D}}+\\frac{1}{\\sqrt{B C}} $$", "solution": "13. Let us construct the circles $\\sigma_{1}$ with center $A$ and radius $R_{1}=A D, \\sigma_{2}$ with center $B$ and radius $R_{2}=B C$, and $\\sigma_{3}$ with center $P$ and radius $x$. The points $C$ and $D$ lie on $\\sigma_{2}$ and $\\sigma_{1}$ respectively, and $C D$ is tangent to $\\sigma_{3}$. From this it is plain that the greatest value of $x$ occurs when $C D$ is also tangent to $\\sigma_{1}$ and $\\sigma_{2}$. We shall show that in this case the required inequality is really an equality, i.e., that $\\frac{1}{\\sqrt{x}}=\\frac{1}{\\sqrt{A D}}+\\frac{1}{\\sqrt{B C}}$. Then the inequality will immediately follow. Denote the point of tangency of $C D$ with $\\sigma_{3}$ by $M$. By the Pythagorean theorem we have $C D=\\sqrt{\\left(R_{1}+R_{2}\\right)^{2}-\\left(R_{1}-R_{2}\\right)^{2}}=2 \\sqrt{R_{1} R_{2}}$. On the other hand, $C D=C M+M D=2 \\sqrt{R_{2} x}+2 \\sqrt{R_{1} x}$. Hence, we obtain $\\frac{1}{\\sqrt{x}}=\\frac{1}{\\sqrt{R_{1}}}+\\frac{1}{\\sqrt{R_{2}}}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (IND 2) A bicentric quadrilateral is one that is both inscribable in and circumscribable about a circle. Show that for such a quadrilateral, the centers of the two associated circles are collinear with the point of intersection of the diagonals.", "solution": "14. Lemma 1. In a quadrilateral $A B C D$ circumscribed about a circle, with points of tangency $P, Q, R, S$ on $D A, A B, B C, C D$ respectively, the lines $A C, B D, P R, Q S$ concur. Proof. Follows immediately, for example, from Brianchon's theorem. Lemma 2. Let a variable chord $X Y$ of a circle $C(I, r)$ subtend a right angle at a fixed point $Z$ within the circle. Then the locus of the midpoint $P$ of $X Y$ is a circle whose center is at the midpoint $M$ of $I Z$ and whose radius is $\\sqrt{r^{2} / 2-I Z^{2} / 4}$. Proof. From $\\angle X Z Y=90^{\\circ}$ follows $\\overrightarrow{Z X} \\cdot \\overrightarrow{Z Y}=(\\overrightarrow{I X}-\\overrightarrow{I Z}) \\cdot(\\overrightarrow{I Y}-\\overrightarrow{I Z})=0$. Therefore,  $$ \\begin{aligned} \\overrightarrow{M P}^{2} & =(\\overrightarrow{M I}+\\overrightarrow{I P})^{2}=\\frac{1}{4}(-\\overrightarrow{I Z}+\\overrightarrow{I X}+\\overrightarrow{I Y})^{2} \\\\ & =\\frac{1}{4}\\left(I X^{2}+I Y^{2}-I Z^{2}+2(\\overrightarrow{I X}-\\overrightarrow{I Z}) \\cdot(\\overrightarrow{I Y}-\\overrightarrow{I Z})\\right) \\\\ & =\\frac{1}{2} r^{2}-\\frac{1}{4} I Z^{2} \\end{aligned} $$  Lemma 3. Using notation as in Lemma 1, if $A B C D$ is cyclic, $P R$ is perpendicular to $Q S$. Proof. Consider the inversion in $C(I, r)$, mapping $A$ to $A^{\\prime}$ etc. $(P, Q, R, S$ are fixed). As is easily seen, $A^{\\prime}, B^{\\prime}, C^{\\prime}, D^{\\prime}$ will lie at the midpoints of $P Q, Q R, R S, S P$, respectively. $A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime}$ is a parallelogram, but also cyclic, since inversion preserves circles; thus it must be a rectangle, and so $P R \\perp Q S$. Now we return to the main result. Let $I$ and $O$ be the incenter and circumcenter, $Z$ the intersection of the diagonals, and $P, Q, R, S, A^{\\prime}, B^{\\prime}, C^{\\prime}, D^{\\prime}$ points as defined in Lemmas 1 and 3. From Lemma 3, the chords $P Q, Q R, R S, S P$ subtend $90^{\\circ}$ at $Z$. Therefore by Lemma 2 the points $A^{\\prime}, B^{\\prime}, C^{\\prime}, D^{\\prime}$ lie on a circle whose center is the midpoint $Y$ of $I Z$. Since this circle is the image of the circle $A B C D$ under the considered inversion (centered at $I$ ), it follows that $I, O, Y$ are collinear, and hence so are $I, O, Z$. Remark. This is the famous Newton's theorem for bicentric quadrilaterals.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (IRE 1) Let $a, b, c, d, m, n$ be positive integers such that $a^{2}+b^{2}+c^{2}+d^{2}=$ 1989, $a+b+c+d=m^{2}$, and the largest of $a, b, c, d$ is $n^{2}$. Determine, with proof, the values of $m$ and $n$.", "solution": "15. By Cauchy's inequality, $44<\\sqrt{1989}<a+b+c+d \\leq \\sqrt{2 \\cdot 1989}<90$. Since $m^{2}=a+b+c+d$ is of the same parity as $a^{2}+b^{2}+c^{2}+d^{2}=1989$, $m^{2}$ is either 49 or 81 . Let $d=\\max \\{a, b, c, d\\}$. Suppose that $m^{2}=49$. Then $(49-d)^{2}=(a+b+c)^{2}>a^{2}+b^{2}+c^{2}=$ $1989-d^{2}$, and so $d^{2}-49 d+206>0$. This inequality does not hold for $5 \\leq d \\leq 44$. Since $d \\geq \\sqrt{1989 / 4}>22$, $d$ must be at least 45 , which is impossible because $45^{2}>1989$. Thus we must have $m^{2}=81$ and $m=9$. Now, $4 d>81$ implies $d \\geq 21$. On the other hand, $d<\\sqrt{1989}$, and hence $d=25$ or $d=36$. Suppose that $d=25$ and put $a=25-p, b=25-q$, $c=25-r$ with $p, q, r \\geq 0$. From $a+b+c=56$ it follows that $p+q+r=19$, which, together with $(25-p)^{2}+(25-q)^{2}+(25-r)^{2}=1364$, gives us $p^{2}+q^{2}+r^{2}=439>361=(p+q+r)^{2}$, a contradiction. Therefore $d=36$ and $n=6$. Remark. A little more calculation yields the unique solution $a=12$, $b=15, c=18, d=36$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (ISR 1) The set $\\left\\{a_{0}, a_{1}, \\ldots, a_{n}\\right\\}$ of real numbers satisfies the following conditions: (i) $a_{0}=a_{n}=0$; (ii) for $1 \\leq k \\leq n-1$,  $$ a_{k}=c+\\sum_{i=k}^{n-1} a_{i-k}\\left(a_{i}+a_{i+1}\\right) . $$  Prove that $c \\leq \\frac{1}{4 n}$.", "solution": "16. Define $S_{k}=\\sum_{i=0}^{k} a_{i}(k=0,1, \\ldots, n)$ and $S_{-1}=0$. We note that $S_{n-1}=$ $S_{n}$. Hence  $$ \\begin{aligned} S_{n} & =\\sum_{k=0}^{n-1} a_{k}=n c+\\sum_{k=0}^{n-1} \\sum_{i=k}^{n-1} a_{i-k}\\left(a_{i}+a_{i+1}\\right) \\\\ & =n c+\\sum_{i=0}^{n-1} \\sum_{k=0}^{i} a_{i-k}\\left(a_{i}+a_{i+1}\\right)=n c+\\sum_{i=0}^{n-1}\\left(a_{i}+a_{i+1}\\right) \\sum_{k=0}^{i} a_{i-k} \\\\ & =n c+\\sum_{i=0}^{n-1}\\left(S_{i+1}-S_{i-1}\\right) S_{i}=n c+S_{n}^{2} \\end{aligned} $$  i.e., $S_{n}^{2}-S_{n}+n c=0$. Since $S_{n}$ is real, the discriminant of the quadratic equation must be positive, and hence $c \\leq \\frac{1}{4 n}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (MON 1) Given seven points in the plane, some of them are connected by segments so that: (i) among any three of the given points, two are connected by a segment; (i) the number of segments is minimal.  How many segments does a figure satisfying (i) and (ii) contain? Give an example of such a figure.", "solution": "17. A figure consisting of 9 lines is shown below. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-531.jpg?height=237&width=318&top_left_y=1210&top_left_x=627)  Now we show that 8 lines are not sufficient. Assume the opposite. By the pigeonhole principle, there is a vertex, say $A$, that is joined to at most 2 other vertices. Let $B, C, D, E$ denote the vertices to which $A$ is not joined, and $F, G$ the other two vertices. Then any two vertices of $B, C, D, E$ must be mutually joined for an edge to exist within the triangle these two points form with A. This accounts for 6 segments. Since only two segments remain, among $A, F$, and $G$ at least two are not joined. Taking these two and one of $B, C, D, E$ that is not joined to any of them (it obviously exists), we get a triple of points, no two of which are joined; a contradiction.  Second solution. Since (a) is equivalent to the fact that no three points make a \"blank triangle,\" by Turan's theorem the number of \"blank edges\" cannot exceed $\\left[7^{2} / 4\\right]=12$, leaving at least $7 \\cdot 6 / 2-12=9$ segments. For general $n$, the answer is $[(n-1) / 2]^{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (MON 4) Given a convex polygon $A_{1} A_{2} \\ldots A_{n}$ with area $S$, and a point $M$ in the same plane, determine the area of polygon $M_{1} M_{2} \\ldots M_{n}$, where $M_{i}$ is the image of $M$ under rotation $\\mathcal{R}_{A_{i}}^{\\alpha}$ around $A_{i}$ by $\\alpha, i=1,2, \\ldots, n$.", "solution": "18. Consider the triangle $M A_{i} M_{i}$. Obviously, the point $M_{i}$ is the image of $A_{i}$ under the composition $C$ of rotation $R_{M}^{\\alpha / 2-90^{\\circ}}$ and homothety $H_{M}^{2 \\sin (\\alpha / 2)}$. Therefore, the polygon $M_{1} M_{2} \\ldots M_{n}$ is obtained as the image of $A_{1} A_{2} \\ldots A_{n}$ under the rotational homothety $C$ with coefficient $2 \\sin (\\alpha / 2)$. Therefore $S_{M_{1} M_{2} \\ldots M_{n}}=4 \\sin ^{2}(\\alpha / 2) \\cdot S$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (MON 6) A positive integer is written in each square of an $m \\times n$ board. The allowed move is to add an integer $k$ to each of two adjacent numbers in such a way that no negative numbers are obtained. (Two squares are adjacent if they have a common side.) Find a necessary and sufficient condition for it to be possible for all the numbers to be zero by a finite sequence of moves.", "solution": "19. Let us color the board in a chessboard fashion. Denote by $S_{b}$ and $S_{w}$ respectively the sum of numbers in the black and in the white squares. It is clear that every allowed move leaves the difference $S_{b}-S_{w}$ unchanged. Therefore a necessary condition for annulling all the numbers is $S_{b}=S_{w}$. We now show it is sufficient. Assuming $S_{b}=S_{w}$ let us observe a triple of (different) cells $a, b, c$ with respective values $x_{a}, x_{b}, x_{c}$ where $a$ and $c$ are both adjacent to $b$. We first prove that we can reduce $x_{a}$ to be 0 if $x_{a}>0$. If $x_{a} \\leq x_{b}$, we subtract $x_{a}$ from both $a$ and $b$. If $x_{a}>x_{b}$, we add $x_{a}-x_{b}$ to $b$ and $c$ and proceed as in the previous case. Applying the reduction in sequence, along the entire board, we reduce all cells except two neighboring cells to be 0 . Since $S_{b}=S_{w}$ is invariant, the two cells must have equal values and we can thus reduce them both to 0 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (AUS 3) Ali Barber, the carpet merchant, has a rectangular piece of carpet whose dimensions are unknown. Unfortunately, his tape measure is broken and he has no other measuring instruments. However, he finds that if he lays it flat on the floor of either of his storerooms, then each corner of the carpet touches a different wall of that room. If the two rooms have dimensions of 38 feet by 55 feet and 50 feet by 55 feet, what are the carpet dimensions?", "solution": "2. Let the carpet have width $x$, length $y$. Suppose that the carpet $E F G H$ lies in a room $A B C D, E$ being on $A B, F$ on $B C, G$ on $C D$, and $H$ on $D A$. Then $\\triangle A E H \\equiv \\triangle C G F \\sim \\triangle B F E \\equiv \\triangle D H G$. Let $\\frac{y}{x}=k, A E=a$ and $A H=b$. In that case $B E=k b$ and $D H=k a$. Thus $a+k b=50, k a+b=55$, whence $a=\\frac{55 k-50}{k^{2}-1}$ and $b=\\frac{50 k-55}{k^{2}-1}$. Hence $x^{2}=a^{2}+b^{2}=\\frac{5525 k^{2}-11000 k+5525}{\\left(k^{2}-1\\right)^{2}}$, i.e.,  $$ x^{2}\\left(k^{2}-1\\right)^{2}=5525 k^{2}-11000 k+5525 $$  Similarly, from the equations for the second storeroom, we get  $$ x^{2}\\left(k^{2}-1\\right)^{2}=4469 k^{2}-8360 k+4469 . $$  Combining the two equations, we get $5525 k^{2}-11000 k+5525=4469 k^{2}-$ $8360 k+4469$, which implies $k=2$ or $1 / 2$. Without loss of generality we have $y=2 x$ and $a+2 b=50,2 a+b=55$; hence $a=20, b=15$, $x=\\sqrt{15^{2}+20^{2}}=25$, and $y=50$. We have thus shown that the carpet is 25 feet by 50 feet.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. (NET 1) ${ }^{\\text {IMO3 }}$ Given a set $S$ in the plane containing $n$ points and satisfying the conditions: (i) no three points of $S$ are collinear, (ii) for every point $P$ of $S$ there exist at least $k$ points in $S$ that have the same distance to $P$, prove that the following inequality holds:  $$ k<\\frac{1}{2}+\\sqrt{2 n} $$", "solution": "20. Suppose $k \\geq 1 / 2+\\sqrt{2 n}$. Consider a point $P$ in $S$. There are at least $k$ points in $S$ having all the same distance to $P$, so there are at least $\\binom{k}{2}$ pairs of points $A, B$ with $A P=B P$. Since this is true for every point $P \\in S$, there are at least $n\\binom{k}{2}$ triples of points $(A, B, P)$ for which $A P=B P$ holds. However,  $$ \\begin{aligned} n\\binom{k}{2} & =n \\frac{k(k-1)}{2} \\geq \\frac{n}{2}\\left(\\sqrt{2 n}+\\frac{1}{2}\\right)\\left(\\sqrt{2 n}-\\frac{1}{2}\\right) \\\\ & =\\frac{n}{2}\\left(2 n-\\frac{1}{4}\\right)>n(n-1)=2\\binom{n}{2} \\end{aligned} $$  Since $\\binom{n}{2}$ is the number of all possible pairs $(A, B)$ with $A, B \\in S$, there must exist a pair of points $A, B$ with more than two points $P_{i}$ such that $A P_{i}=B P_{i}$. These points $P_{i}$ are collinear (they lie on the perpendicular bisector of $A B$ ), contradicting condition (1).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. (NET 2) Prove that the intersection of a plane and a regular tetrahedron can be an obtuse-angled triangle and that the obtuse angle in any such triangle is always smaller than $120^{\\circ}$.", "solution": "21. In order to obtain a triangle as the intersection we must have three points $P, Q, R$ on three sides of the tetrahedron passing through one vertex, say $T$. It is clear that we may suppose w.l.o.g. that $P$ is a vertex, and $Q$ and $R$ lie on the edges $T P_{1}$ and $T P_{2}\\left(P_{1}, P_{2}\\right.$ are vertices) or on their extensions respectively. Suppose that $\\overrightarrow{T Q}=\\lambda \\overrightarrow{T P_{1}}$ and $\\overrightarrow{T R}=\\mu \\overrightarrow{T P_{2}}$, where $\\lambda, \\mu>0$. Then  $$ \\cos \\angle Q P R=\\frac{\\overrightarrow{P Q} \\cdot \\overrightarrow{P R}}{\\overline{P Q} \\cdot \\overline{P R}}=\\frac{(\\lambda-1)(\\mu-1)+1}{2 \\sqrt{\\lambda^{2}-\\lambda+1} \\sqrt{\\mu^{2}-\\mu+1}} $$  In order to obtain an obtuse angle (with $\\cos <0$ ) we must choose $\\mu<1$ and $\\lambda>\\frac{2-\\mu}{1-\\mu}>1$. Since $\\sqrt{\\lambda^{2}-\\lambda+1}>\\lambda-1$ and $\\sqrt{\\mu^{2}-\\mu+1}>1-\\mu$, we get that for $(\\lambda-1)(\\mu-1)+1<0$,  $$ \\cos \\angle Q P R>\\frac{1-(1-\\mu)(\\lambda-1)}{2(1-\\mu)(\\lambda-1)}>-\\frac{1}{2} ; \\quad \\text { hence } \\angle Q P R<120^{\\circ} $$  Remark. After obtaining the formula for $\\cos \\angle Q P R$, the official solution was as follows: For fixed $\\mu_{0}<1$ and $\\lambda>1, \\cos \\angle Q P R$ is a decreasing function of $\\lambda$ : indeed,  $$ \\frac{\\partial \\cos \\angle Q P R}{\\partial \\lambda}=\\frac{\\mu-(3-\\mu) \\lambda}{4\\left(\\lambda^{2}-\\lambda+1\\right)^{3 / 2}\\left(\\mu^{2}-\\mu+1\\right)^{1 / 2}}<0 $$  Similarly, for a fixed, sufficiently large $\\lambda_{0}, \\cos \\angle Q P R$ is decreasing for $\\mu$ decreasing to 0 . Since $\\lim _{\\lambda \\rightarrow 0, \\mu \\rightarrow 0+} \\cos \\angle Q P R=-1 / 2$, we conclude that $\\angle Q P R<120^{\\circ}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "22", "problem_type": null, "exam": "IMO", "problem": "22. (PHI 1) ${ }^{\\text {IMO1 }}$ Prove that the set $\\{1,2, \\ldots, 1989\\}$ can be expressed as the disjoint union of 17 subsets $A_{1}, A_{2}, \\ldots, A_{17}$ such that: (i) each $A_{i}$ contains the same number of elements; (ii) the sum of all elements of each $A_{i}$ is the same for $i=1,2, \\ldots, 17$.", "solution": "22. The statement remains valid if 17 is replaced by any divisor $k$ of $1989=3^{2}$. $13 \\cdot 17,1<k<1989$, so let $k$ be one such divisor. The set $\\{1,2, \\ldots, 1989\\}$ can be partitioned as $\\{1,2, \\ldots, 3 k\\} \\cup \\bigcup_{j=1}^{L}\\{(2 j+1) k+1,(2 j+1) k+$ $2, \\ldots,(2 j+1) k+2 k\\}=X \\cup Y_{1} \\cup \\cdots \\cup Y_{L}$, where $L=(1989-3 k) / 2 k$. The required statement will be an obvious consequence of the following two claims. Claim 1. The set $X=\\{1,2, \\ldots, 3 k\\}$ can be partitioned into $k$ disjoint subsets, each having 3 elements and the same sum. Proof. Since $k$ is odd, let $t=k-1 / 2$ and $X=\\{1,2, \\ldots, 6 t+3\\}$. For $l=1,2, \\ldots, t$, define  $$ \\begin{aligned} X_{2 l-1} & =\\{l, 3 t+1+l, 6 t+5-2 l\\} \\\\ X_{2 l} & =\\{t+1+l, 2 t+1+l, 6 t+4-2 l\\} \\\\ X_{2 t+1} & =X_{k}=\\{t+1,4 t+2,4 t+3\\} \\end{aligned} $$  It is easily seen that these three subsets are disjoint and that the sum of elements in each set is $9 t+6$. Claim 2. Each $Y_{j}=\\{(2 j+1) k+1, \\ldots,(2 j+1) k+2 k\\}$ can be partitioned into $k$ disjoint subsets, each having 2 elements and the same sum. Proof. The obvious partitioning works:  $$ Y_{j}=\\{(2 j+1) k+1,(2 j+1) k+2 k\\} \\cup \\cdots \\cup\\{(2 j+1) k+k,(2 j+1) k+(k+1)\\} . $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "23", "problem_type": null, "exam": "IMO", "problem": "23. (POL 2) ${ }^{\\mathrm{IMO} 6} \\mathrm{We}$ consider permutations $\\left(x_{1}, \\ldots, x_{2 n}\\right)$ of the set $\\{1, \\ldots$, $2 n\\}$ such that $\\left|x_{i}-x_{i+1}\\right|=n$ for at least one $i \\in\\{1, \\ldots, 2 n-1\\}$. For every natural number $n$, find out whether permutations with this property are more or less numerous than the remaining permutations of $\\{1, \\ldots, 2 n\\}$.", "solution": "23. Two numbers $x, y \\in\\{1, \\ldots, 2 n\\}$ will be called twins if $|x-y|=n$. Then the set $\\{1, \\ldots, 2 n\\}$ splits into $n$ pairs of twins. A permutation $\\left(x_{1}, \\ldots, x_{2 n}\\right)$ of this set is said to be of type $T_{k}$ if $\\left|x_{i}-x_{i+1}\\right|=n$ holds for exactly $k$ indices $i$ (thus a permutation of type $T_{0}$ contains no pairs of neighboring twins). Denote by $F_{k}(n)$ the number of $T_{k}$-type permutations of $\\{1, \\ldots, 2 n\\}$. Let $\\left(x_{1}, \\ldots, x_{2 n}\\right)$ be a permutation of type $T_{0}$. Removing $x_{2 n}$ and its twin, we obtain a permutation of $2 n-2$ elements consisting of $n-1$ pairs of twins. This new permutation is of one of the following types: (i) type $T_{0}: x_{2 n}$ can take $2 n$ values, and its twin can take any of $2 n-2$ positions; (ii) type $T_{1}: x_{2 n}$ can take any one of $2 n$ values, but its twin must be placed to separate the unique pair of neighboring twins in the new permutation. The recurrence formula follows:  $$ F_{0}(n)=2 n\\left[(2 n-2) F_{0}(n-1)+F_{1}(n-1)\\right] $$  Now let $\\left(x_{1}, \\ldots, x_{2 n}\\right)$ be a permutation of type $T_{1}$, and let $\\left(x_{j}, x_{j+1}\\right)$ be the unique neighboring twin pair. Similarly, on removing this pair we get a permutation of $2 n-2$ elements, either of type $T_{0}$ or of type $T_{1}$. The pair $\\left(x_{j}, x_{j+1}\\right)$ is chosen out of $n$ twin pairs and can be arranged in two ways. Also, in the first case it can be placed anywhere ( $2 n-1$ possible positions), but in the second case it must be placed to separate the unique pair of neighboring twins. Hence,  $$ F_{1}(n)=2 n\\left[(2 n-1) F_{0}(n-1)+F_{1}(n-1)\\right]=F_{0}(n)+2 n F_{0}(n-1) $$  This implies that $F_{0}(n)<F_{1}(n)$. Therefore the permutations with at least one neighboring twin pair are more numerous than those with no such pairs. Remark 1. As in the official solution, formulas (1) and (2) together give for $F_{0}$ the recurrence  $$ F_{0}(n)=2 n\\left[(2 n-1) F_{0}(n-1)+(2 n-2) F_{0}(n-2)\\right] $$  For the ratio $p_{n}=F_{0}(n) /(2 n)$ !, simple algebraic manipulation yields $p_{n}=$ $p_{n-1}+\\frac{p_{n-2}}{(2 n-3)(2 n-1)}$. Since $p_{1}=0$, we get $p_{n}<p_{n-1}+\\frac{1}{(2 n-3)(2 n-1)}=p_{n-1}+\\frac{1}{2(2 n-3)}-\\frac{1}{2(2 n-1)}<\\cdots<\\frac{1}{2}$. Remark 2. Using the inclusion-exclusion principle, the following formula can be obtained:  $$ \\begin{aligned} F_{0}(n)= & 2^{0}\\binom{n}{0}(2 n)!-2^{1}\\binom{n}{1}(2 n-1)!+2^{2}\\binom{n}{2}(2 n-2)!-\\cdots \\\\ & \\cdots+(-1)^{n-1} 2^{n}\\binom{n}{n} n! \\end{aligned} $$  One consequence is that in fact, $\\lim _{n \\rightarrow \\infty} p_{n}=1 / e$. Second solution. Let $f: T_{0} \\rightarrow T_{1}$ be the mapping defined as follows: if $\\left(x_{1}, x_{2}, \\ldots, x_{2 n}\\right) \\in T_{0}$ and $x_{k}, k>2$, is the twin of $x_{1}$, then  $$ f\\left(x_{1}, x_{2}, \\ldots, x_{2 n}\\right)=\\left(x_{2}, \\ldots, x_{k-1}, x_{1}, x_{k}, \\ldots, x_{2 n}\\right) $$  The mapping $f$ is injective, but not surjective. Thus $F_{0}(n)<F_{1}(n)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "24", "problem_type": null, "exam": "IMO", "problem": "24. (POL 5) For points $A_{1}, \\ldots, A_{5}$ on the sphere of radius 1 , what is the maximum value that $\\min _{1 \\leq i, j \\leq 5} A_{i} A_{j}$ can take? Determine all configurations for which this maximum is attained. (Or: determine the diameter of any set $\\left\\{A_{1}, \\ldots, A_{5}\\right\\}$ for which this maximum is attained.)", "solution": "24. Instead of Euclidean distance, we will use the angles $\\angle A_{i} O A_{j}, O$ denoting the center of the sphere. Let $\\left\\{A_{1}, \\ldots, A_{5}\\right\\}$ be any set for which $\\min _{i \\neq j} \\angle A_{i} O A_{j} \\geq \\pi / 2$ (such a set exists: take for example five vertices of an octagon). We claim that two of the $A_{i}$ 's must be antipodes, thus implying that $\\min _{i \\neq j} \\angle A_{i} O A_{j}$ is exactly equal to $\\pi / 2$, and consequently that $\\min _{i \\neq j} A_{i} A_{j}=\\sqrt{2}$. Suppose no two of the five points are antipodes. Visualize $A_{5}$ as the south pole. Then $A_{1}, \\ldots, A_{4}$ lie in the northern hemisphere, including the equator (but excluding the north pole). No two of $A_{1}, \\ldots, A_{4}$ can lie in the interior of a quarter of this hemisphere, which means that any two of them differ in longitude by at least $\\pi / 2$. Hence, they are situated on four meridians that partition the sphere into quarters. Finally, if one of them does not lie on the equator, its two neighbors must. Hence, in any case there will exist an antipodal pair, giving us a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "25", "problem_type": null, "exam": "IMO", "problem": "25. (KOR 3) Let $a, b$ be integers that are not perfect squares. Prove that if  $$ x^{2}-a y^{2}-b z^{2}+a b w^{2}=0 $$  has a nontrivial solution in integers, then so does  $$ x^{2}-a y^{2}-b z^{2}=0 $$", "solution": "25. We may assume w.l.o.g. that $a>0$ (because $a, b<0$ is impossible, and $a, b \\neq 0$ from the condition of the problem). Let $\\left(x_{0}, y_{0}, z_{0}, w_{0}\\right) \\neq$ $(0,0,0,0)$ be a solution of $x^{2}-a y^{2}-b z^{2}+a b w^{2}$. Then  $$ x_{0}^{2}-a y_{0}^{2}=b\\left(z_{0}^{2}-a w_{0}^{2}\\right) $$  Multiplying both sides by $\\left(z_{0}^{2}-a w_{0}^{2}\\right)$, we get  $$ \\begin{gathered} \\left(x_{0}^{2}-a y_{0}^{2}\\right)\\left(z_{0}^{2}-a w_{0}^{2}\\right)-b\\left(z_{0}^{2}-a w_{0}^{2}\\right)^{2}=0 \\\\ \\Leftrightarrow\\left(x_{0} z_{0}-a y_{0} w_{0}\\right)^{2}-a\\left(y_{0} z_{0}-x_{0} w_{0}\\right)^{2}-b\\left(z_{0}^{2}-a w_{0}^{2}\\right)^{2}=0 . \\end{gathered} $$  Hence, for $x_{1}=x_{0} z_{0}-a y_{0} w_{0}, \\quad y_{1}=y_{0} z_{0}-x_{0} w_{0}, \\quad z_{1}=z_{0}^{2}-a w_{0}^{2}$, we have  $$ x_{1}^{2}-a y_{1}^{2}-b z_{1}^{2}=0 . $$  If $\\left(x_{1}, y_{1}, z_{1}\\right)$ is the trivial solution, then $z_{1}=0$ implies $z_{0}=w_{0}=0$ and similarly $x_{0}=y_{0}=0$ because $a$ is not a perfect square. This contradicts the initial assumption.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "26", "problem_type": null, "exam": "IMO", "problem": "26. (KOR 4) Let $n$ be a positive integer and let $a, b$ be given real numbers. Determine the range of $x_{0}$ for which  $$ \\sum_{i=0}^{n} x_{i}=a \\quad \\text { and } \\quad \\sum_{i=0}^{n} x_{i}^{2}=b $$  where $x_{0}, x_{1}, \\ldots, x_{n}$ are real variables.", "solution": "26. By the Cauchy-Schwarz inequality,  $$ \\left(\\sum_{i=1}^{n} x_{i}\\right)^{2} \\leq n \\sum_{i=1}^{n} x_{i}^{2} $$  Since $\\sum_{i=1}^{n} x_{i}=a-x_{0}$ and $\\sum_{i=1}^{n} x_{i}^{2}=b-x_{0}^{2}$, we have $\\left(a-x_{0}\\right)^{2} \\leq$ $n\\left(b-x_{0}^{2}\\right)$, i.e.,  $$ (n+1) x_{0}^{2}-2 a x_{0}+\\left(a^{2}-n b\\right) \\leq 0 . $$  The discriminant of this quadratic is $D=4 n(n+1)\\left[b-a^{2} /(n+1)\\right]$, so we conclude that (i) if $a^{2}>(n+1) b$, then such an $x_{0}$ does not exist; (ii) if $a^{2}=(n+1) b$, then $x_{0}=a / n+1 ; \\quad$ and (iii) if $a^{2}<(n+1) b$, then $\\frac{a-\\sqrt{D} / 2}{n+1} \\leq x_{0} \\leq \\frac{a+\\sqrt{D} / 2}{n+1}$.  It is easy to see that these conditions for $x_{0}$ are also sufficient.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "27", "problem_type": null, "exam": "IMO", "problem": "27. (ROM 1) Let $m$ be a positive odd integer, $m \\geq 2$. Find the smallest positive integer $n$ such that $2^{1989}$ divides $m^{n}-1$.", "solution": "27. Let $n$ be the required exponent, and suppose $n=2^{k} q$, where $q$ is an odd integer. Then we have  $$ m^{n}-1=\\left(m^{2^{k}}-1\\right)\\left[\\left(m^{2^{k}(q-1)}+\\cdots+m^{2^{k}}+1\\right]=\\left(m^{2^{k}}-1\\right) A\\right. $$  where $A$ is odd. Therefore $m^{n}-1$ and $m^{2^{k}}-1$ are divisible by the same power of 2 , and so $n=2^{k}$. Next, we observe that  $$ \\begin{aligned} m^{2^{k}}-1 & =\\left(m^{2^{k-1}}-1\\right)\\left(m^{2^{k-1}}+1\\right)=\\ldots \\\\ & =\\left(m^{2}-1\\right)\\left(m^{2}+1\\right)\\left(m^{4}+1\\right) \\cdots\\left(m^{2^{k-1}}+1\\right) \\end{aligned} $$  Let $s$ be the maximal positive integer for which $m \\equiv \\pm 1\\left(\\bmod 2^{s}\\right)$. Then $m^{2}-1$ is divisible by $2^{s+1}$ and not divisible by $2^{s+2}$. All the numbers $m^{2}+1, m^{4}+1, \\ldots, m^{2^{k-1}}+1$ are divisible by 2 and not by 4 . Hence $m^{2^{k}}-1$ is divisible by $2^{s+k}$ and not by $2^{s+k+1}$. It follows from the above consideration that the smallest exponent $n$ equals $2^{1989-s}$ if $s \\leq 1989$, and $n=1$ if $s>1989$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "28", "problem_type": null, "exam": "IMO", "problem": "28. (ROM 2) Consider in a plane $\\Pi$ the points $O, A_{1}, A_{2}, A_{3}, A_{4}$ such that $\\sigma\\left(O A_{i} A_{j}\\right) \\geq 1$ for all $i, j=1,2,3,4, i \\neq j$. Prove that there is at least one pair $i_{0}, j_{0} \\in\\{1,2,3,4\\}$ such that $\\sigma\\left(O A_{i_{0}} A_{j_{0}}\\right) \\geq \\sqrt{2}$. (We have denoted by $\\sigma\\left(O A_{i} A_{j}\\right)$ the area of triangle $O A_{i} A_{j}$.)", "solution": "28. Assume w.l.o.g. that the rays $O A_{1}, O A_{2}, O A_{3}, O A_{4}$ are arranged clockwise. Setting $O A_{1}=a, O A_{2}=b, O A_{3}=c, O A_{4}=d$, and $\\angle A_{1} O A_{2}=x$, $\\angle A_{2} O A_{3}=y, \\angle A_{3} O A_{4}=z$, we have  $$ \\begin{aligned} & S_{1}=\\sigma\\left(O A_{1} A_{2}\\right)=\\frac{1}{2} a b|\\sin x|, S_{2}=\\sigma\\left(O A_{1} A_{3}\\right)=\\frac{1}{2} a c|\\sin (x+y)|, \\\\ & S_{3}=\\sigma\\left(O A_{1} A_{4}\\right)=\\frac{1}{2} a d|\\sin (x+y+z)|, S_{4}=\\sigma\\left(O A_{2} A_{3}\\right)=\\frac{1}{2} b c|\\sin y|, \\\\ & S_{5}=\\sigma\\left(O A_{2} A_{4}\\right)=\\frac{1}{2} b d|\\sin (y+z)|, S_{6}=\\sigma\\left(O A_{3} A_{4}\\right)=\\frac{1}{2} c d|\\sin z| . \\end{aligned} $$  Since $\\sin (x+y+z) \\sin y+\\sin x \\sin z=\\sin (x+y) \\sin (y+z)$, it follows that there exists a choice of $k, l \\in\\{0,1\\}$ such that  $$ S_{1} S_{6}+(-1)^{k} S_{2} S_{5}+(-1)^{l} S_{3} S_{4}=0 $$  For example (w.l.o.g.), if $S_{3} S_{4}=S_{1} S_{6}+S_{2} S_{5}$, we have  $$ \\left(\\max _{1 \\leq i \\leq 6} S_{i}\\right)^{2} \\geq S_{3} S_{4}=S_{1} S_{6}+S_{2} S_{5} \\geq 1+1=2 $$  i.e., $\\max _{1 \\leq i \\leq 6} S_{i} \\geq \\sqrt{2}$ as claimed.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "29", "problem_type": null, "exam": "IMO", "problem": "29. (ROM 4) A flock of 155 birds sit down on a circle $C$. Two birds $P_{i}, P_{j}$ are mutually visible if $m\\left(P_{i} P_{j}\\right) \\leq 10^{\\circ}$. Find the smallest number of mutually visible pairs of birds. (One assumes that a position (point) on $C$ can be occupied simultaneously by several birds.)", "solution": "29. Let $P_{i}$, sitting at the place $A$, and $P_{j}$ sitting at $B$, be two birds that can see each other. Let $k$ and $l$ respectively be the number of birds visible from $B$ but not from $A$, and the number of those visible from $A$ but not from $B$. Assume that $k \\geq l$. Then if all birds from $B$ fly to $A$, each of them will see $l$ new birds, but won't see $k$ birds anymore. Hence the total number of mutually visible pairs does not increase, while the number of distinct positions occupied by at least one bird decreases by one. Repeating this operation as many times as possible one can arrive at a situation in which two birds see each other if and only if they are in the same position. The number of such distinct positions is at most 35 , while the total number of mutually visible pairs is not greater than at the beginning. Thus the problem is equivalent to the following one: (1) If $x_{i} \\geq 0$ are integers with $\\sum_{j=1}^{35} x_{j}=155$, find the least possible value of $\\sum_{j=1}^{35}\\left(x_{j}^{2}-x_{j}\\right) / 2$. If $x_{j} \\geq x_{i}+2$ for some $i, j$, then the sum of $\\left(x_{j}^{2}-x_{j}\\right) / 2$ decreases (for $x_{j}-x_{i}-2$ ) if $x_{i}, x_{j}$ are replaced with $x_{i}+1, x_{j}-1$. Consequently, our sum attains its minimum when the $x_{i}$ 's differ from each other by at most 1 . In this case, all the $x_{i}$ 's are equal to either $[155 / 35]=4$ or $[155 / 35]+1=5$, where $155=20 \\cdot 4+15 \\cdot 5$. It follows that the (minimum possible) number of mutually visible pairs is $20 \\cdot \\frac{4 \\cdot 3}{2}+15 \\cdot \\frac{5 \\cdot 4}{2}=270$. Second solution for (1). Considering the graph consisting of birds as vertices and pairs of mutually nonvisible birds as edges, we see that there is no complete 36 -subgraph. Turan's theorem gives the answer immediately. (See problem (SL89-17).)", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (AUS 4) Ali Barber, the carpet merchant, has a rectangular piece of carpet whose dimensions are unknown. Unfortunately, his tape measure is broken and he has no other measuring instruments. However, he finds that if he lays it flat on the floor of either of his storerooms, then each corner of the carpet touches a different wall of that room. He knows that the sides of the carpet are integral numbers of feet and that his two storerooms have the same (unknown) length, but widths of 38 feet and 50 feet respectively. What are the carpet dimensions?", "solution": "3. Let the carpet have width $x$, length $y$. Let the length of the storerooms be $q$. Let $y / x=k$. Then, as in the previous problem, $(k q-50)^{2}+(50 k-q)^{2}=$ $(k q-38)^{2}+(38 k-q)^{2}$, i.e.,  $$ k q=22\\left(k^{2}+1\\right) $$  Also, as before, $x^{2}=\\left(\\frac{k q-50}{k^{2}-1}\\right)^{2}+\\left(\\frac{50 k-q}{k^{2}-1}\\right)^{2}$, i.e.,  $$ x^{2}\\left(q^{2}-1\\right)^{2}=\\left(k^{2}+1\\right)\\left(q^{2}-1900\\right) $$  which, together with (1), yields  $$ x^{2} k^{2}\\left(k^{2}-1\\right)^{2}=\\left(k^{2}+1\\right)\\left(484 k^{4}-932 k^{2}+484\\right) $$  Since $k$ is rational, let $k=c / d$, where $c$ and $d$ are integers with $\\operatorname{gcd}(c, d)=$ 1. Then we obtain  $$ x^{2} c^{2}\\left(c^{2}-d^{2}\\right)^{2}=c^{2}\\left(484 c^{4}-448 c^{2} d^{2}-448 d^{4}\\right)+484 d^{6} $$  We thus have $c^{2} \\mid 484 d^{6}$, but since $(c, d)=1$, we have $c^{2}|484 \\Rightarrow c| 22$. Analogously, $d \\mid 22$; thus $k=1,2,11,22, \\frac{1}{2}, \\frac{1}{11}, \\frac{1}{22}, \\frac{2}{11}, \\frac{11}{2}$. Since reciprocals lead to the same solution, we need only consider $k \\in\\left\\{1,2,11,22, \\frac{11}{2}\\right\\}$, yielding $q=44,55,244,485,125$, respectively. We can test these values by substituting them into (2). Only $k=2$ gives us an integer solution, namely $x=25, y=50$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "30", "problem_type": null, "exam": "IMO", "problem": "30. (SWE 2) ${ }^{\\mathrm{IMO} 5}$ For which positive integers $n$ does there exist a positive integer $N$ such that none of the integers $1+N, 2+N, \\ldots, n+N$ is the power of a prime number?", "solution": "30. For all $n$ such $N$ exists. For a given $n$ choose $N=(n+1)!^{2}+1$. Then $1+j$ is a proper factor of $N+j$ for $1 \\leq j \\leq n$. So if $N+j=p^{m}$ is a power of a prime $p$, then $1+j=p^{r}$ for some integer $r, 1 \\leq r<m$. But then $p^{r+1}$ divides both $(n+1)!^{2}=N-1$ and $p^{m}=N+j$, implying that $p^{r+1} \\mid 1+j$, which is impossible. Thus none of $N+1, N+2, \\ldots, N+n$ is a power of a prime. Second solution. Let $p_{1}, p_{2}, \\ldots, p_{2 n}$ be distinct primes. By the Chinese remainder theorem, there exists a natural number $N$ such that $p_{1} p_{2}$ $N+1, p_{3} p_{4}\\left|N+2, \\ldots, p_{2 n-1} p_{2 n}\\right| N+n$, and then obviously none of the numbers $N+1, \\ldots, N+n$ can be a power of a prime.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "31", "problem_type": null, "exam": "IMO", "problem": "31. (SWE 3) Let $a_{1} \\geq a_{2} \\geq a_{3}$ be given positive integers and let $N\\left(a_{1}, a_{2}, a_{3}\\right)$ be the number of solutions $\\left(x_{1}, x_{2}, x_{3}\\right)$ of the equation  $$ \\frac{a_{1}}{x_{1}}+\\frac{a_{2}}{x_{2}}+\\frac{a_{3}}{x_{3}}=1 $$  where $x_{1}, x_{2}$, and $x_{3}$ are positive integers. Show that  $$ N\\left(a_{1}, a_{2}, a_{3}\\right) \\leq 6 a_{1} a_{2}\\left(3+\\ln \\left(2 a_{1}\\right)\\right) $$", "solution": "31. Let us denote by $N_{p q r}$ the number of solutions for which $a_{p} / x_{p} \\geq a_{q} / x_{q} \\geq$ $a_{r} / x_{r}$, where $(p, q, r)$ is one of six permutations of $(1,2,3)$. It is clearly enough to prove that $N_{p q r}+N_{q p r} \\leq 2 a_{1} a_{2}\\left(3+\\ln \\left(2 a_{1}\\right)\\right)$. First, from  $$ \\frac{3 a_{p}}{x_{p}} \\geq \\frac{a_{p}}{x_{p}}+\\frac{a_{q}}{x_{q}}+\\frac{a_{r}}{x_{r}}=1 \\quad \\text { and } \\quad \\frac{a_{p}}{x_{p}}<1 $$  we get $a_{p}+1 \\leq x_{p} \\leq 3 a_{p}$. Similarly, for fixed $x_{p}$ we have  $$ \\frac{2 a_{q}}{x_{q}} \\geq \\frac{a_{q}}{x_{q}}+\\frac{a_{r}}{x_{r}}=1-\\frac{a_{p}}{x_{p}} \\quad \\text { and } \\quad \\frac{a_{q}}{x_{q}} \\leq \\min \\left(\\frac{a_{p}}{x_{p}}, 1-\\frac{a_{p}}{x_{p}}\\right), $$  which gives $\\max \\left\\{a_{q} \\cdot x_{p} / a_{p}, a_{q} \\cdot x_{p} /\\left(x_{p}-a_{p}\\right)\\right\\} \\leq x_{q} \\leq 2 a_{q} \\cdot x_{p} /\\left(x_{p}-a_{p}\\right)$, i.e., if $a_{p}+1 \\leq x_{p} \\leq 2 a_{p}$ there are at most $a_{q} \\cdot x_{p} /\\left(x_{p}-a_{p}\\right)+1 / 2$ possible values for $x_{q}$ (because there are $[2 x]-[x]=[x+1 / 2]$ integers between $x$ and $2 x)$, and if $2 a_{p}+1 \\leq x_{p} \\leq 3 a_{p}$, at most $2 a_{q} \\cdot x_{p} /\\left(x_{p}-a_{p}\\right)-a_{q} \\cdot x_{p} / a_{p}+$ 1 possible values. Given $x_{p}$ and $x_{q}, x_{r}$ is uniquely determined. Hence  $$ \\begin{aligned} N_{p q r} & \\leq \\sum_{x_{p}=a_{p}+1}^{2 a_{p}}\\left(\\frac{a_{q} \\cdot x_{p}}{x_{p}-a_{p}}+\\frac{1}{2}\\right)+\\sum_{x_{p}=2 a_{p}+1}^{3 a_{p}}\\left(\\frac{2 a_{q} \\cdot x_{p}}{x_{p}-a_{p}}-\\frac{a_{q} \\cdot x_{p}}{a_{p}}+1\\right) \\\\ & =\\frac{3 a_{p}}{2}+a_{q} \\sum_{k=1}^{a_{p}}\\left[\\frac{k+a_{p}}{k}+\\left(\\frac{2\\left(k+2 a_{p}\\right)}{k+a_{p}}-\\frac{k+2 a_{p}}{a_{p}}\\right)\\right] \\\\ & =\\frac{3 a_{p}}{2}+a_{q} \\sum_{k=1}^{a_{p}}\\left[1-\\frac{k}{a_{p}}+a_{p}\\left(\\frac{1}{k}+\\frac{2}{k+a_{p}}\\right)\\right] \\\\ & =\\frac{3 a_{p}}{2}-\\frac{a_{q}}{2}+a_{p} a_{q}\\left(\\frac{1}{2}+\\sum_{k=1}^{a_{p}}\\left(\\frac{1}{k}+\\frac{2}{k+a_{p}}\\right)\\right) \\\\ & \\leq a_{p} a_{q}\\left(\\frac{3}{2 a_{q}}-\\frac{1}{2 a_{p}}+\\ln \\left(2 a_{p}\\right)+\\frac{5}{2}-\\ln 2\\right), \\end{aligned} $$  where we have used $\\sum_{k=1}^{n}(1 / k+2 /(k+n)) \\leq \\ln (2 n)+2-\\ln 2$ (this can be proved by induction). Hence, $N_{p q r}+N_{q p r} \\leq 2 a_{p} a_{q}\\left(1+0.5+\\ln \\left(2 a_{p}\\right)+2-\\ln 2\\right)<2 a_{1} a_{2}\\left(2.81+\\ln \\left(2 a_{1}\\right)\\right)$. Remark. The official solution was somewhat simpler, but used that the interval $(x, 2 x]$, for real $x$, cannot contain more than $x$ integers, which is false in general. Thus it could give only a weaker estimate $N \\leq 6 a_{1} a_{2}\\left(9 / 2-\\ln 2+\\ln \\left(2 a_{1}\\right)\\right)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "32", "problem_type": null, "exam": "IMO", "problem": "32. (USA 3) The vertex $A$ of the acute triangle $A B C$ is equidistant from the circumcenter $O$ and the orthocenter $H$. Determine all possible values for the measure of angle $A$.", "solution": "32. Let $C C^{\\prime}$ be an altitude, and $R$ the circumradius. Then, since $A H=R$, we have $A C^{\\prime}=|R \\sin B|$ and hence (1) $C C^{\\prime}=|R \\sin B \\tan A|$. On the other hand, $C C^{\\prime}=|B C \\sin B|=2|R \\sin A \\sin B|$, which together with (1) yields $2|\\sin A|=|\\tan A| \\Rightarrow|\\cos A|=1 / 2$. Hence, $\\angle A$ is $60^{\\circ}$. (Without the condition that the triangle is acute, $\\angle A$ could also be $120^{\\circ}$.) Second Solution. For a point $X$, let $\\bar{X}$ denote the vector $O X$. Then $|\\bar{A}|=|\\bar{B}|=|\\bar{C}|=R$ and $\\bar{H}=\\bar{A}+\\bar{B}+\\bar{C}$, and moreover,  $$ R^{2}=(\\bar{H}-\\bar{A})^{2}=(\\bar{B}+\\bar{C})^{2}=2 \\bar{B}^{2}+2 \\bar{C}^{2}-(\\bar{B}-\\bar{C})^{2}=4 R^{2}-B C^{2} $$  It follows that $\\sin A=\\frac{B C}{2 R}=\\sqrt{3} / 2$, i.e., that $\\angle A=60^{\\circ}$. Third Solution. Let $A_{1}$ be the midpoint of $B C$. It is well known that $A H=2 O A_{1}$, and since $A H=A O=B O$, it means that in the rightangled triangle $B O A_{1}$ the relation $B O=2 O A_{1}$ holds. Thus $\\angle B O A_{1}=$ $\\angle A=60^{\\circ}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (BUL 3) Prove that for every integer $n>1$ the equation  $$ \\frac{x^{n}}{n!}+\\frac{x^{n-1}}{(n-1)!}+\\cdots+\\frac{x^{2}}{2!}+\\frac{x}{1!}+1=0 $$  has no rational roots.", "solution": "4. First we note that for every integer $k>0$ and prime number $p, p^{k}$ doesn't divide $k!$. This follows from the fact that the highest exponent $r$ of $p$ for which $p^{r} \\mid k$ ! is  $$ r=\\left[\\frac{k}{p}\\right]+\\left[\\frac{k}{p^{2}}\\right]+\\cdots<\\frac{k}{p}+\\frac{k}{p^{2}}+\\cdots=\\frac{k}{p-1}<k . $$  Now suppose that $\\alpha$ is a rational root of the given equation. Then  $$ \\alpha^{n}+\\frac{n!}{(n-1)!} \\alpha^{n-1}+\\cdots+\\frac{n!}{2!} \\alpha^{2}+\\frac{n!}{1!} \\alpha+n!=0 $$  from which we can conclude that $\\alpha$ must be an integer, not equal to $\\pm 1$. Let $p$ be a prime divisor of $n$ and let $r$ be the highest exponent of $p$ for which $p^{r} \\mid n$ !. Then $p \\mid \\alpha$. Since $p^{k} \\mid \\alpha^{k}$ and $p^{k} \\nmid k$ !, we obtain that $p^{r+1} \\mid n!\\alpha^{k} / k$ ! for $k=1,2, \\ldots, n$. But then it follows from (1) that $p^{r+1} \\mid n$ !, a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (COL 1) Consider the polynomial $p(x)=x^{n}+n x^{n-1}+a_{2} x^{n-2}+\\cdots+a_{n}$ having all real roots. If $r_{1}^{16}+r_{2}^{16}+\\cdots+r_{n}^{16}=n$, where the $r_{j}$ are the roots of $p(x)$, find all such roots.", "solution": "5. According to the Cauchy-Schwarz inequality,  $$ \\left(\\sum_{i=1}^{n} a_{i}\\right)^{2} \\leq\\left(\\sum_{i=1}^{n} a_{i}^{2}\\right)\\left(\\sum_{i=1}^{n} 1^{2}\\right)=n\\left(\\sum_{i=1}^{n} a_{i}^{2}\\right) $$  Since $r_{1}+\\cdots+r_{n}=-n$, applying this inequality we obtain $r_{1}^{2}+\\ldots+r_{n}^{2} \\geq n$, and applying it three more times, we obtain  $$ r_{1}^{16}+\\cdots+r_{n}^{16} \\geq n $$  with equality if and only if $r_{1}=r_{2}=\\ldots=r_{n}=-1$ and $p(x)=(x+1)^{n}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (CZS 1) For a triangle $A B C$, let $k$ be its circumcircle with radius $r$. The bisectors of the inner angles $A, B$, and $C$ of the triangle intersect respectively the circle $k$ again at points $A^{\\prime}, B^{\\prime}$, and $C^{\\prime}$. Prove the inequality  $$ 16 Q^{3} \\geq 27 r^{4} P $$  where $Q$ and $P$ are the areas of the triangles $A^{\\prime} B^{\\prime} C^{\\prime}$ and $A B C$ respectively.", "solution": "6. Let us denote the measures of the inner angles of the triangle $A B C$ by $\\alpha, \\beta, \\gamma$. Then $P=r^{2}(\\sin 2 \\alpha+\\sin 2 \\beta+\\sin 2 \\gamma) / 2$. Since the inner angles of the triangle $A^{\\prime} B^{\\prime} C^{\\prime}$ are $(\\beta+\\gamma) / 2,(\\gamma+\\alpha) / 2,(\\alpha+\\beta) / 2$, we also have $Q=r^{2}[\\sin (\\beta+\\gamma)+\\sin (\\gamma+\\alpha)+\\sin (\\alpha+\\beta)] / 2$. Applying the AM-GM mean inequality, we now obtain  $$ \\begin{aligned} 16 Q^{3} & =\\frac{16}{8} r^{6}(\\sin (\\beta+\\gamma)+\\sin (\\gamma+\\alpha)+\\sin (\\alpha+\\beta))^{3} \\\\ & \\geq 54 r^{6} \\sin (\\beta+\\gamma) \\sin (\\gamma+\\alpha) \\sin (\\alpha+\\beta) \\\\ & =27 r^{6}[\\cos (\\alpha-\\beta)-\\cos (\\alpha+\\beta+2 \\gamma)] \\sin (\\alpha+\\beta) \\\\ & =27 r^{6}[\\cos (\\alpha-\\beta)+\\cos \\gamma] \\sin (\\alpha+\\beta) \\\\ & =\\frac{27}{2} r^{6}[\\sin (\\alpha+\\beta+\\gamma)+\\sin (\\alpha+\\beta-\\gamma)+\\sin 2 \\alpha+\\sin 2 \\beta] \\\\ & =\\frac{27}{2} r^{6}[\\sin (2 \\gamma)+\\sin 2 \\alpha+\\sin 2 \\beta]=27 r^{4} P . \\end{aligned} $$  This completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (FIN 1) Show that any two points lying inside a regular $n$-gon $E$ can be joined by two circular arcs lying inside $E$ and meeting at an angle of at least $\\left(1-\\frac{2}{n}\\right) \\pi$.", "solution": "7. Assume that $P_{1}$ and $P_{2}$ are points inside $E$, and that the line $P_{1} P_{2}$ intersects the perimeter of $E$ at $Q_{1}$ and $Q_{2}$. If we prove the statement for $Q_{1}$ and $Q_{2}$, we are done, since these arcs can be mapped homothetically to join $P_{1}$ and $P_{2}$. Let $V_{1}, V_{2}$ be two vertices of $E$. Then applying two homotheties to the inscribed circle of $E$ one can find two arcs (one of them may be a side of $E)$ joining these two points, both tangent to the sides of $E$ that meet at $V_{1}$ and $V_{2}$. If $A$ is any point of the side $V_{2} V_{3}$, two homotheties with center $V_{1}$ take the arcs joining $V_{1}$ to $V_{2}$ and $V_{3}$ into arcs joining $V_{1}$ to $A$; their angle of incidence at $A$ remains $(1-2 / n) \\pi$. Next, for two arbitrary points $Q_{1}$ and $Q_{2}$ on two different sides $V_{1} V_{2}$ and $V_{3} V_{4}$, we join $V_{1}$ and $V_{2}$ to $Q_{2}$ with pairs of arcs that meet at $Q_{2}$ and have an angle of incidence $(1-2 / n) \\pi$. The two arcs that meet the line $Q_{1} Q_{2}$ again outside $E$ meet at $Q_{2}$ at an angle greater than or equal to $(1-2 / n) \\pi$. Two homotheties with center $Q_{2}$ carry these arcs to ones meeting also at $Q_{1}$ with the same angle of incidence.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (FRA 2) Let $R$ be a rectangle that is the union of a finite number of rectangles $R_{i}, 1 \\leq i \\leq n$, satisfying the following conditions: (i) The sides of every rectangle $R_{i}$ are parallel to the sides of $R$. (ii) The interiors of any two different $R_{i}$ are disjoint. (iii) Every $R_{i}$ has at least one side of integral length. Prove that $R$ has at least one side of integral length.", "solution": "8. Let $A, B, C, D$ denote the vertices of $R$. We consider the set $\\mathcal{S}$ of all points $E$ of the plane that are vertices of at least one rectangle, and its subset $\\mathcal{S}^{\\prime}$ consisting of those points in $\\mathcal{S}$ that have both coordinates integral in the orthonormal coordinate system with point $A$ as the origin and lines $A B, A D$ as axes. First, to each $E \\in \\mathcal{S}$ we can assign an integer $n_{E}$ as the number of rectangles $R_{i}$ with one vertex at $E$. It is easy to check that $n_{E}=1$ if $E$ is one of the vertices $A, B, C, D$; in all other cases $n_{E}$ is either 2 or 4. Furthermore, for each rectangle $R_{i}$ we define $f\\left(R_{i}\\right)$ as the number of vertices of $R_{i}$ that belong to $\\mathcal{S}^{\\prime}$. Since every $R_{i}$ has at least one side of integer length, $f\\left(R_{i}\\right)$ can take only values 0,2 , or 4 . Therefore we have  $$ \\sum_{i=1}^{n} f\\left(R_{i}\\right) \\equiv 0(\\bmod 2) $$  On the other hand, $\\sum_{i=1}^{n} f\\left(R_{i}\\right)$ is equal to $\\sum_{E \\in \\mathcal{S}^{\\prime}} n_{E}$, implying that  $$ \\sum_{E \\in \\mathcal{S}^{\\prime}} n_{E} \\equiv 0(\\bmod 2) $$  However, since $n_{A}=1$, at least one other $n_{E}$, where $E \\in \\mathcal{S}^{\\prime}$, must be odd, and that can happen only for $E$ being $B, C$, or $D$. We conclude that at least one of the sides of $R$ has integral length. Second solution. Consider the coordinate system introduced above. If $D$ is a rectangle whose sides are parallel to the axes of the system, it is easy to prove that  $$ \\int_{D} \\sin 2 \\pi x \\sin 2 \\pi y d x d y=0 $$  if and only if at least one side of $D$ has integral length. This holds for all $R_{i}$ 's, so that adding up these equalities we get $\\int_{R} \\sin 2 \\pi x \\sin 2 \\pi y d x d y=0$. Thus, $R$ also has a side of integral length.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1989", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (FRA 4) For all integers $n, n \\geq 0$, there exist uniquely determined integers $a_{n}, b_{n}, c_{n}$ such that  $$ (1+4 \\sqrt[3]{2}-4 \\sqrt[3]{4})^{n}=a_{n}+b_{n} \\sqrt[3]{2}+c_{n} \\sqrt[3]{4} $$  Prove that $c_{n}=0$ implies $n=0$.", "solution": "9. From $a_{n+1}+b_{n+1} \\sqrt[3]{2}+c_{n+1} \\sqrt[3]{4}=\\left(a_{n}+b_{n} \\sqrt[3]{2}+c_{n} \\sqrt[3]{4}\\right)(1+4 \\sqrt[3]{2}-4 \\sqrt[3]{4})$ we obtain $a_{n+1}=a_{n}-8 b_{n}+8 c_{n}$. Since $a_{0}=1, a_{n}$ is odd for all $n$. For an integer $k>0$, we can write $k=2^{l} k^{\\prime}, k^{\\prime}$ being odd and $l$ a nonnegative integer. Let us set $v(k)=l$, and define $\\beta_{n}=v\\left(b_{n}\\right), \\gamma_{n}=v\\left(c_{n}\\right)$. We prove the following lemmas: Lemma 1. For every integer $p \\geq 0, b_{2^{p}}$ and $c_{2^{p}}$ are nonzero, and $\\beta_{2^{p}}=$ $\\gamma_{2^{p}}=p+2$. Proof. By induction on $p$. For $p=0, b_{1}=4$ and $c_{1}=-4$, so the assertion is true. Suppose that it holds for $p$. Then  $$ (1+4 \\sqrt[3]{2}-4 \\sqrt[3]{4})^{2^{p+1}}=\\left(a+2^{p+2}\\left(b^{\\prime} \\sqrt[3]{2}+c^{\\prime} \\sqrt[3]{4}\\right)\\right)^{2} \\text { with } a, b^{\\prime}, \\text { and } c^{\\prime} \\text { odd. } $$  Then we easily obtain that $(1+4 \\sqrt[3]{2}-4 \\sqrt[3]{4})^{2^{p+1}}=A+2^{p+3}(B \\sqrt[3]{2}+$ $C \\sqrt[3]{4}$ ), where $A, B=a b^{\\prime}+2^{p+1} E, C=a c^{\\prime}+2^{p+1} F$ are odd. Therefore Lemma 1 holds for $p+1$. Lemma 2. Suppose that for integers $n, m \\geq 0, \\beta_{n}=\\gamma_{n}=\\lambda>\\beta_{m}=$ $\\gamma_{m}=\\mu$. Then $b_{n+m}, c_{n+m}$ are nonzero and $\\beta_{n+m}=\\gamma_{n+m}=\\mu$. Proof. Calculating $\\left(a^{\\prime}+2^{\\lambda}\\left(b^{\\prime} \\sqrt[3]{2}+c^{\\prime} \\sqrt[3]{4}\\right)\\right)\\left(a^{\\prime \\prime}+2^{\\mu}\\left(b^{\\prime \\prime} \\sqrt[3]{2}+c^{\\prime \\prime} \\sqrt[3]{4}\\right)\\right)$, with $a^{\\prime}, b^{\\prime}, c^{\\prime}, a^{\\prime \\prime}, b^{\\prime \\prime}, c^{\\prime \\prime}$ odd, we easily obtain the product $A+2^{\\mu}(B \\sqrt[3]{2}+$ $C \\sqrt[3]{4})$, where $A, B=a^{\\prime} b^{\\prime \\prime}+2^{\\lambda-\\mu} E$, and $C=a^{\\prime} c^{\\prime \\prime}+2^{\\lambda-\\mu} F$ are odd, which proves Lemma 2. Since every integer $n>0$ can be written as $n=2^{p_{r}}+\\cdots+2^{p_{1}}$, with $0 \\leq p_{1}<\\cdots<p_{r}$, from Lemmas 1 and 2 it follows that $c_{n}$ is nonzero, and that $\\gamma_{n}=p_{1}+2$. Remark. $b_{1989}$ and $c_{1989}$ are divisible by 4 , but not by 8 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (AUS 3) The integer 9 can be written as a sum of two consecutive integers: $9=4+5$. Moreover, it can be written as a sum of (more than one) consecutive positive integers in exactly two ways: $9=4+5=2+3+4$. Is there an integer that can be written as a sum of 1990 consecutive integers and that can be written as a sum of (more than one) consecutive positive integers in exactly 1990 ways?", "solution": "1. Let $N$ be a number that can be written as a sum of 1990 consecutive integers and as a sum of consecutive positive integers in exactly 1990 ways. The former requirement gives us $N=m+(m+1)+\\cdots+(m+1989)=$ $995(2 m+1989)$ for some $m$. Thus $2 \\nmid N, 5 \\mid N$, and $199 \\mid N$. The latter requirement tells us that there are exactly 1990 ways to express $N$ as $n+(n+1)+\\cdots+(n+k)$, or equivalently, express $2 N$ as $(k+1)(2 n+k)$. Since $N$ is odd, it follows that one of the factors $k+1$ and $2 n+k$ is odd and the other is divisible by 2 , but not by 4 . Evidently $k+1<2 n+k$. On the other hand, every factorization $2 N=a b, 1<a<b$, corresponds to a single pair $(n, k)$, where $n=\\frac{b-a+1}{2}$ (which is an integer) and $k=a-1$. The number of such factorizations is equal to $d(2 N) / 2-1$ because $a=b$ is impossible (here $d(x)$ denotes the number of positive divisors of an $x \\in \\mathbb{N}$ ). Hence we must have $d(2 N)=2 \\cdot 1991=3982$. Now let $2 N=$ $2 \\cdot 5^{e_{1}} \\cdot 199^{e_{2}} \\cdot p_{3}^{e_{3}} \\cdots p_{r}^{e_{r}}$ be a factorization of $2 N$ into prime numbers, where $p_{3}, \\ldots, p_{r}$ are distinct primes other than 2,5 , and 199 and $e_{1}, \\cdots, e_{r}$ are positive integers. Then $d(2 N)=2\\left(e_{1}+1\\right)\\left(e_{2}+1\\right) \\cdots\\left(e_{r}+1\\right)$, from which we deduce $\\left(e_{1}+1\\right)\\left(e_{2}+1\\right) \\cdots\\left(e_{r}+1\\right)=1991=11 \\cdot 181$. We thus get $\\left\\{e_{1}, e_{2}\\right\\}=\\{10,180\\}$ and $e_{3}=\\cdots=e_{r}=0$. Hence $N=5^{10} \\cdot 199^{180}$ and $N=5^{180} \\cdot 199^{10}$ are the only possible solutions. These numbers indeed satisfy the desired properties.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (ICE 2) A plane cuts a right circular cone into two parts. The plane is tangent to the circumference of the base of the cone and passes through the midpoint of the altitude. Find the ratio of the volume of the smaller part to the volume of the whole cone.", "solution": "10. Let $r$ be the radius of the base and $h$ the height of the cone. We may assume w.l.o.g. that $r=1$. Let $A$ be the top of the cone, $B C$ the diameter of the circumference of the base such that the plane touches the circumference at $B, O$ the center of the base, and $H$ the midpoint of $O A$ (also belonging to the plane). Let $B H$ cut the sheet of the cone at $D$. By applying Menelaus's theorem to $\\triangle A O C$ and $\\triangle B H O$, we conclude that $\\frac{A D}{D C}=\\frac{C B}{B O} \\cdot \\frac{O H}{H A}=\\frac{1}{2}$ and $\\frac{H D}{D B}=\\frac{H A}{A O} \\cdot \\frac{O C}{C B}=\\frac{1}{4}$. The plane cuts the cone in an ellipse whose major axis is $B D$. Let $E$ be the center of this ellipse and $F G$ its minor axis. We have $\\frac{B E}{E D}=\\frac{1}{2}$. Let $E^{\\prime}, F^{\\prime}, G^{\\prime}$ be radial projections of $E, F, G$ from $A$ onto the base of the cone. Then $E$ sits on $B C$. Let $h(X)$ denote the height of a point $X$ with respect to the base of the cone. We have $h(E)=h(D) / 2=h / 3$.  Hence $E F=2 E^{\\prime} F^{\\prime} / 3$. Applying Menelaus's theorem to $\\triangle B H O$ we get $\\frac{O E^{\\prime}}{E^{\\prime} B}=\\frac{B E}{E H} \\cdot \\frac{H A}{A O}=1$. Hence $E F=\\frac{2}{3} \\frac{\\sqrt{3}}{2}=\\frac{1}{\\sqrt{3}}$. Let $d$ denote the distance from $A$ to the plane. Let $V_{1}$ and $V$ denote the volume of the cone above the plane (on the same side of the plane as $A$ ) and the total volume of the cone. We have  $$ \\begin{aligned} \\frac{V_{1}}{V} & =\\frac{B E \\cdot E F \\cdot d}{h}=\\frac{(2 B H / 3)(1 / \\sqrt{3})\\left(2 S_{A H B} / B H\\right)}{h} \\\\ & =\\frac{(2 / 3)(1 / \\sqrt{3})(h / 2)}{h}=\\frac{1}{3 \\sqrt{3}} . \\end{aligned} $$  Since this ratio is smaller than $1 / 2$, we have indeed selected the correct volume for our ratio.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (IND $\\left.3^{\\prime}\\right)^{\\mathrm{IMO}}$ Given a circle with two chords $A B, C D$ that meet at $E$, let $M$ be a point of chord $A B$ other than $E$. Draw the circle through $D, E$, and $M$. The tangent line to the circle $D E M$ at $E$ meets the lines $B C, A C$ at $F, G$, respectively. Given $\\frac{A M}{A B}=\\lambda$, find $\\frac{G E}{E F}$.", "solution": "11. Assume $\\mathcal{B}(A, E, M, B)$. Since $A, B, C, D$ lie on a circle, we have $\\angle G C E=$ $\\angle M B D$ and $\\angle M A D=\\angle F C E$. Since $F D$ is tangent to the circle around $\\triangle E M D$ at $E$, we have $\\angle M D E=\\angle F E B=\\angle A E G$. Consequently, $\\angle C E F=180^{\\circ}-\\angle C E A-\\angle F E B=180^{\\circ}-\\angle M E D-\\angle M D E=\\angle E M D$ and $\\angle C E G=180^{\\circ}-\\angle C E F=180^{\\circ}-\\angle E M D=\\angle D M B$. It follows that $\\triangle C E F \\sim \\triangle A M D$ and $\\triangle C E G \\sim$ $\\triangle B M D$. From the first similarity we obtain $C E \\cdot M D=A M \\cdot E F$, and from the second we obtain $C E$. $M D=B M \\cdot E G$. Hence  $$ \\begin{gathered} A M \\cdot E F=B M \\cdot E G \\Longrightarrow \\\\ \\frac{G E}{E F}=\\frac{A M}{B M}=\\frac{\\lambda}{1-\\lambda} . \\end{gathered} $$  ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-543.jpg?height=371&width=353&top_left_y=1023&top_left_x=894)  If $\\mathcal{B}(A, M, E, B)$, interchanging the roles of $A$ and $B$ we similarly obtain $\\frac{G E}{E F}=\\frac{\\lambda}{1-\\lambda}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (IRE 1) Let $A B C$ be a triangle and $L$ the line through $C$ parallel to the side $A B$. Let the internal bisector of the angle at $A$ meet the side $B C$ at $D$ and the line $L$ at $E$ and let the internal bisector of the angle at $B$ meet the side $A C$ at $F$ and the line $L$ at $G$. If $G F=D E$, prove that $A C=B C$.", "solution": "12. Let $d(X, l)$ denote the distance of a point $X$ from a line $l$. Using the elementary facts that $A F: F C=c: a$ and $B D: D C=c: b$, we obtain $d(F, L)=\\frac{a}{a+c} h_{c}$ and $d(D, L)=\\frac{b}{b+c} h_{c}$, where $h_{a}$ is the altitude of $\\triangle A B C$ from $A$. We also have $\\angle F G C=\\beta / 2, \\angle D E C=\\alpha / 2$. It follows that  $$ D E=\\frac{d(D, L)}{\\sin (\\alpha / 2)} \\quad \\text { and } \\quad F G=\\frac{d(F, L)}{\\sin (\\beta / 2)} $$  Now suppose that $a>b$. Since the function $f(x)=\\frac{x}{x+c}$ is strictly increasing, we deduce $d(F, L)>d(D, L)$. Furthermore, $\\sin (\\alpha / 2)>\\sin (\\beta / 2)$, so we get from (1) that $F G>D E$. Similarly, $a<b$ implies $F G<D E$. Hence we must have $a=b$, i.e., $A C=B C$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (IRE 2) An eccentric mathematician has a ladder with $n$ rungs that he always ascends and descends in the following way: When he ascends, each step he takes covers $a$ rungs of the ladder, and when he descends, each step he takes covers $b$ rungs of the ladder, where $a$ and $b$ are fixed positive integers. By a sequence of ascending and descending steps he can climb from ground level to the top rung of the ladder and come back down to ground level again. Find, with proof, the minimum value of $n$, expressed in terms of $a$ and $b$.", "solution": "13. We will call the ground the \"zeroth\" rung. We will prove that the minimum $n$ is $n=a+b-(a, b)$. It is plain that if $(a, b)=k>1$, the scientist can climb only onto the rungs divisible by $k$ and we can just observe these rungs to obtain the situation equivalent to $a^{\\prime}=a / k, b^{\\prime}=b / k$, and $n^{\\prime}=a^{\\prime}+b^{\\prime}-1$. Thus let us assume that $(a, b)=1$ and show that $n=a+b-1$. We obviously have $n>a$. Consider $n=a+b-k, k \\geq 1$, and let us assume without loss of generality that $a>b$ (otherwise, we can reverse the problem starting from the top rung in our round trip). Then we can uniquely define the numbers $r_{i}, 0 \\leq r_{i}<b$, by $r_{i} \\equiv i a(\\bmod b)$. We now describe the only possible sequence of moves. From a position $0 \\leq p \\leq b-k$ we can move only $a$ rungs upward and for $p>b-1$ we can move only $b$ rungs downward. If we end up at $b-k<p \\leq b-1$, we are stuck. Hence, given that we are at $r_{i}$, if $r_{i} \\leq b-k$, we can move to $a+r_{i}$, and when we descend as far as we can go we will end up at $r_{i+1} \\equiv a+r_{i}(\\bmod b)$. If the mathematician climbs to the highest rung and then comes back to $r_{i}=0$, then we deduce $b \\mid i a$, so $i \\geq b$. But since $(a, b)=1$, there exists $0<j<b$ such that $r_{j} \\equiv j a \\equiv b-1(\\bmod b)$. Thus the mathematician has visited the position $b-1$. For him not to get stuck we must have $k \\leq 1$ and $n \\geq a+b-1$. For $n=a+b-1$ by induction he can come to any position $r_{i}, i \\geq 0$, so he eventually comes to $r_{j}=b-1$, climbs to the highest rung, and then continues until he gets to $r_{b}=0$. Hence the answer to the problem is $n=a+b-1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (JAP 2) In the coordinate plane a rectangle with vertices $(0,0),(m, 0)$, $(0, n),(m, n)$ is given where both $m$ and $n$ are odd integers. The rectangle is partitioned into triangles in such a way that (i) each triangle in the partition has at least one side (to be called a \"good\" side) that lies on a line of the form $x=j$ or $y=k$, where $j$ and $k$ are integers, and the altitude on this side has length 1 ; (ii) each \"bad\" side (i.e., a side of any triangle in the partition that is not a \"good\" one) is a common side of two triangles in the partition. Prove that there exist at least two triangles in the partition each of which has two good sides.", "solution": "14. Let $V$ be the set of all midpoints of bad sides, and $E$ the set of segments connecting two points in $V$ that belong to the same triangle. Each edge in $E$ is parallel to exactly one good side and thus is parallel to the coordinate grid and has half-integer coordinates. Thus, the edges of $E$ are a subset of the grid formed by joining the centers of the squares in the original grid to each other. Let $G$ be a graph whose set of vertices is $V$ and set of edges is $E$. The degree of each vertex $X$, denoted by $d(X)$, is 0 , 1 , or 2 . We observe the following cases: (i) $d(X)=0$ for some $X$. Then both triangles containing $X$ have two good sides. (ii) $d(X)=1$ for some $X$. Since $\\sum_{X \\in V} d(X)=2|E|$ is even, it follows that at least another vertex $Y$ has the degree 1. Hence both $X$ and $Y$ belong to triangles having two good sides. (iii) $d(X)=2$ for all $X \\in V$. We will show that this case cannot occur. We prove first that centers of all the squares of the $m \\times n$ board belong to $V \\cup E$. A bad side contains no points with half-integer coordinates in its interior other than its midpoint. Therefore either a point $X$ is in $V$, or it lies on the segment connecting the midpoints of the two bad sides. Evidently, the graph $G$ can be partitioned into disjoint cycles. Each center of a square is passed exactly once in exactly one cycle. Let us color the board black and white in a standard chessboard fashion. Each cycle passes through centers that must alternate in color, and hence it contains an equal number of black and white centers. Consequently, the numbers of black and white squares on the entire board must be equal, contradicting the condition that $m$ and $n$ are odd. Our proof is thus completed.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (MEX 2) Determine for which positive integers $k$ the set  $$ X=\\{1990,1990+1,1990+2, \\ldots, 1990+k\\} $$  can be partitioned into two disjoint subsets $A$ and $B$ such that the sum of the elements of $A$ is equal to the sum of the elements of $B$.", "solution": "15. Let $S(Z)$ denote the sum of all the elements of a set $Z$. We have $S(X)=$ $(k+1) \\cdot 1990+\\frac{k(k+1)}{2}$. To partition the set into two parts with equal sums, $S(X)$ must be even and hence $\\frac{k(k+1)}{2}$ must be even. Hence $k$ is of the form $4 r$ or $4 r+3$, where $r$ is an integer. For $k=4 r+3$ we can partition $X$ into consecutive fourtuplets $\\{1990+$ $4 l, 1990+4 l+1,1990+4 l+2,1990+4 l+3\\}$ for $0 \\leq l \\leq r$ and put $1990+4 l, 1990+4 l+3 \\in A$ and $1990+4 l+1,1990+4 l+2 \\in B$ for all $l$. This would give us $S(A)=S(B)=(3980+4 r+3)(r+1)$. For $k=4 r$ the numbers of elements in $A$ and $B$ must differ. Let us assume without loss of generality $|A|<|B|$. Then $S(A) \\leq(1990+2 r+1)+(1990+$ $2 r+2)+\\cdots+(1990+4 r)$ and $S(B) \\geq 1990+1991+\\cdots+(1990+2 r)$. Plugging these inequalities into the condition $S(A)=S(B)$ gives us $r \\geq 23$ and consequently $k \\geq 92$. We note that $B=\\{1990,1991, \\ldots, 2034,2052,2082\\}$ and $A=\\{2035,2036, \\ldots, 2051,2053, \\ldots, 2081\\}$ is a partition for $k=92$ that satisfies $S(A)=S(B)$. To construct a partition out of higher $k=4 r$ we use the $k=92$ partition for the first 93 elements and construct for the remaining elements as was done for $k=4 r+3$. Hence we can construct a partition exactly for the integers $k$ of the form $k=4 r+3, r \\geq 0$, and $k=4 r, r \\geq 23$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (NET 1) ${ }^{\\text {IMO6 }}$ Is there a 1990-gon with the following properties (i) and (ii)? (i) All angles are equal; (ii) The lengths of the 1990 sides are a permutation of the numbers $1^{2}, 2^{2}, \\ldots, 1989^{2}, 1990^{2}$.", "solution": "16. Let $A_{0} A_{1} \\ldots A_{1989}$ be the desired 1990-gon. We also define $A_{1990}=A_{0}$. Let $O$ be an arbitrary point. For $1 \\leq i \\leq 1990$ let $B_{i}$ be a point such that $\\overrightarrow{O B_{i}}=\\overrightarrow{A_{i-1} A_{i}}$. We define $B_{0}=B_{1990}$. The points $B_{i}$ must satisfy the following properties: $\\angle B_{i} O B_{i+1}=\\frac{2 \\pi}{1990}, 0 \\leq i \\leq 1989$, lengths of $O B_{i}$ are a permutation of $1^{2}, 2^{2}, \\ldots, 1989^{2}, 1990^{2}$, and $\\sum_{i=0}^{1989} \\overrightarrow{O B_{i}}=\\overrightarrow{0}$. Conversely, any such set of points $B_{i}$ corresponds to a desired 1990-gon. Hence, our goal is to construct vectors $\\overrightarrow{O B_{i}}$ satisfying all the stated properties. Let us group vectors of lengths $(2 n-1)^{2}$ and $(2 n)^{2}$ into pairs and put them diametrically opposite each other. The length of the resulting vectors is $4 n-1$. The problem thus reduces to arranging vectors of lengths $3,7,11, \\ldots, 3979$ at mutual angles of $\\frac{2 \\pi}{995}$ such that their sum is $\\overrightarrow{0}$. We partition the 995 directions into 199 sets of five directions at mutual angles $\\frac{2 \\pi}{5}$. The directions when intersected with a unit circle form a regular pentagon. We group the set of lengths of vectors $3,7, \\ldots, 3979$ into 199 sets of five consecutive elements of the set. We place each group of lengths on directions belonging to the same group of directions, thus constructing five vectors. We use that $\\overrightarrow{O C_{1}}+\\cdots+\\overrightarrow{O C_{n}}=0$ where $O$ is the center of a regular $n$-gon $C_{1} \\ldots C_{n}$. In other words, vectors of equal lengths along directions that form a regular $n$-gon cancel each other out. Such are the groups of five directions. Hence, we can assume for each group of five lengths for its lengths to be $\\{0,4,8,12,16\\}$. We place these five lengths in a random fashion on a single group of directions. We then rotate the configuration clockwise by $\\frac{2 \\pi}{199}$ to cover other groups of directions and repeat until all groups of directions are exhausted. It follows that all vectors of each of the lengths $\\{0,4,8,12,16\\}$ will form a regular 199-gon and will thus cancel each other out. We have thus constructed a way of obtaining points $B_{i}$ and have hence shown the existence of the 1990-gon satisfying $(i)$ and (ii).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (NET 3) Unit cubes are made into beads by drilling a hole through them along a diagonal. The beads are put on a string in such a way that they can move freely in space under the restriction that the vertices of two neighboring cubes are touching. Let $A$ be the beginning vertex and $B$ be the end vertex. Let there be $p \\times q \\times r$ cubes on the string $(p, q, r \\geq 1)$. (a) Determine for which values of $p, q$, and $r$ it is possible to build a block with dimensions $p, q$, and $r$. Give reasons for your answers. (b) The same question as (a) with the extra condition that $A=B$.", "solution": "17. Let us set a coordinate system denoting the vertices of the block. The vertices of the unit cubes of the block can be described as $\\{(x, y, z) \\mid 0 \\leq$ $x \\leq p, 0 \\leq y \\leq q, 0 \\leq z \\leq r\\}$, and we restrict our attention to only these points. Suppose the point $A$ is fixed at $(a, b, c)$. Then for every other necklace point $(x, y, z)$ numbers $x-a, y-b$, and $z-c$ must be of equal parity. Conversely, every point $(x, y, z)$ such that $x-a, y-b$, and $z-c$ are of the same parity has to be a necklace point. Consider the graph $G$ whose vertices are all such points and edges are all diagonals of the unit cubes through these points. In part (a) we are looking for an open or closed Euler path, while in part (b) we are looking for a closed Euler path. Necklace points in the interior of the $(p, q, r)$ box have degree 8, points on the surface have degree 4 , points on the edge have degree 2 , and points on the corner have degree 1. A closed Euler path can be formed if and only if all vertices are of an even degree, while an open Euler path can be formed if and only if exactly two vertices have an odd degree. Hence the problem in part (a) amounts to being able to choose a point $A$ such that 0 or 2 corner vertices are necklace vertices, whereas in part (b) no corner points can be necklace vertices. We distinguish two cases. (i) At least two of $p, q, r$, say $p, q$, are even. We can choose $a=1, b=c=$ 0 . In this case none of the corners is a necklace point. Hence a closed Euler path exists. (ii) At most one of $p, q, r$ is even. However one chooses $A$, exactly two necklace points are at the corners. Hence, an open Euler path exists, but it is impossible to form a closed path. Hence, in part (a), a box can be made of all ( $p, q, r)$ and in part (b) only those $(p, q, r)$ where at least two of the numbers are even.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (NOR) Let $a, b$ be natural numbers with $1 \\leq a \\leq b$, and $M=\\left[\\frac{a+b}{2}\\right]$. Define the function $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ by  $$ f(n)= \\begin{cases}n+a, & \\text { if } n<M \\\\ n-b, & \\text { if } n \\geq M\\end{cases} $$  Let $f^{1}(n)=f(n), f^{i+1}(n)=f\\left(f^{i}(n)\\right), i=1,2, \\ldots$. Find the smallest natural number $k$ such that $f^{k}(0)=0$.", "solution": "18. Clearly, it suffices to consider the case $(a, b)=1$. Let $S$ be the set of integers such that $M-b \\leq x \\leq M+a-1$. Then $f(S) \\subseteq S$ and $0 \\in S$. Consequently, $f^{k}(0) \\in S$. Let us assume for $k>0$ that $f^{k}(0)=0$. Since $f(m)=m+a$ or $f(m)=m-b$, it follows that $k$ can be written as $k=r+s$, where $r a-s b=0$. Since $a$ and $b$ are relatively prime, it follows that $k \\geq a+b$. Let us now prove that $f^{a+b}(0)=0$. In this case $a+b=r+s$ and hence $f^{a+b}(0)=(a+b-s) a-s b=(a+b)(a-s)$. Since $a+b \\mid f^{a+b}(0)$ and $f^{a+b}(0) \\in S$, it follows that $f^{a+b}(0)=0$. Thus for $(a, b)=1$ it follows that $k=a+b$. For other $a$ and $b$ we have $k=\\frac{a+b}{(a, b)}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (POL 1) Let $P$ be a point inside a regular tetrahedron $T$ of unit volume. The four planes passing through $P$ and parallel to the faces of $T$ partition $T$ into 14 pieces. Let $f(P)$ be the joint volume of those pieces that are neither a tetrahedron nor a parallelepiped (i.e., pieces adjacent to an edge but not to a vertex). Find the exact bounds for $f(P)$ as $P$ varies over $T$.", "solution": "19. Let $d_{1}, d_{2}, d_{3}, d_{4}$ be the distances of the point $P$ to the tetrahedron. Let $d$ be the height of the regular tetrahedron. Let $x_{i}=d_{i} / d$. Clearly, $x_{1}+$ $x_{2}+x_{3}+x_{4}=1$, and given this condition, the parameters vary freely as we vary $P$ within the tetrahedron. The four tetrahedra have volumes $x_{1}^{3}, x_{2}^{3}, x_{3}^{3}$, and $x_{4}^{3}$, and the four parallelepipeds have volumes of $6 x_{2} x_{3} x_{4}$, $6 x_{1} x_{3} x_{4}, 6 x_{1} x_{2} x_{4}$, and $6 x_{1} x_{2} x_{3}$. Hence, using $x_{1}+x_{2}+x_{3}+x_{4}=1$ and setting $g(x)=x^{2}(1-x)$, we directly verify that  $$ \\begin{aligned} f(P) & =f\\left(x_{1}, x_{2}, x_{3}, x_{4}\\right)=1-\\sum_{i=1}^{4} x_{i}^{3}-6 \\sum_{1 \\leq i<j<k \\leq 4} x_{i} x_{j} x_{k} \\\\ & =3\\left(g\\left(x_{1}\\right)+g\\left(x_{2}\\right)+g\\left(x_{3}\\right)+g\\left(x_{4}\\right)\\right) . \\end{aligned} $$  We note that $g(0)=0$ and $g(1)=0$. Hence, as $x_{1}$ tends to 1 and other variables tend to $0, f\\left(x_{1}, x_{2}, x_{3}, x_{4}\\right)=0$. Thus $f(P)$ is sharply bounded downwards at 0 . We now find an upper bound. We note that  $$ \\begin{aligned} g\\left(x_{i}+x_{j}\\right) & =\\left(x_{i}+x_{j}\\right)^{2}\\left(1-x_{1}-x_{2}\\right) \\\\ & =g\\left(x_{i}\\right)+g\\left(x_{j}\\right)+2 x_{i} x_{j}\\left(1-\\frac{3}{2}\\left(x_{i}+x_{j}\\right)\\right) \\end{aligned} $$  thus for $x_{i}+x_{j} \\leq 2 / 3$ and $x_{i}, x_{j}>0$ we have $g\\left(x_{i}+x_{j}\\right)+g(0) \\geq$ $g\\left(x_{i}\\right)+g\\left(x_{j}\\right)$. Equality holds only when $x_{i}+x_{j}=2 / 3$. Assuming without loss of generality $x_{1} \\geq x_{2} \\geq x_{3} \\geq x_{4}$, we have $g\\left(x_{1}\\right)+$ $g\\left(x_{2}\\right)+g\\left(x_{3}\\right)+g\\left(x_{4}\\right)<g\\left(x_{1}\\right)+g\\left(x_{2}\\right)+g\\left(x_{3}+x_{4}\\right)$. Assuming $y_{1}+y_{2}+y_{3}=1$ and $y_{1} \\geq y_{2} \\geq y_{3}$, we have $g\\left(y_{1}\\right)+g\\left(y_{2}\\right)+g\\left(y_{3}\\right) \\leq g\\left(y_{1}\\right)+g\\left(y_{2}+y_{3}\\right)$. Hence $g\\left(x_{1}\\right)+g\\left(x_{2}\\right)+g\\left(x_{3}\\right)+g\\left(x_{4}\\right)<g(x)+g(1-x)$ for some $x$. We also have $g(x)+g(1-x)=x(1-x) \\leq 1 / 4$. Hence $f(P) \\leq 3 / 4$. Equality holds for $x_{1}=x_{2}=1 / 2, x_{3}=x_{4}=0$ (corresponding to the midpoint of an edge), and as the variables converge to these values, $f(P)$ converges to $3 / 4$. Hence the bounds for $f(P)$ are  $$ 0<f(P)<\\frac{3}{4} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (CAN 1) Given $n$ countries with three representatives each, $m$ committees $A(1), A(2), \\ldots A(m)$ are called $a$ cycle if (i) each committee has $n$ members, one from each country; (ii) no two committees have the same membership; (iii) for $i=1,2, \\ldots, m$, committee $A(i)$ and committee $A(i+1)$ have no member in common, where $A(m+1)$ denotes $A(1)$; (iv) if $1<|i-j|<m-1$, then committees $A(i)$ and $A(j)$ have at least one member in common. Is it possible to have a cycle of 1990 committees with 11 countries?", "solution": "2. We will call a cycle with $m$ committees and $n$ countries an $(m, n)$ cycle. We will number the delegates from each country with numbers $1,2,3$ and denote committees by arrays of these integers (of length $n$ ) defining which of the delegates from each country is in the committee. We will first devise methods of constructing larger cycles out of smaller cycles. Let $A_{1}, \\ldots, A_{m}$ be an $(m, n)$ cycle, where $m$ is odd. Then the following is a $(2 m, n+1)$ cycle:  $$ \\left(A_{1}, 1\\right),\\left(A_{2}, 2\\right), \\ldots,\\left(A_{m}, 1\\right),\\left(A_{1}, 2\\right),\\left(A_{2}, 1\\right), \\ldots,\\left(A_{m}, 2\\right) $$  Also, let $A_{1}, \\ldots, A_{m}$ be an $(m, n)$ cycle and $k \\leq m$ an even integer. Then the cycle  $$ \\begin{gathered} \\left(A_{1}, 3\\right),\\left(A_{2}, 1\\right),\\left(A_{3}, 2\\right), \\ldots,\\left(A_{k-2}, 1\\right),\\left(A_{k-1}, 2\\right) \\\\ \\left(A_{k}, 3\\right),\\left(A_{k-1}, 1\\right),\\left(A_{k-2}, 2\\right), \\ldots,\\left(A_{2}, 2\\right) \\end{gathered} $$  is a $(2(k-1), n+1)$ cycle. Starting from the $((1),(2),(3))$ cycle with parameters $(3,1)$ we can sequentially construct larger cycles using the shown methods. The obtained cycles have parameters as follows:  $$ (6,2),(10,3), \\ldots,\\left(2^{k}+2, k\\right), \\ldots,(1026,10),(1990,11) $$  Thus there exists a cycle of 1990 committees with 11 countries.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. (POL 3) Prove that every integer $k$ greater than 1 has a multiple that is less than $k^{4}$ and can be written in the decimal system with at most four different digits.", "solution": "20. Let $n$ be the unique integer such that $2^{n-1} \\leq k<2^{n}$. Let $S(n)$ be the set of numbers less than $10^{n}$ that are written with only the digits $\\{0,1\\}$ in the decimal system. Evidently $|S(n)|=2^{n}>k$ and hence there exist two numbers $x, y \\in S(n)$ such that $k \\mid x-y$. Let us show that $w=|x-y|$ is the desired number. By definition $k \\mid w$. We also have  $$ w<1.2 \\cdot 10^{n-1} \\leq 1.2 \\cdot\\left(2^{3} \\sqrt{2}\\right)^{n-1} \\leq 1.2 \\cdot k^{3} \\sqrt{k} \\leq k^{4} $$  Finally, since $x, y \\in S(n)$, it follows that $w=|x-y|$ can be written using only the digits $\\{0,1,8,9\\}$. This completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. (ROM $\\mathbf{1}^{\\prime}$ ) Let $n$ be a composite natural number and $p$ a proper divisor of $n$. Find the binary representation of the smallest natural number $N$ such that $\\frac{\\left(1+2^{p}+2^{n-p}\\right) N-1}{2^{n}}$ is an integer.", "solution": "21. We must solve the congruence $\\left(1+2^{p}+2^{n-p}\\right) N \\equiv 1\\left(\\bmod 2^{n}\\right)$. Since $(1+$ $2^{p}+2^{n-p}$ ) and $2^{n}$ are coprime, there clearly exists a unique $N$ satisfying this equation and $0<N<2^{n}$. Let us assume $n=m p$. Then we have $\\left(1+2^{p}\\right)\\left(\\sum_{j=0}^{m-1}(-1)^{j} 2^{j p}\\right) \\equiv$ $1\\left(\\bmod 2^{n}\\right)$ and $\\left(1+2^{n-p}\\right)\\left(1-2^{n-p}\\right) \\equiv 1\\left(\\bmod 2^{n}\\right)$. By multiplying the two congruences we obtain  $$ \\left(1+2^{p}\\right)\\left(1+2^{n-p}\\right)\\left(1-2^{n-p}\\right)\\left(\\sum_{j=0}^{m-1}(-1)^{j} 2^{j p}\\right) \\equiv 1\\left(\\bmod 2^{n}\\right) $$  Since $\\left(1+2^{p}\\right)\\left(1+2^{n-p}\\right) \\equiv\\left(1+2^{p}+2^{n-p}\\right)\\left(\\bmod 2^{n}\\right)$, it follows that $N \\equiv$ $\\left(1-2^{n-p}\\right)\\left(\\sum_{j=0}^{m-1}(-1)^{j} 2^{j p}\\right)\\left(\\bmod 2^{n}\\right)$. The integer $N=\\sum_{j=0}^{m-1}(-1)^{j} 2^{j p}-$ $2^{n-p}+2^{n}$ satisfies the congruence and $0<N \\leq 2^{n}$. Using that for $a>b$ we have in binary representation  $$ 2^{a}-2^{b}=\\underbrace{11 \\ldots 11}_{a-b \\text { times }} \\underbrace{00 \\ldots 00}_{b \\text { times }} $$  the binary representation of $N$ is calculated as follows:  $$ N=\\left\\{\\begin{array}{cc} \\underbrace{11 \\ldots 11}_{p \\text { times }} \\underbrace{11 \\ldots 11}_{p \\text { times }} \\underbrace{00 \\ldots 00}_{p \\text { times }} \\ldots \\underbrace{11 \\ldots 11}_{p \\text { times }} \\underbrace{00 \\ldots 00}_{p-1 \\text { times }} 1, & 2 \\nmid \\frac{n}{p} \\\\ \\underbrace{11 \\ldots 11}_{p-1 \\text { times }} \\underbrace{00 \\ldots 00}_{p+1} \\underbrace{11 \\ldots 11}_{p \\text { times }} \\underbrace{00 \\ldots 00}_{p \\text { times }} \\ldots \\underbrace{11 \\ldots 11}_{p \\text { times }} \\underbrace{00 \\ldots 00}_{p-1 \\text { times }} 1, & 2 \\left\\lvert\\, \\frac{n}{p}\\right. \\end{array}\\right. $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "22", "problem_type": null, "exam": "IMO", "problem": "22. (ROM 4) Ten localities are served by two international airlines such that there exists a direct service (without stops) between any two of these localities and all airline schedules offer round-trip service between the cities they serve. Prove that at least one of the airlines can offer two disjoint round trips each containing an odd number of landings.", "solution": "22. We can assume without loss of generality that each connection is serviced by only one airline and the problem reduces to finding two disjoint monochromatic cycles of the same color and of odd length on a complete graph of 10 points colored by two colors. We use the following two standard lemmas: Lemma 1. Given a complete graph on six points whose edges are colored with two colors there exists a monochromatic triangle. $\\operatorname{Proof}$. Let us denote the vertices by $c_{1}, c_{2}, c_{3}, c_{4}, c_{5}, c_{6}$. By the pigeonhole principle at least three vertices out of $c_{1}$, say $c_{2}, c_{3}, c_{4}$, are of the same color, let us call it red. Assuming that at least one of the edges connecting points $c_{2}, c_{3}, c_{4}$ is red, the connected points along with $c_{1}$ form a red triangle. Otherwise, edges connecting $c_{2}, c_{3}, c_{4}$ are all of the opposite color, let us call it blue, and hence in all cases we have a monochromatic triangle. Lemma 2. Given a complete graph on five points whose edges are colored two colors there exists a monochromatic triangle or a monochromatic cycle of length five. Proof. Let us denote the vertices by $c_{1}, c_{2}, c_{3}, c_{4}, c_{5}$. Assume that out of a point $c_{i}$ three vertices are of the same color. We can then proceed as in Lemma 1 to obtain a monochromatic triangle. Otherwise, each point is connected to other points with exactly two red and two blue vertices. Hence, we obtain monochromatic cycles starting from a single point and moving along the edges of the same color. Since each cycle must be of length at least three (i.e., we cannot have more than one cycle of one color), it follows that for both red and blue we must have one cycle of length five of that color. We now apply the lemmas. Let us denote the vertices by $c_{1}, c_{2}, \\ldots, c_{10}$. We apply Lemma 1 to vertices $c_{1}, \\ldots, c_{6}$ to obtain a monochromatic triangle. Out of the seven remaining vertices we select 6 and again apply Lemma 1 to obtain another monochromatic triangle. If they are of the same color, we are done. Otherwise, out of the nine edges connecting the two triangles of opposite color at least 5 are of the same color, we can assume blue w.l.o.g., and hence a vertex of a red triangle must contain at least two blue edges whose endpoints are connected with a blue edge. Hence there exist two triangles of different colors joined at a vertex. These take up five points. Applying Lemma 2 on the five remaining points, we obtain a monochromatic cycle of odd length that is of the same color as one of the two joined triangles and disjoint from both of them.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "23", "problem_type": null, "exam": "IMO", "problem": "23. (ROM 5) ${ }^{\\mathrm{IMO} 3}$ Find all positive integers $n$ having the property that $\\frac{2^{n}+1}{n^{2}}$ is an integer.", "solution": "23. Let us assume $n>1$. Obviously $n$ is odd. Let $p \\geq 3$ be the smallest prime divisor of $n$. In this case $(p-1, n)=1$. Since $2^{n}+1 \\mid 2^{2 n}-1$, we have that $p \\mid 2^{2 n}-1$. Thus it follows from Fermat's little theorem and elementary number theory that $p \\mid\\left(2^{2 n}-1,2^{p-1}-1\\right)=2^{(2 n, p-1)}-1$. Since $(2 n, p-1) \\leq 2$, it follows that $p \\mid 3$ and hence $p=3$. Let us assume now that $n$ is of the form $n=3^{k} d$, where $2,3 \\nmid d$. We first prove that $k=1$. Lemma. If $2^{m}-1$ is divisible by $3^{r}$, then $m$ is divisible by $3^{r-1}$. Proof. This is the lemma from (SL97-14) with $p=3, a=2^{2}, k=m$, $\\alpha=1$, and $\\beta=r$. Since $3^{2 k}$ divides $n^{2} \\mid 2^{2 n}-1$, we can apply the lemma to $m=2 n$ and $r=2 k$ to conclude that $3^{2 k-1} \\mid n=3^{k} d$. Hence $k=1$. Finally, let us assume $d>1$ and let $q$ be the smallest prime factor of $d$. Obviously $q$ is odd, $q \\geq 5$, and $(n, q-1) \\in\\{1,3\\}$. We then have $q \\mid 2^{2 n}-1$ and $q \\mid 2^{q-1}-1$. Consequently, $q \\mid 2^{(2 n, q-1)}-1=2^{2(n, q-1)}-1$, which divides $2^{6}-1=63=3^{2} \\cdot 7$, so we must have $q=7$. However, in that case we obtain $7|n| 2^{n}+1$, which is a contradiction, since powers of two can only be congruent to 1,2 and 4 modulo 7 . It thus follows that $d=1$ and $n=3$. Hence $n>1 \\Rightarrow n=3$. It is easily verified that $n=1$ and $n=3$ are indeed solutions. Hence these are the only solutions.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "24", "problem_type": null, "exam": "IMO", "problem": "24. (THA 2) Let $a, b, c, d$ be nonnegative real numbers such that $a b+b c+$ $c d+d a=1$. Show that  $$ \\frac{a^{3}}{b+c+d}+\\frac{b^{3}}{a+c+d}+\\frac{c^{3}}{a+b+d}+\\frac{d^{3}}{a+b+c} \\geq \\frac{1}{3} $$  ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-262.jpg?height=62&width=1190&top_left_y=791&top_left_x=173) a function $f: \\mathbb{Q}^{+} \\rightarrow \\mathbb{Q}^{+}$such that  $$ f(x f(y))=\\frac{f(x)}{y}, \\quad \\text { for all } x, y \\text { in } \\mathbb{Q}^{+} $$", "solution": "24. Let us denote $A=b+c+d, B=a+c+d, C=a+b+d, D=a+b+c$. Since $a b+b c+c d+d a=1$ the numbers $A, B, C, D$ are all positive. By trivially applying the AM-GM inequality we have:  $$ a^{2}+b^{2}+c^{2}+d^{2} \\geq a b+b c+c d+d a=1 . $$  We will prove the inequality assuming only that $A, B, C, D$ are positive and $a^{2}+b^{2}+c^{2}+d^{2} \\geq 1$. In this case we may assume without loss of generality that $a \\geq b \\geq c \\geq d \\geq 0$. Hence $a^{3} \\geq b^{3} \\geq c^{3} \\geq d^{3} \\geq 0$ and $\\frac{1}{A} \\geq \\frac{1}{B} \\geq \\frac{1}{C} \\geq \\frac{1}{D}>0$. Using the Chebyshev and Cauchy inequalities we obtain:  $$ \\begin{aligned} & \\frac{a^{3}}{A}+\\frac{b^{3}}{B}+\\frac{c^{3}}{C}+\\frac{d^{3}}{D} \\\\ & \\geq \\frac{1}{4}\\left(a^{3}+b^{3}+c^{3}+d^{3}\\right)\\left(\\frac{1}{A}+\\frac{1}{B}+\\frac{1}{C}+\\frac{1}{D}\\right) \\\\ & \\geq \\frac{1}{16}\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)(a+b+c+d)\\left(\\frac{1}{A}+\\frac{1}{B}+\\frac{1}{C}+\\frac{1}{D}\\right) \\\\ & =\\frac{1}{48}\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)(A+B+C+D)\\left(\\frac{1}{A}+\\frac{1}{B}+\\frac{1}{C}+\\frac{1}{D}\\right) \\geq \\frac{1}{3} \\end{aligned} $$  This completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "26", "problem_type": null, "exam": "IMO", "problem": "26. (USA 2) Let $P$ be a cubic polynomial with rational coefficients, and let $q_{1}, q_{2}, q_{3}, \\ldots$ be a sequence of rational numbers such that $q_{n}=P\\left(q_{n+1}\\right)$ for all $n \\geq 1$. Prove that there exists $k \\geq 1$ such that for all $n \\geq 1, q_{n+k}=q_{n}$.", "solution": "26. We note that $|P(x) / x| \\rightarrow \\infty$. Hence, there exists an integer number $M$ such that $M>\\left|q_{1}\\right|$ and $|P(x)| \\leq|x| \\Rightarrow|x|<M$. It follows that $\\left|q_{i}\\right|<M$ for all $i \\in \\mathbb{N}$ because assuming $\\left|q_{i}\\right| \\geq M$ for some $i$ we get $\\left|q_{i-1}\\right|=$ $\\left|P\\left(q_{i}\\right)\\right|>\\left|q_{i}\\right| \\geq M$ and this ultimately contradicts $\\left|q_{1}\\right|<M$. Let us define $q_{1}=\\frac{r}{s}$ and $P(x)=\\frac{a x^{3}+b x^{2}+c x+d}{e}$ where $r, s, a, b, c, d, e$ are all integers. For $N=s a$ we shall prove by induction that $N q_{i}$ is an integer for all $i \\in \\mathbb{N}$. By definition $N \\neq 0$.  For $i=1$ this obviously holds. Assume it holds for some $i \\in \\mathbb{N}$. Then using $q_{i}=P\\left(q_{i+1}\\right)$ we have that $N q_{i+1}$ is a zero of the polynomial  $$ \\begin{aligned} Q(x) & =\\frac{e}{a} N^{3}\\left(P\\left(\\frac{x}{N}\\right)-q_{i}\\right) \\\\ & =x^{3}+(s b) x^{2}+\\left(s^{2} a c\\right) x+\\left(s^{3} a^{2} d-s^{2} a e\\left(N q_{i}\\right)\\right) . \\end{aligned} $$  Since $Q(x)$ is a monic polynomial with integer coefficients (a conclusion for which we must assume the induction hypothesis) and $N q_{i+1}$ is rational it follows by the rational root theorem that $N q_{i+1}$ is an integer. It follows that all $q_{i}$ are multiples of $1 / N$. Since $-M<q_{i}<M$ we conclude that $q_{i}$ can take less than $T=2 M|N|$ distinct values. Therefore for each $j$ there are $m_{j}$ and $m_{j}+k_{j}\\left(k_{j}>0\\right)$ both belonging to the set $\\{j T+1, j T+2, \\ldots, j T+T\\}$ such that $q_{m_{j}}=q_{m_{j}+k_{j}}$. Since $k_{j}<T$ for all $k_{j}$ it follows that there exists a positive integer $k$ which appears an infinite number of times in the sequence $k_{j}$, i.e. there exist infinitely many integers $m$ such that $q_{m}=q_{m+k}$. Moreover, $q_{m}=q_{m+k}$ clearly implies $q_{n}=q_{n+k}$ for all $n \\leq m$. Hence $q_{n}=q_{n+k}$ holds for all $n$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "27", "problem_type": null, "exam": "IMO", "problem": "27. (USS 1) Find all natural numbers $n$ for which every natural number whose decimal representation has $n-1$ digits 1 and one digit 7 is prime.", "solution": "27. Let us denote by $A_{n}(k)$ the $n$-digit number which consists of $n-1$ ones and one digit seven in the $k+1$-th rightmost position $(0 \\leq k<n)$. Then $A_{n}(k)=\\left(10^{n}+54 \\cdot 10^{k}-1\\right) / 9$. We note that if $3 \\mid n$ we have that $3 \\mid A_{n}(k)$ for all $k$. Hence $n$ cannot be divisible by 3 . Now let $3 \\nmid n$. We claim that for each such $n \\geq 5$, there exists $k<n$ for which $7 \\mid A_{n}(k)$. We see that $A_{n}(k)$ is divisible by 7 if and only if $10^{n}-1 \\equiv 2 \\cdot 10^{k}(\\bmod 7)$. There are several cases.  $$ \\begin{aligned} & n \\equiv 1(\\bmod 6) \\text {. Then } 10^{n}-1 \\equiv 2 \\equiv 2 \\cdot 10^{0} \\text {, so } 7 \\mid A_{n}(0) \\text {. } \\\\ & n \\equiv 2(\\bmod 6) \\text {. Then } 10^{n}-1 \\equiv 1 \\equiv 2 \\cdot 10^{4} \\text {, so } 7 \\mid A_{n}(4) . \\\\ & n \\equiv 4(\\bmod 6) . \\text { Then } 10^{n}-1 \\equiv 3 \\equiv 2 \\cdot 10^{5} \\text {, so } 7 \\mid A_{n}(5) \\text {. } \\\\ & n \\equiv 5(\\bmod 6) \\text {. Then } 10^{n}-1 \\equiv 4 \\equiv 2 \\cdot 10^{2} \\text {, so } 7 \\mid A_{n}(2) . \\end{aligned} $$  The remaining cases are $n=1,2,4$. For $n=4$ the number $1711=29 \\cdot 59$ is composite, while it is easily checked that $n=1$ and $n=2$ are solutions. Hence the answer is $n=1,2$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "28", "problem_type": null, "exam": "IMO", "problem": "28. (USS 3) Prove that on the coordinate plane it is impossible to draw a closed broken line such that (i) the coordinates of each vertex are rational; (ii) the length each of its edges is 1 ; (iii) the line has an odd number of vertices.", "solution": "28. Let us first prove the following lemma.  Lemma. Let $\\left(b^{\\prime} / a^{\\prime}, d^{\\prime} / c^{\\prime}\\right)$ and $\\left(b^{\\prime \\prime} / a^{\\prime \\prime}, d^{\\prime \\prime} / c^{\\prime \\prime}\\right)$ be two points with rational coordinates where the fractions given are irreducible. If both $a^{\\prime}$ and $c^{\\prime}$ are odd and the distance between the two points is 1 then it follows that $a^{\\prime \\prime}$ and $c^{\\prime \\prime}$ are odd, and that $b^{\\prime}+d^{\\prime}$ and $b^{\\prime \\prime}+d^{\\prime \\prime}$ are of a different parity. Proof. Let $b / a$ and $d / c$ be irreducible fractions such that $b^{\\prime} / a^{\\prime}-b^{\\prime \\prime} / a^{\\prime \\prime}=$ $b / a$ and $d^{\\prime} / c^{\\prime}-d^{\\prime \\prime} / c^{\\prime \\prime}=d / c$. Then it follows that $b^{2} / a^{2}+d^{2} / c^{2}=$ $1 \\Rightarrow b^{2} c^{2}+a^{2} d^{2}=a^{2} c^{2}$. Since $(a, b)=1$ and $(c, d)=1$ it follows that $a|c, c| a$ and hence $a=c$. Consequently $b^{2}+d^{2}=a^{2}$. Since $a$ is mutually co-prime to $b$ and $d$ it follows that $a$ and $b+d$ are odd. From $b^{\\prime \\prime} / a^{\\prime \\prime}=b / a+b^{\\prime} / a^{\\prime}$ we get that $a^{\\prime \\prime} \\mid a a^{\\prime}$, so $a^{\\prime \\prime}$ is odd. Similarly, $c^{\\prime \\prime}$ is odd as well. Now it follows that $b^{\\prime \\prime} \\equiv b+b^{\\prime}$ and similarly $d^{\\prime \\prime} \\equiv d+d^{\\prime}$ $(\\bmod 2)$. Hence $b^{\\prime \\prime}+d^{\\prime \\prime} \\equiv b^{\\prime}+d^{\\prime}+b+d \\equiv b^{\\prime}+d^{\\prime}+1(\\bmod 2)$, from which it follows that $b^{\\prime}+d^{\\prime}$ and $b^{\\prime \\prime}+d^{\\prime \\prime}$ are of a different parity. Without loss of generality we start from the origin of the coordinate system $(0 / 1,0 / 1)$. Initially $b+d=0$ and after moving to each subsequent point along the broken line $b+d$ changes parity by the lemma. Hence it will not be possible to return to the origin after an odd number of steps since $b+d$ will be odd.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. $(\\mathbf{C Z S} 1)^{\\mathrm{IMO} 2}$ On a circle, $2 n-1(n \\geq 3)$ different points are given. Find the minimal natural number $N$ with the property that whenever $N$ of the given points are colored black, there exist two black points such that the interior of one of the corresponding arcs contains exactly $n$ of the given $2 n-1$ points.", "solution": "3. A segment connecting two points which divides the given circle into two arcs one of which contains exactly $n$ points in its interior we will call a good segment. Good segments determine one or more closed polygonal lines that we will call stars. Let us compute the number of stars. Note first that $\\operatorname{gcd}(n+1,2 n-1)=\\operatorname{gcd}(n+1,3)$. (i) Suppose that $3 \\nmid n+1$. Then the good segments form a single star. Among any $n$ points, two will be adjacent vertices of the star. On the other hand, we can select $n-1$ alternate points going along the star, and in this case no two points lie on a good segment. Hence $N=n$. (ii) If $3 \\mid n+1$, we obtain three stars of $\\left[\\frac{2 n-1}{3}\\right]$ vertices. If more than $\\left[\\frac{2 n-1}{6}\\right]=\\frac{n-2}{3}$ points are chosen on any of the stars, then two of them will be connected with a good segment. On the other hand, we can select $\\frac{n-2}{3}$ alternate points on each star, which adds up to $n-2$ points in total, no two of which lie on a good segment. Hence $N=n-1$. To sum up, $N=n$ for $3 \\nmid 2 n-1$ and $N=n-1$ for $3 \\mid 2 n-1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (CZS 2) Assume that the set of all positive integers is decomposed into $r$ (disjoint) subsets $A_{1} \\cup A_{2} \\cup \\cdots A_{r}=\\mathbb{N}$. Prove that one of them, say $A_{i}$, has the following property: There exists a positive $m$ such that for any $k$ one can find numbers $a_{1}, a_{2}, \\ldots, a_{k}$ in $A_{i}$ with $0<a_{j+1}-a_{j} \\leq m$ $(1 \\leq j \\leq k-1)$.", "solution": "4. Assuming that $A_{1}$ is not such a set $A_{i}$, it follows that for every $m$ there exist $m$ consecutive numbers not in $A_{1}$. It follows that $A_{2} \\cup A_{3} \\cup \\cdots \\cup A_{r}$ contains arbitrarily long sequences of numbers. Inductively, let us assume that $A_{j} \\cup A_{j+1} \\cup \\cdots \\cup A_{r}$ contains arbitrarily long sequences of consecutive numbers and none of $A_{1}, A_{2}, \\ldots, A_{j-1}$ is the desired set $A_{i}$. Let us assume that $A_{j}$ is also not $A_{i}$. Hence for each $m$ there exists $k(m)$ such that among $k(m)$ elements of $A_{j}$ there exist two consecutive elements that differ by at least $m$. Let us consider $m \\cdot k(m)$ consecutive numbers in $A_{j} \\cup \\cdots \\cup A_{r}$, which exist by the induction hypothesis. Then either $A_{j}$ contains fewer than $k(m)$ of these integers, in which case $A_{j+1} \\cup \\cdots \\cup A_{r}$ contains $m$ consecutive integers by the pigeonhole principle or $A_{j}$ contains $k(m)$ integers among which there exists a gap of length $m$ of consecutive integers that belong to $A_{j+1} \\cup \\cdots \\cup A_{r}$. Hence we have proven that $A_{j+1} \\cup \\cdots \\cup A_{r}$ contains sequences of integers of arbitrary length. By induction, assuming that $A_{1}, A_{2}, \\ldots, A_{r-1}$ do not satisfy the conditions to be the set $A_{i}$, it follows that $A_{r}$ contains sequences of consecutive integers of arbitrary length and hence satisfies the conditions necessary for it to be the set $A_{i}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (FRA 1) Given $\\triangle A B C$ with no side equal to another side, let $G, K$, and $H$ be its centroid, incenter, and orthocenter, respectively. Prove that $\\angle G K H>90^{\\circ}$.", "solution": "5. Let $O$ be the circumcenter of $A B C, E$ the midpoint of $O H$, and $R$ and $r$ the radii of the circumcircle and incircle respectively. We use the following facts from elementary geometry: $\\overrightarrow{O H}=3 \\overrightarrow{O G}, O K^{2}=R^{2}-2 R r$, and $K E=\\frac{R}{2}-r$. Hence $\\overrightarrow{K H}=2 \\overrightarrow{K E}-\\overrightarrow{K O}$ and $\\overrightarrow{K G}=\\frac{2 \\overrightarrow{K E}+\\overrightarrow{K O}}{3}$. We then obtain  $$ \\overrightarrow{K H} \\cdot \\overrightarrow{K G}=\\frac{1}{3}\\left(4 K E^{2}-K O^{2}\\right)=-\\frac{2}{3} r(R-2 r)<0 . $$  Hence $\\cos \\angle G K H<0 \\Rightarrow \\angle G K H>90^{\\circ}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (FRG 2) ${ }^{\\mathrm{IMO} 5}$ Two players $A$ and $B$ play a game in which they choose numbers alternately according to the following rule: At the beginning, an initial natural number $n_{0}>1$ is given. Knowing $n_{2 k}$, player $A$ may choose any $n_{2 k+1} \\in \\mathbb{N}$ such that  $$ n_{2 k} \\leq n_{2 k+1} \\leq n_{2 k}^{2} $$  Then player $B$ chooses a number $n_{2 k+2} \\in \\mathbb{N}$ such that  $$ \\frac{n_{2 k+1}}{n_{2 k+2}}=p^{r} $$  where $p$ is a prime number and $r \\in \\mathbb{N}$. It is stipulated that player $A$ wins the game if he (she) succeeds in choosing the number 1990, and player $B$ wins if he (she) succeeds in choosing 1.  For which natural numbers $n_{0}$ can player $A$ manage to win the game, for which $n_{0}$ can player $B$ manage to win, and for which $n_{0}$ can players $A$ and $B$ each force a tie?", "solution": "6. Let $W$ denote the set of all $n_{0}$ for which player $A$ has a winning strategy, $L$ the set of all $n_{0}$ for which player $B$ has a winning strategy, and $T$ the set of all $n_{0}$ for which a tie is ensured.  Lemma. Assume $\\{m, m+1, \\ldots 1990\\} \\subseteq W$ and that there exists $s \\leq 1990$ such that $s / p^{r} \\geq m$, where $p^{r}$ is the largest degree of a prime that divides $s$. Then all integers $x$ such that $\\sqrt{s} \\leq x<m$ also belong in W.  Proof. Starting from $x$, player $A$ can choose $s$, and by definition of $s$, player $B$ cannot choose a number smaller than $m$. This ensures player $A$ the victory. We now have trivially that since $45^{2}=2025>1990$, it follows that for $n_{0} \\in\\{45, \\ldots, 1990\\}$ player $A$ can choose 1990 in the first move. Hence $\\{45, \\ldots, 1990\\} \\subseteq W$. Using $m=45$ and selecting $s=420=2^{2} \\cdot 3 \\cdot 5 \\cdot 7$ we apply the lemma to get that all integers $x$ such that $\\sqrt{420}<21 \\leq x \\leq 1990$ are in $W$. Again, using $m=21$ and selecting $s=168=2^{3} \\cdot 3 \\cdot 7$ we apply the lemma to get that all integers $x$ such that $\\sqrt{168}<13 \\leq x \\leq 1990$ are in $W$. Selecting $s=105$ we obtain the new value for $m$ at $m=11$. Selecting $s=60$ we obtain $m=8$. Thus $\\{8, \\ldots, 1990\\} \\subseteq W$. For $n_{0}>1990$ there exists $r \\in N$ such that $2^{r} \\cdot 3^{2}<n_{0} \\leq 2^{r+1} \\cdot 3^{2}<n_{0}^{2}$. Player $A$ can take $n_{1}=2^{r+1} \\cdot 3^{2}$. The number player $B$ selects has to satisfy $8 \\leq n_{2}<n_{0}$. After finitely many steps he will select $8 \\leq n_{2 r} \\leq 1990$, and $A$ will have a winning strategy. Hence all $m \\geq 8$ belong to $W$. Now let us consider the case $n_{0} \\leq 5$. Since the smallest number divisible by three different primes is 30 and $n_{0}^{2} \\leq 5^{2}=25<30$, it follows that $n_{1}$ is of the form $n_{1}=p^{r}$ or $n_{1}=p^{r} \\cdot q^{s}$, where $p$ and $q$ are two different primes. In the first case player $B$ can choose 1 and win, while in the second case he can select the smaller of $p^{r}, q^{s}$, which is also smaller than $\\sqrt{n_{1}} \\leq n_{0}$. Thus player $B$ can eventually reach $n_{2 k}=1$. Thus $\\{2,3,4,5\\} \\subseteq L$. Finally, for $n_{0}=6$ or $n_{0}=7$ player $A$ must select a number divisible by at least three primes, which must be $30=2 \\cdot 3 \\cdot 5$ or $42=2 \\cdot 3 \\cdot 7$; otherwise, $B$ can select a degree of a prime smaller than $n_{0}$, yielding $n_{2}<6$ and victory for $B$. Player $B$ must select a number smaller than 8 . Hence, he has to select 6 in both cases. Afterwards, to avoid losing the game, player $A$ will always choose 30 and player $B$ always 6 . In this case we would have a tie. Hence $T \\subseteq\\{6,7\\}$. Considering that we have accounted for all integers $n_{0}>1$, the final solution is $L=\\{2,3,4,5\\}, T=\\{6,7\\}$, and $W=\\{x \\in \\mathbb{N} \\mid x \\geq 8\\}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (GRE 2) Let $f(0)=f(1)=0$ and  $$ f(n+2)=4^{n+2} f(n+1)-16^{n+1} f(n)+n \\cdot 2^{n^{2}}, \\quad n=0,1,2,3, \\ldots $$  Show that the numbers $f(1989), f(1990), f(1991)$ are divisible by 13.", "solution": "7. Let $f(n)=g(n) 2^{n^{2}}$ for all $n$. The recursion then transforms into $g(n+$ 2) $-2 g(n+1)+g(n)=n \\cdot 16^{-n-1}$ for $n \\in \\mathbb{N}_{0}$. By summing this equation from 0 to $n-1$, we get  $$ g(n+1)-g(n)=\\frac{1}{15^{2}} \\cdot\\left(1-(15 n+1) 16^{-n}\\right) $$  By summing up again from 0 to $n-1$ we get $g(n)=\\frac{1}{15^{3}} \\cdot(15 n-32+$ $\\left.(15 n+2) 16^{-n+1}\\right)$. Hence  $$ f(n)=\\frac{1}{15^{3}} \\cdot\\left(15 n+2+(15 n-32) 16^{n-1}\\right) \\cdot 2^{(n-2)^{2}} $$  Now let us look at the values of $f(n)$ modulo 13:  $$ f(n) \\equiv 15 n+2+(15 n-32) 16^{n-1} \\equiv 2 n+2+(2 n-6) 3^{n-1} $$  We have $3^{3} \\equiv 1(\\bmod 13)$. Plugging in $n \\equiv 1(\\bmod 13)$ and $n \\equiv 1(\\bmod$ 3 ) for $n=1990$ gives us $f(1990) \\equiv 0(\\bmod 13)$. We similarly calculate $f(1989) \\equiv 0$ and $f(1991) \\equiv 0(\\bmod 13)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (HUN 1) For a given positive integer $k$ denote the square of the sum of its digits by $f_{1}(k)$ and let $f_{n+1}(k)=f_{1}\\left(f_{n}(k)\\right)$. Determine the value of $f_{1991}\\left(2^{1990}\\right)$.", "solution": "8. Since $2^{1990}<8^{700}<10^{700}$, we have $f_{1}\\left(2^{1990}\\right)<(9 \\cdot 700)^{2}<4 \\cdot 10^{7}$. We then have $f_{2}\\left(2^{1990}\\right)<(3+9 \\cdot 7)^{2}<4900$ and finally $f_{3}\\left(2^{1990}\\right)<(3+9 \\cdot 3)^{2}=30^{2}$. It is easily shown that $f_{k}(n) \\equiv f_{k-1}(n)^{2}(\\bmod 9)$. Since $2^{6} \\equiv 1(\\bmod 9)$, we have $2^{1990} \\equiv 2^{4} \\equiv 7$ (all congruences in this problem will be $\\bmod 9$ ). It follows that $f_{1}\\left(2^{1990}\\right) \\equiv 7^{2} \\equiv 4$ and $f_{2}\\left(2^{1990}\\right) \\equiv 4^{2} \\equiv 7$. Indeed, it follows that $f_{2 k}\\left(2^{1990}\\right) \\equiv 7$ and $f_{2 k+1}\\left(2^{1990}\\right) \\equiv 4$ for all integer $k>0$. Thus $f_{3}\\left(2^{1990}\\right)=r^{2}$ where $r<30$ is an integer and $r \\equiv f_{2}\\left(2^{1990}\\right) \\equiv 7$. It follows that $r \\in\\{7,16,25\\}$ and hence $f_{3}\\left(2^{1990}\\right) \\in\\{49,256,625\\}$. It follows that $f_{4}\\left(2^{1990}\\right)=169, f_{5}\\left(2^{1990}\\right)=256$, and inductively $f_{2 k}\\left(2^{1990}\\right)=169$ and $f_{2 k+1}\\left(2^{1990}\\right)=256$ for all integer $k>1$. Hence $f_{1991}\\left(2^{1990}\\right)=256$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1990", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (HUN 3) The incenter of the triangle $A B C$ is $K$. The midpoint of $A B$ is $C_{1}$ and that of $A C$ is $B_{1}$. The lines $C_{1} K$ and $A C$ meet at $B_{2}$, the lines $B_{1} K$ and $A B$ at $C_{2}$. If the areas of the triangles $A B_{2} C_{2}$ and $A B C$ are equal, what is the measure of angle $\\angle C A B$ ?", "solution": "9. Let $a, b, c$ be the lengths of the sides of $\\triangle A B C, s=\\frac{a+b+c}{2}, r$ the inradius of the triangle, and $c_{1}$ and $b_{1}$ the lengths of $A B_{2}$ and $A C_{2}$ respectively. As usual we will denote by $S(X Y Z)$ the area of $\\triangle X Y Z$. We have  $$ \\begin{gathered} S\\left(A C_{1} B_{2}\\right)=\\frac{A C_{1} \\cdot A B_{2}}{A C \\cdot A B} S(A B C)=\\frac{c_{1} r s}{2 b} \\\\ S\\left(A K B_{2}\\right)=\\frac{c_{1} r}{2}, \\quad S\\left(A C_{1} K\\right)=\\frac{c r}{4} \\end{gathered} $$  From $S\\left(A C_{1} B_{2}\\right)=S\\left(A K B_{2}\\right)+S\\left(A C_{1} K\\right)$ we get $\\frac{c_{1} r s}{2 b}=\\frac{c_{1} r}{2}+\\frac{c r}{4}$; therefore $(a-b+c) c_{1}=b c$. By looking at the area of $\\triangle A B_{1} C_{2}$ we similarly obtain $(a+b-c) b_{1}=b c$. From these two equations and from $S(A B C)=S\\left(A B_{2} C_{2}\\right)$, from which we have $b_{1} c_{1}=b c$, we obtain  $$ a^{2}-(b-c)^{2}=b c \\Rightarrow \\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\\cos (\\angle B A C)=\\frac{1}{2} \\Rightarrow \\angle B A C=60^{\\circ} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (PHI 3) Let $A B C$ be any triangle and $P$ any point in its interior. Let $P_{1}, P_{2}$ be the feet of the perpendiculars from $P$ to the two sides $A C$ and $B C$. Draw $A P$ and $B P$, and from $C$ drop perpendiculars to $A P$ and $B P$. Let $Q_{1}$ and $Q_{2}$ be the feet of these perpendiculars. Prove that the lines $Q_{1} P_{2}, Q_{2} P_{1}$, and $A B$ are concurrent.", "solution": "1. All the angles $\\angle P P_{1} C, \\angle P P_{2} C, \\angle P Q_{1} C, \\angle P Q_{2} C$ are right, hence $P_{1}, P_{2}$, $Q_{1}, Q_{2}$ lie on the circle with diameter $P C$. The result now follows immediately from Pascal's theorem applied to the hexagon $P_{1} P P_{2} Q_{1} C Q_{2}$. It tells us that the points of intersection of the three pairs of lines $P_{1} C, P Q_{1}$ (intersection $A$ ), $P_{1} Q_{2}, P_{2} Q_{1}$ (intersection ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-553.jpg?height=333&width=335&top_left_y=376&top_left_x=919) $X)$ and $P Q_{2}, P_{2} C$ (intersection $B$ ) are collinear.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (USA 5) ${ }^{\\mathrm{IMO} 4}$ Suppose $G$ is a connected graph with $n$ edges. Prove that it is possible to label the edges of $G$ from 1 to $n$ in such a way that in every vertex $v$ of $G$ with two or more incident edges, the set of numbers labeling those edges has no common divisor greater than 1.", "solution": "10. We start at some vertex $v_{0}$ and walk along distinct edges of the graph, numbering them $1,2, \\ldots$ in the order of appearance, until this is no longer possible without reusing an edge. If there are still edges which are not numbered, one of them has a vertex which has already been visited (else $G$ would not be connected). Starting from this vertex, we continue to walk along unused edges resuming the numbering, until we eventually get stuck. Repeating this procedure as long as possible, we shall number all the edges. Let $v$ be a vertex which is incident with $e \\geq 2$ edges. If $v=v_{0}$, then it is on the edge 1 , so the gcd at $v$ is 1 . If $v \\neq v_{0}$, suppose that it was reached for the first time by the edge $r$. At that time there was at least one unused edge incident with $v$ (as $e \\geq 2$ ), hence one of them was labelled by $r+1$. The gcd at $v$ again equals $\\operatorname{gcd}(r, r+1)=1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (AUS 4) Prove that  $$ \\sum_{m=0}^{995} \\frac{(-1)^{m}}{1991-m}\\binom{1991-m}{m}=\\frac{1}{1991} $$", "solution": "11. To start with, observe that $\\frac{1}{n-m}\\binom{n-m}{m}=\\frac{1}{n}\\left[\\binom{n-m}{m}+\\binom{n-m-1}{m-1}\\right]$.  For $n=1,2, \\ldots$ set $S_{n}=\\sum_{m=0}^{[n / 2]}(-1)^{m}\\binom{n-m}{m}$. Using the identity $\\binom{m}{k}=$ $\\binom{m-1}{k}+\\binom{m-1}{k-1}$ we obtain the following relation for $S_{n}$ :  $$ \\begin{aligned} S_{n+1} & =\\sum_{m}(-1)^{m}\\binom{n-m+1}{m} \\\\ & =\\sum_{m}(-1)^{m}\\binom{n-m}{m}+\\sum_{m}(-1)^{m}\\binom{n-m}{m-1}=S_{n}-S_{n-1} . \\end{aligned} $$  Since the initial members of the sequence $S_{n}$ are $1,1,0,-1,-1,0,1,1, \\ldots$, we thus find that $S_{n}$ is periodic with period 6 . Now the sum from the problem reduces to  $$ \\begin{gathered} \\frac{1}{1991}\\binom{1991}{0}-\\frac{1}{1991}\\left[\\binom{1990}{1}+\\binom{1989}{0}\\right]+\\cdots-\\frac{1}{1991}\\left[\\binom{996}{995}+\\binom{995}{994}\\right] \\\\ =\\frac{1}{1991}\\left(S_{1991}-S_{1989}\\right)=\\frac{1}{1991}(0-(-1))=\\frac{1}{1991} . \\end{gathered} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. $(\\mathbf{C H N} 3)^{\\mathrm{IMO} 3}$ Let $S=\\{1,2,3, \\ldots, 280\\}$. Find the minimal natural number $n$ such that in any $n$-element subset of $S$ there are five numbers that are pairwise relatively prime.", "solution": "12. Let $A_{m}$ be the set of those elements of $S$ which are divisible by $m$. By the inclusion-exclusion principle, the number of elements divisible by $2,3,5$ or 7 equals  $$ \\begin{aligned} & \\left|A_{2} \\cup A_{3} \\cup A_{5} \\cup A_{7}\\right| \\\\ & =\\left|A_{2}\\right|+\\left|A_{3}\\right|+\\left|A_{5}\\right|+\\left|A_{7}\\right|-\\left|A_{6}\\right|-\\left|A_{10}\\right|-\\left|A_{14}\\right|-\\left|A_{15}\\right| \\\\ & \\quad-\\left|A_{21}\\right|-\\left|A_{35}\\right|+\\left|A_{30}\\right|+\\left|A_{42}\\right|+\\left|A_{70}\\right|+\\left|A_{105}\\right|-\\left|A_{210}\\right| \\\\ & =140+93+56+40-46-28-20-18 \\\\ & \\quad-13-8+9+6+4+2-1=216 . \\end{aligned} $$  Among any five elements of the set $A_{2} \\cup A_{3} \\cup A_{5} \\cup A_{7}$, one of the sets $A_{2}, A_{3}, A_{5}, A_{7}$ contains at least two, and those two are not relatively prime. Therefore $n>216$.  We claim that the answer is $n=217$. First notice that the set $A_{2} \\cup A_{3} \\cup$ $A_{5} \\cup A_{7}$ consists of four prime $(2,3,5,7)$ and 212 composite numbers. The set $S \\backslash A$ contains exactly 8 composite numbers: namely, $11^{2}, 11 \\cdot 13,11$. $17,11 \\cdot 19,11 \\cdot 23,13^{2}, 13 \\cdot 17,13 \\cdot 19$. Thus $S$ consists of the unity, 220 composite numbers and 59 primes. Let $A$ be a 217 -element subset of $S$, and suppose that there are no five pairwise relatively prime numbers in $A$. Then $A$ can contain at most 4 primes (or unity and three primes) and at least 213 composite numbers. Hence the set $S \\backslash A$ contains at most 7 composite numbers. Consequently, at least one of the following 8 five-element sets is disjoint with $S \\backslash A$, and is thus entirely contained in $A$ :  $$ \\begin{array}{ll} \\{2 \\cdot 23,3 \\cdot 19,5 \\cdot 17,7 \\cdot 13,11 \\cdot 11\\}, & \\{2 \\cdot 29,3 \\cdot 23,5 \\cdot 19,7 \\cdot 17,11 \\cdot 13\\}, \\\\ \\{2 \\cdot 31,3 \\cdot 29,5 \\cdot 23,7 \\cdot 19,11 \\cdot 17\\}, & \\{2 \\cdot 37,3 \\cdot 31,5 \\cdot 29,7 \\cdot 23,11 \\cdot 19\\}, \\\\ \\{2 \\cdot 41,3 \\cdot 37,5 \\cdot 31,7 \\cdot 29,11 \\cdot 23\\}, & \\{2 \\cdot 43,3 \\cdot 41,5 \\cdot 37,7 \\cdot 31,13 \\cdot 17\\}, \\\\ \\{2 \\cdot 47,3 \\cdot 43,5 \\cdot 41,7 \\cdot 37,13 \\cdot 19\\}, & \\{2 \\cdot 2,3 \\cdot 3,5 \\cdot 5,7 \\cdot 7,13 \\cdot 13\\} . \\end{array} $$  As each of these sets consists of five numbers relatively prime in pairs, the claim is proved.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (POL 4) Given any integer $n \\geq 2$, assume that the integers $a_{1}, a_{2}, \\ldots, a_{n}$ are not divisible by $n$ and, moreover, that $n$ does not divide $a_{1}+a_{2}+$ $\\cdots+a_{n}$. Prove that there exist at least $n$ different sequences $\\left(e_{1}, e_{2}, \\cdots, e_{n}\\right)$ consisting of zeros or ones such that $e_{1} a_{1}+e_{2} a_{2}+\\cdots+e_{n} a_{n}$ is divisible by $n$.", "solution": "13. Call a sequence $e_{1}, \\ldots, e_{n}$ good if $e_{1} a_{1}+\\cdots+e_{n} a_{n}$ is divisible by $n$. Among the sums $s_{0}=0, s_{1}=a_{1}, s_{2}=a_{1}+a_{2}, \\ldots, s_{n}=a_{1}+\\cdots+a_{n}$, two give the same remainder modulo $n$, and their difference corresponds to a good sequence. To show that, permuting the $a_{i}$ 's, we can find $n-1$ different sequences, we use the following Lemma. Let $A$ be a $k \\times n(k \\leq n-2)$ matrix of zeros and ones, whose every row contains at least one 0 and at least two 1 's. Then it is possible to permute columns of $A$ is such a way that in any row 1 's do not form a block. Proof. We will use the induction on $k$. The case $k=1$ and arbitrary $n \\geq 3$ is trivial. Suppose that $k \\geq 2$ and that for $k-1$ and any $n \\geq k+1$ the lemma is true. Consider a $k \\times n$ matrix $A, n \\geq k+2$. We mark an element $a_{i j}$ if either it is the only zero in the $i$-th row, or one of the 1 's in the row if it contains exactly two 1 's. Since $n \\geq 4$, every row contains at most two marked elements, which adds up to at most $2 k<2 n$ marked elements in total. It follows that there is a column with at most one marked element. Assume w.l.o.g. that it is the first column and that $a_{1 j}$ isn't marked for $j>1$. The matrix $B$, obtained by omitting the first row and first column from $A$, satisfies the conditions of the lemma. Therefore, we can permute columns of $B$ and get the required form. Considered as a permutation of column of $A$, this permutation may leave a block of 1's only in the first row of $A$. In the case that it is so, if $a_{11}=1$ we put the first column in the last place, otherwise we put it between any two columns having 1's in the first row. The obtained matrix has the required property. Suppose now that we have got $k$ different nontrivial good sequences $e_{1}^{i}, \\ldots, e_{n}^{i}, i=1, \\ldots, k$, and that $k \\leq n-2$. The matrix $A=\\left(e_{j}^{i}\\right)$ fulfils the conditions of Lemma, hence there is a permutation $\\sigma$ from Lemma. Now among the sums $s_{0}=0, s_{1}=a_{\\sigma(1)}, s_{2}=a_{\\sigma(1)}+a_{\\sigma(2)}$, $\\ldots, s_{n}=a_{\\sigma(1)}+\\cdots+a_{\\sigma(n)}$, two give the same remainder modulo $n$. Let $s_{p} \\equiv s_{q}(\\bmod n), p<q$. Then $n \\mid s_{q}-s_{p}=a_{\\sigma(p+1)}+\\cdots+a_{\\sigma(q)}$, and this yields a good sequence $e_{1}, \\ldots, e_{n}$ with $e_{\\sigma(p+1)}=\\cdots=e_{\\sigma(q)}=1$ and other $e$ 's equal to zero. Since from the construction we see that none of the sequences $e_{\\sigma(j)^{i}}$ has all 1's in a block, in this way we have got a new nontrivial good sequence, and we can continue this procedure until there are $n-1$ sequences. Together with the trivial $0, \\ldots, 0$ sequence, we have found $n$ good sequences.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (POL 3) Let $a, b, c$ be integers and $p$ an odd prime number. Prove that if $f(x)=a x^{2}+b x+c$ is a perfect square for $2 p-1$ consecutive integer values of $x$, then $p$ divides $b^{2}-4 a c$.", "solution": "14. Suppose that $f\\left(x_{0}\\right), f\\left(x_{0}+1\\right), \\ldots, f\\left(x_{0}+2 p-2\\right)$ are squares. If $p \\mid a$ and $p \\nmid b$, then $f(x) \\equiv b x+c(\\bmod p)$ for $x=x_{0}, \\ldots, x_{0}+p-1$ form a complete system of residues modulo $p$. However, a square is always congruent to exactly one of the $\\frac{p+1}{2}$ numbers $0,1^{2}, 2^{2}, \\ldots,\\left(\\frac{p-1}{2}\\right)^{2}$ and thus cannot give every residue modulo $p$. Also, if $p \\mid a$ and $p \\mid b$, then $p \\mid b^{2}-4 a c$. We now assume $p \\nmid a$. The following identities hold for any quadric polynomial:  $$ 4 a \\cdot f(x)=(2 a x+b)^{2}-\\left(b^{2}-4 a c\\right) $$  and  $$ f(x+p)-f(x)=p(2 a x+b)+p^{2} a . $$  Suppose that there is an $y, x_{0} \\leq y \\leq x_{0}+p-2$, for which $f(y)$ is divisible by $p$. Then both $f(y)$ and $f(y+p)$ are squares divisible by $p$, and therefore both are divisible by $p^{2}$. But relation (2) implies that $p \\mid 2 a y+b$, and hence by (1) $b^{2}-4 a c$ is divisible by $p$ as well. Therefore it suffices to show that such an $y$ exists, and for that aim we prove that there are two such $y$ in $\\left[x_{0}, x_{0}+p-1\\right]$. Assume the opposite. Since for $x=x_{0}, x_{0}+1, \\ldots, x_{0}+p-1 f(x)$ is congruent modulo $p$ to one of the $\\frac{p-1}{2}$ numbers $1^{2}, 2^{2}, \\ldots,\\left(\\frac{p-1}{2}\\right)^{2}$, it follows by the pigeon-hole principle that for some mutually distinct $u, v, w \\in\\left\\{x_{0}, \\ldots, x_{0}+p-1\\right\\}$ we have $f(u) \\equiv f(v) \\equiv f(w)(\\bmod p)$. Consequently the difference $f(u)-f(v)=$ $(u-v)(a(u+v)+b)$ is divisible by $p$, but it is clear that $p \\nmid u-v$, hence $a(u+v) \\equiv-b(\\bmod p)$. Similarly $a(u+w) \\equiv-b(\\bmod p)$, which together with the previous congruence yields $p|a(v-w) \\Rightarrow p| v-w$ which is clearly impossible. It follows that $p \\mid f\\left(y_{1}\\right)$ for at least one $y_{1}$, $x_{0} \\leq y_{1}<x_{0}+p$. If $y_{2}, x_{0} \\leq y_{2}<x_{0}+p$ is such that $a\\left(y_{1}+y_{2}\\right)+b \\equiv 0(\\bmod p)$, we have $p\\left|f\\left(y_{1}\\right)-f\\left(y_{2}\\right) \\Rightarrow p\\right| f\\left(y_{2}\\right)$. If $y_{1}=y_{2}$, then by (1) $p \\mid b^{2}-4 a c$. Otherwise, among $y_{1}, y_{2}$ one belongs to $\\left[x_{0}, x_{0}+p-2\\right]$ as required. Second solution. Using Legendre's symbols $\\left(\\frac{a}{p}\\right)$ for quadratic residues we can prove a stronger statement for $p \\geq 5$. It can be shown that  $$ \\sum_{x=0}^{p-1}\\left(\\frac{a x^{2}+b x+c}{p}\\right)=-\\left(\\frac{a}{p}\\right) \\quad \\text { if } \\quad p \\nmid b^{2}-4 a c $$  hence for at most $\\frac{p+3}{2}$ values of $x$ between $x_{0}$ and $x_{0}+p-1$ inclusive, $a x^{2}+b x+c$ is a quadratic residue or 0 modulo $p$. Therefore, if $p \\geq 5$ and $f(x)$ is a square for $\\frac{p+5}{2}$ consecutive values, then $p \\mid b^{2}-4 a c$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (USS 2) Let $a_{n}$ be the last nonzero digit in the decimal representation of the number $n$ !. Does the sequence $a_{1}, a_{2}, \\ldots, a_{n}, \\ldots$ become periodic after a finite number of terms?", "solution": "15. Assume that the sequence has the period $T$. We can find integers $k>m>$ 0 , as large as we like, such that $10^{k} \\equiv 10^{m}(\\bmod T)$, using for example Euler's theorem. It is obvious that $a_{10^{k}-1}=a_{10^{k}}$ and hence, taking $k$ sufficiently large and using the periodicity, we see that  $$ a_{2 \\cdot 10^{k}-10^{m}-1}=a_{10^{k}-1}=a_{10^{k}}=a_{2 \\cdot 10^{k}-10^{m}} $$  Since $\\left(2 \\cdot 10^{k}-10^{m}\\right)!=\\left(2 \\cdot 10^{k}-10^{m}\\right)\\left(2 \\cdot 10^{k}-10^{m}-1\\right)!$ and the last nonzero digit of $2 \\cdot 10^{k}-10^{m}$ is nine, we must have $a_{2 \\cdot 10^{k}-10^{m}-1}=5$ (if $s$ is a digit, the last digit of $9 s$ is $s$ only if $s=5$ ). But this means that 5 divides $n$ ! with a greater power than 2 does, which is impossible. Indeed, if the exponents of these powers are $\\alpha_{2}, \\alpha_{5}$ respectively, then $\\alpha_{5}=$ $[n / 5]+\\left[n / 5^{2}\\right]+\\cdots \\leq \\alpha_{2}=[n / 2]+\\left[n / 2^{2}\\right]+\\cdots$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (ROM 1) ${ }^{\\mathrm{IMO} 2}$ Let $n>6$ and $a_{1}<a_{2}<\\cdots<a_{k}$ be all natural numbers that are less than $n$ and relatively prime to $n$. Show that if $a_{1}, a_{2}, \\ldots, a_{k}$ is an arithmetic progression, then $n$ is a prime number or a natural power of two.", "solution": "16. Let $p$ be the least prime number that does not divide $n$ : thus $a_{1}=1$ and $a_{2}=p$. Since $a_{2}-a_{1}=a_{3}-a_{2}=\\cdots=r$, the $a_{i}$ 's are $1, p, 2 p-1,3 p-2, \\ldots$ We have the following cases: $p=2$. Then $r=1$ and the numbers $1,2,3, \\ldots, n-1$ are relatively prime to $n$, hence $n$ is a prime. $p=3$. Then $r=2$, so every odd number less than $n$ is relatively prime to $n$, from which we deduce that $n$ has no odd divisors. Therefore $n=2^{k}$ for some $k \\in \\mathbb{N}$. $p>3$. Then $r=p-1$ and $a_{k+1}=a_{1}+k(p-1)=1+k(p-1)$. Since $n-1$ also must belong to the progression, we have $p-1 \\mid n-2$. Let $q$ be any prime divisor of $p-1$. Then also $q \\mid n-2$. On the other hand, since $q<p$, it must divide $n$ too, therefore $q \\mid 2$, i.e. $q=2$. This means that $p-1$ has no prime divisors other than 2 and thus $p=2^{l}+1$ for some $l \\geq 2$. But in order for $p$ to be prime, $l$ must be even (because $3 \\mid 2^{l}+1$ for $l$ odd). Now we recall that $2 p-1$ is also relatively prime to $n$; but $2 p-1=2^{l+1}+1$ is divisible by 3 , which is a contradiction because $3 \\mid n$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (HKG 4) Find all positive integer solutions $x, y, z$ of the equation $3^{x}+$ $4^{y}=5^{z}$ 。", "solution": "17. Taking the equation $3^{x}+4^{y}=5^{z}(x, y, z>0)$ modulo 3 , we get that $5^{z} \\equiv 1(\\bmod 3)$, hence $z$ is even, say $z=2 z_{1}$. The equation then becomes $3^{x}=5^{2 z_{1}}-4^{y}=\\left(5^{z_{1}}-2^{y}\\right)\\left(5^{z_{1}}+2^{y}\\right)$. Each factor $5^{z_{1}}-2^{y}$ and $5^{z_{1}}+2^{y}$ is a power of 3 , for which the only possibility is $5^{z_{1}}+2^{y}=3^{x}$ and $5^{z_{1}}-2^{y}=$ 1. Again modulo 3 these equations reduce to $(-1)^{z_{1}}+(-1)^{y}=0$ and $(-1)^{z_{1}}-(-1)^{y}=1$, implying that $z_{1}$ is odd and $y$ is even. Particularly, $y \\geq 2$. Reducing the equation $5^{z_{1}}+2^{y}=3^{x}$ modulo 4 we get that $3^{x} \\equiv 1$, hence $x$ is even. Now if $y>2$, modulo 8 this equation yields $5 \\equiv 5^{z_{1}} \\equiv$ $3^{x} \\equiv 1$, a contradiction. Hence $y=2, z_{1}=1$. The only solution of the original equation is $x=y=z=2$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (BUL 1) Find the highest degree $k$ of 1991 for which $1991^{k}$ divides the number  $$ 1990^{1991^{1992}}+1992^{1991^{1990}} $$", "solution": "18. For integers $a>0, n>0$ and $\\alpha \\geq 0$, we shall write $a^{\\alpha} \\| n$ when $a^{\\alpha} \\mid n$ and $a^{\\alpha+1} \\nmid n$. Lemma. For every odd number $a \\geq 3$ and an integer $n \\geq 0$ it holds that  $$ a^{n+1} \\|(a+1)^{a^{n}}-1 \\quad \\text { and } \\quad a^{n+1} \\|(a-1)^{a^{n}}+1 $$  Proof. We shall prove the first relation by induction (the second is analogous). For $n=0$ the statement is obvious. Suppose that it holds for some $n$, i.e. that $(1+a)^{a^{n}}=1+N a^{n+1}, a \\nmid N$. Then  $$ (1+a)^{a^{n+1}}=\\left(1+N a^{n+1}\\right)^{a}=1+a \\cdot N a^{n+1}+\\binom{a}{2} N^{2} a^{2 n+2}+M a^{3 n+3} $$  for some integer $M$. Since $\\binom{a}{2}$ is divisible by $a$ for $a$ odd, we deduce that the part of the above sum behind $1+a \\cdot N a^{n+1}$ is divisible by $a^{n+3}$. Hence $(1+a)^{a^{n+1}}=1+N^{\\prime} a^{n+2}$, where $a \\nmid N^{\\prime}$. It follows immediately from Lemma that  $$ 1991^{1993} \\| 1990^{1991^{1992}}+1 \\quad \\text { and } \\quad 1991^{1991} \\| 1992^{1991^{1990}}-1 $$  Adding these two relations we obtain immediately that $k=1991$ is the desired value.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (IRE 5) Let $a$ be a rational number with $0<a<1$ and suppose that  $$ \\cos 3 \\pi a+2 \\cos 2 \\pi a=0 $$  (Angle measurements are in radians.) Prove that $a=2 / 3$.", "solution": "19. Set $x=\\cos (\\pi a)$. The given equation is equivalent to $4 x^{3}+4 x^{2}-3 x-2=0$, which factorizes as $(2 x+1)\\left(2 x^{2}+x-2\\right)=0$. The case $2 x+1=0$ yields $\\cos (\\pi a)=-1 / 2$ and $a=2 / 3$. It remains to show that if $x$ satisfies $2 x^{2}+x-2=0$ then $a$ is not rational. The polynomial equation $2 x^{2}+x-2=0$ has two real roots, $x_{1,2}=\\frac{-1 \\pm \\sqrt{17}}{4}$, and since $|x| \\leq 1$ we must have $x=\\cos \\pi a=\\frac{-1+\\sqrt{17}}{4}$. We now prove by induction that, for every integer $n \\geq 0, \\cos \\left(2^{n} \\pi a\\right)=$ $\\frac{a_{n}+b_{n} \\sqrt{17}}{4}$ for some odd integers $a_{n}, b_{n}$. The case $n=0$ is trivial. Also, if $\\cos \\left(2^{n} \\pi a\\right)=\\frac{a_{n}+b_{n} \\sqrt{17}}{4}$, then  $$ \\begin{aligned} \\cos \\left(2^{n+1} \\pi a\\right) & =2 \\cos ^{2}\\left(2^{n} \\pi a\\right)-1 \\\\ & =\\frac{1}{4}\\left(\\frac{a_{n}^{2}+17 b_{n}^{2}-8}{2}+a_{n} b_{n} \\sqrt{17}\\right)=\\frac{a_{n+1}+b_{n+1} \\sqrt{17}}{4} . \\end{aligned} $$  By the inductive step that $a_{n}, b_{n}$ are odd, it is obvious that $a_{n+1}, b_{n+1}$ are also odd. This proves the claim. Note also that, since $a_{n+1}=\\frac{1}{2}\\left(a_{n}^{2}+17 b_{n}^{2}-8\\right)>a_{n}$, the sequence $\\left\\{a_{n}\\right\\}$ is strictly increasing. Hence the set of values of $\\cos \\left(2^{n} \\pi a\\right), n=0,1,2, \\ldots$, is infinite (because $\\sqrt{17}$ is irrational). However, if $a$ were rational, then the set of values of $\\cos m \\pi a, m=1,2, \\ldots$, would be finite, a contradiction. Therefore the only possible value for $a$ is $2 / 3$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (JAP 5) For an acute triangle $A B C, M$ is the midpoint of the segment $B C, P$ is a point on the segment $A M$ such that $P M=B M, H$ is the foot of the perpendicular line from $P$ to $B C, Q$ is the point of intersection of segment $A B$ and the line passing through $H$ that is perpendicular to $P B$, and finally, $R$ is the point of intersection of the segment $A C$ and the line passing through $H$ that is perpendicular to $P C$. Show that the circumcircle of $\\triangle Q H R$ is tangent to the side $B C$ at point $H$.", "solution": "2. Let $H Q$ meet $P B$ at $Q^{\\prime}$ and $H R$ meet $P C$ at $R^{\\prime}$. From $M P=M B=M C$ we have $\\angle B P C=90^{\\circ}$. So $P R^{\\prime} H Q^{\\prime}$ is a rectangle. Since $P H$ is perpendicular to $B C$, it follows that the circle with diameter $P H$, through $P, R^{\\prime}, H, Q^{\\prime}$, is tangent to $B C$. It is now sufficient to show that $Q R$ is parallel to $Q^{\\prime} R^{\\prime}$. Let $C P$ meet $A B$ at $X$, and $B P$ meet $A C$ at $Y$. Since $P$ is on the median, it follows (for ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-553.jpg?height=403&width=432&top_left_y=806&top_left_x=877) example, by Ceva's theorem) that $A X / X B=A Y / Y C$, i.e. that $X Y$ is parallel to $B C$. Consequently, $P Y / B P=P X / C P$. Since $H Q$ is parallel to $C X$, we have $Q Q^{\\prime} / H Q^{\\prime}=$ $P X / C P$ and similarly $R R^{\\prime} / H R^{\\prime}=P Y / B P$. It follows that $Q Q^{\\prime} / H Q^{\\prime}=$ $R R^{\\prime} / H R^{\\prime}$, hence $Q R$ is parallel to $Q^{\\prime} R^{\\prime}$ as required. Second solution. It suffices to show that $\\angle R H C=\\angle R Q H$, or equivalently $R H: Q H=P C: P B$. We assume $P C: P B=1: x$. Let $X \\in A B$ and $Y \\in A C$ be points such that $M X \\perp P B$ and $M Y \\perp P C$. Since $M X$ bisects $\\angle A M B$ and $M Y$ bisects $A M C$, we deduce  $$ \\begin{aligned} & A X: X B=A M: M B=A Y: Y C \\Rightarrow X Y \\| B C \\Rightarrow \\\\ & \\quad \\Rightarrow \\triangle X Y M \\sim \\triangle C B P \\Rightarrow X M: M Y=1: x . \\end{aligned} $$  Now from $C H: H B=1: x^{2}$ we obtain $R H: M Y=C H: C M=1: \\frac{1+x^{2}}{2}$ and $Q H: M X=B H: B M=x^{2}: \\frac{1+x^{2}}{2}$. Therefore  $$ R H: Q H=\\frac{2}{1+x^{2}} M Y: \\frac{2 x^{2}}{1+x^{2}} M X=1: x $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. (IRE 3) Let $\\alpha$ be the positive root of the equation $x^{2}=1991 x+1$. For natural numbers $m, n$ define  $$ m * n=m n+[\\alpha m][\\alpha n] $$  where $[x]$ is the greatest integer not exceeding $x$. Prove that for all natural numbers $p, q, r$,  $$ (p * q) * r=p *(q * r) $$", "solution": "20. We prove the result with 1991 replaced by any positive integer $k$. For natural numbers $p, q$, let $\\epsilon=(\\alpha p-[\\alpha p])(\\alpha q-[\\alpha q])$. Then $0<\\epsilon<1$ and  $$ \\epsilon=\\alpha^{2} p q-\\alpha(p[\\alpha q]+q[\\alpha p])+[\\alpha p][\\alpha q] . $$  Multiplying this equality by $\\alpha-k$ and using $\\alpha^{2}=k \\alpha+1$, i.e. $\\alpha(\\alpha-k)=1$, we get  $$ (\\alpha-k) \\epsilon=\\alpha(p q+[\\alpha p][\\alpha q])-(p[\\alpha q]+q[\\alpha p]+k[\\alpha p][\\alpha q]) $$  Since $0<(\\alpha-k) \\epsilon<1$, we have $[\\alpha(p * q)]=p[\\alpha q]+q[\\alpha p]+k[\\alpha p][\\alpha q]$. Now  $$ \\begin{aligned} (p * q) * r & =(p * q) r+[\\alpha(p * q)][\\alpha r]= \\\\ & =p q r+[\\alpha p][\\alpha q] r+[\\alpha q][\\alpha r] p+[\\alpha r][\\alpha p] q+k[\\alpha p][\\alpha q][\\alpha r] . \\end{aligned} $$  Since the last expression is symmetric, the same formula is obtained for $p *(q * r)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. (HKG 6) Let $f(x)$ be a monic polynomial of degree 1991 with integer coefficients. Define $g(x)=f^{2}(x)-9$. Show that the number of distinct integer solutions of $g(x)=0$ cannot exceed 1995.", "solution": "21. The polynomial $g(x)$ factorizes as $g(x)=f(x)^{2}-9=(f(x)-3)(f(x)+3)$. If one of the equations $f(x)+3=0$ and $f(x)-3=0$ has no integer solutions, then the number of integer solutions of $g(x)=0$ clearly does not exceed 1991. Suppose now that both $f(x)+3=0$ and $f(x)-3=0$ have integer solutions. Let $x_{1}, \\ldots, x_{k}$ be distinct integer solutions of the former, and $x_{k+1}, \\ldots, x_{k+l}$ be distinct integer solutions of the latter equation. There exist monic polynomials $p(x), q(x)$ with integer coefficients such that $f(x)+3=\\left(x-x_{1}\\right)\\left(x-x_{2}\\right) \\ldots\\left(x-x_{k}\\right) p(x)$ and $f(x)-3=$ $\\left(x-x_{k+1}\\right)\\left(x-x_{k+2}\\right) \\ldots\\left(x-x_{k+l}\\right) q(x)$. Thus we obtain $\\left(x-x_{1}\\right)\\left(x-x_{2}\\right) \\ldots\\left(x-x_{k}\\right) p(x)-\\left(x-x_{k+1}\\right)\\left(x-x_{k+2}\\right) \\ldots\\left(x-x_{k+l}\\right) q(x)=6$. Putting $x=x_{k+1}$ we get $\\left(x_{k+1}-x_{1}\\right)\\left(x_{k+1}-x_{2}\\right) \\cdots\\left(x_{k+1}-x_{k}\\right) \\mid 6$, and since the product of more than four distinct integers cannot divide 6 , this implies $k \\leq 4$. Similarly $l \\leq 4$; hence $g(x)=0$ has at most 8 distinct integer solutions. Remark. The proposer provided a solution for the upper bound of 1995 roots which was essentially the same as that of (IMO74-6).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "22", "problem_type": null, "exam": "IMO", "problem": "22. (USA 4) Real constants $a, b, c$ are such that there is exactly one square all of whose vertices lie on the cubic curve $y=x^{3}+a x^{2}+b x+c$. Prove that the square has sides of length $\\sqrt[4]{72}$.", "solution": "22. Suppose w.l.o.g. that the center of the square is at the origin $O(0,0)$. We denote the curve $y=f(x)=x^{3}+a x^{2}+b x+c$ by $\\gamma$ and the vertices of the square by $A, B, C, D$ in this order. At first, the symmetry with respect to the point $O$ maps $\\gamma$ into the curve $\\bar{\\gamma}\\left(y=f(-x)=x^{3}-a x^{2}+b x-c\\right)$. Obviously $\\bar{\\gamma}$ also passes through $A, B, C, D$, and thus has four different intersection points with $\\gamma$. Then $2 a x^{2}+2 c$ has at least four distinct solution, which implies $a=c=0$. Particularly, $\\gamma$ passes through $O$ and intersects all quadrants, and hence $b<0$. Further, the curve $\\gamma^{\\prime}$, obtained by rotation of $\\gamma$ around $O$ for $90^{\\circ}$, has an equation $-x=f(y)$ and also contains the points $A, B, C, D$ and $O$. The intersection points $(x, y)$ of $\\gamma \\cap \\gamma^{\\prime}$ are determined by $-x=f(f(x))$, and hence they are roots of a polynomial $p(x)=f(f(x))+x$ of 9-th degree.  But the number of times that one cubic actually crosses the other in each quadrant is in the general case even (draw the picture!), and since $A B C D$ is the only square lying on $\\gamma \\cap \\gamma^{\\prime}$, the intersection points $A, B, C, D$ must be double. It follows that  $$ p(x)=x[(x-r)(x+r)(x-s)(x+s)]^{2}, $$  where $r, s$ are the $x$-coordinates of $A$ and $B$. On the other hand, $p(x)$ is defined by $\\left(x^{3}+b x\\right)^{3}+b\\left(x^{3}+b x\\right)+x$, and therefore equating of coefficients with (1) yields  $$ \\begin{array}{cc} 3 b=-2\\left(r^{2}+s^{2}\\right), & 3 b^{2}=\\left(r^{2}+s^{2}\\right)^{2}+2 r^{2} s^{2}, \\\\ b\\left(b^{2}+1\\right)=-2 r^{2} s^{2}\\left(r^{2}+s^{2}\\right), & b^{2}+1=r^{4} s^{4} . \\end{array} $$  Straightforward solving this system of equations gives $b=-\\sqrt{8}$ and $r^{2}+$ $s^{2}=\\sqrt{18}$. The line segment from $O$ to $(r, s)$ is half a diagonal of the square, and thus a side of the square has length $a=\\sqrt{2\\left(r^{2}+s^{2}\\right)}=\\sqrt[4]{72}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "23", "problem_type": null, "exam": "IMO", "problem": "23. (IND 2) Let $f$ and $g$ be two integer-valued functions defined on the set of all integers such that (a) $f(m+f(f(n)))=-f(f(m+1)-n$ for all integers $m$ and $n$; (b) $g$ is a polynomial function with integer coefficients and $g(n)=g(f(n))$ for all integers $n$. Determine $f(1991)$ and the most general form of $g$.", "solution": "23. From (i), replacing $m$ by $f(f(m))$, we get  $$ f(f(f(m))+f(f(n)))=-f(f(f(f(m))+1))-n $$  analogously $\\quad f(f(f(n))+f(f(m)))=-f(f(f(f(n))+1))-m$. From these relations we get $f(f(f(f(m))+1))-f(f(f(f(n))+1))=m-n$. Again from (i),  $$ \\begin{aligned} & f(f(f(f(m))+1))=f(-m-f(f(2))) \\\\ & \\text { and } \\quad f(f(f(f(n))+1))=f(-n-f(f(2))) . \\end{aligned} $$  Setting $f(f(2))=k$ we obtain $f(-m-k)-f(-n-k)=m-n$ for all integers $m, n$. This implies $f(m)=f(0)-m$. Then also $f(f(m))=m$, and using this in (i) we finally get  $$ f(n)=-n-1 \\quad \\text { for all integers } n $$  Particularly $f(1991)=-1992$. From (ii) we obtain $g(n)=g(-n-1)$ for all integers $n$. Since $g$ is a polynomial, it must also satisfy $g(x)=g(-x-1)$ for all real $x$. Let us now express $g$ as a polynomial on $x+1 / 2: g(x)=h(x+1 / 2)$. Then $h$ satisfies $h(x+1 / 2)=h(-x-1 / 2)$, i.e. $h(y)=h(-y)$, hence it is a polynomial in $y^{2}$; thus $g$ is a polynomial in $(x+1 / 2)^{2}=x^{2}+x+1 / 4$. Hence $g(n)=p\\left(n^{2}+n\\right)$ (for some polynomial $p$ ) is the most general form of $g$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "24", "problem_type": null, "exam": "IMO", "problem": "24. (IND 1) An odd integer $n \\geq 3$ is said to be \"nice\" if there is at least one permutation $a_{1}, a_{2}, \\ldots, a_{n}$ of $1,2, \\ldots, n$ such that the $n$ sums $a_{1}-a_{2}+$ $a_{3}-\\cdots-a_{n-1}+a_{n}, a_{2}-a_{3}+a_{4}-\\cdots-a_{n}+a_{1}, a_{3}-a_{4}+a_{5}-\\cdots-a_{1}+$ $a_{2}, \\ldots, a_{n}-a_{1}+a_{2}-\\cdots-a_{n-2}+a_{n-1}$ are all positive. Determine the set of all \"nice\" integers.", "solution": "24. Let $y_{k}=a_{k}-a_{k+1}+a_{k+2}-\\cdots+a_{k+n-1}$ for $k=1,2, \\ldots, n$, where we define $x_{i+n}=x_{i}$ for $1 \\leq i \\leq n$. We then have $y_{1}+y_{2}=2 a_{1}, y_{2}+y_{3}=$ $2 a_{2}, \\ldots, y_{n}+y_{1}=2 a_{n}$. (i) Let $n=4 k-1$ for some integer $k>0$. Then for each $i=1,2, \\ldots, n$ we have that $y_{i}=\\left(a_{i}+a_{i+1}+\\cdots+a_{i-1}\\right)-2\\left(a_{i+1}+a_{i+3}+\\cdots+a_{i-2}\\right)=1+$ $2+\\cdots+(4 k-1)-2\\left(a_{i+1}+a_{i+3}+\\cdots+a_{i-2}\\right)$ is even. Suppose now that $a_{1}, \\ldots, a_{n}$ is a good permutation. Then each $y_{i}$ is positive and even, so $y_{i} \\geq 2$. But for some $t \\in\\{1, \\ldots, n\\}$ we must have $a_{t}=1$, and thus $y_{t}+y_{t+1}=2 a_{t}=2$ which is impossible. Hence the numbers $n=4 k-1$ are not good. (ii) Let $n=4 k+1$ for some integer $k>0$. Then $2,4, \\ldots, 4 k, 4 k+1,4 k-$ $1, \\ldots, 3,1$ is a permutation with the desired property. Indeed, in this case $y_{1}=y_{4 k+1}=1, y_{2}=y_{4 k}=3, \\ldots, y_{2 k}=y_{2 k+2}=4 k-1$, $y_{2 k+1}=4 k+1$. Therefore all nice numbers are given by $4 k+1, k \\in \\mathbb{N}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "25", "problem_type": null, "exam": "IMO", "problem": "25. (USA 1) Suppose that $n \\geq 2$ and $x_{1}, x_{2}, \\ldots, x_{n}$ are real numbers between 0 and 1 (inclusive). Prove that for some index $i$ between 1 and $n-1$ the inequality  $$ x_{i}\\left(1-x_{i+1}\\right) \\geq \\frac{1}{4} x_{1}\\left(1-x_{n}\\right) $$  holds.", "solution": "25. Since replacing $x_{1}$ by 1 can only reduce the set of indices $i$ for which the desired inequality holds, we may assume $x_{1}=1$. Similarly we may assume $x_{n}=0$. Now we can let $i$ be the largest index such that $x_{i}>1 / 2$. Then $x_{i+1} \\leq 1 / 2$, hence  $$ x_{i}\\left(1-x_{i+1}\\right) \\geq \\frac{1}{4}=\\frac{1}{4} x_{1}\\left(1-x_{n}\\right) . $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "26", "problem_type": null, "exam": "IMO", "problem": "26. (CZS 1) Let $n \\geq 2$ be a natural number and let the real numbers $p, a_{1}, a_{2}, \\ldots, a_{n}, b_{1}, b_{2}, \\ldots, b_{n}$ satisfy $1 / 2 \\leq p \\leq 1,0 \\leq a_{i}, 0 \\leq b_{i} \\leq p$, $i=1, \\ldots, n$, and $\\sum_{i=1}^{n} a_{i}=\\sum_{i=1}^{n} b_{i}=1$. Prove the inequality  $$ \\sum_{i=1}^{n} b_{i} \\prod_{\\substack{j=1 \\\\ j \\neq i}}^{n} a_{j} \\leq \\frac{p}{(n-1)^{n-1}} $$", "solution": "26. Without loss of generality we can assume $b_{1} \\geq b_{2} \\geq \\cdots \\geq b_{n}$. We denote by $A_{i}$ the product $a_{1} a_{2} \\ldots a_{i-1} a_{i+1} \\ldots a_{n}$. If for some $i<j$ holds $A_{i}<A_{j}$, then $b_{i} A_{i}+b_{j} A_{j} \\leq b_{i} A_{j}+b_{j} A_{i}$ (or equivalently $\\left(b_{i}-b_{j}\\right)\\left(A_{i}-A_{j}\\right) \\leq 0$ ). Therefore the sum $\\sum_{i=1}^{n} b_{i} A_{i}$ does not decrease when we rearrange the numbers $a_{1}, \\ldots, a_{n}$ so that $A_{1} \\geq \\cdots \\geq A_{n}$, and consequently $a_{1} \\leq \\cdots \\leq$ $a_{n}$. Further, for fixed $a_{i}$ 's and $\\sum b_{i}=1$, the sum $\\sum_{i=1}^{n} b_{i} A_{i}$ is maximal when $b_{1}$ takes the largest possible value, i.e. $b_{1}=p, b_{2}$ takes the remaining largest possible value $b_{2}=1-p$, whereas $b_{3}=\\cdots=b_{n}=0$. In this case  $$ \\begin{aligned} \\sum_{i=1}^{n} b_{i} A_{i} & =p A_{1}+(1-p) A_{2}=a_{3} \\ldots a_{n}\\left(p a_{2}+(1-p) a_{1}\\right) \\\\ & \\leq p\\left(a_{1}+a_{2}\\right) a_{3} \\ldots a_{n} \\leq \\frac{p}{(n-1)^{n-1}} \\end{aligned} $$  using the inequality between the geometric and arithmetic means for $a_{3}, \\ldots, a_{n}, a_{1}+a_{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "27", "problem_type": null, "exam": "IMO", "problem": "27. (POL 2) Determine the maximum value of the sum  $$ \\sum_{i<j} x_{i} x_{j}\\left(x_{i}+x_{j}\\right) $$  over all $n$-tuples $\\left(x_{1}, \\ldots, x_{n}\\right)$, satisfying $x_{i} \\geq 0$ and $\\sum_{i=1}^{n} x_{i}=1$.", "solution": "27. Write $F\\left(x_{1}, \\ldots, x_{n}\\right)=\\sum_{i<j} x_{i} x_{j}\\left(x_{i}+x_{j}\\right)$. Choose an $n$-tuple $\\left(x_{1}, \\ldots, x_{n}\\right)$, $\\sum_{i=1}^{n} x_{i}=1, x_{i} \\geq 0$ with at least three nonzero components, and assume w.l.o.g. that $x_{1} \\geq \\cdots \\geq x_{k-1} \\geq x_{k} \\geq x_{k+1}=\\cdots=x_{n}=0$. We claim that replacing $x_{k-1}, x_{k}$ with $x_{k-1}+x_{k}, 0$ the value of $F$ increases. Write for brevity $x_{k-1}=a, x_{k}=b$. Then  $$ \\begin{gathered} F(\\ldots, a+b, 0,0, \\ldots)-F(\\ldots, a, b, 0, \\ldots) \\\\ =\\sum_{i=1}^{k-2} x_{i}(a+b)\\left(x_{i}+a+b\\right)-\\sum_{i=1}^{k-2}\\left[x_{i} a\\left(x_{i}+a\\right)+x_{i} b\\left(x_{i}+b\\right)\\right]-a b(a+b) \\end{gathered} $$  $$ =a b\\left(2 \\sum_{i=1}^{k-2} x_{i}-a-b\\right)=a b(2-3(a+b))>0 $$  because $x_{k-1}+x_{k} \\leq \\frac{2}{3}\\left(x_{1}+x_{k-1}+x_{k-2}\\right) \\leq \\frac{2}{3}$. Repeating this procedure we can reduce the number of nonzero $x_{i}$ 's to two, increasing the value of $F$ in each step. It remains to maximize $F$ over $n$-tuples $\\left(x_{1}, x_{2}, 0, \\ldots, 0\\right)$ with $x_{1}, x_{2} \\geq 0, x_{1}+x_{2}=1$ : in this case $F$ equals $x_{1} x_{2}$ and attains its maximum value $\\frac{1}{4}$ when $x_{1}=x_{2}=\\frac{1}{2}, x_{3}=\\ldots, x_{n}=0$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "28", "problem_type": null, "exam": "IMO", "problem": "28. (NET 1) ${ }^{\\mathrm{IMO}}$ Given a real number $a>1$, construct an infinite and bounded sequence $x_{0}, x_{1}, x_{2}, \\ldots$ such that for all natural numbers $i$ and $j, i \\neq j$, the following inequality holds:  $$ \\left|x_{i}-x_{j}\\right||i-j|^{a} \\geq 1 $$", "solution": "28. Let $x_{n}=c(n \\sqrt{2}-[n \\sqrt{2}])$ for some constant $c>0$. For $i>j$, putting $p=[i \\sqrt{2}]-[j \\sqrt{2}]$, we have $\\left|x_{i}-x_{j}\\right|=c|(i-j) \\sqrt{2}-p|=\\frac{\\left|2(i-j)^{2}-p^{2}\\right| c}{(i-j) \\sqrt{2}+p} \\geq \\frac{c}{(i-j) \\sqrt{2}+p} \\geq \\frac{c}{4(i-j)}$, because $p<(i-j) \\sqrt{2}+1$. Taking $c=4$, we obtain that for any $i>j$, $(i-j)\\left|x_{i}-x_{j}\\right| \\geq 1$. Of course, this implies $(i-j)^{a}\\left|x_{i}-x_{j}\\right| \\geq 1$ for any $a>1$ 。 Remark. The constant 4 can be replaced with $3 / 2+\\sqrt{2}$. Second solution. Another example of a sequence $\\left\\{x_{n}\\right\\}$ is constructed in the following way: $x_{1}=0, x_{2}=1, x_{3}=2$ and $x_{3^{k} i+m}=x_{m}+\\frac{i}{3^{k}}$ for $i=1,2$ and $1 \\leq m \\leq 3^{k}$. It is easily shown that $|i-j| \\cdot\\left|x_{i}-x_{j}\\right| \\geq 1 / 3$ for any $i \\neq j$. Third solution. If $n=b_{0}+2 b_{1}+\\cdots+2^{k} b_{k}, b_{i} \\in\\{0,1\\}$, then one can set $x_{n}$ to be $=b_{0}+2^{-a} b_{1}+\\cdots+2^{-k a} b_{k}$. In this case it holds that $|i-j|^{a}\\left|x_{i}-x_{j}\\right| \\geq$ $\\frac{2^{a}-2}{2^{a}-1}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "29", "problem_type": null, "exam": "IMO", "problem": "29. (FIN 2) We call a set $S$ on the real line $\\mathbb{R}$ superinvariant if for any stretching $A$ of the set by the transformation taking $x$ to $A(x)=x_{0}+$ $a\\left(x-x_{0}\\right)$ there exists a translation $B, B(x)=x+b$, such that the images of $S$ under $A$ and $B$ agree; i.e., for any $x \\in S$ there is a $y \\in S$ such that $A(x)=B(y)$ and for any $t \\in S$ there is a $u \\in S$ such that $B(t)=A(u)$. Determine all superinvariant sets. Remark. It is assumed that $a>0$.", "solution": "29. One easily observes that the following sets are super-invariant: one-point set, its complement, closed and open half-lines or their complements, and the whole real line. To show that these are the only possibilities, we first observe that $S$ is super-invariant if and only if for each $a>0$ there is a $b$ such that $x \\in S \\Leftrightarrow a x+b \\in S$. (i) Suppose that for some $a$ there are two such $b$ 's: $b_{1}$ and $b_{2}$. Then $x \\in$ $S \\Leftrightarrow a x+b_{1} \\in S$ and $x \\in S \\Leftrightarrow a x+b_{2} \\in S$, which implies that $S$ is periodic: $y \\in S \\Leftrightarrow y+\\frac{b_{1}-b_{2}}{a} \\in S$. Since $S$ is identical to a translate of any stretching of $S$, all positive numbers are periods of $S$. Therefore $S \\equiv \\mathbb{R}$. (ii) Assume that, for each $a, b=f(a)$ is unique. Then for any $a_{1}$ and $a_{2}$,  $$ \\begin{aligned} x \\in S & \\Leftrightarrow a_{1} x+f\\left(a_{1}\\right) \\in S \\Leftrightarrow a_{1} a_{2} x+a_{2} f\\left(a_{1}\\right)+f\\left(a_{2}\\right) \\in S \\\\ & \\Leftrightarrow a_{2} x+f\\left(a_{2}\\right) \\in S \\Leftrightarrow a_{1} a_{2} x+a_{1} f\\left(a_{2}\\right)+f\\left(a_{1}\\right) \\in S . \\end{aligned} $$  As above it follows that $a_{1} f\\left(a_{2}\\right)+f\\left(a_{1}\\right)=a_{2} f\\left(a_{1}\\right)+f\\left(a_{2}\\right)$, or equivalently $f\\left(a_{1}\\right)\\left(a_{2}-1\\right)=f\\left(a_{2}\\right)\\left(a_{1}-1\\right)$. Hence (for some $\\left.c\\right), f(a)=c(a-1)$ for all $a$. Now $x \\in S \\Leftrightarrow a x+c(a-1) \\in S$ actually means that $y-c \\in S \\Leftrightarrow a y-c \\in S$ for all $a$. Then it is easy to conclude that $\\{y-c \\mid y \\in S\\}$ is either a half-line or the whole line, and so is $S$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (PRK 1) Let $S$ be any point on the circumscribed circle of $\\triangle P Q R$. Then the feet of the perpendiculars from $S$ to the three sides of the triangle lie on the same straight line. Denote this line by $l(S, P Q R)$. Suppose that the hexagon $A B C D E F$ is inscribed in a circle. Show that the four lines $l(A, B D F), l(B, A C E), l(D, A B F)$, and $l(E, A B C)$ intersect at one point if and only if $C D E F$ is a rectangle.", "solution": "3. Consider the problem with the unit circle on the complex plane. For convenience, we use the same letter for a point in the plane and its corresponding complex number. Lemma 1. Line $l(S, P Q R)$ contains the point $Z=\\frac{P+Q+R+S}{2}$.  Proof. Suppose $P^{\\prime}, Q^{\\prime}, R^{\\prime}$ are the feet of perpendiculars from $S$ to $Q R$, $R P, P Q$ respectively. It suffices to show that $P^{\\prime}, Q^{\\prime}, R^{\\prime}, Z$ are on the same line. Let us first represent $P^{\\prime}$ by $Q, R, S$. Since $P^{\\prime} \\in Q R$, we have $\\frac{P^{\\prime}-Q}{R-Q}=\\overline{\\left(\\frac{P^{\\prime}-Q}{R-Q}\\right)}$, that is,  $$ \\left(P^{\\prime}-Q\\right)(\\bar{R}-\\bar{Q})=\\left(\\overline{P^{\\prime}}-\\bar{Q}\\right)(R-Q) $$  On the other hand, since $S P^{\\prime} \\perp Q R$, the ratio $\\frac{P^{\\prime}-S}{R-Q}$ is purely imaginary. Thus  $$ \\left(P^{\\prime}-S\\right)(\\bar{R}-\\bar{Q})=-\\left(\\overline{P^{\\prime}}-\\bar{S}\\right)(R-Q) $$  Eliminating $\\overline{P^{\\prime}}$ from (1) and (2) and using the fact that $\\bar{X}=X^{-1}$ for $X$ on the unit circle, we obtain $P^{\\prime}=(Q+R+S-Q R / S) / 2$ and analogously $Q^{\\prime}=(P+R+S-P R / S) / 2$ and $R^{\\prime}=(P+Q+S-$ $P Q / S) / 2$. Hence $Z-P^{\\prime}=(P+Q R / S) / 2, Z-Q^{\\prime}=(Q+P R / S) / 2$ and $Z-R^{\\prime}=(R+P Q / S) / 2$. Setting $P=p^{2}, Q=q^{2}, R=r^{2}$, $S=s^{2}$ we obtain $Z-P^{\\prime}=\\frac{p q r}{2 s}\\left(\\frac{p s}{q r}+\\frac{q r}{p s}\\right), Z-Q^{\\prime}=\\frac{p q r}{2 s}\\left(\\frac{q s}{p r}+\\frac{p r}{q s}\\right)$ and $Z-P^{\\prime}=\\frac{p q r}{2 s}\\left(\\frac{r s}{p q}+\\frac{p q}{r s}\\right)$. Since $x+x^{-1}=2 \\operatorname{Re} x$ is real for all $x$ on the unit circle, it follows that the ratio of every pair of these differences is real, which means that $Z, P^{\\prime}, Q^{\\prime}, R^{\\prime}$ belong to the same line. Lemma 2. If $P, Q, R, S$ are four different points on a circle, then the lines $l(P, Q R S), l(Q, R S P), l(R, S P Q), l(S, P Q R)$ intersect at one point. Proof. By Lemma 1, they all pass through $\\frac{P+Q+R+S}{2}$. Now we can find the needed conditions for $A, B, \\ldots, F$. In fact, the lines $l(A, B D F), l(D, A B F)$ meet at $Z_{1}=\\frac{A+B+D+F}{2}$, and $l(B, A C E)$, $l(E, A B C)$ meet at $Z_{2}=\\frac{A+B+C+E}{2}$. Hence, $Z_{1} \\equiv Z_{2}$ if and only if $D-C=E-F \\Leftrightarrow C D E F$ is a rectangle. Remark. The line $l(S, P Q R)$ is widely known as Simson line; the proof that the feet of perpendiculars are collinear is straightforward. The key claim, Lemma 1, is a known property of Simson lines, and can be shown elementarily:  * $l(S, P Q R)$ passes through the midpoint $X$ of $H S$, where $H$ is the orthocenter of $P Q R$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "30", "problem_type": null, "exam": "IMO", "problem": "30. (BUL 3) Two students $A$ and $B$ are playing the following game: Each of them writes down on a sheet of paper a positive integer and gives the sheet to the referee. The referee writes down on a blackboard two integers, one of which is the sum of the integers written by the players. After that, the referee asks student $A$ : \"Can you tell the integer written by the other student?\" If $A$ answers \"no,\" the referee puts the same question to student $B$. If $B$ answers \"no,\" the referee puts the question back to $A$, and so on. Assume that both students are intelligent and truthful. Prove that after a finite number of questions, one of the students will answer \"yes.\"", "solution": "30. Let $a$ and $b$ be the integers written by $A$ and $B$ respectively, and let $x<y$ be the two integers written by the referee. Suppose that none of $A$ and $B$ ever answers \"yes\". Initially, regardless of $a, A$ knows that $0 \\leq b \\leq y$ and answers \"no\". In the second step, $B$ knows that $A$ obtained $0 \\leq b \\leq y$, but if $a$ were greater than $x, A$ would know that $a+b=y$ and would thus answer \"yes\". So $B$ concludes $0 \\leq a \\leq x$ but answers \"no\". The process continues. Suppose that, in the $n$-th step, $A$ knows that $B$ obtained $r_{n-1} \\leq a \\leq s_{n-1}$. If $b>x-r_{n-1}, B$ would know that $a+b>x$ and hence $a+b=y$, while if $b<y-s_{n-1}, B$ would know that $a+b<y$, i.e. $a+b=x$ : in both cases he would be able to guess $a$. However, $B$ answered \"no\", from which $A$ concludes $y-s_{n-1} \\leq b \\leq x-r_{n-1}$. Put $r_{n}=y-s_{n-1}$ and $s_{n}=x-r_{n-1}$. Similarly, in the next step $B$ knows that $A$ obtained $r_{n} \\leq b \\leq s_{n}$ and, since $A$ answered \"no\", concludes $y-s_{n} \\leq a \\leq x-r_{n}$. Put $r_{n+1}=y-s_{n}$ and $s_{n+1}=x-r_{n}$. Notice that in both cases $s_{i+1}-r_{i+1}=s_{i}-r_{i}-(y-x)$. Since $y-x>0$, there exists an $m$ for which $s_{m}-r_{m}<0$, a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (FRA 2) ${ }^{\\mathrm{IMO5}}$ Let $A B C$ be a triangle and $M$ an interior point in $A B C$. Show that at least one of the angles $\\measuredangle M A B, \\measuredangle M B C$, and $\\measuredangle M C A$ is less than or equal to $30^{\\circ}$.", "solution": "4. Assume the contrary, that $\\angle M A B, \\angle M B C, \\angle M C A$ are all greater than $30^{\\circ}$. By the sine Ceva theorem, it holds that  $$ \\begin{aligned} & \\sin \\angle M A C \\sin \\angle M B A \\sin \\angle M C B \\\\ = & \\sin \\angle M A B \\sin \\angle M B C \\sin \\angle M C A>\\sin ^{3} 30^{\\circ}=\\frac{1}{8} . \\end{aligned} $$  On the other hand, since $\\angle M A C+\\angle M B A+\\angle M C B<180^{\\circ}-3 \\cdot 30^{\\circ}=90^{\\circ}$, Jensen's inequality applied on the concave function $\\ln \\sin x(x \\in[0, \\pi])$ gives us $\\sin \\angle M A C \\sin \\angle M B A \\sin \\angle M C B<\\sin ^{3} 30^{\\circ}$, contradicting (*).  Second solution. Denote the intersections of $P A, P B, P C$ with $B C, C A$, $A B$ by $A_{1}, B_{1}, C_{1}$, respectively. Suppose that each of the angles $\\angle P A B$, $\\angle P B C, \\angle P C A$ is greater than $30^{\\circ}$ and denote $P A=2 x, P B=2 y, P C=$ $2 z$. Then $P C_{1}>x, P A_{1}>y, P B_{1}>z$. On the other hand, we know that  $$ \\frac{P C_{1}}{P C+P C_{1}}+\\frac{P A_{1}}{P A+P A_{1}}+\\frac{P B_{1}}{P B+P B_{1}}=\\frac{S_{A B P}}{S_{A B C}}+\\frac{S_{P B C}}{S_{A B C}}+\\frac{S_{A P C}}{S_{A B C}}=1 . $$  Since the function $\\frac{t}{p+t}$ is increasing, we obtain $\\frac{x}{2 z+x}+\\frac{y}{2 x+y}+\\frac{z}{2 y+z}<1$. But on the contrary, Cauchy-Schwartz inequality (or alternatively Jensen's inequality) yields  $$ \\frac{x}{2 z+x}+\\frac{y}{2 x+y}+\\frac{z}{2 y+z} \\geq \\frac{(x+y+z)^{2}}{x(2 z+x)+y(2 x+y)+z(2 y+z)}=1 . $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (SPA 4) In the triangle $A B C$, with $\\measuredangle A=60^{\\circ}$, a parallel $I F$ to $A C$ is drawn through the incenter $I$ of the triangle, where $F$ lies on the side $A B$. The point $P$ on the side $B C$ is such that $3 B P=B C$. Show that $\\measuredangle B F P=\\measuredangle B / 2$.", "solution": "5. Let $P_{1}$ be the point on the side $B C$ such that $\\angle B F P_{1}=\\beta / 2$. Then $\\angle B P_{1} F=180^{\\circ}-3 \\beta / 2$, and the sine law gives us $\\frac{B F}{B P_{1}}=\\frac{\\sin (3 \\beta / 2)}{\\sin (\\beta / 2)}=$ $3-4 \\sin ^{2}(\\beta / 2)=1+2 \\cos \\beta$. Now we calculate $\\frac{B F}{B P}$. We have $\\angle B I F=120^{\\circ}-\\beta / 2, \\angle B F I=60^{\\circ}$ and $\\angle B I C=120^{\\circ}, \\angle B C I=\\gamma / 2=60^{\\circ}-\\beta / 2$. By the sine law,  $$ B F=B I \\frac{\\sin \\left(120^{\\circ}-\\beta / 2\\right)}{\\sin 60^{\\circ}}, \\quad B P=\\frac{1}{3} B C=B I \\frac{\\sin 120^{\\circ}}{3 \\sin \\left(60^{\\circ}-\\beta / 2\\right)} . $$  It follows that $\\frac{B F}{B P}=\\frac{3 \\sin \\left(60^{\\circ}-\\beta / 2\\right) \\sin \\left(60^{\\circ}+\\beta / 2\\right)}{\\sin ^{2} 60^{\\circ}}=4 \\sin \\left(60^{\\circ}-\\beta / 2\\right) \\sin \\left(60^{\\circ}+\\right.$ $\\beta / 2)=2\\left(\\cos \\beta-\\cos 120^{\\circ}\\right)=2 \\cos \\beta+1=\\frac{B F}{B P_{1}}$. Therefore $P \\equiv P_{1}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (USS 4) ${ }^{\\mathrm{IMO}}$ Prove for each triangle $A B C$ the inequality  $$ \\frac{1}{4}<\\frac{I A \\cdot I B \\cdot I C}{l_{A} l_{B} l_{C}} \\leq \\frac{8}{27} $$  where $I$ is the incenter and $l_{A}, l_{B}, l_{C}$ are the lengths of the angle bisectors of $A B C$.", "solution": "6. Let $a, b, c$ be sides of the triangle. Let $A_{1}$ be the intersection of line $A I$ with $B C$. By the known fact, $B A_{1}: A_{1} C=c: b$ and $A I: I A_{1}=A B: B A_{1}$, hence $B A_{1}=\\frac{a c}{b+c}$ and $\\frac{A I}{I A_{1}}=\\frac{A B}{B A_{1}}=\\frac{b+c}{a}$. Consequently $\\frac{A I}{l_{A}}=\\frac{b+c}{a+b+c}$. Put $a=n+p, b=p+m, c=m+n$ : it is obvious that $m, n, p$ are positive. Our inequality becomes  $$ 2<\\frac{(2 m+n+p)(m+2 n+p)(m+n+2 p)}{(m+n+p)^{3}} \\leq \\frac{64}{27} $$  The right side inequality immediately follows from the inequality between arithmetic and geometric means applied on $2 m+n+p, m+2 n+p$ and $m+n+2 p$. For the left side inequality, denote by $T=m+n+p$. Then we can write $(2 m+n+p)(m+2 n+p)(m+n+2 p)=(T+m)(T+n)(T+p)$ and $(T+m)(T+n)(T+p)=T^{3}+(m+n+p) T^{2}+(m n+n p+p n) T+m n p>2 T^{3}$. Remark. The inequalities cannot be improved. In fact, $\\frac{A I \\cdot B I \\cdot C I}{l_{A} l_{B} l_{C}}$ is equal to $8 / 27$ for $a=b=c$, while it can be arbitrarily close to $1 / 4$ if $a=b$ and $c$ is sufficiently small.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (CHN 2) Let $O$ be the center of the circumsphere of a tetrahedron $A B C D$. Let $L, M, N$ be the midpoints of $B C, C A, A B$ respectively, and assume that $A B+B C=A D+C D, B C+C A=B D+A D$, and $C A+A B=$ $C D+B D$. Prove that $\\angle L O M=\\angle M O N=\\angle N O L$.", "solution": "7. The given equations imply $A B=C D, A C=B D, A D=B C$. Let $L_{1}$, $M_{1}, N_{1}$ be the midpoints of $A D, B D, C D$ respectively. Then the above equalities yield  $$ \\begin{gathered} L_{1} M_{1}=A B / 2=L M \\\\ L_{1} M_{1}\\|A B\\| L M \\\\ L_{1} M=C D / 2=L M_{1} \\\\ L_{1} M\\|C D\\| L M_{1} \\end{gathered} $$  Thus $L, M, L_{1}, M_{1}$ are coplanar and $L M L_{1} M_{1}$ is a rhombus as well as ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-556.jpg?height=317&width=410&top_left_y=248&top_left_x=888) $M N M_{1} N_{1}$ and $L N L_{1} N_{1}$. Then the segments $L L_{1}, M M_{1}, N N_{1}$ have the common midpoint $Q$ and $Q L \\perp Q M$, $Q L \\perp Q N, Q M \\perp Q N$. We also infer that the line $N N_{1}$ is perpendicular to the plane $L M L_{1} M_{1}$ and hence to the line $A B$. Thus $Q A=Q B$, and similarly, $Q B=Q C=Q D$, hence $Q$ is just the center $O$, and $\\angle L O M=$ $\\angle M O N=\\angle N O L=90^{\\circ}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (NET 1) Let $S$ be a set of $n$ points in the plane. No three points of $S$ are collinear. Prove that there exists a set $P$ containing $2 n-5$ points satisfying the following condition: In the interior of every triangle whose three vertices are elements of $S$ lies a point that is an element of $P$.", "solution": "8. Let $P_{1}\\left(x_{1}, y_{1}\\right), P_{2}\\left(x_{2}, y_{2}\\right), \\ldots, P_{n}\\left(x_{n}, y_{n}\\right)$ be the $n$ points of $S$ in the coordinate plane. We may assume $x_{1}<x_{2}<\\cdots<x_{n}$ (choosing adequate axes and renumbering the points if necessary). Define $d$ to be half the minimum distance of $P_{i}$ from the line $P_{j} P_{k}$, where $i, j, k$ go through all possible combinations of mutually distinct indices. First we define a set $T$ containing $2 n-4$ points:  $$ T=\\left\\{\\left(x_{i}, y_{i}-d\\right),\\left(x_{i}, y_{i}+d\\right) \\mid i=2,3, \\ldots, n-1\\right\\} $$  Consider any triangle $P_{k} P_{l} P_{m}$, where $k<l<m$. Its interior contains at least one of the two points $\\left(x_{l}, y_{l} \\pm d\\right)$, so $T$ is a set of $2 n-4$ points with the required property. However, at least one of the points of $T$ is useless. The convex hull of $S$ is a polygon with at least three points in $S$ as vertices. Let $P_{j}$ be a vertex of that hull distinct from $P_{1}$ and $P_{n}$. Clearly one of the points $\\left(x_{j}, y_{j} \\pm d\\right)$ lies outside the convex hull, and thus can be left out. The remaining set of $2 n-5$ points satisfies the conditions.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1991", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (FRA 3) In the plane we are given a set $E$ of 1991 points, and certain pairs of these points are joined with a path. We suppose that for every point of $E$, there exist at least 1593 other points of $E$ to which it is joined by a path. Show that there exist six points of $E$ every pair of which are joined by a path. Alternative version. Is it possible to find a set $E$ of 1991 points in the plane and paths joining certain pairs of the points in $E$ such that every point of $E$ is joined with a path to at least 1592 other points of $E$, and in every subset of six points of $E$ there exist at least two points that are not joined?", "solution": "9. Let $A_{1}, A_{2}$ be two points of $E$ which are joined. In $E \\backslash\\left\\{A_{1}, A_{2}\\right\\}$, there are at most 397 points to which $A_{1}$ is not joined, and at most as much to which $A_{2}$ is not joined. Consequently, there exists a point $A_{3}$ which is joined to both $A_{1}$ and $A_{2}$. There are at most $3 \\cdot 397=1191$ points of $E \\backslash\\left\\{A_{1}, A_{2}, A_{3}\\right\\}$ to which at least one of $A_{1}, A_{2}, A_{3}$ is not joined, hence it is possible to choose a point $A_{4}$ joined to $A_{1}, A_{2}, A_{3}$. Similarly, there exists a point $A_{5}$ which is joined to all $A_{1}, A_{2}, A_{3}, A_{4}$. Finally, among the remaining 1986 points, there are at most $5 \\cdot 397=1985$ which are not joined to one of the points $A_{1}, \\ldots, A_{5}$. Thus there is at least one point $A_{6}$ joined to all $A_{1}, \\ldots, A_{5}$. It is clear that $A_{1}, \\ldots, A_{6}$ are pairwise joined. Solution of the alternative version. Let be given 1991 points instead. Number the points from 1 to 1991, and join $i$ and $j$ if and only if $i-j$ is not a multiple of 5 . Then each $i$ is joined to 1592 or 1593 other points, and obviously among any six points there are two which are not joined.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (AUS 2) Prove that for any positive integer $m$ there exist an infinite number of pairs of integers $(x, y)$ such that (i) $x$ and $y$ are relatively prime; (ii) $y$ divides $x^{2}+m$; (iii) $x$ divides $y^{2}+m$.", "solution": "1. Assume that a pair $(x, y)$ with $x<y$ satisfies the required conditions. We claim that the pair $\\left(y, x_{1}\\right)$ also satisfies the conditions, where $x_{1}=\\frac{y^{2}+m}{x}$ (note that $x_{1}>y$ is a positive integer). This will imply the desired result, since starting from the pair $(1,1)$ we can obtain arbitrarily many solutions. First, we show that $\\operatorname{gcd}\\left(x_{1}, y\\right)=1$. Suppose to the contrary that $\\operatorname{gcd}\\left(x_{1}, y\\right)$ $=d>1$. Then $d\\left|x_{1}\\right| y^{2}+m \\Rightarrow d \\mid m$, which implies $d|y| x^{2}+m \\Rightarrow d \\mid x$. But this last is impossible, since $\\operatorname{gcd}(x, y)=1$. Thus it remains to show that $x_{1} \\mid y^{2}+m$ and $y \\mid x_{1}^{2}+m$. The former relation is obvious. Since $\\operatorname{gcd}(x, y)=1$, the latter is equivalent to $y \\mid\\left(x x_{1}\\right)^{2}+m x^{2}=y^{4}+2 m y^{2}+$ $m^{2}+m x^{2}$, which is true because $y \\mid m\\left(m+x^{2}\\right)$ by the assumption. Hence $\\left(y, x_{1}\\right)$ indeed satisfies all the required conditions. Remark. The original problem asked to prove the existence of a pair $(x, y)$ of positive integers satisfying the given conditions such that $x+y \\leq m+1$. The problem in this formulation is trivial, since the pair $x=y=1$ satisfies the conditions. Moreover, this is sometimes the only solution with $x+y \\leq m+1$. For example, for $m=3$ the least nontrivial solution is $\\left(x_{0}, y_{0}\\right)=(1,4)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (ITA 1) ${ }^{\\mathrm{IMO5}}$ Let $V$ be a finite subset of Euclidean space consisting of points $(x, y, z)$ with integer coordinates. Let $S_{1}, S_{2}, S_{3}$ be the projections of $V$ onto the $y z, x z, x y$ planes, respectively. Prove that  $$ |V|^{2} \\leq\\left|S_{1}\\right|\\left|S_{2}\\right|\\left|S_{3}\\right| $$  ( $|X|$ denotes the number of elements of $X$ ).", "solution": "10. Let us set $S(x)=\\{(y, z) \\mid(x, y, z) \\in V\\}, S_{y}(x)=\\left\\{z \\mid(x, z) \\in S_{y}\\right\\}$ and $S_{z}(x)=\\left\\{y \\mid(x, y) \\in S_{z}\\right\\}$. Clearly $S(x) \\subset S_{x}$ and $S(x) \\subset S_{y}(x) \\times S_{z}(x)$. It follows that  $$ \\begin{aligned} |V| & =\\sum_{x}|S(x)| \\leq \\sum_{x} \\sqrt{\\left|S_{x}\\right|\\left|S_{y}(x)\\right|\\left|S_{z}(x)\\right|} \\\\ & =\\sqrt{\\left|S_{x}\\right|} \\sum_{x} \\sqrt{\\left|S_{y}(x)\\right|\\left|S_{z}(x)\\right|} \\end{aligned} $$  Using the Cauchy-Schwarz inequality we also get  $$ \\sum_{x} \\sqrt{\\left|S_{y}(x)\\right|\\left|S_{z}(x)\\right|} \\leq \\sqrt{\\sum_{x}\\left|S_{y}(x)\\right|} \\sqrt{\\sum_{x}\\left|S_{z}(x)\\right|}=\\sqrt{\\left|S_{y}\\right|\\left|S_{z}\\right|} $$  Now (1) and (2) together yield $|V| \\leq \\sqrt{\\left|S_{x}\\right|\\left|S_{y}\\right|\\left|S_{z}\\right|}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (JAP 2) In a triangle $A B C$, let $D$ and $E$ be the intersections of the bisectors of $\\angle A B C$ and $\\angle A C B$ with the sides $A C, A B$, respectively. Determine the angles $\\angle A, \\angle B, \\angle C$ if  $$ \\measuredangle B D E=24^{\\circ}, \\quad \\measuredangle C E D=18^{\\circ} . $$", "solution": "11. Let $I$ be the incenter of $\\triangle A B C$. Since $90^{\\circ}+\\alpha / 2=\\angle B I C=\\angle D I E=$ $138^{\\circ}$, we obtain that $\\angle A=96^{\\circ}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-571.jpg?height=295&width=1043&top_left_y=1086&top_left_x=260)  Let $D^{\\prime}$ and $E^{\\prime}$ be the points symmetric to $D$ and $E$ with respect to $C E$ and $B D$ respectively, and let $S$ be the intersection point of $E D^{\\prime}$ and $B D$. Then $\\angle B D E^{\\prime}=24^{\\circ}$ and $\\angle D^{\\prime} D E^{\\prime}=\\angle D^{\\prime} D E-\\angle E^{\\prime} D E=24^{\\circ}$, which means that $D E^{\\prime}$ bisects the angle $S D D^{\\prime}$. Moreover, $\\angle E^{\\prime} S B=\\angle E S B=$ $\\angle E D S+\\angle D E S=60^{\\circ}$ and hence $S E^{\\prime}$ bisects the angle $D^{\\prime} S B$. It follows that $E^{\\prime}$ is the excenter of $\\triangle D^{\\prime} D S$ and consequently $\\angle D^{\\prime} D C=\\angle D D^{\\prime} C=$ $\\angle S D^{\\prime} E^{\\prime}=\\left(180^{\\circ}-72^{\\circ}\\right) / 2=54^{\\circ}$. Finally, $\\angle C=180^{\\circ}-2 \\cdot 54^{\\circ}=72^{\\circ}$ and $\\angle B=12^{\\circ}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (NET 1) Let $f, g$, and $a$ be polynomials with real coefficients, $f$ and $g$ in one variable and $a$ in two variables. Suppose  $$ f(x)-f(y)=a(x, y)(g(x)-g(y)) \\quad \\text { for all } x, y \\in \\mathbb{R} $$  Prove that there exists a polynomial $h$ with $f(x)=h(g(x))$ for all $x \\in \\mathbb{R}$.", "solution": "12. Let us set $\\operatorname{deg} f=n$ and $\\operatorname{deg} g=m$. We shall prove the result by induction on $n$. If $n<m$, then $\\operatorname{deg}_{x}[f(x)-f(y)]<\\operatorname{deg}_{x}[g(x)-g(y)]$, which implies that $f(x)-f(y)=0$, i.e., that $f$ is constant. The statement trivially holds. Assume now that $n \\geq m$. Transition to $f_{1}(x)=f(x)-f(0)$ and $g_{1}(x)=$ $g(x)-g(0)$ allows us to suppose that $f(0)=g(0)=0$. Then the given condition for $y=0$ gives us $f(x)=f_{1}(x) g(x)$, where $f_{1}(x)=a(x, 0)$ and $\\operatorname{deg} f_{1}=n-m$. We now have  $$ \\begin{aligned} a(x, y)(g(x)-g(y)) & =f(x)-f(y)=f_{1}(x) g(x)-f_{1}(y) g(y) \\\\ & =\\left[f_{1}(x)-f_{1}(y)\\right] g(x)+f_{1}(y)[g(x)-g(y)] . \\end{aligned} $$  Since $g(x)$ is relatively prime to $g(x)-g(y)$, it follows that $f_{1}(x)-f_{1}(y)=$ $b(x, y)(g(x)-g(y))$ for some polynomial $b(x, y)$. By the induction hypothesis there exists a polynomial $h_{1}$ such that $f_{1}(x)=h_{1}(g(x))$ and consequently $f(x)=g(x) \\cdot h_{1}(g(x))=h(g(x))$ for $h(t)=t h_{1}(t)$. Thus the induction is complete.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (NZL 1) ${ }^{\\mathrm{IMO1}}$ Find all integer triples $(p, q, r)$ such that $1<p<q<r$ and $(p-1)(q-1)(r-1)$ is a divisor of $(p q r-1)$.", "solution": "13. Let us define  $$ \\begin{aligned} F(p, q, r)= & \\frac{(p q r-1)}{(p-1)(q-1)(r-1)} \\\\ = & 1+\\frac{1}{p-1}+\\frac{1}{q-1}+\\frac{1}{r-1} \\\\ & +\\frac{1}{(p-1)(q-1)}+\\frac{1}{(q-1)(r-1)}+\\frac{1}{(r-1)(p-1)} \\end{aligned} $$  Obviously $F$ is a decreasing function of $p, q, r$. Suppose that $1<p<q<r$ are integers for which $F(p, q, r)$ is an integer. Observe that $p, q, r$ are either all even or all odd. Indeed, if for example $p$ is odd and $q$ is even, then $p q r-1$ is odd while $(p-1)(q-1)(r-1)$ is even, which is impossible. Also, if $p, q, r$ are even then $F(p, q, r)$ is odd. If $p \\geq 4$, then $1<F(p, q, r) \\leq F(4,6,8)=191 / 105<2$, which is impossible. Hence $p \\leq 3$. Let $p=2$. Then $q, r$ are even and $1<F(2, q, r) \\leq F(2,4,6)=47 / 15<4$. Therefore $F(2, q, r)=3$. This equality reduces to $(q-3)(r-3)=5$, with the unique solution $q=4, r=8$. Let $p=3$. Then $q, r$ are odd and $1<F(3, q, r) \\leq F(3,5,7)=104 / 48<3$. Therefore $F(3, q, r)=2$. This equality reduces to $(q-4)(r-4)=11$, which leads to $q=5, r=15$. Hence the only solutions $(p, q, r)$ of the problem are $(2,4,8)$ and $(3,5,15)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (POL 2) For any positive integer $x$ define  $$ \\begin{aligned} g(x) & =\\text { greatest odd divisor of } x, \\\\ f(x) & =\\left\\{\\begin{array}{l} x / 2+x / g(x), \\text { if } x \\text { is even; } \\\\ 2^{(x+1) / 2}, \\text { if } x \\text { is odd. } \\end{array}\\right. \\end{aligned} $$  Construct the sequence $x_{1}=1, x_{n+1}=f\\left(x_{n}\\right)$. Show that the number 1992 appears in this sequence, determine the least $n$ such that $x_{n}=1992$, and determine whether $n$ is unique.", "solution": "14. We see that $x_{1}=2^{0}$. Suppose that for some $m, r \\in \\mathbb{N}$ we have $x_{m}=2^{r}$. Then inductively $x_{m+i}=2^{r-i}(2 i+1)$ for $i=1,2, \\ldots, r$ and $x_{m+r+1}=$ $2^{r+1}$. Since every natural number can be uniquely represented as the product of an odd number and a power of two, we conclude that every natural number occurs in our sequence exactly once. Moreover, it follows that $2 k-1=x_{k(k+1) / 2}$. Thus $x_{n}=1992=2^{3} \\cdot 249$ implies that $x_{n+3}=255=2 \\cdot 128-1=x_{128 \\cdot 129 / 2}=x_{8256}$. Hence $n=8253$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (PRK 2) Does there exist a set $M$ with the following properties? (i) The set $M$ consists of 1992 natural numbers. (ii) Every element in $M$ and the sum of any number of elements have the form $m^{k}(m, k \\in \\mathbb{N}, k \\geq 2)$.", "solution": "15. The result follows from the following lemma by taking $n=\\frac{1992 \\cdot 1993}{2}$ and $M=\\{d, 2 d, \\ldots, 1992 d\\}$. Lemma. For every $n \\in \\mathbb{N}$ there exists a natural number $d$ such that all the numbers $d, 2 d, \\ldots, n d$ are of the form $m^{k}(m, k \\in \\mathbb{N}, k \\geq 2)$. Proof. Let $p_{1}, p_{2}, \\ldots, p_{n}$ be distinct prime numbers. We shall find $d$ in the form $d=2^{\\alpha_{2}} 3^{\\alpha_{3}} \\cdots n^{\\alpha_{n}}$, where $\\alpha_{i} \\geq 0$ are integers such that $k d$ is a perfect $p_{k}$ th power. It is sufficient to find $\\alpha_{i}, i=2,3, \\ldots, n$, such that $\\alpha_{i} \\equiv 0\\left(\\bmod p_{j}\\right)$ if $i \\neq j$ and $\\alpha_{i} \\equiv-1\\left(\\bmod p_{j}\\right)$ if $i=j$. But the existence of such $\\alpha_{i}$ 's is an immediate consequence of the Chinese remainder theorem.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (KOR 2) Prove that $N=\\frac{5^{125}-1}{5^{25}-1}$ is a composite number.", "solution": "16. Observe that $x^{4}+x^{3}+x^{2}+x+1=\\left(x^{2}+3 x+1\\right)^{2}-5 x(x+1)^{2}$. Thus for $x=5^{25}$ we have  $$ \\begin{aligned} N & =x^{4}+x^{3}+x^{2}+x+1 \\\\ & =\\left(x^{2}+3 x+1-5^{13}(x+1)\\right)\\left(x^{2}+3 x+1+5^{13}(x+1)\\right)=A \\cdot B . \\end{aligned} $$  Clearly, both $A$ and $B$ are positive integers greater than 1.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (SWE 1) Let $\\alpha(n)$ be the number of digits equal to one in the binary representation of a positive integer $n$. Prove that: (a) the inequality $\\alpha\\left(n^{2}\\right) \\leq \\frac{1}{2} \\alpha(n)(\\alpha(n)+1)$ holds; (b) the above inequality is an equality for infinitely many positive integers; (c) there exists a sequence $\\left(n_{i}\\right)_{1}^{\\infty}$ such that $\\alpha\\left(n_{i}^{2}\\right) / \\alpha\\left(n_{i}\\right) \\rightarrow 0$ as $i \\rightarrow \\infty$.  Alternative parts: Prove that there exists a sequence $\\left(n_{i}\\right)_{1}^{\\infty}$ such that $\\alpha\\left(n_{i}^{2}\\right) / \\alpha\\left(n_{i}\\right)$ tends to (d) $\\infty$; (e) an arbitrary real number $\\gamma \\in(0,1)$; (f) an arbitrary real number $\\gamma \\geq 0$.", "solution": "17. (a) Let $n=\\sum_{i=1}^{k} 2^{a_{i}}$, so that $\\alpha(n)=k$. Then  $$ n^{2}=\\sum_{i} 2^{2 a_{i}}+\\sum_{i<j} 2^{a_{i}+a_{j}+1} $$  has at most $k+\\binom{k}{2}=\\frac{k(k+1)}{2}$ binary ones. (b) The above inequality is an equality for all numbers $n_{k}=2^{k}$. (c) Put $n_{m}=2^{2^{m}-1}-\\sum_{j=1}^{m} 2^{2^{m}-2^{j}}$, where $m>1$. It is easy to see that $\\alpha\\left(n_{m}\\right)=2^{m}-m$. On the other hand, squaring and simplifying yields $n_{m}^{2}=1+\\sum_{i<j} 2^{2^{m+1}+1-2^{i}-2^{j}}$. Hence $\\alpha\\left(n_{m}^{2}\\right)=1+\\frac{m(m+1)}{2}$ and thus  $$ \\frac{\\alpha\\left(n_{m}^{2}\\right)}{\\alpha\\left(n_{m}\\right)}=\\frac{2+m(m+1)}{2\\left(2^{m}-m\\right)} \\rightarrow 0 \\quad \\text { as } m \\rightarrow \\infty $$  Solution to the alternative parts. (1) Let $n=\\sum_{i=1}^{n} 2^{2^{i}}$. Then $n^{2}=\\sum_{i=1}^{n} 2^{2^{i+1}}+\\sum_{i<j} 2^{2^{i}+2^{j}+1}$ has exactly $\\frac{k(k+1)}{2}$ binary ones, and therefore $\\frac{\\alpha\\left(n^{2}\\right)}{\\alpha(n)}=\\frac{2 k}{k(k+1)} \\rightarrow \\infty$. (2) Consider the sequence $n_{i}$ constructed in part (c). Let $\\theta>1$ be a constant to be chosen later, and let $N_{i}=2^{m_{i}} n_{i}-1$ where $m_{i}>\\alpha\\left(n_{i}\\right)$ is such that $m_{i} / \\alpha\\left(n_{i}\\right) \\rightarrow \\theta$ as $i \\rightarrow \\infty$. Then $\\alpha\\left(N_{i}\\right)=\\alpha\\left(n_{i}\\right)+m_{i}-1$, whereas $N_{i}^{2}=2^{2 m_{i}} n_{i}^{2}-2^{m_{i}+1} n_{i}+1$ and $\\alpha\\left(N_{i}^{2}\\right)=\\alpha\\left(n_{i}^{2}\\right)-\\alpha\\left(n_{i}\\right)+m_{i}$. It follows that  $$ \\lim _{i \\rightarrow \\infty} \\frac{\\alpha\\left(N_{i}^{2}\\right)}{\\alpha\\left(N_{i}\\right)}=\\lim _{i \\rightarrow \\infty} \\frac{\\alpha\\left(n_{i}^{2}\\right)+(\\theta-1) \\alpha\\left(n_{i}\\right)}{(1+\\theta) \\alpha\\left(n_{i}\\right)}=\\frac{\\theta-1}{\\theta+1} $$  which is equal to $\\gamma \\in[0,1]$ for $\\theta=\\frac{1+\\gamma}{1-\\gamma}$ (for $\\gamma=1$ we set $m_{i} / \\alpha\\left(n_{i}\\right) \\rightarrow$ $\\infty)$. (3) Let be given a sequence $\\left(n_{i}\\right)_{i=1}^{\\infty}$ with $\\alpha\\left(n_{i}^{2}\\right) / \\alpha\\left(n_{i}\\right) \\rightarrow \\gamma$. Taking $m_{i}>$ $\\alpha\\left(n_{i}\\right)$ and $N_{i}=2^{m_{i}} n_{i}+1$ we easily find that $\\alpha\\left(N_{i}\\right)=\\alpha\\left(n_{i}\\right)+1$ and $\\alpha\\left(N_{i}^{2}\\right)=\\alpha\\left(n_{i}^{2}\\right)+\\alpha\\left(n_{i}\\right)+1$. Hence $\\alpha\\left(N_{i}^{2}\\right) / \\alpha\\left(N_{i}\\right)=\\gamma+1$. Continuing this procedure we can construct a sequence $t_{i}$ such that $\\alpha\\left(t_{i}^{2}\\right) / \\alpha\\left(t_{i}\\right)=$ $\\gamma+k$ for an arbitrary $k \\in \\mathbb{N}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (USA 2) Let $[x]$ denote the greatest integer less than or equal to $x$. Pick any $x_{1}$ in $[0,1)$ and define the sequence $x_{1}, x_{2}, x_{3}, \\ldots$ by $x_{n+1}=0$ if $x_{n}=0$ and $x_{n+1}=1 / x_{n}-\\left[1 / x_{n}\\right]$ otherwise. Prove that  $$ x_{1}+x_{2}+\\cdots+x_{n}<\\frac{F_{1}}{F_{2}}+\\frac{F_{2}}{F_{3}}+\\cdots+\\frac{F_{n}}{F_{n+1}} $$  where $F_{1}=F_{2}=1$ and $F_{n+2}=F_{n+1}+F_{n}$ for $n \\geq 1$.", "solution": "18. Let us define inductively $f^{1}(x)=f(x)=\\frac{1}{x+1}$ and $f^{n}(x)=f\\left(f^{n-1}(x)\\right)$, and let $g_{n}(x)=x+f(x)+f^{2}(x)+\\cdots+f^{n}(x)$. We shall prove first the following statement.  Lemma. The function $g_{n}(x)$ is strictly increasing on $[0,1]$, and $g_{n-1}(1)=$ $F_{1} / F_{2}+F_{2} / F_{3}+\\cdots+F_{n} / F_{n+1}$. Proof. Since $f(x)-f(y)=\\frac{y-x}{(1+x)(1+y)}$ is smaller in absolute value than $x-y$, it follows that $x>y$ implies $f^{2 k}(x)>f^{2 k}(y)$ and $f^{2 k+1}(x)<$ $f^{2 k+1}(y)$, and moreover that for every integer $k \\geq 0$,  $$ \\left[f^{2 k}(x)-f^{2 k}(y)\\right]+\\left[f^{2 k+1}(x)-f^{2 k+1}(y)\\right]>0 $$  Hence if $x>y$, we have $g_{n}(x)-g_{n}(y)=(x-y)+[f(x)-f(y)]+\\cdots+$ $\\left[f^{n}(x)-f^{n}(y)\\right]>0$, which yields the first part of the lemma. The second part follows by simple induction, since $f^{k}(1)=F_{k+1} / F_{k+2}$. If some $x_{i}=0$ and consequently $x_{j}=0$ for all $j \\geq i$, then the problem reduces to the problem with $i-1$ instead of $n$. Thus we may assume that all $x_{1}, \\ldots, x_{n}$ are different from 0 . If we write $a_{i}=\\left[1 / x_{i}\\right]$, then $x_{i}=\\frac{1}{a_{i}+x_{i+1}}$. Thus we can regard $x_{i}$ as functions of $x_{n}$ depending on $a_{1}, \\ldots, a_{n-1}$. Suppose that $x_{n}, a_{n-1}, \\ldots, a_{3}, a_{2}$ are fixed. Then $x_{2}, x_{3}, \\ldots, x_{n}$ are all fixed, and $x_{1}=\\frac{1}{a_{1}+x_{2}}$ is maximal when $a_{1}=1$. Hence the sum $S=$ $x_{1}+x_{2}+\\cdots+x_{n}$ is maximized for $a_{1}=1$. We shall show by induction on $i$ that $S$ is maximized for $a_{1}=a_{2}=\\cdots=$ $a_{i}=1$. In fact, assuming that the statement holds for $i-1$ and thus $a_{1}=$ $\\cdots=a_{i-1}=1$, having $x_{n}, a_{n-1}, \\ldots, a_{i+1}$ fixed we have that $x_{n}, \\ldots, x_{i+1}$ are also fixed, and that $x_{i-1}=f\\left(x_{i}\\right), \\ldots, x_{1}=f^{i-1}\\left(x_{i}\\right)$. Hence by the lemma, $S=g_{i-1}\\left(x_{i}\\right)+x_{i+1}+\\cdots+x_{n}$ is maximal when $x_{i}=\\frac{1}{a_{i}+x_{i+1}}$ is maximal, that is, for $a_{i}=1$. Thus the induction is complete. It follows that $x_{1}+\\cdots+x_{n}$ is maximal when $a_{1}=\\cdots=a_{n-1}=1$, so that $x_{1}+\\cdots+x_{n}=g_{n-1}\\left(x_{1}\\right)$. By the lemma, the latter does not exceed $g_{n-1}(1)$. This completes the proof. Remark. The upper bound is the best possible, because it is approached by taking $x_{n}$ close to 1 and inductively (in reverse) defining $x_{i-1}=\\frac{1}{1+x_{i}}=$ $\\frac{1}{a_{i}+x_{i}}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (IRE 1) Let $f(x)=x^{8}+4 x^{6}+2 x^{4}+28 x^{2}+1$. Let $p>3$ be a prime and suppose there exists an integer $z$ such that $p$ divides $f(z)$. Prove that there exist integers $z_{1}, z_{2}, \\ldots, z_{8}$ such that if  $$ g(x)=\\left(x-z_{1}\\right)\\left(x-z_{2}\\right) \\cdots\\left(x-z_{8}\\right) $$  then all coefficients of $f(x)-g(x)$ are divisible by $p$.", "solution": "19. Observe that $f(x)=\\left(x^{4}+2 x^{2}+3\\right)^{2}-8\\left(x^{2}-1\\right)^{2}=\\left[x^{4}+2(1-\\sqrt{2}) x^{2}+\\right.$ $3+2 \\sqrt{2}]\\left[x^{4}+2(1+\\sqrt{2}) x^{2}+3-2 \\sqrt{2}\\right]$. Now it is easy to find that the roots of $f$ are  $$ x_{1,2,3,4}= \\pm i(i \\sqrt[4]{2} \\pm 1) \\quad \\text { and } \\quad x_{5,6,7,8}= \\pm i(\\sqrt[4]{2} \\pm 1) $$  In other words, $x_{k}=\\alpha_{i}+\\beta_{j}$, where $\\alpha_{i}^{2}=-1$ and $\\beta_{j}^{4}=2$. We claim that any root of $f$ can be obtained from any other using rational functions. In fact, we have  $$ \\begin{aligned} & x^{3}=-\\alpha_{i}-3 \\beta_{j}+3 \\alpha_{i} \\beta_{j}^{2}+\\beta_{j}^{3} \\\\ & x^{5}=11 \\alpha_{i}+7 \\beta_{j}-10 \\alpha_{i} \\beta_{j}^{2}-10 \\beta_{j}^{3} \\\\ & x^{7}=-71 \\alpha_{i}-49 \\beta_{j}+35 \\alpha_{i} \\beta_{j}^{2}+37 \\beta_{j}^{3} \\end{aligned} $$  from which we easily obtain that $\\alpha_{i}=24^{-1}\\left(127 x+5 x^{3}+19 x^{5}+5 x^{7}\\right), \\quad \\beta_{j}=24^{-1}\\left(151 x+5 x^{3}+19 x^{5}+5 x^{7}\\right)$. Since all other values of $\\alpha$ and $\\beta$ can be obtained as rational functions of $\\alpha_{i}$ and $\\beta_{j}$, it follows that all the roots $x_{l}$ are rational functions of a particular root $x_{k}$. We now note that if $x_{1}$ is an integer such that $f\\left(x_{1}\\right)$ is divisible by $p$, then $p>3$ and $x_{1} \\in \\mathbb{Z}_{p}$ is a root of the polynomial $f$. By the previous consideration, all remaining roots $x_{2}, \\ldots, x_{8}$ of $f$ over the field $\\mathbb{Z}_{p}$ are rational functions of $x_{1}$, since 24 is invertible in $Z_{p}$. Then $f(x)$ factors as  $$ f(x)=\\left(x-x_{1}\\right)\\left(x-x_{2}\\right) \\cdots\\left(x-x_{8}\\right), $$  and the result follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (CHN 1) Let $\\mathbb{R}^{+}$be the set of all nonnegative real numbers. Given two positive real numbers $a$ and $b$, suppose that a mapping $f: \\mathbb{R}^{+} \\rightarrow \\mathbb{R}^{+}$ satisfies the functional equation  $$ f(f(x))+a f(x)=b(a+b) x $$  Prove that there exists a unique solution of this equation.", "solution": "2. Let us define $x_{n}$ inductively as $x_{n}=f\\left(x_{n-1}\\right)$, where $x_{0} \\geq 0$ is a fixed real number. It follows from the given equation in $f$ that $x_{n+2}=-a x_{n+1}+$ $b(a+b) x_{n}$. The general solution to this equation is of the form  $$ x_{n}=\\lambda_{1} b^{n}+\\lambda_{2}(-a-b)^{n}, $$  where $\\lambda_{1}, \\lambda_{2} \\in \\mathbb{R}$ satisfy $x_{0}=\\lambda_{1}+\\lambda_{2}$ and $x_{1}=\\lambda_{1} b-\\lambda_{2}(a+b)$. In order to have $x_{n} \\geq 0$ for all $n$ we must have $\\lambda_{2}=0$. Hence $x_{0}=\\lambda_{1}$ and $f\\left(x_{0}\\right)=x_{1}=\\lambda_{1} b=b x_{0}$. Since $x_{0}$ was arbitrary, we conclude that $f(x)=b x$ is the only possible solution of the functional equation. It is easily verified that this is indeed a solution.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. (FRA 1) ${ }^{\\mathrm{IMO} 4}$ In the plane, let there be given a circle $C$, a line $l$ tangent to $C$, and a point $M$ on $l$. Find the locus of points $P$ that have the following property: There exist two points $Q$ and $R$ on $l$ such that $M$ is the midpoint of $Q R$ and $C$ is the incircle of $P Q R$.", "solution": "20. Denote by $U$ the point of tangency of the circle $C$ and the line $l$. Let $X$ and $U^{\\prime}$ be the points symmetric to $U$ with respect to $S$ and $M$ respectively; these points do not depend on the choice of $P$. Also, let $C^{\\prime}$ be the excircle of $\\triangle P Q R$ corresponding to $P, S^{\\prime}$ the center of $C^{\\prime}$, and $W, W^{\\prime}$ the points of tangency of $C$ and $C^{\\prime}$ with the line $P Q$ respectively. Obviously, $\\triangle W S P \\sim \\triangle W^{\\prime} S^{\\prime} P$. Since $S X \\| S^{\\prime} U^{\\prime}$ and $S X: S^{\\prime} U^{\\prime}=$ $S W: S^{\\prime} W^{\\prime}=S P: S^{\\prime} P$, we deduce that $\\Delta S X P \\sim \\Delta S^{\\prime} U^{\\prime} P$, and consequently that $P$ lies on the line $X U^{\\prime}$. On the other hand, it is easy to show that each point $P$ of the ray $U^{\\prime} X$ over $X$ satisfies the required condition. Thus the desired locus is ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-575.jpg?height=543&width=539&top_left_y=908&top_left_x=799) the extension of $U^{\\prime} X$ over $X$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. (GBR 2) ${ }^{\\mathrm{IMO} 6}$ For each positive integer $n$, denote by $s(n)$ the greatest integer such that for all positive integers $k \\leq s(n), n^{2}$ can be expressed as a sum of squares of $k$ positive integers. (a) Prove that $s(n) \\leq n^{2}-14$ for all $n \\geq 4$. (b) Find a number $n$ such that $s(n)=\\overline{n^{2}}-14$. (c) Prove that there exist infinitely many positive integers $n$ such that $s(n)=n^{2}-14$.", "solution": "21. (a) Representing $n^{2}$ as a sum of $n^{2}-13$ squares is equivalent to representing 13 as a sum of numbers of the form $x^{2}-1, x \\in \\mathbb{N}$, such as $0,3,8,15, \\ldots$ But it is easy to check that this is impossible, and hence $s(n) \\leq n^{2}-14$. (b) Let us prove that $s(13)=13^{2}-14=155$. Observe that  $$ \\begin{aligned} 13^{2} & =8^{2}+8^{2}+4^{2}+4^{2}+3^{2} \\\\ & =8^{2}+8^{2}+4^{2}+4^{2}+2^{2}+2^{2}+1^{2} \\\\ & =8^{2}+8^{2}+4^{2}+3^{2}+3^{2}+2^{2}+1^{2}+1^{2}+1^{2} \\end{aligned} $$  Given any representation of $n^{2}$ as a sum of $m$ squares one of which is even, we can construct a representation as a sum of $m+3$ squares by dividing the odd square into four equal squares. Thus the first equality enables us to construct representations with $5,8,11, \\ldots, 155$ squares, the second to construct ones with $7,10,13, \\ldots, 154$ squares, and the third with $9,12, \\ldots, 153$ squares. It remains only to represent $13^{2}$ as a sum of $k=2,3,4,6$ squares. This can be done as follows:  $$ \\begin{aligned} 13^{2} & =12^{2}+5^{2}=12^{2}+4^{2}+3^{2} \\\\ & =11^{2}+4^{2}+4^{2}+4^{2}=12^{2}+3^{2}+2^{2}+2^{2}+2^{2}+2^{2} \\end{aligned} $$  (c) We shall prove that whenever $s(n)=n^{2}-14$ for some $n \\geq 13$, it also holds that $s(2 n)=(2 n)^{2}-14$. This will imply that $s(n)=n^{2}-14$ for any $n=2^{t} \\cdot 13$. If $n^{2}=x_{1}^{2}+\\cdots+x_{r}^{2}$, then we have $(2 n)^{2}=\\left(2 x_{1}\\right)^{2}+\\cdots+\\left(2 x_{r}\\right)^{2}$. Replacing $\\left(2 x_{i}\\right)^{2}$ with $x_{i}^{2}+x_{i}^{2}+x_{i}^{2}+x_{i}^{2}$ as long as it is possible we can obtain representations of $(2 n)^{2}$ consisting of $r, r+3, \\ldots, 4 r$ squares. This gives representations of $(2 n)^{2}$ into $k$ squares for any $k \\leq 4 n^{2}-62$. Further, we observe that each number $m \\geq 14$ can be written as a sum of $k \\geq m$ numbers of the form $x^{2}-1, x \\in \\mathbb{N}$, which is easy to verify. Therefore if $k \\leq 4 n^{2}-14$, it follows that $4 n^{2}-k$ is a sum of $k$ numbers of the form $x^{2}-1$ (since $k \\geq 4 n^{2}-k \\geq 14$ ), and consequently $4 n^{2}$ is a sum of $k$ squares. Remark. One can find exactly the value of $f(n)$ for each $n$ :  $$ f(n)= \\begin{cases}1, & \\text { if } n \\text { has a prime divisor congruent to } 3 \\bmod 4 \\\\ 2, & \\text { if } n \\text { is of the form } 5 \\cdot 2^{k}, k \\text { a positive integer; } \\\\ n^{2}-14, & \\text { otherwise. }\\end{cases} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (CHN 2) The diagonals of a quadrilateral $A B C D$ are perpendicular: $A C \\perp B D$. Four squares, $A B E F, B C G H, C D I J, D A K L$, are erected externally on its sides. The intersection points of the pairs of straight lines $C L, D F ; D F, A H ; A H, B J ; B J, C L$ are denoted by $P_{1}, Q_{1}, R_{1}, S_{1}$, respectively, and the intersection points of the pairs of straight lines $A I, B K$; $B K, C E ; C E, D G ; D G, A I$ are denoted by $P_{2}, Q_{2}, R_{2}, S_{2}$, respectively. Prove that $P_{1} Q_{1} R_{1} S_{1} \\cong P_{2} Q_{2} R_{2} S_{2}$.", "solution": "3. Consider two squares $A B^{\\prime} C D^{\\prime}$ and $A^{\\prime} B C^{\\prime} D$. Since $A C \\perp B D$, these two squares are homothetic, which implies that the lines $A A^{\\prime}, B B^{\\prime}, C C^{\\prime}, D D^{\\prime}$ are concurrent at a certain point $O$. Since the rotation about $A$ by $90^{\\circ}$ takes $\\triangle A B K$ into $\\triangle A F D$, it follows that $B K \\perp D F$. Denote by $T$ the intersection of $B K$ and $D F$. The rotation about some point $X$ by $90^{\\circ}$ maps $B K$ into $D F$ if and only if $T X$ bisects an angle between $B K$ and $D F$. Therefore $\\angle F T A=$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-567.jpg?height=342&width=470&top_left_y=1659&top_left_x=847) $\\angle A T K=45^{\\circ}$. Moreover, the quadrilateral $B A^{\\prime} D T$ is cyclic, which implies that $\\angle B T A^{\\prime}=B D A^{\\prime}=45^{\\circ}$ and consequently that the points $A, T, A^{\\prime}$ are collinear. It follows that the point $O$ lies on a bisector of $\\angle B T D$ and therefore the rotation $\\mathcal{R}$ about $O$ by $90^{\\circ}$ takes $B K$ into $D F$. Analogously, $\\mathcal{R}$ maps the lines $C E, D G, A I$ into $A H, B J, C L$. Hence the quadrilateral $P_{1} Q_{1} R_{1} S_{1}$ is the image of the quadrilateral $P_{2} Q_{2} R_{2} S_{2}$, and the result follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. $(\\mathbf{C H N} 3)^{\\mathrm{IMO}}$ Given nine points in space, no four of which are coplanar, find the minimal natural number $n$ such that for any coloring with red or blue of $n$ edges drawn between these nine points there always exists a triangle having all edges of the same color.", "solution": "4. There are 36 possible edges in total. If not more than 3 edges are left undrawn, then we can choose 6 of the given 9 points no two of which are connected by an undrawn edge. These 6 points together with the edges between them form a two-colored complete graph, and thus by a wellknown result there exists at least one monochromatic triangle. It follows that $n \\leq 33$. In order to show that $n=33$, we shall give an example of a graph with 32 edges that does not contain a monochromatic triangle. Let us start with a complete graph $C_{5}$ with 5 vertices. Its edges can be colored in two colors so that there is no monochromatic triangle (Fig. 1). Furthermore, given a graph $\\mathcal{H}$ with $k$ vertices without monochromatic triangles, we can add to it a new vertex, join it to all vertices of $\\mathcal{H}$ except $A$, and color each edge $B X$ in the same way as $A X$. The obtained graph obviously contains no monochromatic triangles. Applying this construction four times to the graph $C_{5}$ we get an example like that of Fig. 2. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-568.jpg?height=338&width=396&top_left_y=1069&top_left_x=326)  Fig. 1 ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-568.jpg?height=326&width=396&top_left_y=1079&top_left_x=868)  Fig. 2  Second solution. For simplicity, we call the colors red and blue. Let $r(k, l)$ be the least positive integer $r$ such that each complete $r$-graph whose edges are colored in red and blue contains either a complete red $k$-graph or a complete blue $l$-graph. Also, let $t(n, k)$ be the greatest possible number of edges in a graph with $n$ vertices that does not contain a complete $k$-graph. These numbers exist by the theorems of Ramsey and Turán. Let us assume that $r(k, l)<n$. Every graph with $n$ vertices and $t(n, r(k, l))$ +1 edges contains a complete subgraph with $r(k, l)$ vertices, and this subgraph contains either a red complete $k$-graph or a blue complete $l$ graph. We claim that $t(n, r(k, l))+1$ is the smallest number of edges with the above property. By the definition of $r(k, l)$ there exists a coloring of the complete graph $H$ with $r(k, l)-1$ vertices in two colors such that no red complete $k$-graph or blue complete $l$-graph exists. Let $c_{i j}$ be the color in which the edge $(i, j)$ of $H$ is colored, $1 \\leq i<j \\leq r(k, l)-1$. Consider a complete $r(k, l)$-1-partite graph $G$ with $n$ vertices and exactly $t(n, r(k, l))$ edges and denote its partitions by $P_{i}, i=1, \\ldots, r(k, l)-1$. If we color each edge of $H$ between $P_{i}$ and $P_{j}(j<i)$ in the color $c_{i j}$, we obviously obtain a graph with $n$ vertices and $t(n, r(k, l))$ edges in two colors that contains neither a red complete $k$-graph nor a blue complete $l$-graph. Therefore the answer to our problem is $t(9, r(3,3))+1=t(9,6)+1=33$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (COL 3) Let $A B C D$ be a convex quadrilateral such that $A C=$ $B D$. Equilateral triangles are constructed on the sides of the quadrilateral. Let $O_{1}, O_{2}, O_{3}, O_{4}$ be the centers of the triangles constructed on $A B, B C, C D, D A$ respectively. Show that $O_{1} O_{3}$ is perpendicular to $O_{2} O_{4}$.", "solution": "5. Denote by $K, L, M$, and $N$ the midpoints of the sides $A B, B C, C D$, and $D A$, respectively. The quadrilateral $K L M N$ is a rhombus. We shall prove that $O_{1} O_{3} \\| K M$. Similarly, $O_{2} O_{4} \\| L N$, and the desired result follows immediately. We have $\\overrightarrow{O_{1} O_{3}}=\\overrightarrow{K M}+\\left(\\overrightarrow{O_{1} K}+\\overrightarrow{M O_{3}}\\right)$. Assume that $A B C D$ is positively oriented. A rotational homothety $\\mathcal{R}$ with angle $-90^{\\circ}$ and coefficient $1 / \\sqrt{3}$ takes the vectors $\\overrightarrow{B K}$ and $\\overrightarrow{C M}$ into $\\overrightarrow{O_{1} K}$ and $\\overrightarrow{M O_{3}}$ respectively. Therefore  $$ \\begin{aligned} \\overrightarrow{O_{1} O_{3}} & =\\overrightarrow{K M}+\\left(\\overrightarrow{O_{1} K}+\\overrightarrow{M O_{3}}\\right)=\\overrightarrow{K M}+\\mathcal{R}(\\overrightarrow{B K}+\\overrightarrow{C M}) \\\\ & =\\overrightarrow{K M}+\\frac{1}{2} \\mathcal{R}(\\overrightarrow{B A}+\\overrightarrow{C D})=\\overrightarrow{K M}+\\mathcal{R}(\\overrightarrow{L N}) \\end{aligned} $$  Since $L N \\perp K M$, it follows that $\\mathcal{R}(L N)$ is parallel to $K M$ and so is $\\mathrm{O}_{1} \\mathrm{O}_{3}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (IND 2) ${ }^{\\mathrm{IMO} 2}$ Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that  $$ f\\left(x^{2}+f(y)\\right)=y+f(x)^{2} \\quad \\text { for all } x, y \\text { in } \\mathbb{R} $$", "solution": "6. It is easy to see that $f$ is injective and surjective. From $f\\left(x^{2}+f(y)\\right)=$ $f\\left((-x)^{2}+f(y)\\right)$ it follows that $f(x)^{2}=(f(-x))^{2}$, which implies $f(-x)=$ $-f(x)$ because $f$ is injective. Furthermore, there exists $z \\in \\mathbb{R}$ such that $f(z)=0$. From $f(-z)=-f(z)=0$ we deduce that $z=0$. Now we have $f\\left(x^{2}\\right)=f\\left(x^{2}+f(0)\\right)=0+(f(x))^{2}=f(x)^{2}$, and consequently $f(x)=f(\\sqrt{x})^{2}>0$ for all $x>0$. It also follows that $f(x)<0$ for $x<0$. In other words, $f$ preserves sign. Now setting $x>0$ and $y=-f(x)$ in the given functional equation we obtain  $$ f(x-f(x))=f\\left(\\sqrt{x}^{2}+f(-x)\\right)=-x+f(\\sqrt{x})^{2}=-(x-f(x)) . $$  But since $f$ preserves sign, this implies that $f(x)=x$ for $x>0$. Moreover, since $f(-x)=-f(x)$, it follows that $f(x)=x$ for all $x$. It is easily verified that this is indeed a solution.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (IND 4) Circles $G, G_{1}, G_{2}$ are three circles related to each other as follows: Circles $G_{1}$ and $G_{2}$ are externally tangent to one another at a point $W$ and both these circles are internally tangent to the circle $G$. Points $A, B, C$ are located on the circle $G$ as follows: Line $B C$ is a direct common tangent to the pair of circles $G_{1}$ and $G_{2}$, and line $W A$ is the transverse common tangent at $W$ to $G_{1}$ and $G_{2}$, with $W$ and $A$ lying on the same side of the line $B C$. Prove that $W$ is the incenter of the triangle $A B C$.", "solution": "7. Let $G_{1}, G_{2}$ touch the chord $B C$ at $P, Q$ and touch the circle $G$ at $R, S$ respectively. Let $D$ be the midpoint of the complementary $\\operatorname{arc} B C$ of $G$. The homothety centered at $R$ mapping $G_{1}$ onto $G$ also maps the line $B C$ onto a tangent of $G$ parallel to $B C$. It follows that this line touches $G$ at point $D$, which is therefore the image of $P$ under the homothety. Hence $R, P$, and $D$ are collinear. Since $\\angle D B P=\\angle D C B=\\angle D R B$, it follows that $\\triangle D B P \\sim \\triangle D R B$ and consequently that $D P \\cdot D R=D B^{2}$. Similarly, points $S, Q, D$ are collinear and satisfy $D Q \\cdot D S=D B^{2}=D P \\cdot D R$.  Hence $D$ lies on the radical axis of the circles $G_{1}$ and $G_{2}$, i.e., on their common tangent $A W$, which also implies that $A W$ bisects the angle $B A D$. Furthermore, since $D B=D C=D W=\\sqrt{D P \\cdot D R}$, it follows from the lemma of (SL99-14) that $W$ is the incenter of $\\triangle A B C$. Remark. According to the third solution of (SL93-3), both $P W$ and $Q W$ contain the incenter of $\\triangle A B C$, and the result is immediate. The problem can also be solved by inversion centered at $W$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (IND 5) Show that in the plane there exists a convex polygon of 1992 sides satisfying the following conditions: (i) its side lengths are $1,2,3, \\ldots, 1992$ in some order; (ii) the polygon is circumscribable about a circle.  Alternative formulation. Does there exist a 1992-gon with side lengths $1,2,3, \\ldots, 1992$ circumscribed about a circle? Answer the same question for a 1990-gon.", "solution": "8. For simplicity, we shall write $n$ instead of 1992.  Lemma. There exists a tangent $n$-gon $A_{1} A_{2} \\ldots A_{n}$ with sides $A_{1} A_{2}=a_{1}$, $A_{2} A_{3}=a_{2}, \\ldots, A_{n} A_{1}=a_{n}$ if and only if the system  $$ x_{1}+x_{2}=a_{1}, x_{2}+x_{3}=a_{2}, \\ldots, x_{n}+x_{1}=a_{n} $$  has a solution $\\left(x_{1}, \\ldots, x_{n}\\right)$ in positive reals. Proof. Suppose that such an $n$-gon $A_{1} A_{2} \\ldots A_{n}$ exists. Let the side $A_{i} A_{i+1}$ touch the inscribed circle at point $P_{i}$ (where $\\left.A_{n+1}=A_{1}\\right)$. Then $x_{1}=$ $A_{1} P_{n}=A_{1} P_{1}, x_{2}=A_{2} P_{1}=A_{2} P_{2}, \\ldots, x_{n}=A_{n} P_{n-1}=A_{n} P_{n}$ is clearly a positive solution of (1). Now suppose that the system (1) has a positive real solution $\\left(x_{1}, \\ldots\\right.$, $x_{n}$ ). Let us draw a polygonal line $A_{1} A_{2} \\ldots A_{n+1}$ touching a circle of radius $r$ at points $P_{1}, P_{2}, \\ldots, P_{n}$ respectively such that $A_{1} P_{1}=$ $A_{n+1} P_{n}=x_{1}$ and $A_{i} P_{i}=A_{i} P_{i-1}=x_{i}$ for $i=2, \\ldots, n$. Observe that $O A_{1}=O A_{n+1}=\\sqrt{x_{1}^{2}+r^{2}}$ and the function $f(r)=\\angle A_{1} O A_{2}+$ $\\angle A_{2} O A_{3}+\\cdots+\\angle A_{n} O A_{n+1}=$ $2\\left(\\arctan \\frac{x_{1}}{r}+\\cdots+\\arctan \\frac{x_{n}}{r}\\right)$ is continuous. Thus $A_{1} A_{2} \\ldots A_{n+1}$ is a closed simple polygonal line if and only if $f(r)=360^{\\circ}$. But such an $r$ exists, since $f(r) \\rightarrow 0$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-570.jpg?height=331&width=444&top_left_y=1156&top_left_x=887) when $r \\rightarrow \\infty$ and $f(r) \\rightarrow \\infty$ when $r \\rightarrow 0$. This proves the second direction of the lemma. For $n=4 k$, the system (1) is solvable in positive reals if $a_{i}=i$ for $i \\equiv 1,2$ $(\\bmod 4), a_{i}=i+1$ for $i \\equiv 3$ and $a_{i}=i-1$ for $i \\equiv 0(\\bmod 4)$. Indeed, one solution is given by $x_{i}=1 / 2$ for $i \\equiv 1, x_{i}=3 / 2$ for $i \\equiv 3$ and $x_{i}=i-3 / 2$ for $i \\equiv 0,2(\\bmod 4)$. Remark. For $n=4 k+2$ there is no such $n$-gon. In fact, solvability of the system (1) implies $a_{1}+a_{3}+\\cdots=a_{2}+a_{4}+\\cdots$, while in the case $n=4 k+2$ the sum $a_{1}+a_{2}+\\cdots+a_{n}$ is odd.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1992", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (IRN 1) Let $f(x)$ be a polynomial with rational coefficients and $\\alpha$ be a real number such that $\\alpha^{3}-\\alpha=f(\\alpha)^{3}-f(\\alpha)=33^{1992}$. Prove that for each $n \\geq 1$,  $$ \\left(f^{(n)}(\\alpha)\\right)^{3}-f^{(n)}(\\alpha)=33^{1992} $$  where $f^{(n)}(x)=f(f(\\ldots f(x)))$, and $n$ is a positive integer.", "solution": "9. Since the equation $x^{3}-x-c=0$ has only one real root for every $c>$ $2 /(3 \\sqrt{3}), \\alpha$ is the unique real root of $x^{3}-x-33^{1992}=0$. Hence $f^{n}(\\alpha)=$ $f(\\alpha)=\\alpha$. Remark. Consider any irreducible polynomial $g(x)$ in the place of $x^{3}-$ $x-33^{1992}$. The problem amounts to proving that if $\\alpha$ and $f(\\alpha)$ are roots of $g$, then any $f^{(n)}(\\alpha)$ is also a root of $g$. In fact, since $g(f(x))$ vanishes at $x=\\alpha$, it must be divisible by the minimal polynomial of $\\alpha$, that is, $g(x)$. It follows by induction that $g\\left(f^{(n)}(x)\\right)$ is divisible by $g(x)$ for all $n \\in \\mathbb{N}$, and hence $g\\left(f^{(n)}(\\alpha)\\right)=0$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (BRA 1) Show that there exists a finite set $A \\subset \\mathbb{R}^{2}$ such that for every $X \\in A$ there are points $Y_{1}, Y_{2}, \\ldots, Y_{1993}$ in $A$ such that the distance between $X$ and $Y_{i}$ is equal to 1 , for every $i$.", "solution": "1. First we notice that for a rational point $O$ (i.e., with rational coordinates), there exist 1993 rational points in each quadrant of the unit circle centered at $O$. In fact, it suffices to take  $$ X=\\left\\{\\left.O+\\left( \\pm \\frac{t^{2}-1}{t^{2}+1}, \\pm \\frac{2 t}{t^{2}+1}\\right) \\right\\rvert\\, t=1,2, \\ldots, 1993\\right\\} $$  Now consider the set $A=\\{(i / q, j / q) \\mid i, j=0,1, \\ldots, 2 q\\}$, where $q=$ $\\prod_{i=1}^{1993}\\left(t^{2}+1\\right)$. We claim that $A$ gives a solution for the problem. Indeed, for any $P \\in A$ there is a quarter of the unit circle centered at $P$ that is contained in the square $[0,2] \\times[0,2]$. As explained above, there are 1993 rational points on this quarter circle, and by definition of $q$ they all belong to $A$. Remark. Substantially the same problem was proposed by Bulgaria for IMO 71: see (SL71-2), where we give another possible construction of a set $A$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (IND 5) A natural number $n$ is said to have the property $P$ if whenever $n$ divides $a^{n}-1$ for some integer $a, n^{2}$ also necessarily divides $a^{n}-1$. (a) Show that every prime number has property $P$. (b) Show that there are infinitely many composite numbers $n$ that possess property $P$.", "solution": "10. (a) Let $n=p$ be a prime and let $p \\mid a^{p}-1$. By Fermat's theorem $p \\mid$ $a^{p-1}-1$, so that $p \\mid a^{\\operatorname{gcd}(p, p-1)}-1=a-1$, i.e., $a \\equiv 1(\\bmod p)$. Since then $a^{i} \\equiv 1(\\bmod p)$, we obtain $p \\mid a^{p-1}+\\cdots+a+1$ and hence $p^{2} \\mid a^{p}-1=(a-1)\\left(a^{p-1}+\\cdots+a+1\\right)$. (b) Let $n=p_{1} \\cdots p_{k}$ be a product of distinct primes and let $n \\mid a^{n}-1$. Then from $p_{i} \\mid a^{n}-1=\\left(a^{\\left(n / p_{i}\\right)}\\right)^{p_{i}}-1$ and part (a) we conclude that $p_{i}^{2} \\mid a^{n}-1$. Since this is true for all indices $i$, we also have $n^{2} \\mid a^{n}-1$; hence $n$ has the property $P$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (IRE 1) ${ }^{\\mathrm{IMO1}}$ Let $n>1$ be an integer and let $f(x)=x^{n}+5 x^{n-1}+3$. Prove that there do not exist polynomials $g(x), h(x)$, each having integer coefficients and degree at least one, such that $f(x)=g(x) h(x)$.", "solution": "11. Due to the extended Eisenstein criterion, $f$ must have an irreducible factor of degree not less than $n-1$. Since $f$ has no integral zeros, it must be irreducible. Second solution. The proposer's solution was as follows. Suppose that $f(x)=g(x) h(x)$, where $g, h$ are nonconstant polynomials with integer coefficients. Since $|f(0)|=3$, either $|g(0)|=1$ or $|h(0)|=1$. We may assume $|g(0)|=1$ and that $g(x)=\\left(x-\\alpha_{1}\\right) \\cdots\\left(x-\\alpha_{k}\\right)$. Then $\\left|\\alpha_{1} \\cdots \\alpha_{k}\\right|=$ 1. Since $\\alpha_{i}^{n-1}\\left(\\alpha_{i}+5\\right)=-3$, taking the product over $i=1,2, \\ldots, k$ yields $\\left|\\left(\\alpha_{1}+5\\right) \\cdots\\left(\\alpha_{k}+5\\right)\\right|=|g(-5)|=3^{k}$. But $f(-5)=g(-5) h(-5)=3$, so the only possibility is $\\operatorname{deg} g=k=1$. This is impossible, because $f$ has no integral zeros. Remark. Generalizing this solution, it can be shown that if $a, m, n$ are positive integers and $p<a-1$ is a prime, then $F(x)=x^{m}(x-a)^{n}+p$ is irreducible. The details are left to the reader.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (IRE 2) Let $n, k$ be positive integers with $k \\leq n$ and let $S$ be a set containing $n$ distinct real numbers. Let $T$ be the set of all real numbers of the form $x_{1}+x_{2}+\\cdots+x_{k}$, where $x_{1}, x_{2}, \\ldots, x_{k}$ are distinct elements of $S$. Prove that $T$ contains at least $k(n-k)+1$ distinct elements.", "solution": "12. Let $x_{1}<x_{2}<\\cdots<x_{n}$ be the elements of $S$. We use induction on $n$. The result is trivial for $k=1$ or $n=k$, so assume that it is true for $n-1$ numbers. Then there exist $m=(k-1)(n-k)+1$ distinct sums of $k-1$ numbers among $x_{2}, \\ldots, x_{n}$; call these sums $S_{i}, S_{1}<S_{2}<\\cdots<S_{m}$. Then $x_{1}+S_{1}, x_{1}+S_{2}, \\ldots, x_{1}+S_{m}$ are distinct sums of $k$ of the numbers $x_{1}, x_{2}, \\ldots, x_{n}$. However, the biggest of these sums is  $$ x_{1}+S_{m} \\leq x_{1}+x_{n-k+2}+x_{n-k+3}+\\cdots+x_{n} $$  hence we can find $n-k$ sums that are greater and thus not included here: $x_{2}+x_{n-k+2}+\\cdots+x_{n}, x_{3}+x_{n-k+2}+\\cdots+x_{n}, \\ldots, x_{n-k+1}+x_{n-k+2}+\\cdots+x_{n}$. This counts for $k(n-k)+1$ sums in total. Remark. Equality occurs if $S$ is an arithmetic progression.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (IRE 3) Let $S$ be the set of all pairs $(m, n)$ of relatively prime positive integers $m, n$ with $n$ even and $m<n$. For $s=(m, n) \\in S$ write $n=2^{k} n_{0}$, where $k, n_{0}$ are positive integers with $n_{0}$ odd and define $f(s)=\\left(n_{0}, m+\\right.$ $n-n_{0}$ ). Prove that $f$ is a function from $S$ to $S$ and that for each $s=(m, n) \\in S$, there exists a positive integer $t \\leq \\frac{m+n+1}{4}$ such that $f^{t}(s)=s$, where  $$ f^{t}(s)=\\underbrace{(f \\circ f \\circ \\cdots \\circ f)}_{t \\text { times }}(s) . $$  If $m+n$ is a prime number that does not divide $2^{k}-1$ for $k=1,2, \\ldots, m+$ $n-2$, prove that the smallest value of $t$ that satisfies the above conditions is $\\left[\\frac{m+n+1}{4}\\right]$, where $[x]$ denotes the greatest integer less than or equal to $x$.", "solution": "13. For an odd integer $N>1$, let $S_{N}=\\{(m, n) \\in S \\mid m+n=N\\}$. If $f(m, n)=\\left(m_{1}, n_{1}\\right)$, then $m_{1}+n_{1}=m+n$ with $m_{1}$ odd and $m_{1} \\leq \\frac{n}{2}<$ $\\frac{N}{2}<n_{1}$, so $f$ maps $S_{N}$ to $S_{N}$. Also $f$ is bijective, since if $f(m, n)=$ $\\left(m_{1}, n_{1}\\right)$, then $n$ is uniquely determined as the even number of the form $2^{k} m_{1}$ that belongs to the interval $\\left[\\frac{N+1}{2}, N\\right]$, and this also determines $m$. Note that $S_{N}$ has at most $\\left[\\frac{N+1}{4}\\right]$ elements, with equality if and only if $N$ is prime. Thus if $(m, n) \\in S_{N}$, there exist $s, r$ with $1 \\leq s<r \\leq\\left[\\frac{N+5}{4}\\right]$ such that $f^{s}(m, n)=f^{r}(m, n)$. Consequently $f^{t}(m, n)=(m, n)$, where $t=r-s, 0<t \\leq\\left[\\frac{N+1}{4}\\right]=\\left[\\frac{m+n+1}{4}\\right]$. Suppose that $(m, n) \\in S_{N}$ and $t$ is the least positive integer with $f^{t}(m, n)=(m, n)$. We write $(m, n)=\\left(m_{0}, n_{0}\\right)$ and $f^{i}(m, n)=\\left(m_{i}, n_{i}\\right)$ for $i=1, \\ldots, t$. Then there exist positive integers $a_{i}$ such that $2^{a_{i}} m_{i}=n_{i-1}$, $i=1, \\ldots, t$. Since $m_{t}=m_{0}$, multiplying these equalities gives  $$ \\begin{aligned} 2^{a_{1}+a_{2}+\\cdots+a_{t}} m_{0} m_{1} \\cdots m_{t-1} & =n_{0} n_{1} \\cdots n_{t-1} \\\\ & \\equiv(-1)^{t} m_{0} m_{1} \\cdots m_{t-1}(\\bmod N) . \\end{aligned} $$  It follows that $N \\mid 2^{k} \\pm 1$ and consequently $N \\mid 2^{2 k}-1$, where $k=$ $a_{1}+\\cdots+a_{t}$. On the other hand, it also follows that $2^{k}\\left|n_{0} n_{1} \\cdots n_{t-1}\\right|$ $(N-1)(N-3) \\cdots(N-2[N / 4])$. But since  $$ \\frac{(N-1)(N-3) \\cdots\\left(N-2\\left[\\frac{N}{4}\\right]\\right)}{1 \\cdot 3 \\cdots\\left(2\\left[\\frac{N-2}{4}\\right]+1\\right)}=\\frac{2 \\cdot 4 \\cdots(N-1)}{1 \\cdot 2 \\cdots \\frac{N-1}{2}}=2^{\\frac{N-1}{2}} $$  we conclude that $0<k \\leq \\frac{N-1}{2}$, where equality holds if and only if $\\left\\{n_{1}, \\ldots, n_{t}\\right\\}$ is the set of all even integers from $\\frac{N+1}{2}$ to $N-1$, and consequently $t=\\frac{N+1}{4}$. Now if $N \\nmid 2^{h}-1$ for $1 \\leq h<N-1$, we must have $2 k=N-1$. Therefore $t=\\frac{N+1}{4}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (ISR 1) The vertices $D, E, F$ of an equilateral triangle lie on the sides $B C, C A, A B$ respectively of a triangle $A B C$. If $a, b, c$ are the respective lengths of these sides, and $S$ the area of $A B C$, prove that  $$ D E \\geq \\frac{2 \\sqrt{2} S}{\\sqrt{a^{2}+b^{2}+c^{2}+4 \\sqrt{3} S}} $$", "solution": "14. Consider any point $T$ inside the triangle $A B C$ or on its boundary. Since  $$ \\begin{aligned} 2 S & =2\\left(S_{A E T F}+S_{B F T D}+S_{C D T E}\\right) \\\\ & \\leq A T \\cdot E F+B T \\cdot F D+C T \\cdot D E=(A T+B T+C T) D E \\end{aligned} $$  it suffices to find a point $T$ such that  $$ (A T+B T+C T)^{2} \\geq \\frac{a^{2}+b^{2}+c^{2}+4 S \\sqrt{3}}{2} . $$  We distinguish two cases: (i) If all angles of $\\triangle A B C$ are less than $120^{\\circ}$, then the sum $A T+B T+C T$ attains its minimum when $T$ is the Torricelli point, i.e., such that $\\angle A T B=\\angle B T C=\\angle C T A=120^{\\circ}$. In this case, by the cosine theorem we get  $$ \\begin{aligned} A T^{2}+A T \\cdot B T+B T^{2} & =c^{2}, \\\\ B T^{2}+B T \\cdot C T+C T^{2} & =a^{2}, \\\\ C T^{2}+C T \\cdot A T+A T^{2} & =b^{2}, \\\\ 3(A T \\cdot B T+B T \\cdot C T+C T \\cdot A T) & =4 \\sqrt{3}\\left(S_{A T B}+S_{B T C}+S_{C T A}\\right) \\\\ & =4 \\sqrt{3} S . \\end{aligned} $$  Adding these four equalities, we obtain $2(A T+B T+C T)^{2}=a^{2}+$ $b^{2}+c^{2}+4 \\sqrt{3} S$. (ii) Let $\\angle A C B \\geq 120^{\\circ}$. We claim that $T=C$ satisfies the requirements. Indeed, $a^{2}+b^{2}+c^{2}+4 \\sqrt{3} S=a^{2}+b^{2}+\\left(a^{2}+b^{2}-2 a b \\cos \\angle C\\right)+$ $2 \\sqrt{3} a b \\sin \\angle C=2\\left(a^{2}+b^{2}\\right)+2 a b(\\sqrt{3} \\sin \\angle C-\\cos \\angle C)=2\\left(a^{2}+b^{2}\\right)+$ $4 a b \\sin \\left(\\angle C-30^{\\circ}\\right) \\leq 2(a+b)^{2}$, which proves the desired inequality.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (MCD 1) ${ }^{\\mathrm{IMO} 4}$ For three points $A, B, C$ in the plane we define $m(A B C)$ to be the smallest length of the three altitudes of the triangle $A B C$, where in the case of $A, B, C$ collinear, $m(A B C)=0$. Let $A, B, C$ be given points in the plane. Prove that for any point $X$ in the plane,  $$ m(A B C) \\leq m(A B X)+m(A X C)+m(X B C) $$", "solution": "15. Denote by $d(P Q R)$ the diameter of a triangle $P Q R$. It is clear that $d(P Q R) \\cdot m(P Q R)=2 S_{P Q R}$. So if the point $X$ lies inside the triangle $A B C$ or on its boundary, we have $d(A B X), d(B C X), d(C A X) \\leq d(A B C)$, which implies  $$ \\begin{aligned} m(A B X)+m(B C X)+m(C A X) & =\\frac{2 S_{A B X}}{d(A B X)}+\\frac{2 S_{B C X}}{d(B C X)}+\\frac{2 S_{C A X}}{d(C A X)} \\\\ & \\geq \\frac{2 S_{A B X}+2 S_{B C X}+2 S_{C A X}}{d(A B C)} \\\\ & =\\frac{2 S_{A B C}}{d(A B C)}=m(A B C) \\end{aligned} $$  If $X$ is outside $\\triangle A B C$ but inside the angle $B A C$, consider the point $Y$ of intersection of $A X$ and $B C$. Then $m(A B X)+m(B C X)+m(C A X) \\geq$ $m(A B Y)+m(B C Y)+m(C A Y) \\geq m(A B C)$. Also, if $X$ is inside the opposite angle of $\\angle B A C$ (i.e., $\\angle D A E$, where $\\mathcal{B}(D, A, B)$ and $\\mathcal{B}(E, A, C)$ ), then $m(A B X)+m(B C X)+m(C A X) \\geq m(B C X) \\geq m(A B C)$. Since these are substantially all possible different positions of point $X$, we have finished the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (MCD 3) Let $n \\in \\mathbb{N}, n \\geq 2$, and $A_{0}=\\left(a_{01}, a_{02}, \\ldots, a_{0 n}\\right)$ be any $n$-tuple of natural numbers such that $0 \\leq a_{0 i} \\leq i-1$, for $i=1, \\ldots, n$. The $n$-tuples $A_{1}=\\left(a_{11}, a_{12}, \\ldots, a_{1 n}\\right), A_{2}=\\left(a_{21}, a_{22}, \\ldots, a_{2 n}\\right), \\ldots$ are defined by $a_{i+1, j}=\\operatorname{Card}\\left\\{a_{i, l} \\mid 1 \\leq l \\leq j-1, a_{i, l} \\geq a_{i, j}\\right\\}, \\quad$ for $i \\in \\mathbb{N}$ and $j=1, \\ldots, n$. Prove that there exists $k \\in \\mathbb{N}$, such that $A_{k+2}=A_{k}$.", "solution": "16. Let $S_{n}=\\left\\{A=\\left(a_{1}, \\ldots, a_{n}\\right) \\mid 0 \\leq a_{i}<i\\right\\}$. For $A=\\left(a_{1}, \\ldots, a_{n}\\right)$, let $A^{\\prime}=\\left(a_{1}, \\ldots, a_{n-1}\\right)$, so that we can write $A=\\left(A^{\\prime}, a_{n}\\right)$. The proof of the statement from the problem will be given by induction on $n$. For $n=2$ there are two possibilities for $A_{0}$, so one directly checks that $A_{2}=A_{0}$. Now assume that $n \\geq 3$ and that $A_{0}=\\left(A_{0}^{\\prime}, a_{0 n}\\right) \\in S_{n}$. It is clear that then any $A_{i}$ is in $S_{n}$ too. By the induction hypothesis there exists $k \\in \\mathbb{N}$ such that $A_{k}^{\\prime}=A_{k+2}^{\\prime}=A_{k+4}^{\\prime}=\\cdots$ and $A_{k+1}^{\\prime}=A_{k+3}^{\\prime}=\\cdots$. Observe that if we increase (decrease) $a_{k n}, a_{k+1, n}$ will decrease (respectively increase), and this will also increase (respectively decrease) $a_{k+2, n}$. Hence $a_{k n}, a_{k+2, n}, a_{k+4, n}, \\ldots$ is monotonically increasing or decreasing, and since it is bounded (by 0 and $n-1$ ), it follows that we will eventually have $a_{k+2 i, n}=a_{k+2 i+2, n}=\\cdots$. Consequently $A_{k+2 i}=A_{k+2 i+2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (NET 2) ${ }^{\\mathrm{IMO} 6}$ Let $n$ be an integer greater than 1 . In a circular arrangement of $n$ lamps $L_{0}, \\ldots, L_{n-1}$, each one of that can be either ON or OFF, we start with the situation where all lamps are ON, and then carry out a sequence of steps, $S_{t e p_{0}}, S t e p_{1}, \\ldots$. If $L_{j-1}(j$ is taken $\\bmod n)$ is ON, then $S_{t e p}^{j}$ changes the status of $L_{j}$ (it goes from ON to OFF or from OFF to ON) but does not change the status of any of the other lamps. If $L_{j-1}$ is OFF, then $S_{t e p}^{j}$ does not change anything at all. Show that: (a) There is a positive integer $M(n)$ such that after $M(n)$ steps all lamps are ON again. (b) If $n$ has the form $2^{k}$, then all lamps are ON after $n^{2}-1$ steps. (c) If $n$ has the form $2^{k}+1$, then all lamps are ON after $n^{2}-n+1$ steps.", "solution": "17. We introduce the rotation operation Rot to the left by one, so that $S_{t e p_{j}}=$ Rot $^{-j} \\circ$ Step $_{0} \\circ$ Rot $^{j}$. Now writing Step ${ }^{*}={\\text { Rot } \\circ \\text { Step }_{0} \\text {, the problem is }}^{-20}$ transformed into the question whether there is an $M(n)$ such that all lamps are $O N$ again after $M(n)$ successive applications of Step*. We operate in the field $\\mathbb{Z}_{2}$, representing $O F F$ by 0 and $O N$ by 1 . So if the status of $L_{j}$ at some moment is given by $v_{j} \\in \\mathbb{Z}_{2}$, the effect of $S t e p_{j}$ is that $v_{j}$ is replaced by $v_{j}+v_{j-1}$. With the $n$-tuple $v_{0}, \\ldots, v_{n-1}$ we associate the polynomial  $$ P(x)=v_{n-1} x^{n-1}+v_{0} x^{n-2}+v_{1} x^{n-3}+\\cdots+v_{n-2} . $$  By means of Step*, this polynomial is transformed into the polynomial $Q(x)$ over $\\mathbb{Z}$ of degree less than $n$ that satisfies $Q(x) \\equiv x P(x)(\\bmod$ $\\left.x^{n}+x^{n-1}+1\\right)$. From now on, the sign $\\equiv$ always stands for congruence with this modulus. (i) It suffices to show the existence of $M(n)$ with $x^{M(n)} \\equiv 1$. Because the number of residue classes is finite, there are $r, q, r<q$ such that $x^{q} \\equiv x^{r}$, i.e., $x^{r}\\left(x^{q-r}-1\\right)=0$. One can take $M(n)=q-r$. (Or simply note that there are only finitely many possible configurations; since each operation is bijective, the configuration that reappears first must be $O N, O N, \\ldots, O N$.) (ii) We shall prove that if $n=2^{k}$, then $x^{n^{2}-1} \\equiv 1$. We have $x^{n^{2}} \\equiv$ $\\left(x^{n-1}+1\\right)^{n} \\equiv x^{n^{2}-n}+1$, because all binomial coefficients of order $n=2^{k}$ are even, apart from the first one and the last one. Since also $x^{n^{2}} \\equiv x^{n^{2}-1}+x^{n^{2}-n}$, this is what we wanted. (iii) Now if $n=2^{k}+1$, we prove that $x^{n^{2}-n+1} \\equiv 1$. We have $x^{n^{2}-1} \\equiv$ $\\left(x^{n+1}\\right)^{n-1} \\equiv\\left(x+x^{n}\\right)^{n-1} \\equiv x^{n-1}+x^{n^{2}-n}$ (again by evenness of binomial coefficients of order $n-1=2^{k}$ ). Together with $x^{n^{2}} \\equiv x^{n^{2}-1}+$ $x^{n^{2}-n}$, this leads to $x^{n^{2}} \\equiv x^{n-1}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (POL 1) Let $S_{n}$ be the number of sequences $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$, where $a_{i} \\in$ $\\{0,1\\}$, in which no six consecutive blocks are equal. Prove that $S_{n} \\rightarrow \\infty$ as $n \\rightarrow \\infty$.", "solution": "18. Let $B_{n}$ be the set of sequences with the stated property $\\left(S_{n}=\\left|B_{n}\\right|\\right)$. We shall prove by induction on $n$ that $S_{n} \\geq \\frac{3}{2} S_{n-1}$ for every $n$. Suppose that for every $i \\leq n, S_{i} \\geq \\frac{3}{2} S_{i-1}$, and consequently $S_{i} \\leq$ $\\left(\\frac{2}{3}\\right)^{n-i} S_{n}$. Let us consider the $2 S_{n}$ sequences obtained by putting 0 or 1 at the end of any sequence from $B_{n}$. If some sequence among them does not belong to $B_{n+1}$, then for some $k \\geq 1$ it can be obtained by extending some sequence from $B_{n+1-6 k}$ by a sequence of $k$ terms repeated six times. The number of such sequences is $2^{k} S_{n+1-6 k}$. Hence the number of sequences not satisfying our condition is not greater than  $$ \\sum_{k \\geq 1} 2^{k} S_{n+1-6 k} \\leq \\sum_{k \\geq 1} 2^{k}\\left(\\frac{2}{3}\\right)^{6 k-1} S_{n}=\\frac{3}{2} S_{n} \\frac{2(2 / 3)^{6}}{1-2(2 / 3)^{6}}=\\frac{192}{601} S_{n}<\\frac{1}{2} S_{n} $$  Therefore $S_{n+1}$ is not smaller than $2 S_{n}-\\frac{1}{2} S_{n}=\\frac{3}{2} S_{n}$. Thus we have $S_{n} \\geq\\left(\\frac{3}{2}\\right)^{n}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (ROM 2) Let $a, b, n$ be positive integers, $b>1$ and $b^{n}-1 \\mid a$. Show that the representation of the number $a$ in the base $b$ contains at least $n$ digits different from zero.", "solution": "19. Let $s$ be the minimum number of nonzero digits that can appear in the $b$ adic representation of any number divisible by $b^{n}-1$. Among all numbers divisible by $b^{n}-1$ and having $s$ nonzero digits in base $b$, we choose the number $A$ with the minimum sum of digits. Let $A=a_{1} b^{n_{1}}+\\cdots+a_{s} b^{n_{s}}$, where $0<a_{i} \\leq b-1$ and $n_{1}>n_{2}>\\cdots>n_{s}$. First, suppose that $n_{i} \\equiv n_{j}(\\bmod n), i \\neq j$. Consider the number  $$ B=A-a_{i} b^{n_{i}}-a_{j} b^{n_{j}}+\\left(a_{i}+a_{j}\\right) b^{n_{j}+k n} $$  with $k$ chosen large enough so that $n_{j}+k n>n_{1}$ : this number is divisible by $b^{n}-1$ as well. But if $a_{i}+a_{j}<b$, then $B$ has $s-1$ digits in base $b$, which is impossible; on the other hand, $a_{i}+a_{j} \\geq b$ is also impossible, for otherwise $B$ would have sum of digits less for $b-1$ than that of $A$ (because $B$ would have digits 1 and $a_{i}+a_{j}-b$ in the positions $\\left.n_{j}+k n+1, n_{j}+k n\\right)$. Therefore $n_{i} \\not \\equiv n_{j}$ if $i \\neq j$. Let $n_{i} \\equiv r_{i}$, where $r_{i} \\in\\{0,1, \\ldots, n-1\\}$ are distinct. The number $C=$ $a_{1} b^{r_{1}}+\\cdots+a_{s} b^{r_{s}}$ also has $s$ digits and is divisible by $b^{n}-1$. But since $C<b^{n}$, the only possibility is $C=b^{n}-1$ which has exactly $n$ digits in base $b$. It follows that $s=n$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (CAN 2) Let triangle $A B C$ be such that its circumradius $R$ is equal to 1. Let $r$ be the inradius of $A B C$ and let $p$ be the inradius of the orthic triangle $A^{\\prime} B^{\\prime} C^{\\prime}$ of triangle $A B C$. Prove that $p \\leq 1-\\frac{1}{3}(1+r)^{2}$. Remark. The orthic triangle is the triangle whose vertices are the feet of the altitudes of $A B C$.", "solution": "2. It is well known that $r \\leq \\frac{1}{2} R$. Therefore $\\frac{1}{3}(1+r)^{2} \\leq \\frac{1}{3}\\left(1+\\frac{1}{2}\\right)^{2}=\\frac{3}{4}$.  It remains only to show that $p \\leq \\frac{1}{4}$. We note that $p$ does not exceed one half of the circumradius of $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$. However, by the theorem on the nine-point circle, this circumradius is equal to $\\frac{1}{2} R$, and the conclusion follows. Second solution. By a well-known relation we have $\\cos A+\\cos B+\\cos C=$ $1+\\frac{r}{R}(=1+r$ when $R=1)$. Next, recalling that the incenter of $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ is at the orthocenter of $\\triangle A B C$, we easily obtain $p=2 \\cos A \\cos B \\cos C$. Cosines of angles of a triangle satisfy the identity $\\cos ^{2} A+\\cos ^{2} B+\\cos ^{2} C+$ $2 \\cos A \\cos B \\cos C=1$ (the proof is straightforward: see (SL81-11)). Thus  $$ \\begin{aligned} p+\\frac{1}{3}(1+r)^{2} & =2 \\cos A \\cos B \\cos C+\\frac{1}{3}(\\cos A+\\cos B+\\cos C)^{2} \\\\ & \\leq 2 \\cos A \\cos B \\cos C+\\cos ^{2} A+\\cos ^{2} B+\\cos ^{2} C=1 \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. (ROM 3) Let $c_{1}, \\ldots, c_{n} \\in \\mathbb{R}(n \\geq 2)$ such that $0 \\leq \\sum_{i=1}^{n} c_{i} \\leq n$. Show that we can find integers $k_{1}, \\ldots, k_{n}$ such that $\\sum_{i=1}^{n} k_{i}=0$ and  $$ 1-n \\leq c_{i}+n k_{i} \\leq n \\quad \\text { for every } i=1, \\ldots, n $$", "solution": "20. For every real $x$ we shall denote by $\\lfloor x\\rfloor$ and $\\lceil x\\rceil$ the greatest integer less than or equal to $x$ and the smallest integer greater than or equal to $x$ respectively. The condition $c_{i}+n k_{i} \\in[1-n, n]$ is equivalent to $k_{i} \\in I_{i}=$ $\\left[\\frac{1-c_{i}}{n}-1,1-\\frac{c_{i}}{n}\\right]$. For every $c_{i}$, this interval contains two integers (not necessarily distinct), namely $p_{i}=\\left\\lceil\\frac{1-c_{i}}{n}-1\\right\\rceil \\leq q_{i}=\\left\\lfloor 1-\\frac{c_{i}}{n}\\right\\rfloor$. In order to show that there exist integers $k_{i} \\in I_{i}$ with $\\sum_{i=1}^{n} k_{i}=0$, it is sufficient to show that $\\sum_{i=1}^{n} p_{i} \\leq 0 \\leq \\sum_{i=1}^{n} q_{i}$. Since $p_{i}<\\frac{1-c_{i}}{n}$, we have  $$ \\sum_{i=1}^{n} p_{i}<1-\\sum_{i=1}^{n} \\frac{c_{i}}{n} \\leq 1 $$  and consequently $\\sum_{i=1}^{n} p_{i} \\leq 0$ because the $p_{i}$ 's are integers. On the other hand, $q_{i}>-\\frac{c_{i}}{n}$ implies  $$ \\sum_{i=1}^{n} q_{i}>-\\sum_{i=1}^{n} \\frac{c_{i}}{n} \\geq-1 $$  which leads to $\\sum_{i=1}^{n} q_{i} \\geq 0$. The proof is complete.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. (GBR 1) A circle $S$ is said to cut a circle $\\Sigma$ diametrally if their common chord is a diameter of $\\Sigma$. Let $S_{A}, S_{B}, S_{C}$ be three circles with distinct centers $A, B, C$ respectively. Prove that $A, B, C$ are collinear if and only if there is no unique circle $S$ that cuts each of $S_{A}, S_{B}, S_{C}$ diametrally. Prove further that if there exists more than one circle $S$ that cuts each of $S_{A}, S_{B}, S_{C}$ diametrally, then all such circles pass through two fixed points. Locate these points in relation to the circles $S_{A}, S_{B}, S_{C}$.", "solution": "21. Assume that $S$ is a circle with center $O$ that cuts $S_{i}$ diametrically in points $P_{i}, Q_{i}, i \\in\\{A, B, C\\}$, and denote by $r_{i}, r$ the radii of $S_{i}$ and $S$ respectively. Since $O A$ is perpendicular to $P_{A} Q_{A}$, it follows by Pythagoras's theorem that $O A^{2}+A P_{A}^{2}=O P_{A}^{2}$, i.e., $r_{A}^{2}+O A^{2}=r^{2}$. Analogously $r_{B}^{2}+O B^{2}=r^{2}$ and $r_{C}^{2}+O C^{2}=r^{2}$. Thus if $O_{A}, O_{B}, O_{C}$ are the feet of perpendiculars from $O$ to $B C, C A, A B$ respectively, then $O_{C} A^{2}-O_{C} B^{2}=r_{B}^{2}-r_{A}^{2}$. Since the left-hand side is a monotonic function of $O_{C} \\in A B$, the point $O_{C}$ is uniquely determined by the imposed conditions. The same holds for $O_{A}$ and $O_{B}$. If $A, B, C$ are not collinear, then the positions of $O_{A}, O_{B}, O_{C}$ uniquely determine the point $O$, and therefore the circle $S$ also. On the other hand, if $A, B, C$ are collinear, all one can deduce is that $O$ lies on the lines $l_{A}, l_{B}, l_{C}$ through $O_{A}, O_{B}, O_{C}$, perpendicular to $B C, C A, A B$ respectively. By this, $l_{A}, l_{B}, l_{C}$ are parallel, so $O$ can be either anywhere on the line if these lines coincide, or ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-586.jpg?height=480&width=544&top_left_y=1409&top_left_x=810) nowhere if they don't coincide. So if there exists more than one circle $S$, $A, B, C$ lie on a line and the foot $O^{\\prime}$ of the perpendicular from $O$ to the line $A B C$ is fixed. If $X, Y$ are the intersection points of $S$ and the line $A B C$, then $r^{2}=O X^{2}=O A^{2}+r_{A}^{2}$ and consequently $O^{\\prime} X^{2}=O^{\\prime} A^{2}+r_{A}^{2}$, which implies that $X, Y$ are fixed.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "22", "problem_type": null, "exam": "IMO", "problem": "22. (GBR 2) ${ }^{\\mathrm{IMO} 2} A, B, C, D$ are four points in the plane, with $C, D$ on the same side of the line $A B$, such that $A C \\cdot B D=A D \\cdot B C$ and $\\measuredangle A D B=$ $90^{\\circ}+\\measuredangle A C B$. Find the ratio  $$ \\frac{A B \\cdot C D}{A C \\cdot B D} $$  and prove that circles $A C D, B C D$ are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicular.)", "solution": "22. Let $M$ be the point inside $\\angle A D B$ that satisfies $D M=D B$ and $D M \\perp$ $D B$. Then $\\angle A D M=\\angle A C B$ and $A D / D M=A C / C B$. It follows that the triangles $A D M, A C B$ are similar; hence $\\angle C A D=\\angle B A M$ (because $\\angle C A B=\\angle D A M$ ) and $A B / A M=A C / A D$. Consequently the triangles $C A D, B A M$ are similar and therefore $\\frac{A C}{A B}=\\frac{C D}{B M}=$ $\\frac{C D}{\\sqrt{2} B D}$. Hence $\\frac{A B \\cdot C D}{A C \\cdot B D}=\\sqrt{2}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-587.jpg?height=380&width=391&top_left_y=239&top_left_x=882)  Let $C T, C U$ be the tangents at $C$ to the circles $A C D, B C D$ respectively. Then (in oriented angles) $\\angle T C U=\\angle T C D+\\angle D C U=\\angle C A D+\\angle C B D=$ $90^{\\circ}$, as required. Second solution to the first part. Denote by $E, F, G$ the feet of the perpendiculars from $D$ to $B C, C A, A B$. Consider the pedal triangle $E F G$. Since $F G=A D \\sin \\angle A$, from the sine theorem we have $F G: G E: E F=$ $(C D \\cdot A B):(B D \\cdot A C):(A D \\cdot B C)$. Thus $E G=F G$. On the other hand, $\\angle E G F=\\angle E G D+\\angle D G F=\\angle C B D+\\angle C A D=90^{\\circ}$ implies that $E F: E G=\\sqrt{2}: 1$; hence the required ratio is $\\sqrt{2}$. Third solution to the first part. Under inversion centered at $C$ and with power $r^{2}=C A \\cdot C B$, the triangle $D A B$ maps into a right-angled isosceles triangle $D^{*} A^{*} B^{*}$, where  $$ D^{*} A^{*}=\\frac{A D \\cdot B C}{C D}, D^{*} B^{*}=\\frac{A C \\cdot B D}{C D}, A^{*} B^{*}=\\frac{A B \\cdot C D}{C D} . $$  Thus $D^{*} B^{*}: A^{*} B^{*}=\\sqrt{2}$, and this is the required ratio.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "23", "problem_type": null, "exam": "IMO", "problem": "23. (GBR 3) A finite set of (distinct) positive integers is called a \" $D S$-set\" if each of the integers divides the sum of them all. Prove that every finite set of positive integers is a subset of some $D S$-set.", "solution": "23. Let the given numbers be $a_{1}, \\ldots, a_{n}$. Put $s=a_{1}+\\cdots+a_{n}$ and $m=$ $\\operatorname{lcm}\\left(a_{1}, \\ldots, a_{n}\\right)$ and write $m=2^{k} r$ with $k \\geq 0$ and $r$ odd. Let the binary expansion of $r$ be $r=2^{k_{0}}+2^{k_{1}}+\\cdots+2^{k_{t}}$, with $0=k_{0}<\\cdots<k_{t}$. Adjoin to the set $\\left\\{a_{1}, \\ldots, a_{n}\\right\\}$ the numbers $2^{k_{i}} s, i=1,2, \\ldots, t$. The sum of the enlarged set is $r s$. Finally, adjoin $r s, 2 r s, 2^{2} r s, \\ldots, 2^{l-1} r s$ for $l=$ $\\max \\left\\{k, k_{t}\\right\\}$. The resulting set has sum $2^{l} r s$, which is divisible by $m$ and so by each of $a_{j}$, and also by the $2^{i} s$ above and by $r s, 2 r s, \\ldots, 2^{l-1} r s$. Therefore this is a $D S$-set. Second solution. We show by induction that there is a $D S$-set containing 1 and $n$. For $n=2,3$, take $\\{1,2,3\\}$. Assume that $\\left\\{1, n, b_{1}, \\ldots, b_{k}\\right\\}$ is a $D S$ set. Then $\\left\\{1, n+1, n, 2(n+1) n, 2(n+1) b_{1}, \\ldots, 2(n+1) b_{k}\\right\\}$ is a $D S$-set too. For given $a_{1}, \\ldots, a_{n}$ let $m$ be a sufficiently large common multiple of the $a_{i}$ 's such that $u=m-\\left(a_{1}+\\cdots+a_{n}\\right) \\neq a_{i}$ for all $i$. There exist $b_{1}, \\ldots, b_{k}$ such that $\\left\\{1, u, b_{1}, \\ldots, b_{k}\\right\\}$ is a $D S$-set. It is clear that $\\left\\{a_{1}, \\ldots, a_{n}, u, m u, m b_{1}, \\ldots, m b_{k}\\right\\}$ is a $D S$-set containing $a_{1}, \\ldots, a_{n}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "24", "problem_type": null, "exam": "IMO", "problem": "24. (USA 3) Prove that  $$ \\frac{a}{b+2 c+3 d}+\\frac{b}{c+2 d+3 a}+\\frac{c}{d+2 a+3 b}+\\frac{d}{a+2 b+3 c} \\geq \\frac{2}{3} $$  for all positive real numbers $a, b, c, d$.", "solution": "24. By the Cauchy-Schwarz inequality, if $x_{1}, x_{2}, \\ldots, x_{n}$ and $y_{1}, y_{2}, \\ldots, y_{n}$ are positive numbers, then  $$ \\left(\\sum_{i=1}^{n} \\frac{x_{i}}{y_{i}}\\right)\\left(\\sum_{i=1}^{n} x_{i} y_{i}\\right) \\geq\\left(\\sum_{i=1}^{n} x_{i}\\right)^{2} $$  Applying this to the numbers $a, b, c, d$ and $b+2 c+3 d, c+2 d+3 a, d+2 a+$ $3 b, a+2 b+3 c$ (here $n=4$ ), we obtain  $$ \\begin{gathered} \\frac{a}{b+2 c+3 d}+\\frac{b}{c+2 d+3 a}+\\frac{c}{d+2 a+3 b}+\\frac{d}{a+2 b+3 c} \\\\ \\geq \\frac{(a+b+c+d)^{2}}{4(a b+a c+a d+b c+b d+c d)} \\geq \\frac{2}{3} \\end{gathered} $$  The last inequality follows, for example, from $(a-b)^{2}+(a-c)^{2}+\\cdots+$ $(c-d)^{2} \\geq 0$. Equality holds if and only if $a=b=c=d$. Second solution. Putting $A=b+2 c+3 d, B=c+2 d+3 a, C=d+2 a+3 b$, $D=a+2 b+3 c$, our inequality transforms into  $$ \\begin{aligned} & \\frac{-5 A+7 B+C+D}{24 A}+\\frac{-5 B+7 C+D+A}{24 B} \\\\ & +\\frac{-5 C+7 D+A+B}{24 C}+\\frac{-5 D+7 A+B+C}{24 D} \\geq \\frac{2}{3} \\end{aligned} $$  This follows from the arithmetic-geometric mean inequality, since $\\frac{B}{A}+\\frac{C}{B}+$ $\\frac{D}{C}+\\frac{A}{D} \\geq 4$, etc.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "25", "problem_type": null, "exam": "IMO", "problem": "25. (VIE 1) Solve the following system of equations, in which $a$ is a given number satisfying $|a|>1$ :  $$ \\begin{aligned} x_{1}^{2} & =a x_{2}+1, \\\\ x_{2}^{2} & =a x_{3}+1 \\\\ \\cdots & \\cdots \\\\ x_{999}^{2} & =a x_{1000}+1 \\\\ x_{1000}^{2} & =a x_{1}+1 \\end{aligned} $$", "solution": "25. We need only consider the case $a>1$ (since the case $a<-1$ is reduced to $a>1$ by taking $a^{\\prime}=-a, x_{i}^{\\prime}=-x_{i}$ ). Since the left sides of the equations are nonnegative, we have $x_{i} \\geq-\\frac{1}{a}>-1, i=1, \\ldots, 1000$. Suppose w.l.o.g. that $x_{1}=\\max \\left\\{x_{i}\\right\\}$. In particular, $x_{1} \\geq x_{2}, x_{3}$. If $x_{1} \\geq 0$, then we deduce that $x_{1000}^{2} \\geq 1 \\Rightarrow x_{1000} \\geq 1$; further, from this we deduce that $x_{999}>1$ etc., so either $x_{i}>1$ for all $i$ or $x_{i}<0$ for all $i$. (i) $x_{i}>1$ for every $i$. Then $x_{1} \\geq x_{2}$ implies $x_{1}^{2} \\geq x_{2}^{2}$, so $x_{2} \\geq x_{3}$. Thus $x_{1} \\geq x_{2} \\geq \\cdots \\geq x_{1000} \\geq x_{1}$, and consequently $x_{1}=\\cdots=x_{1000}$. In this case the only solution is $x_{i}=\\frac{1}{2}\\left(a+\\sqrt{a^{2}+4}\\right)$ for all $i$. (ii) $x_{i}<0$ for every $i$. Then $x_{1} \\geq x_{3}$ implies $x_{1}^{2} \\leq x_{3}^{2} \\Rightarrow x_{2} \\leq x_{4}$. Similarly, this leads to $x_{3} \\geq x_{5}$, etc. Hence $x_{1} \\geq x_{3} \\geq x_{5} \\geq \\cdots \\geq x_{999} \\geq x_{1}$ and $x_{2} \\leq x_{4} \\leq \\cdots \\leq x_{2}$, so we deduce that $x_{1}=x_{3}=\\cdots$ and $x_{2}=x_{4}=$ $\\cdots$. Therefore the system is reduced to $x_{1}^{2}=a x_{2}+1, x_{2}^{2}=a x_{1}+1$. Subtracting these equations, one obtains $\\left(x_{1}-x_{2}\\right)\\left(x_{1}+x_{2}+a\\right)=0$. There are two possibilities: (1) If $x_{1}=x_{2}$, then $x_{1}=x_{2}=\\cdots=\\frac{1}{2}\\left(a-\\sqrt{a^{2}+4}\\right)$. (2) $x_{1}+x_{2}+a=0$ is equivalent to $x_{1}^{2}+a x_{1}+\\left(a^{2}-1\\right)=0$. The discriminant of the last equation is $4-3 a^{2}$. Therefore if $a>\\frac{2}{\\sqrt{3}}$, this case yields no solutions, while if $a \\leq \\frac{2}{\\sqrt{3}}$, we obtain $x_{1}=$ $\\frac{1}{2}\\left(-a-\\sqrt{4-3 a^{2}}\\right), x_{2}=\\frac{1}{2}\\left(-a+\\sqrt{4-3 a^{2}}\\right)$, or vice versa.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "26", "problem_type": null, "exam": "IMO", "problem": "26. (VIE 2) Let $a, b, c, d$ be four nonnegative numbers satisfying $a+b+c+d=$ 1. Prove the inequality  $$ a b c+b c d+c d a+d a b \\leq \\frac{1}{27}+\\frac{176}{27} a b c d $$", "solution": "26. Set  $$ \\begin{aligned} f(a, b, c, d) & =a b c+b c d+c d a+d a b-\\frac{176}{27} a b c d \\\\ & =a b(c+d)+c d\\left(a+b-\\frac{176}{27} a b\\right) . \\end{aligned} $$  If $a+b-\\frac{176}{a} b \\leq 0$, by the arithmetic-geometric inequality we have $f(a, b, c, d) \\leq a b(c+d) \\leq \\frac{1}{27}$. On the other hand, if $a+b-\\frac{176}{a} b>0$, the value of $f$ increases if $c, d$ are replaced by $\\frac{c+d}{2}, \\frac{c+d}{2}$. Consider now the following fourtuplets:  $$ \\begin{gathered} P_{0}(a, b, c, d), P_{1}\\left(a, b, \\frac{c+d}{2}, \\frac{c+d}{2}\\right), P_{2}\\left(\\frac{a+b}{2}, \\frac{a+b}{2}, \\frac{c+d}{2}, \\frac{c+d}{2}\\right), \\\\ P_{3}\\left(\\frac{1}{4}, \\frac{a+b}{2}, \\frac{c+d}{2}, \\frac{1}{4}\\right), P_{4}\\left(\\frac{1}{4}, \\frac{1}{4}, \\frac{1}{4}, \\frac{1}{4}\\right) \\end{gathered} $$  From the above considerations we deduce that for $i=0,1,2,3$ either $f\\left(P_{i}\\right) \\leq f\\left(P_{i+1}\\right)$, or directly $f\\left(P_{i}\\right) \\leq 1 / 27$. Since $f\\left(P_{4}\\right)=1 / 27$, in every case we are led to  $$ f(a, b, c, d)=f\\left(P_{0}\\right) \\leq \\frac{1}{27} $$  Equality occurs only in the cases $(0,1 / 3,1 / 3,1 / 3)$ (with permutations) and ( $1 / 4,1 / 4,1 / 4,1 / 4)$. Remark. Lagrange multipliers also work. On the boundary of the set one of the numbers $a, b, c, d$ is 0 , and the inequality immediately follows, while for an extremum point in the interior, among $a, b, c, d$ there are at most two distinct values, in which case one easily verifies the inequality.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (SPA 1) Consider the triangle $A B C$, its circumcircle $k$ with center $O$ and radius $R$, and its incircle with center $I$ and radius $r$. Another circle $k_{c}$ is tangent to the sides $C A, C B$ at $D, E$, respectively, and it is internally tangent to $k$. Show that the incenter $I$ is the midpoint of $D E$.", "solution": "3. Let $O_{1}$ and $\\rho$ be the center and radius of $k_{c}$. It is clear that $C, I, O_{1}$ are collinear and $C I / C O_{1}=r / \\rho$. By Stewart's theorem applied to $\\triangle O C O_{1}$,  $$ O I^{2}=\\frac{r}{\\rho} O O_{1}^{2}+\\left(1-\\frac{r}{\\rho}\\right) O C^{2}-C I \\cdot I O_{1} . $$  Since $O O_{1}=R-\\rho, O C=R$ and by Euler's formula $O I^{2}=R^{2}-2 R r$, substituting these values in (1) gives $C I \\cdot I O_{1}=r \\rho$, or equivalently $C O_{1}$. $I O_{1}=\\rho^{2}=D O_{1}^{2}$. Hence the triangles $C O_{1} D$ and $D O_{1} I$ are similar, implying $\\angle D I O_{1}=90^{\\circ}$. Since $C D=C E$ and the line $C O_{1}$ bisects the segment $D E$, it follows that $I$ is the midpoint of $D E$. Second solution. Under the inversion with center $C$ and power $a b, k_{c}$ is transformed into the excircle of $\\widehat{A} \\widehat{B} C$ corresponding to $C$. Thus $C D=$ $\\frac{a b}{s}$, where $s$ is the common semiperimeter of $\\triangle A B C$ and $\\triangle \\widehat{A} \\widehat{B} C$, and consequently the distance from $D$ to $B C$ is $\\frac{a b}{s} \\sin C=\\frac{2 S_{A B C}}{s}=2 r$. The statement follows immediately. Third solution. We shall prove a stronger statement: Let $A B C D$ be a convex quadrilateral inscribed in a circle $k$, and $k^{\\prime}$ the circle that is tangent to segments $B O, A O$ at $K, L$ respectively (where $O=B D \\cap A C$ ), and internally to $k$ at $M$. Then $K L$ contains the incenters $I, J$ of $\\triangle A B C$ and $\\triangle A B D$. Let $K^{\\prime}, K^{\\prime \\prime}, L^{\\prime}, L^{\\prime \\prime}, N$ denote the midpoints of arcs $B C, B D, A C, A D, A B$ that don't contain $M ; X^{\\prime}, X^{\\prime \\prime}$ the points on $k$ defined by $X^{\\prime} N=N X^{\\prime \\prime}=$ $K^{\\prime} K^{\\prime \\prime}=L^{\\prime} L^{\\prime \\prime}$ (as oriented arcs); and set $S=A K^{\\prime} \\cap B L^{\\prime \\prime}, \\bar{M}=N S \\cap k$, $\\bar{K}=K^{\\prime \\prime} M \\cap B O, \\bar{L}=L^{\\prime} M \\cap A O$. It is clear that $I=A K^{\\prime} \\cap B L^{\\prime}, J=A K^{\\prime \\prime} \\cap B L^{\\prime \\prime}$. Furthermore, $X^{\\prime} \\bar{M}$ contains $I$ (to see this, use the fact that for $A, B, C, D, E, F$ on $k$, lines $A D, B E, C F$ are concurrent if and only if $A B \\cdot C D \\cdot E F=B C \\cdot D E \\cdot F A$, and then express $A \\bar{M} / \\bar{M} B$ by applying this rule to $A M B K^{\\prime} N L^{\\prime \\prime}$ and show that $A K^{\\prime}, \\bar{M} X^{\\prime}, B L^{\\prime}$ are concurrent). Analogously, $X^{\\prime \\prime} \\bar{M}$ contains $J$. Now the points $B, \\bar{K}, I, S, \\bar{M}$ lie on a circle $(\\angle B \\overline{K M}=\\angle B I \\bar{M}=\\angle B S \\bar{M})$, and points $A, \\bar{L}, J, S, \\bar{M}$ do so as well. Lines $I \\bar{K}, J \\bar{L}$ are parallel to $K^{\\prime \\prime} L^{\\prime}$ (because $\\angle \\overline{M K} I=\\angle \\bar{M} B I=$ $\\left.\\angle \\bar{M} K^{\\prime \\prime} L^{\\prime}\\right)$. On the other hand, the quadrilateral $A B I J$ is cyclic, and simple calculation with angles shows that $I J$ is also parallel to $K^{\\prime \\prime} L^{\\prime}$. Hence $\\bar{K}, I, J, \\bar{L}$ are collinear. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-578.jpg?height=496&width=523&top_left_y=967&top_left_x=827)  Finally, $\\bar{K} \\equiv K, \\bar{L} \\equiv L$, and $\\bar{M} \\equiv M$ because the homothety centered at $M$ that maps $k^{\\prime}$ to $k$ sends $K$ to $K^{\\prime \\prime}$ and $L$ to $L^{\\prime}$ (thus $M, K, K^{\\prime \\prime}$, as well as $M, L, L^{\\prime}$, must be collinear). As is seen now, the deciphered picture yields many other interesting properties. Thus, for example, $N, S, M$ are collinear, i.e., $\\angle A M S=\\angle B M S$. Fourth solution. We give an alternative proof of the more general statement in the third solution. Let $W$ be the foot of the perpendicular from $B$ to $A C$. We define $q=C W, h=B W, t=O L=O K, x=A L$, $\\theta=\\measuredangle W B O(\\theta$ is negative if $\\mathcal{B}(O, W, A), \\theta=0$ if $W=O)$, and as usual, $a=B C, b=A C, c=A B$. Let $\\alpha=\\measuredangle K L C$ and $\\beta=\\measuredangle I L C$ (both angles must be acute). Our goal is to prove $\\alpha=\\beta$. We note that $90^{\\circ}-\\theta=2 \\alpha$. One easily gets  $$ \\tan \\alpha=\\frac{\\cos \\theta}{1+\\sin \\theta}, \\quad \\tan \\beta=\\frac{\\frac{2 S_{A B C}}{a+b+c}}{\\frac{b+c-a}{2}-x} $$  Applying Casey's theorem to $A, B, C, k^{\\prime}$, we get $A C \\cdot B K+A L \\cdot B C=$ $A B \\cdot C L$, i.e., $b\\left(\\frac{h}{\\cos \\theta}-t\\right)+x a=c(b-x)$. Using that $t=b-x-q-h \\tan \\theta$ we get  $$ x=\\frac{b(b+c-q)-b h\\left(\\frac{1}{\\cos \\theta}+\\tan \\theta\\right)}{a+b+c} . $$  Plugging (2) into the second equation of (1) and using $b h=2 S_{A B C}$ and $c^{2}=b^{2}+a^{2}-2 b q$, we obtain $\\tan \\alpha=\\tan \\beta$, i.e., $\\alpha=\\beta$, which completes our proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (SPA 2) In the triangle $A B C$, let $D, E$ be points on the side $B C$ such that $\\angle B A D=\\angle C A E$. If $M, N$ are, respectively, the points of tangency with $B C$ of the incircles of the triangles $A B D$ and $A C E$, show that  $$ \\frac{1}{M B}+\\frac{1}{M D}=\\frac{1}{N C}+\\frac{1}{N E} $$", "solution": "4. Let $h$ be the altitude from $A$ and $\\varphi=\\angle B A D$. We have $B M=\\frac{1}{2}(B D+$ $A B-A D)$ and $M D=\\frac{1}{2}(B D-A B+A D)$, so  $$ \\begin{aligned} \\frac{1}{M B}+\\frac{1}{M D} & =\\frac{B D}{M B \\cdot M D}=\\frac{4 B D}{B D^{2}-A B^{2}-A D^{2}+2 A B \\cdot A D} \\\\ & =\\frac{4 B D}{2 A B \\cdot A D(1-\\cos \\varphi)}=\\frac{2 B D \\sin \\varphi}{2 S_{A B D}(1-\\cos \\varphi)} \\\\ & =\\frac{2 B D \\sin \\varphi}{B D \\cdot h(1-\\cos \\varphi)}=\\frac{2}{h \\tan \\frac{\\varphi}{2}} . \\end{aligned} $$  It follows that $\\frac{1}{M B}+\\frac{1}{M D}$ depends only on $h$ and $\\varphi$. Specially, $\\frac{1}{N C}+\\frac{1}{N E}=$ $\\frac{2}{h \\tan (\\varphi / 2)}$ as well.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (FIN 3) ${ }^{\\mathrm{IMO} 3}$ On an infinite chessboard, a solitaire game is played as follows: At the start, we have $n^{2}$ pieces occupying $n^{2}$ squares that form a square of side $n$. The only allowed move is a jump horizontally or vertically over an occupied square to an unoccupied one, and the piece that has been jumped over is removed. For what positive integers $n$ can the game end with only one piece remaining on the board?", "solution": "5. For $n=1$ the game is trivially over. If $n=2$, it can end, for example, in the following way: ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-579.jpg?height=144&width=534&top_left_y=1229&top_left_x=510)  Fig. 1 The sequence of moves shown in Fig. 2 enables us to remove three pieces placed in a $1 \\times 3$ rectangle, using one more piece and one more free cell. In that way, for any $n \\geq 4$ we can reduce an $(n+3) \\times(n+3)$ square to an $n \\times n$ square (Fig. 3). Therefore the game can end for every $n$ that is not divisible by 3 . ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-579.jpg?height=123&width=595&top_left_y=1716&top_left_x=265)  Fig. 2 ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-579.jpg?height=272&width=272&top_left_y=1642&top_left_x=1050)  Fig. 3  Suppose now that one can play the game on a $3 k \\times 3 k$ square so that at the end only one piece remains. Denote the cells by $(i, j), i, j \\in\\{1, \\ldots, 3 k\\}$, and let $S_{0}, S_{1}, S_{2}$ denote the numbers of pieces on those squares $(i, j)$ for which $i+j$ gives remainder $0,1,2$ respectively upon division by 3 . Initially $S_{0}=S_{1}=S_{2}=3 k^{2}$. After each move, two of $S_{0}, S_{1}, S_{2}$ diminish and one increases by one. Thus each move reverses the parity of the $S_{i}$ 's, so that $S_{0}, S_{1}, S_{2}$ are always of the same parity. But in the final position one of the $S_{i}$ 's must be equal to 1 and the other two must be 0 , which is impossible.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (GER 1) ${ }^{\\mathrm{IMO} 5}$ Let $\\mathbb{N}=\\{1,2,3, \\ldots\\}$. Determine whether there exists a strictly increasing function $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ with the following properties:  $$ \\begin{aligned} f(1) & =2 \\\\ f(f(n)) & =f(n)+n \\quad(n \\in \\mathbb{N}) \\end{aligned} $$", "solution": "6. Notice that for $\\alpha=\\frac{1+\\sqrt{5}}{2}, \\alpha^{2} n=\\alpha n+n$ for all $n \\in \\mathbb{N}$. We shall show that $f(n)=\\left[\\alpha n+\\frac{1}{2}\\right]$ (the closest integer to $\\alpha n$ ) satisfies the requirements. Observe that $f$ is strictly increasing and $f(1)=2$. By the definition of $f$, $|f(n)-\\alpha n| \\leq \\frac{1}{2}$ and $f(f(n))-f(n)-n$ is an integer. On the other hand,  $$ \\begin{aligned} |f(f(n))-f(n)-n| & =\\left|f(f(n))-f(n)-\\alpha^{2} n+\\alpha n\\right| \\\\ & =\\left|f(f(n))-\\alpha f(n)+\\alpha f(n)-\\alpha^{2} n-f(n)+\\alpha n\\right| \\\\ & =|(\\alpha-1)(f(n)-\\alpha n)+(f(f(n))-\\alpha f(n))| \\\\ & \\leq(\\alpha-1)|f(n)-\\alpha n|+|f(f(n))-\\alpha f(n)| \\\\ & \\leq \\frac{1}{2}(\\alpha-1)+\\frac{1}{2}=\\frac{1}{2} \\alpha<1, \\end{aligned} $$  which implies that $f(f(n))-f(n)-n=0$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (GEO 3) Let $a, b, c$ be given integers $a>0, a c-b^{2}=P=P_{1} \\cdots P_{m}$ where $P_{1}, \\ldots, P_{m}$ are (distinct) prime numbers. Let $M(n)$ denote the number of pairs of integers $(x, y)$ for which  $$ a x^{2}+2 b x y+c y^{2}=n $$  Prove that $M(n)$ is finite and $M(n)=M\\left(P^{k} \\cdot n\\right)$ for every integer $k \\geq 0$.", "solution": "7. Multiplying by $a$ and $c$ the equation  $$ a x^{2}+2 b x y+c y^{2}=P^{k} n $$  gives $(a x+b y)^{2}+P y^{2}=a P^{k} n$ and $(b x+c y)^{2}+P x^{2}=c P^{k} n$. It follows immediately that $M(n)$ is finite; moreover, $(a x+b y)^{2}$ and $(b x+$ $c y)^{2}$ are divisible by $P$, and consequently $a x+b y, b x+c y$ are divisible by $P$ because $P$ is not divisible by a square greater than 1 . Thus there exist integers $X, Y$ such that $b x+c y=P X, a x+b y=-P Y$. Then $x=-b X-c Y$ and $y=a X+b Y$. Introducing these values into (1) and simplifying the expression obtained we get  $$ a X^{2}+2 b X Y+c Y^{2}=P^{k-1} n $$  Hence $(x, y) \\mapsto(X, Y)$ is a bijective correspondence between integral solutions of (1) and (2), so that $M\\left(P^{k} n\\right)=M\\left(P^{k-1} n\\right)=\\cdots=M(n)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (IND 1) Define a sequence $\\langle f(n)\\rangle_{n=1}^{\\infty}$ of positive integers by $f(1)=1$ and  $$ f(n)= \\begin{cases}f(n-1)-n, & \\text { if } f(n-1)>n ; \\\\ f(n-1)+n, & \\text { if } f(n-1) \\leq n,\\end{cases} $$  for $n \\geq 2$. Let $S=\\{n \\in \\mathbb{N} \\mid f(n)=1993\\}$. (a) Prove that $S$ is an infinite set. (b) Find the least positive integer in $S$. (c) If all the elements of $S$ are written in ascending order as $n_{1}<n_{2}<$ $n_{3}<\\cdots$, show that  $$ \\lim _{i \\rightarrow \\infty} \\frac{n_{i+1}}{n_{i}}=3 . $$", "solution": "8. Suppose that $f(n)=1$ for some $n>0$. Then $f(n+1)=n+2, f(n+$ $2)=2 n+4, f(n+3)=n+1, f(n+4)=2 n+5, f(n+5)=n$, and so by induction $f(n+2 k)=2 n+3+k, f(n+2 k-1)=n+3-k$ for $k=1,2, \\ldots, n+2$. Particularly, $n^{\\prime}=3 n+3$ is the smallest value greater than $n$ for which $f\\left(n^{\\prime}\\right)=1$. It follows that all numbers $n$ with $f(n)=1$ are given by $n=b_{i}$, where $b_{0}=1, b_{n}=3 b_{n-1}+3$. Furthermore, $b_{n}=3+3 b_{n-1}=3+3^{2}+3^{2} b_{n-2}=\\cdots=3+3^{2}+\\cdots+3^{n}+3^{n}=$ $=\\frac{1}{2}\\left(5 \\cdot 3^{n}-3\\right)$. It is seen from above that if $n \\leq b_{i}$, then $f(n) \\leq f\\left(b_{i}-1\\right)=b_{i}+1$. Hence if $f(n)=1993$, then $n \\geq b_{i} \\geq 1992$ for some $i$. The smallest such $b_{i}$ is $b_{7}=5466$, and $f\\left(b_{i}+2 k-1\\right)=b_{i}+3-k=1993$ implies $k=3476$. Thus the least integer in $S$ is $n_{1}=5466+2 \\cdot 3476-1=12417$. All the elements of $S$ are given by $n_{i}=b_{i+6}+2 k-1$, where $b_{i+6}+3-k=$ 1993, i.e., $k=b_{i+6}-1990$. Therefore $n_{i}=3 b_{i+6}-3981=\\frac{1}{2}\\left(5 \\cdot 3^{i+7}-7971\\right)$. Clearly $S$ is infinite and $\\lim _{i \\rightarrow \\infty} \\frac{n_{i+1}}{n_{i}}=3$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1993", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (IND 4) (a) Show that the set $\\mathbb{Q}^{+}$of all positive rational numbers can be partitioned into three disjoint subsets $A, B, C$ satisfying the following conditions:  $$ B A=B, \\quad B^{2}=C, \\quad B C=A, $$  where $H K$ stands for the set $\\{h k \\mid h \\in H, k \\in K\\}$ for any two subsets $H, K$ of $\\mathbb{Q}^{+}$and $H^{2}$ stands for $H H$. (b) Show that all positive rational cubes are in $A$ for such a partition of $\\mathbb{Q}^{+}$. (c) Find such a partition $\\mathbb{Q}^{+}=A \\cup B \\cup C$ with the property that for no positive integer $n \\leq 34$ are both $n$ and $n+1$ in $A$; that is,  $$ \\min \\{n \\in \\mathbb{N} \\mid n \\in A, n+1 \\in A\\}>34 $$", "solution": "9. We shall first complete the \"multiplication table\" for the sets $A, B, C$. It is clear that this multiplication is commutative and associative, so that we have the following relations:  $$ \\begin{aligned} & A C=(A B) B=B B=C \\\\ & A^{2}=A A=(A B) C=B C=A \\\\ & C^{2}=C C=B(B C)=B A=B \\end{aligned} $$  (a) Now put 1 in $A$ and distribute the primes arbitrarily in $A, B, C$. This distribution uniquely determines the partition of $\\mathbb{Q}^{+}$with the stated property. Indeed, if an arbitrary rational number  $$ x=p_{1}^{\\alpha_{1}} \\cdots p_{k}^{\\alpha_{k}} q_{1}^{\\beta_{1}} \\cdots q_{l}^{\\beta_{l}} r_{1}^{\\gamma_{1}} \\cdots r_{m}^{\\gamma_{m}} $$  is given, where $p_{i} \\in A, q_{i} \\in B, r_{i} \\in C$ are primes, it is easy to see that $x$ belongs to $A, B$, or $C$ according as $\\beta_{1}+\\cdots+\\beta_{l}+2 \\gamma_{1}+\\cdots+2 \\gamma_{m}$ is congruent to 0,1 , or $2(\\bmod 3)$. (b) In every such partition, cubes all belong to $A$. In fact, $A^{3}=A^{2} A=$ $A A=A, B^{3}=B^{2} B=C B=A, C^{3}=C^{2} C=B C=A$. (c) By (b) we have $1,8,27 \\in A$. Then $2 \\notin A$, and since the problem is symmetric with respect to $B, C$, we can assume $2 \\in B$ and consequently $4 \\in C$. Also $7 \\notin A$, and also $7 \\notin B$ (otherwise, $28=4 \\cdot 7 \\in A$ and $27 \\in A$ ), so $7 \\in C, 14 \\in A, 28 \\in B$. Further, we see that $3 \\notin A$ (since otherwise $9 \\in A$ and $8 \\in A$ ). Put 3 in $C$. Then $5 \\notin B$ (otherwise $15 \\in A$ and $14 \\in A$ ), so let $5 \\in C$ too. Consequently $6,10 \\in A$. Also $13 \\notin A$, and $13 \\notin C$ because $26 \\notin A$, so $13 \\in B$. Now it is easy to distribute the remaining primes $11,17,19,23,29,31$ : one possibility is  $$ \\begin{aligned} & A=\\{1,6,8,10,14,19,23,27,29,31,33, \\ldots\\}, \\\\ & C=\\{3,4,5,7,18,22,24,26,30,32,34, \\ldots\\} \\\\ & B=\\{2,9,11,12,13,15,16,17,20,21,25,28,35, \\ldots\\} \\end{aligned} $$  Remark. It can be proved that $\\min \\{n \\in \\mathbb{N} \\mid n \\in A, n+1 \\in A\\} \\leq 77$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "1", "problem_type": "Algebra", "exam": "IMO", "problem": "1. A1 (USA) Let $a_{0}=1994$ and $a_{n+1}=\\frac{a_{n}^{2}}{a_{n}+1}$ for each nonnegative integer $n$. Prove that $1994-n$ is the greatest integer less than or equal to $a_{n}$, $0 \\leq n \\leq 998$.", "solution": "1. Obviously $a_{0}>a_{1}>a_{2}>\\cdots$. Since $a_{k}-a_{k+1}=1-\\frac{1}{a_{k}+1}$, we have $a_{n}=a_{0}+\\left(a_{1}-a_{0}\\right)+\\cdots+\\left(a_{n}-a_{n-1}\\right)=1994-n+\\frac{1}{a_{0}+1}+\\cdots+\\frac{1}{a_{n-1}+1}>$ $1994-n$. Also, for $1 \\leq n \\leq 998$,  $$ \\frac{1}{a_{0}+1}+\\cdots+\\frac{1}{a_{n-1}+1}<\\frac{n}{a_{n-1}+1}<\\frac{998}{a_{997}+1}<1 $$  because as above, $a_{997}>997$. Hence $\\left\\lfloor a_{n}\\right\\rfloor=1994-n$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "10", "problem_type": "Combinatorics", "exam": "IMO", "problem": "10. C5 (SWE) At a round table are 1994 girls, playing a game with a deck of $n$ cards. Initially, one girl holds all the cards. In each turn, if at least one girl holds at least two cards, one of these girls must pass a card to each of her two neighbors. The game ends when and only when each girl is holding at most one card. (a) Prove that if $n \\geq 1994$, then the game cannot end. (b) Prove that if $n<1994$, then the game must end.", "solution": "10. (a) The case $n>1994$ is trivial. Suppose that $n=1994$. Label the girls $G_{1}$ to $G_{1994}$, and let $G_{1}$ initially hold all the cards. At any moment give to each card the value $i, i=1, \\ldots, 1994$, if $G_{i}$ holds it. Define the characteristic $C$ of a position as the sum of all these values. Initially $C=1994$. In each move, if $G_{i}$ passes cards to $G_{i-1}$ and $G_{i+1}$ (where $G_{0}=G_{1994}$ and $G_{1995}=G_{1}$ ), $C$ changes for $\\pm 1994$ or does not change, so that it remains divisible by 1994. But if the game ends, the characteristic of the final position will be $C=1+2+\\cdots+1994=$ $997 \\cdot 1995$, which is not divisible by 1994. (b) Whenever a card is passed from one girl to another for the first time, let the girls sign their names on it. Thereafter, if one of them passes a card to her neighbor, we shall assume that the passed card is exactly the one signed by both of them. Thus each signed card is stuck between two neighboring girls, so if $n<1994$, there are two neighbors who never exchange cards. Consequently, there is a girl $G$ who played only a finite number of times. If her neighbor plays infinitely often, then after her last move, $G$ will continue to accumulate cards indefinitely, which is impossible. Hence every girl plays finitely many times.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "11", "problem_type": "Combinatorics", "exam": "IMO", "problem": "11. C6 (FIN) On an infinite square grid, two players alternately mark symbols on empty cells. The first player always marks $X$ 's, the second $O$ 's. One symbol is marked per turn. The first player wins if there are 11 consecutive $X$ 's in a row, column, or diagonal. Prove that the second player can prevent the first from winning.", "solution": "11. Tile the table with dominoes and numbers as shown in the picture. The second player will not lose if whenever the first player plays in a cell of a domino, he plays in the other cell of the domino, and whenever the first player plays on a number, he plays on the same number that is diagonally adjacent. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-593.jpg?height=385&width=399&top_left_y=1025&top_left_x=851)", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "12", "problem_type": "Combinatorics", "exam": "IMO", "problem": "12. C7 (BRA) Prove that for any integer $n \\geq 2$, there exists a set of $2^{n-1}$ points in the plane such that no 3 lie on a line and no $2 n$ are the vertices of a convex $2 n$-gon.", "solution": "12. Define $S_{n}$ recursively as follows: Let $S_{2}=\\{(0,0),(1,1)\\}$ and $S_{n+1}=$ $S_{n} \\cup T_{n}$, where $T_{n}=\\left\\{\\left(x+2^{n-1}, y+M_{n}\\right) \\mid(x, y) \\in S_{n}\\right\\}$, with $M_{n}$ chosen large enough so that the entire set $T_{n}$ lies above every line passing through two points of $S_{n}$. By definition, $S_{n}$ has exactly $2^{n-1}$ points and contains no three collinear points. We claim that no $2 n$ points of this set are the vertices of a convex $2 n$-gon. Consider an arbitrary convex polygon $\\mathcal{P}$ with vertices in $S_{n}$. Join by a diagonal $d$ the two vertices of $\\mathcal{P}$ having the smallest and greatest $x$ coordinates. This diagonal divides $\\mathcal{P}$ into two convex polygons $\\mathcal{P}_{1}, \\mathcal{P}_{2}$, the former lying above $d$. We shall show by induction that both $\\mathcal{P}_{1}, \\mathcal{P}_{2}$ have at most $n$ vertices. Assume to the contrary that $\\mathcal{P}_{1}$ has at least $n+1$ vertices $A_{1}\\left(x_{1}, y_{1}\\right), \\ldots, A_{n+1}\\left(x_{n+1}, y_{n+1}\\right)$ in $S_{n}$, with $x_{1}<\\cdots<x_{n+1}$. It follows that $\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}>\\cdots>\\frac{y_{n+1}-y_{n}}{x_{n+1}-x_{n}}$. By the induction hypothesis, not more than $n-1$ of these vertices belong to $S_{n-1}$ or $T_{n-1}$, so let $A_{k-1}, A_{k} \\in S_{n-1}$, $A_{k+1} \\in T_{n-1}$. But by the construction of $T_{n-1}, \\frac{y_{k+1}-y_{k}}{x_{k+1}-x_{k}}>\\frac{y_{k}-y_{k-1}}{x_{k}-x_{k-1}}$, which gives a contradiction. Similarly, $\\mathcal{P}_{2}$ has no more than $n$ vertices, and therefore $\\mathcal{P}$ itself has at most $2 n-2$ vertices.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "13", "problem_type": "Geometry", "exam": "IMO", "problem": "13. G1 (FRA) A semicircle $\\Gamma$ is drawn on one side of a straight line $l$. $C$ and $D$ are points on $\\Gamma$. The tangents to $\\Gamma$ at $C$ and $D$ meet $l$ at $B$ and $A$ respectively, with the center of the semicircle between them. Let $E$ be the point of intersection of $A C$ and $B D$, and $F$ the point on $l$ such that $E F$ is perpendicular to $l$. Prove that $E F$ bisects $\\angle C F D$.", "solution": "13. Extend $A D$ and $B C$ to meet at $P$, and let $Q$ be the foot of the perpendicular from $P$ to $A B$. Denote by $O$ the center of $\\Gamma$. Since $\\triangle P A Q \\sim \\triangle O A D$ and $\\triangle P B Q \\sim \\triangle O B C$, we obtain $\\frac{A Q}{A D}=\\frac{P Q}{O D}=\\frac{P Q}{O C}=\\frac{B Q}{B C}$. Therefore $\\frac{A Q}{Q B} \\cdot \\frac{B C}{C P} \\cdot \\frac{P D}{D A}=1$, so by the converse Ceva theorem, $A C, B D$, and $P Q$ are concurrent. It follows that $Q \\equiv F$. Finally, since the points $O, C, P, D, F$ are concyclic, we have $\\angle D F P=\\angle D O P=\\angle P O C=\\angle P F C$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "14", "problem_type": "Geometry", "exam": "IMO", "problem": "14. G2 (UKR) $A B C D$ is a quadrilateral with $B C$ parallel to $A D . M$ is the midpoint of $C D, P$ that of $M A$ and $Q$ that of $M B$. The lines $D P$ and $C Q$ meet at $N$. Prove that $N$ is not outside triangle $A B M .{ }^{8}$", "solution": "14. Although it does not seem to have been noticed at the jury, the statement of the problem is false. For $A(0,0), B(0,4), C(1,4), D(7,0)$, we have $M(4,2), P(2,1), Q(2,3)$ and $N(9 / 2,1 / 2) \\notin \\triangle A B M$. The official solution, if it can be called so, actually shows that $N$ lies inside $A B C D$ and goes as follows: The case $A D=B C$ is trivial, so let $A D>B C$. Let $L$ be the midpoint of $A B$. Complete the parallelograms $A D M X$ and $B C M Y$. Now $N=D X \\cap C Y$, so let $C Y$ and $D X$ intersect $A B$ at $K$ and $H$ respectively. From $L X=L Y$ and  $$ \\frac{H L}{L X}=\\frac{H A}{A D}<\\frac{L A}{A D}<\\frac{K B}{A D}<\\frac{K B}{B C}=\\frac{K L}{L Y} $$  we get $H L<K L$, and the statement follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "15", "problem_type": "Geometry", "exam": "IMO", "problem": "15. G3 (RUS) A circle $\\omega$ is tangent to two parallel lines $l_{1}$ and $l_{2}$. A second circle $\\omega_{1}$ is tangent to $l_{1}$ at $A$ and to $\\omega$ externally at $C$. A third circle $\\omega_{2}$ is tangent to $l_{2}$ at $B$, to $\\omega$ externally at $D$, and to $\\omega_{1}$ externally at $E$. $A D$ intersects $B C$ at $Q$. Prove that $Q$ is the circumcenter of triangle $C D E$.", "solution": "15. We shall prove that $A D$ is a common tangent of $\\omega$ and $\\omega_{2}$. Denote by $K, L$ the points of tangency of $\\omega$ with $l_{1}$ and $l_{2}$ respectively. Let $r, r_{1}, r_{2}$ be the radii of $\\omega, \\omega_{1}, \\omega_{2}$ respectively, and set $K A=x, L B=y$. It will be enough if we show that $x y=2 r^{2}$, since this will imply that $\\triangle K L B$ and $\\triangle A K O$ are similar, where $O$ is the center of $\\omega$, and consequently that $O A \\perp K D$ (because $D \\in K B$ ). Now if $O_{1}$ is the center of $\\omega_{1}$, we have $x^{2}=$ $K A^{2}=O O_{1}^{2}-\\left(K O-A O_{1}\\right)^{2}=\\left(r+r_{1}\\right)^{2}-\\left(r-r_{1}\\right)^{2}=4 r r_{1}$ and analogously $y^{2}=4 r r_{2}$. But we also have $\\left(r_{1}+r_{2}\\right)^{2}=O_{1} O_{2}^{2}=(x-y)^{2}+\\left(2 r-r_{1}-r_{2}\\right)^{2}$, so $x^{2}-2 x y+y^{2}=4 r\\left(r_{1}+r_{2}-r\\right)$, from which we obtain $x y=2 r^{2}$ as claimed. Hence $A D$ is tangent to both $\\omega, \\omega_{2}$, and similarly $B C$ is tangent to $\\omega, \\omega_{1}$. It follows that $Q$ lies on the radical axes of pairs of circles $\\left(\\omega, \\omega_{1}\\right)$ and $\\left(\\omega, \\omega_{2}\\right)$. Therefore $Q$ also lies on the radical axis of $\\left(\\omega_{1}, \\omega_{2}\\right)$, i.e., on the common tangent at $E$ of $\\omega_{1}$ and $\\omega_{2}$. Hence $Q C=Q D=Q E$. Second solution. An inversion with center at $D$ maps $\\omega$ and $\\omega_{2}$ to parallel lines, $\\omega_{1}$ and $l_{2}$ to disjoint equal circles touching $\\omega, \\omega_{2}$, and $l_{1}$ to a circle externally tangent to $\\omega_{1}, l_{2}$, and to $\\omega$. It is easy to see that the obtained picture is symmetric (with respect to a diameter of $l_{1}$ ), and that line $A D$ is parallel to the lines $\\omega$ and $\\omega_{2}$. Going back to the initial picture, this means that $A D$ is a common tangent of $\\omega$ and $\\omega_{2}$. The end is like that in the first solution.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "[^6]16. G4 (AUS-ARM) ${ }^{\\mathrm{IMO} 2} N$ is an arbitrary point on the bisector of $\\angle B A C$. $P$ and $O$ are points on the lines $A B$ and $A N$, respectively, such that $\\measuredangle A N P=90^{\\circ}=\\measuredangle A P O . Q$ is an arbitrary point on $N P$, and an arbitrary line through $Q$ meets the lines $A B$ and $A C$ at $E$ and $F$ respectively. Prove that $\\measuredangle O Q E=90^{\\circ}$ if and only if $Q E=Q F$.", "solution": "16. First, assume that $\\angle O Q E=90^{\\circ}$. Extend $P N$ to meet $A C$ at $R$. Then $O E P Q$ and $O R F Q$ are cyclic quadrilaterals; hence we have $\\angle O E Q=$ $\\angle O P Q=\\angle O R Q=\\angle O F Q$. It follows that $\\triangle O E Q \\cong \\triangle O F Q$ and $Q E=Q F$. Now suppose $Q E=Q F$. Let $S$ be the point symmetric to $A$ with respect to $Q$, so that the quadrilateral $A E S F$ is a parallelogram. Draw the line $E^{\\prime} F^{\\prime}$ through $Q$ so that $\\angle O Q E^{\\prime}=90^{\\circ}$ and $E^{\\prime} \\in A B$, $F^{\\prime} \\in A C$. By the first part $Q E^{\\prime}=$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-595.jpg?height=344&width=448&top_left_y=379&top_left_x=851) $Q F^{\\prime}$; hence $A E^{\\prime} S F^{\\prime}$ is also a parallelogram. It follows that $E \\equiv E^{\\prime}, F \\equiv F^{\\prime}$, and $\\angle O Q E=90^{\\circ}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "17", "problem_type": "Geometry", "exam": "IMO", "problem": "17. G5 (CYP) A line $l$ does not meet a circle $\\omega$ with center $O . E$ is the point on $l$ such that $O E$ is perpendicular to $l . M$ is any point on $l$ other than $E$. The tangents from $M$ to $\\omega$ touch it at $A$ and $B . C$ is the point on $M A$ such that $E C$ is perpendicular to $M A . D$ is the point on $M B$ such that $E D$ is perpendicular to $M B$. The line $C D$ cuts $O E$ at $F$. Prove that the location of $F$ is independent of that of $M$.", "solution": "17. We first prove that $A B$ cuts $O E$ in a fixed point $H$. Note that $\\angle O A H=$ $\\angle O M A=\\angle O E A$ (because $O, A, E, M$ lie on a circle); hence $\\triangle O A H \\sim$ $\\triangle O E A$. This implies $O H \\cdot O E=O A^{2}$, i.e., $H$ is fixed. Let the lines $A B$ and $C D$ meet at $K$. Since $E A O B M$ and $E C D M$ are cyclic, we have $\\angle E A K=$ $\\angle E M B=\\angle E C K$, so $E C A K$ is cyclic. Therefore $\\angle E K A=90^{\\circ}$, hence $E K B D$ is also cyclic and $E K \\| O M$. Then $\\angle E K F=$ $\\angle E B D=\\angle E O M=\\angle O E K$, from which we deduce that $K F=F E$. However, since $\\angle E K H=90^{\\circ}$, the ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-595.jpg?height=410&width=451&top_left_y=947&top_left_x=854) point $F$ is the midpoint of $E H$; hence it is fixed.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "18", "problem_type": "Number Theory", "exam": "IMO", "problem": "18. N1 (BUL) $M$ is a subset of $\\{1,2,3, \\ldots, 15\\}$ such that the product of any three distinct elements of $M$ is not a square. Determine the maximum number of elements in $M$.", "solution": "18. Since for each of the subsets $\\{1,4,9\\},\\{2,6,12\\},\\{3,5,15\\}$ and $\\{7,8,14\\}$ the product of its elements is a square and these subsets are disjoint, we have $|M| \\leq 11$. Suppose that $|M|=11$. Then $10 \\in M$ and none of the disjoint subsets $\\{1,4,9\\},\\{2,5\\},\\{6,15\\},\\{7,8,14\\}$ is a subset of $M$. Consequently $\\{3,12\\} \\subset M$, so none of $\\{1\\},\\{4\\},\\{9\\},\\{2,6\\},\\{5,15\\}$, and $\\{7,8,14\\}$ is a subset of $M$ : thus $|M| \\leq 9$, a contradiction. It follows that $|M| \\leq 10$, and this number is attained in the case $M=\\{1,4,5,6,7,10,11,12,13,14\\}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "19", "problem_type": "Number Theory", "exam": "IMO", "problem": "19. N2 (AUS) ${ }^{\\mathrm{IMO} 4}$ Determine all pairs $(m, n)$ of positive integers such that $\\frac{n^{3}+1}{m n-1}$ is an integer.", "solution": "19. Since $m n-1$ and $m^{3}$ are relatively prime, $m n-1$ divides $n^{3}+1$ if and only if it divides $m^{3}\\left(n^{3}+1\\right)=\\left(m^{3} n^{3}-1\\right)+m^{3}+1$. Thus  $$ \\frac{n^{3}+1}{m n-1} \\in \\mathbb{Z} \\Leftrightarrow \\frac{m^{3}+1}{m n-1} \\in \\mathbb{Z} $$  hence we may assume that $m \\geq n$. If $m=n$, then $\\frac{n^{3}+1}{n^{2}-1}=n+\\frac{1}{n-1}$ is an integer, so $m=n=2$. If $n=1$, then $\\frac{2}{m-1} \\in \\mathbb{Z}$, which happens only when $m=2$ or $m=3$. Now suppose $m>n \\geq 2$. Since $m^{3}+1 \\equiv 1$ and $m n-1 \\equiv-1(\\bmod n)$, we deduce $\\frac{n^{3}+1}{m n-1}=k n-1$ for some integer $k>0$. On the other hand, $k n-1<\\frac{n^{3}+1}{n^{2}-1}=n+\\frac{1}{n-1} \\leq 2 n-1$ gives that $k=1$, and therefore $n^{3}+1=(m n-1)(n-1)$. This yields $m=\\frac{n^{2}+1}{n-1}=n+1+\\frac{2}{n-1} \\in \\mathbb{N}$, so $n \\in\\{2,3\\}$ and $m=5$. The solutions with $m<n$ are obtained by symmetry. There are 9 solutions in total: $(1,2),(1,3),(2,1),(3,1),(2,2),(2,5),(3,5)$, $(5,2),(5,3)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "2", "problem_type": "Algebra", "exam": "IMO", "problem": "2. A2 (FRA) ${ }^{\\mathrm{IMO} 1}$ Let $m$ and $n$ be positive integers. The set $A=\\left\\{a_{1}, a_{2}, \\ldots\\right.$, $\\left.a_{m}\\right\\}$ is a subset of $\\{1,2, \\ldots, n\\}$. Whenever $a_{i}+a_{j} \\leq n, 1 \\leq i \\leq j \\leq m$, $a_{i}+a_{j}$ also belongs to $A$. Prove that  $$ \\frac{a_{1}+a_{2}+\\cdots+a_{m}}{m} \\geq \\frac{n+1}{2} . $$", "solution": "2. We may assume that $a_{1}>a_{2}>\\cdots>a_{m}$. We claim that for $i=1, \\ldots, m$, $a_{i}+a_{m+1-i} \\geq n+1$. Indeed, otherwise $a_{i}+a_{m+1-i}, \\ldots, a_{i}+a_{m-1}, a_{i}+a_{m}$ are $i$ different elements of $A$ greater than $a_{i}$, which is impossible. Now by adding for $i=1, \\ldots, m$ we obtain $2\\left(a_{1}+\\cdots+a_{m}\\right) \\geq m(n+1)$, and the result follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "20", "problem_type": "Number Theory", "exam": "IMO", "problem": "20. N3 (FIN) ${ }^{\\text {IMO6 }}$ Find a set $A$ of positive integers such that for any infinite set $P$ of prime numbers, there exist positive integers $m \\in A$ and $n \\notin A$, both the product of the same number of distinct elements of $P$.", "solution": "20. Let $A$ be the set of all numbers of the form $p_{1} p_{2} \\ldots p_{p_{1}}$, where $p_{1}<p_{2}<$ $\\cdots<p_{p_{1}}$ are primes. In other words, $A=\\{2 \\cdot 3,2 \\cdot 5, \\ldots\\} \\cup\\{3 \\cdot 5 \\cdot 7,3 \\cdot 5 \\cdot$ $11, \\ldots\\} \\cup\\{5 \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 17, \\ldots\\} \\cup \\cdots$. This set satisfies the requirements of the problem. Indeed, for any infinite set of primes $P=\\left\\{q_{1}, q_{2}, \\ldots\\right\\}$ (where $q_{1}<q_{2}<\\cdots$ ) we have  $$ m=q_{1} q_{2} \\cdots q_{q_{1}} \\in A \\quad \\text { and } \\quad n=q_{2} q_{3} \\cdots q_{q_{1}+1} \\notin A $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "21", "problem_type": "Number Theory", "exam": "IMO", "problem": "21. N4 (FRA) For any positive integer $x_{0}$, three sequences $\\left\\{x_{n}\\right\\},\\left\\{y_{n}\\right\\}$, and $\\left\\{z_{n}\\right\\}$ are defined as follows: (i) $y_{0}=4$ and $z_{0}=1$; (ii) if $x_{n}$ is even for $n \\geq 0, x_{n+1}=\\frac{x_{n}}{2}, y_{n+1}=2 y_{n}$, and $z_{n+1}=z_{n}$; (iii) if $x_{n}$ is odd for $n \\geq 0, x_{n+1}=x_{n}-\\frac{y_{n}}{2}-z_{n}, y_{n+1}=y_{n}$, and $z_{n+1}=$ $y_{n}+z_{n}$. The integer $x_{0}$ is said to be good if $x_{n}=0$ for some $n \\geq 1$. Find the number of good integers less than or equal to 1994.", "solution": "21. Note first that $y_{n}=2^{k}(k \\geq 2)$ and $z_{k} \\equiv 1(\\bmod 4)$ for all $n$, so if $x_{n}$ is odd, $x_{n+1}$ will be even. Further, it is shown by induction on $n$ that $y_{n}>z_{n}$ when $x_{n-1}$ is even and $2 y_{n}>z_{n}>y_{n}$ when $x_{n-1}$ is odd. In fact, $n=1$ is the trivial case, while if it holds for $n \\geq 1$, then $y_{n+1}=2 y_{n}>z_{n}=z_{n+1}$ if $x_{n}$ is even, and $2 y_{n+1}=2 y_{n}>y_{n}+z_{n}=z_{n+1}$ if $x_{n}$ is odd (since then $x_{n-1}$ is even). If $x_{1}=0$, then $x_{0}=3$ is good. Suppose $x_{n}=0$ for some $n \\geq 2$. Then $x_{n-1}$ is odd and $x_{n-2}$ is even, so that $y_{n-1}>z_{n-1}$. We claim that a pair $\\left(y_{n-1}, z_{n-1}\\right)$, where $2^{k}=y_{n-1}>z_{n-1}>0$ and $z_{n-1} \\equiv 1$ $(\\bmod 4)$, uniquely determines $x_{0}=f\\left(y_{n-1}, z_{n-1}\\right)$. We see that $x_{n-1}=$ $\\frac{1}{2} y_{n-1}+z_{n-1}$, and define $\\left(x_{k}, y_{k}, z_{k}\\right)$ backwards as follows, until we get $\\left(y_{k}, z_{k}\\right)=(4,1)$. If $y_{k}>z_{k}$, then $x_{k-1}$ must have been even, so we define $\\left(x_{k-1}, y_{k-1}, z_{k-1}\\right)=\\left(2 x_{k}, y_{k} / 2, z_{k}\\right)$; otherwise $x_{k-1}$ must have been odd, so we put $\\left(x_{k-1}, y_{k-1}, z_{k-1}\\right)=\\left(x_{k}-y_{k} / 2+z_{k}, y_{k}, z_{k}-y_{k}\\right)$. We eventually arrive at $\\left(y_{0}, z_{0}\\right)=(4,1)$ and a good integer $x_{0}=f\\left(y_{n-1}, z_{n-1}\\right)$, as claimed. Thus for example $\\left(y_{n-1}, z_{n-1}\\right)=(64,61)$ implies $x_{n-1}=93$, $\\left(x_{n-2}, y_{n-2}, z_{n-2}\\right)=(186,32,61)$ etc., and $x_{0}=1953$, while in the case of $\\left(y_{n-1}, z_{n-1}\\right)=(128,1)$ we get $x_{0}=2080$. Note that $y^{\\prime}>y \\Rightarrow f\\left(y^{\\prime}, z^{\\prime}\\right)>f(y, z)$ and $z^{\\prime}>z \\Rightarrow f\\left(y, z^{\\prime}\\right)>f(y, z)$. Therefore there are no $y, z$ for which $1953<f(y, z)<2080$. Hence all good integers less than or equal to 1994 are given as $f(y, z), y=2^{k} \\leq 64$ and $0<z \\equiv 1(\\bmod 4)$, and the number of $\\operatorname{such}(y, z)$ equals $1+2+4+8+16=$ 31. So the answer is 31 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "22", "problem_type": "Number Theory", "exam": "IMO", "problem": "22. N5 (ROM) ${ }^{\\mathrm{IMO} 3}$ For any positive integer $k, A_{k}$ is the subset of $\\{k+1, k+$ $2, \\ldots, 2 k\\}$ consisting of all elements whose digits in base 2 contain exactly three 1's. Let $f(k)$ denote the number of elements in $A_{k}$. (a) Prove that for any positive integer $m, f(k)=m$ has at least one solution. (b) Determine all positive integers $m$ for which $f(k)=m$ has a unique solution.", "solution": "22. (a) Denote by $b(n)$ the number of 1's in the binary representation of $n$. Since $b(2 k+2)=b(k+1)$ and $b(2 k+1)=b(k)+1$, we deduce that  $$ f(k+1)= \\begin{cases}f(k)+1, & \\text { if } b(k)=2 \\\\ f(k), & \\text { otherwise }\\end{cases} $$  The set of $k$ 's with $b(k)=2$ is infinite, so it follows that $f(k)$ is unbounded. Hence $f$ takes all natural values. (b) Since $f$ is increasing, $k$ is a unique solution of $f(k)=m$ if and only if $f(k-1)<f(k)<f(k+1)$. By (1), this inequality is equivalent to $b(k-1)=b(k)=2$. It is easy to see that then $k-1$ must be of the form $2^{t}+1$ for some $t$. In this case, $\\{k+1, \\ldots, 2 k\\}$ contains the number $2^{t+1}+3=10 \\ldots 011_{2}$ and $\\frac{t(t-1)}{2}$ binary $(t+1)$-digit numbers with three 1 's, so $m=f(k)=\\frac{t(t-1)}{2}+1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "23", "problem_type": "Number Theory", "exam": "IMO", "problem": "23. N6 (LAT) Let $x_{1}$ and $x_{2}$ be relatively prime positive integers. For $n \\geq 2$, define $x_{n+1}=x_{n} x_{n-1}+1$. (a) Prove that for every $i>1$, there exists $j>i$ such that $x_{i}^{i}$ divides $x_{j}^{j}$. (b) Is it true that $x_{1}$ must divide $x_{j}^{j}$ for some $j>1$ ?", "solution": "23. (a) Let $p$ be a prime divisor of $x_{i}, i>1$, and let $x_{j} \\equiv u_{j}(\\bmod p)$ where $0 \\leq u_{j} \\leq p-1$ (particularly $u_{i} \\equiv 0$ ). Then $u_{j+1} \\equiv u_{j} u_{j-1}+$ $1(\\bmod p)$. The number of possible pairs $\\left(u_{j}, u_{j+1}\\right)$ is finite, so $u_{j}$ is eventually periodic. We claim that for some $d_{p}>0, u_{i+d_{p}}=0$. Indeed, suppose the contrary and let $\\left(u_{m}, u_{m+1}, \\ldots, u_{m+d-1}\\right)$ be the first period for $m \\geq i$. Then $m \\neq i$. By the assumption $u_{m-1} \\not \\equiv$ $u_{m+d-1}$, but $u_{m-1} u_{m} \\equiv u_{m+1}-1 \\equiv u_{m+d+1}-1 \\equiv u_{m+d-1} u_{m+d} \\equiv$ $u_{m+d-1} u_{m}(\\bmod p)$, which is impossible if $p \\nmid u_{m}$. Hence there is a $d_{p}$ with $u_{i}=u_{i+d_{p}}=0$ and moreover $u_{i+1}=u_{i+d_{p}+1}=1$, so the sequence $u_{j}$ is periodic with period $d_{p}$ starting from $u_{i}$. Let $m$ be the least common multiple of all $d_{p}$ 's, where $p$ goes through all prime divisors of $x_{i}$. Then the same primes divide every $x_{i+k m}, k=1,2, \\ldots$, so for large enough $k$ and $j=i+k m, x_{i}^{i} \\mid x_{j}^{j}$. (b) If $i=1$, we cannot deduce that $x_{i+1} \\equiv 1(\\bmod p)$. The following example shows that the statement from (a) need not be true in this case. Take $x_{1}=22$ and $x_{2}=9$. Then $x_{n}$ is even if and only if $n \\equiv 1(\\bmod$ 3 ), but modulo 11 the sequence $\\left\\{x_{n}\\right\\}$ is $0,9,1,10,0,1,1,2,3,7,0, \\ldots$, so $11 \\mid x_{n}(n>1)$ if and only if $n \\equiv 5(\\bmod 6)$. Thus for no $n>1$ can we have $22 \\mid x_{n}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "24", "problem_type": "Number Theory", "exam": "IMO", "problem": "24. N7 (GBR) A wobbly number is a positive integer whose digits in base 10 are alternately nonzero and zero, the units digit being nonzero. Determine all positive integers that do not divide any wobbly number.", "solution": "24. A multiple of 10 does not divide any wobbly number. Also, if $25 \\mid n$, then every multiple of $n$ ends with $25,50,75$, or 00 ; hence it is not wobbly. We now show that every other number $n$ divides some wobbly number. (i) Let $n$ be odd and not divisible by 5 . For any $k \\geq 1$ there exists $l$ such that $\\left(10^{k}-1\\right) n$ divides $10^{l}-1$, and thus also divides $10^{k l}-1$. Consequently, $v_{k}=\\frac{10^{k l}-1}{10^{k}-1}$ is divisible by $n$, and it is wobbly when $k=2$ (indeed, $v_{2}=101 \\ldots 01$ ). If $n$ is divisible by 5 , one can simply take $5 v_{2}$ instead. (ii) Let $n$ be a power of 2 . We prove by induction on $m$ that $2^{2 m+1}$ has a wobbly multiple $w_{m}$ with exactly $m$ nonzero digits. For $m=1$, take $w_{1}=8$. Suppose that for some $m \\geq 1$ there is a wobbly $w_{m}=$ $2^{2 m+1} d_{m}$. Then the numbers $a \\cdot 10^{2 m}+w_{m}$ are wobbly and divisible by $2^{2 m+1}$ when $a \\in\\{2,4,6,8\\}$. Moreover, one of these numbers is divisible by $2^{2 m+3}$. Indeed, it suffices to choose $a$ such that $\\frac{a}{2}+d_{m}$ is divisible by 4 . This proves the induction step. (iii) Let $n=2^{m} r$, where $m \\geq 1$ and $r$ is odd, $5 \\nmid r$. Then $v_{2 m} w_{m}$ is wobbly and divisible by both $2^{m}$ and $r$ (using notation from (i), $r \\mid v_{2 m}$ ).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "3", "problem_type": "Algebra", "exam": "IMO", "problem": "3. A3 (GBR) ${ }^{\\mathrm{IMO} 5}$ Let $S$ be the set of real numbers greater than -1 . Find all functions $f: S \\rightarrow S$ such that  $$ f(x+f(y)+x f(y))=y+f(x)+y f(x) \\quad \\text { for all } x \\text { and } y \\text { in } S, $$  and $f(x) / x$ is strictly increasing for $-1<x<0$ and for $0<x$.", "solution": "3. The last condition implies that $f(x)=x$ has at most one solution in $(-1,0)$ and at most one solution in $(0, \\infty)$. Suppose that for $u \\in(-1,0)$, $f(u)=u$. Then putting $x=y=u$ in the given functional equation yields $f\\left(u^{2}+2 u\\right)=u^{2}+2 u$. Since $u \\in(-1,0) \\Rightarrow u^{2}+2 u \\in(-1,0)$, we deduce that $u^{2}+2 u=u$, i.e., $u=-1$ or $u=0$, which is impossible. Similarly, if $f(v)=v$ for $v \\in(0, \\infty)$, we are led to the same contradiction. However, for all $x \\in S, f(x+(1+x) f(x))=x+(1+x) f(x)$, so we must have $x+(1+x) f(x)=0$. Therefore $f(x)=-\\frac{x}{1+x}$ for all $x \\in S$. It is directly verified that this function satisfies all the conditions.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "4", "problem_type": "Algebra", "exam": "IMO", "problem": "4. A4 (MON) Let $\\mathbb{R}$ denote the set of all real numbers and $\\mathbb{R}^{+}$the subset of all positive ones. Let $\\alpha$ and $\\beta$ be given elements in $\\mathbb{R}$, not necessarily distinct. Find all functions $f: \\mathbb{R}^{+} \\rightarrow \\mathbb{R}$ such that  $$ f(x) f(y)=y^{\\alpha} f\\left(\\frac{x}{2}\\right)+x^{\\beta} f\\left(\\frac{y}{2}\\right) \\quad \\text { for all } x \\text { and } y \\text { in } \\mathbb{R}^{+} . $$", "solution": "4. Suppose that $\\alpha=\\beta$. The given functional equation for $x=y$ yields $f(x / 2)=x^{-\\alpha} f(x)^{2} / 2$; hence the functional equation can be written as  $$ f(x) f(y)=\\frac{1}{2} x^{\\alpha} y^{-\\alpha} f(y)^{2}+\\frac{1}{2} y^{\\alpha} x^{-\\alpha} f(x)^{2} $$  i.e.,  $$ \\left((x / y)^{\\alpha / 2} f(y)-(y / x)^{\\alpha / 2} f(x)\\right)^{2}=0 $$  Hence $f(x) / x^{\\alpha}=f(y) / y^{\\alpha}$ for all $x, y \\in \\mathbb{R}^{+}$, so $f(x)=\\lambda x^{\\alpha}$ for some $\\lambda$. Substituting into the functional equation we obtain that $\\lambda=2^{1-\\alpha}$ or $\\lambda=0$. Thus either $f(x) \\equiv 2^{1-\\alpha} x^{\\alpha}$ or $f(x) \\equiv 0$. Now let $\\alpha \\neq \\beta$. Interchanging $x$ with $y$ in the given equation and subtracting these equalities from each other, we get $\\left(x^{\\alpha}-x^{\\beta}\\right) f(y / 2)=\\left(y^{\\alpha}-\\right.$ $\\left.y^{\\beta}\\right) f(x / 2)$, so for some constant $\\lambda \\geq 0$ and all $x \\neq 1, f(x / 2)=\\lambda\\left(x^{\\alpha}-x^{\\beta}\\right)$. Substituting this into the given equation, we obtain that only $\\lambda=0$ is possible, i.e., $f(x) \\equiv 0$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "5", "problem_type": "Algebra", "exam": "IMO", "problem": "5. A5 (POL) Let $f(x)=\\frac{x^{2}+1}{2 x}$ for $x \\neq 0$. Define $f^{(0)}(x)=x$ and $f^{(n)}(x)=$ $f\\left(f^{(n-1)}(x)\\right)$ for all positive integers $n$ and $x \\neq 0$. Prove that for all nonnegative integers $n$ and $x \\neq-1,0$, or 1 ,  $$ \\frac{f^{(n)}(x)}{f^{(n+1)}(x)}=1+\\frac{1}{f\\left(\\left(\\frac{x+1}{x-1}\\right)^{2^{n}}\\right)} $$", "solution": "5. If $f^{(n)}(x)=\\frac{p_{n}(x)}{q_{n}(x)}$ for some positive integer $n$ and polynomials $p_{n}, q_{n}$, then  $$ f^{(n+1)}(x)=f\\left(\\frac{p_{n}(x)}{q_{n}(x)}\\right)=\\frac{p_{n}(x)^{2}+q_{n}(x)^{2}}{2 p_{n}(x) q_{n}(x)} $$  Note that $f^{(0)}(x)=x / 1$. Thus $f^{(n)}(x)=\\frac{p_{n}(x)}{q_{n}(x)}$, where the sequence of polynomials $p_{n}, q_{n}$ is defined recursively by  $$ \\begin{gathered} p_{0}(x)=x, \\quad q_{0}(x)=1, \\text { and } \\\\ p_{n+1}(x)=p_{n}(x)^{2}+q_{n}(x)^{2}, q_{n+1}(x)=2 p_{n}(x) q_{n}(x) . \\end{gathered} $$  Furthermore, $p_{0}(x) \\pm q_{0}(x)=x \\pm 1$ and $p_{n+1}(x) \\pm q_{n+1}(x)=p_{n}(x)^{2}+$ $q_{n}(x)^{2} \\pm 2 p_{n}(x) q_{n}(x)=\\left(p_{n}(x) \\pm q_{n}(x)\\right)^{2}$, so $p_{n}(x) \\pm q_{n}(x)=(x \\pm 1)^{2^{n}}$ for all $n$. Hence  $$ p_{n}(x)=\\frac{(x+1)^{2^{n}}+(x-1)^{2^{n}}}{2} \\quad \\text { and } \\quad q_{n}(x)=\\frac{(x+1)^{2^{n}}-(x-1)^{2^{n}}}{2} . $$  Finally,  $$ \\begin{aligned} \\frac{f^{(n)}(x)}{f^{(n+1)}(x)} & =\\frac{p_{n}(x) q_{n+1}(x)}{q_{n}(x) p_{n+1}(x)}=\\frac{2 p_{n}(x)^{2}}{p_{n+1}(x)}=\\frac{\\left((x+1)^{2^{n}}+(x-1)^{2^{n}}\\right)^{2}}{(x+1)^{2^{n+1}}+(x-1)^{2^{n+1}}} \\\\ & =1+\\frac{2\\left(\\frac{x+1}{x-1}\\right)^{2^{n}}}{1+\\left(\\frac{x+1}{x-1}\\right)^{2^{n+1}}}=1+\\frac{1}{f\\left(\\left(\\frac{x+1}{x-1}\\right)^{2^{n}}\\right)} \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "6", "problem_type": "Combinatorics", "exam": "IMO", "problem": "6. C1 (UKR) On a $5 \\times 5$ board, two players alternately mark numbers on empty cells. The first player always marks 1's, the second 0's. One number is marked per turn, until the board is filled. For each of the nine $3 \\times 3$ squares the sum of the nine numbers on its cells is computed. Denote by $A$ the maximum of these sums. How large can the first player make $A$, regardless of the responses of the second player?", "solution": "6. Call the first and second player $M$ and $N$ respectively. $N$ can keep $A \\leq 6$. Indeed, let 10 dominoes be placed as shown in the picture, and whenever $M$ marks a 1 in a cell of some domino, let $N$ mark 0 in the other cell of that domino if it is still empty. Since any $3 \\times 3$ square contains at least three complete domi- ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-591.jpg?height=252&width=249&top_left_y=1193&top_left_x=937) noes, there are at least three 0 's inside. Hence $A \\leq 6$. We now show that $M$ can make $A=6$. Let him start by marking 1 in $c 3$. By symmetry, we may assume that $N$ 's response is made in row 4 or 5 . Then $M$ marks 1 in $c 2$. If $N$ puts 0 in $c 1$, then $M$ can always mark two 1's in $b \\times\\{1,2,3\\}$ as well as three 1's in $\\{a, d\\} \\times\\{1,2,3\\}$. Thus either $\\{a, b, c\\} \\times\\{1,2,3\\}$ or $\\{b, c, d\\} \\times\\{1,2,3\\}$ will contain six 1's. However, if $N$ does not play his second move in $c 1$, then $M$ plays there, and thus he can easily achieve to have six 1's either in $\\{a, b, c\\} \\times\\{1,2,3\\}$ or $\\{c, d, e\\} \\times\\{1,2,3\\}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "7", "problem_type": "Combinatorics", "exam": "IMO", "problem": "7. C2 (COL) In a certain city, age is reckoned in terms of real numbers rather than integers. Every two citizens $x$ and $x^{\\prime}$ either know each other or do not know each other. Moreover, if they do not, then there exists a chain of citizens $x=x_{0}, x_{1}, \\ldots, x_{n}=x^{\\prime}$ for some integer $n \\geq 2$ such that $x_{i-1}$ and $x_{i}$ know each other. In a census, all male citizens declare their ages, and there is at least one male citizen. Each female citizen provides only the information that her age is the average of the ages of all the citizens she knows. Prove that this is enough to determine uniquely the ages of all the female citizens.", "solution": "7. Let $a_{1}, a_{2}, \\ldots, a_{m}$ be the ages of the male citizens $(m \\geq 1)$. We claim that the age of each female citizen can be expressed in the form $c_{1} a_{1}+\\cdots+c_{m} a_{m}$ for some constants $c_{i} \\geq 0$, and we will prove this by induction on the number $n$ of female citizens. The claim is clear if $n=1$. Suppose it holds for $n$ and consider the case of $n+1$ female citizens. Choose any of them, say $A$ of age $x$ who knows $k$ citizens (at least one male). By the induction hypothesis, the age of each of the other $n$ females is expressible as $c_{1} a_{1}+\\cdots+c_{m} a_{m}+c_{0} x$, where $c_{i} \\geq 0$ and $c_{0}+c_{1}+\\cdots+c_{m}=1$. Consequently, the sum of ages of the $k$ citizens who know $A$ is $k x=b_{1} a_{1}+\\cdots+b_{m} a_{m}+b_{0} x$ for some constants $b_{i} \\geq 0$ with sum $k$. But $A$ knows at least one male citizen (who does not contribute to the coefficient of $x)$, so $b_{0} \\leq k-1$. Hence $x=\\frac{b_{1} a_{1}+\\cdots+b_{m} a_{m}}{k-b_{0}}$, and the claim follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "8", "problem_type": "Combinatorics", "exam": "IMO", "problem": "8. C3 (MCD) Peter has three accounts in a bank, each with an integral number of dollars. He is only allowed to transfer money from one account to another so that the amount of money in the latter is doubled. (a) Prove that Peter can always transfer all his money into two accounts. (b) Can Peter always transfer all his money into one account?", "solution": "8. (a) Let $a, b, c, a \\leq b \\leq c$ be the amounts of money in dollars in Peter's first, second, and third account, respectively. If $a=0$, then we are done, so suppose that $a>0$. Let Peter make transfers of money into the first account as follows. Write $b=a q+r$ with $0 \\leq r<a$ and let $q=m_{0}+2 m_{1}+\\cdots+2^{k} m_{k}$ be the binary representation of $q$ $\\left(m_{i} \\in\\{0,1\\}, m_{k}=1\\right)$. In the $i$ th transfer, $i=1,2, \\ldots, k+1$, if $m_{i}=1$ he transfers money from the second account, while if $m_{i}=0$ he does so from the third. In this way he has transferred exactly $\\left(m_{0}+2 m_{1}+\\cdots+2^{k} m_{k}\\right) a$ dollars from the second account, thus leaving $r$ dollars in it, $r<a$. Repeating this procedure, Peter can diminish the amount of money in the smallest account to zero, as required. (b) If Peter has an odd number of dollars, he clearly cannot transfer his money into one account.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1994", "tier": "T0", "problem_label": "9", "problem_type": "Combinatorics", "exam": "IMO", "problem": "9. C4 (EST) There are $n+1$ fixed positions in a row, labeled 0 to $n$ in increasing order from right to left. Cards numbered 0 to $n$ are shuffled and dealt, one in each position. The object of the game is to have card $i$ in the $i$ th position for $0 \\leq i \\leq n$. If this has not been achieved, the following move is executed. Determine the smallest $k$ such that the $k$ th position is occupied by a card $l>k$. Remove this card, slide all cards from the $(k+1)$ st to the $l$ th position one place to the right, and replace the card $l$ in the $l$ th position. (a) Prove that the game lasts at most $2^{n}-1$ moves. (b) Prove that there exists a unique initial configuration for which the game lasts exactly $2^{n}-1$ moves.", "solution": "9. (a) For $i=1, \\ldots, n$, let $d_{i}$ be 0 if the card $i$ is in the $i$ th position, and 1 otherwise. Define $b=d_{1}+2 d_{2}+2^{2} d_{3}+\\cdots+2^{n-1} d_{n}$, so that $0 \\leq b \\leq$ $2^{n}-1$, and $b=0$ if and only if the game is over. After each move some digit $d_{l}$ changes from 1 to 0 while $d_{l+1}, d_{l+2}, \\ldots$ remain unchanged. Hence $b$ decreases after each move, and consequently the game ends after at most $2^{n}-1$ moves. (b) Suppose the game lasts exactly $2^{n}-1$ moves. Then each move decreases $b$ for exactly one, so playing the game in reverse (starting from the final configuration), every move is uniquely determined. It follows that if the configuration that allows a game lasting $2^{n}-1$ moves exists, it must be unique. Consider the initial configuration $0, n, n-1, \\ldots, 2,1$. We prove by induction that the game will last exactly $2^{n}-1$ moves, and that the card 0 will get to the 0 th position only in the last move. This is trivial for $n=1$, so suppose that the claim is true for some $n=m-1 \\geq 1$ and consider the case $n=m$. Obviously the card 0 does not move until the card $m$ gets to the 0 -th position. But if we ignore the card 0 and consider the card $m$ to be the card 0 , the induction hypothesis gives that the card $m$ will move to the 0 th position only after $2^{m-1}-1$ moves. After these $2^{m-1}-1$ moves, we come to the configuration $0, m-1, \\ldots, 2,1, m$. The next move yields $m, 0, m-1, \\ldots, 2,1$, so by the induction hypothesis again we need $2^{m-1}-1$ moves more to finish the game.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "1", "problem_type": "Algebra", "exam": "IMO", "problem": "1. A1 (RUS) ${ }^{\\mathrm{IMO} 2}$ Let $a, b$, and $c$ be positive real numbers such that $a b c=1$. Prove that  $$ \\frac{1}{a^{3}(b+c)}+\\frac{1}{b^{3}(a+c)}+\\frac{1}{c^{3}(a+b)} \\geq \\frac{3}{2} $$", "solution": "1. Let $x=\\frac{1}{a}, y=\\frac{1}{b}, z=\\frac{1}{c}$. Then $x y z=1$ and  $$ S=\\frac{1}{a^{3}(b+c)}+\\frac{1}{b^{3}(c+a)}+\\frac{1}{c^{3}(a+b)}=\\frac{x^{2}}{y+z}+\\frac{y^{2}}{z+x}+\\frac{z^{2}}{x+y} $$  We must prove that $S \\geq \\frac{3}{2}$. From the Cauchy-Schwarz inequality,  $$ [(y+z)+(z+x)+(x+y)] \\cdot S \\geq(x+y+z)^{2} \\Rightarrow S \\geq \\frac{x+y+z}{2} $$  It follows from the A-G mean inequality that $\\frac{x+y+z}{2} \\geq \\frac{3}{2} \\sqrt[3]{x y z}=\\frac{3}{2}$; hence the proof is complete. Equality holds if and only if $x=y=z=1$, i.e., $a=b=c=1$. Remark. After reducing the problem to $\\frac{x^{2}}{y+z}+\\frac{y^{2}}{z+x}+\\frac{z^{2}}{x+y} \\geq \\frac{3}{2}$, we can solve the problem using Jensen's inequality applied to the function $g(u, v)=$ $u^{2} / v$. The problem can also be solved using Muirhead's inequality.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "10", "problem_type": "Geometry", "exam": "IMO", "problem": "10. G4 (UKR) An acute triangle $A B C$ is given. Points $A_{1}$ and $A_{2}$ are taken on the side $B C$ (with $A_{2}$ between $A_{1}$ and $C$ ), $B_{1}$ and $B_{2}$ on the side $A C$ (with $B_{2}$ between $B_{1}$ and $A$ ), and $C_{1}$ and $C_{2}$ on the side $A B$ (with $C_{2}$ between $C_{1}$ and $B$ ) such that  $$ \\angle A A_{1} A_{2}=\\angle A A_{2} A_{1}=\\angle B B_{1} B_{2}=\\angle B B_{2} B_{1}=\\angle C C_{1} C_{2}=\\angle C C_{2} C_{1} . $$  The lines $A A_{1}, B B_{1}$, and $C C_{1}$ form a triangle, and the lines $A A_{2}, B B_{2}$, and $C C_{2}$ form a second triangle. Prove that all six vertices of these two triangles lie on a single circle.", "solution": "10. Let the two triangles be $X_{1} Y_{1} Z_{1}, X_{2} Y_{2} Z_{2}$, with $X_{1}=B B_{1} \\cap C C_{1}, Y_{1}=$ $C C_{1} \\cap A A_{1}, Z_{1}=A A_{1} \\cap B B_{1}$, $X_{2}=B B_{2} \\cap C C_{2}, Y_{2}=C C_{2} \\cap$ $A A_{2}, Z_{2}=A A_{2} \\cap B B_{2}$. First, we observe that $\\angle A B B_{2}=\\angle A C C_{1}$ and $\\angle A B B_{1}=\\angle A C C_{2}$. Consequently $\\angle B Z_{1} A_{1}=\\angle B A A_{1}+$ $\\angle A B B_{1}=\\angle B C C_{2}+\\angle C_{2} C A=$ $\\angle C$ and similarly $\\angle A Z_{2} B_{2}=\\angle C$, $\\angle A Y_{1} C_{1}=\\angle C Y_{2} A_{2}=\\angle B$. Also, $\\triangle A B B_{2} \\sim \\triangle A C C_{1}$; hence ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-601.jpg?height=399&width=475&top_left_y=1280&top_left_x=849) $A C_{1} / A C=A B_{2} / A B$. From the sine formula, we obtain  $$ \\begin{aligned} \\frac{A Z_{1}}{\\sin \\angle A B Z_{1}} & =\\frac{A B}{\\sin \\angle A Z_{1} B}=\\frac{A B}{\\sin \\angle C}=\\frac{A C}{\\sin \\angle B}=\\frac{A C}{\\sin \\angle A Y_{2} C} \\\\ & =\\frac{A Y_{2}}{\\sin \\angle A C Y_{2}} \\Longrightarrow A Z_{1}=A Y_{2} . \\end{aligned} $$  Analogously, $B X_{1}=B Z_{2}$ and $C Y_{1}=C X_{2}$. Furthermore, again from the sine formula,  $$ \\begin{aligned} \\frac{A Y_{1}}{\\sin \\angle A C_{1} Y_{1}} & =\\frac{A C_{1}}{\\sin \\angle A Y_{1} C_{1}}=\\frac{A C_{1}}{A C} \\frac{A C}{\\sin \\angle B} \\\\ & =\\frac{A B_{2}}{A B} \\frac{A B}{\\sin \\angle C}=\\frac{A B_{2}}{\\sin \\angle A Z_{2} B_{2}}=\\frac{A Z_{2}}{\\sin \\angle A B_{2} Z_{2}} . \\end{aligned} $$  Hence, $A Y_{1}=A Z_{2}$ and, analogously, $B Z_{1}=B X_{2}$ and $C X_{1}=C Y_{2}$. We deduce that $Y_{1} Z_{2} \\| B C$ and $Z_{2} X_{1} \\| A C$, which gives us $\\angle Y_{1} Z_{2} X_{1}=$ $180^{\\circ}-\\angle C=180^{\\circ}-\\angle Y_{1} Z_{1} X_{1}$. It follows that $Z_{2}$ lies on the circle circumscribed about $\\triangle X_{1} Y_{1} Z_{1}$. Similarly, so do $X_{2}$ and $Y_{2}$. Second solution. Let $H$ be the orthocenter of $\\triangle A B C$. Triangles $A H B$, $B H C, C H A, A B C$ have the same circumradius $R$. Additionally,  $$ \\angle H A A_{i}=\\angle H B B_{i}=\\angle H C C_{i}=\\theta(i=1,2) . $$  Since $\\angle H B X_{1}=\\angle H C X_{1}=\\theta, B C X_{1} H$ is concyclic and therefore $H X_{1}=$ $2 R \\sin \\theta$. The same holds for $H Y_{1}, H Z_{1}, H X_{2}, H Y_{2}, H Z_{2}$. Hence $X_{i}, Y_{i}, Z_{i}$ $(i=1,2)$ lie on a circle centered at $H$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "11", "problem_type": "Geometry", "exam": "IMO", "problem": "11. G5 (NZL) ${ }^{\\mathrm{IMO} 5}$ Let $A B C D E F$ be a convex hexagon with $A B=B C=$ $C D, D E=E F=F A$, and $\\measuredangle B C D=\\measuredangle E F A=\\pi / 3$ (that is, $60^{\\circ}$ ). Let $G$ and $H$ be two points interior to the hexagon such that angles $A G B$ and $D H E$ are both $2 \\pi / 3$ (that is, $120^{\\circ}$ ). Prove that $A G+G B+G H+D H+$ $H E \\geq C F$.", "solution": "11. Triangles $B C D$ and $E F A$ are equilateral, and hence $B E$ is an axis of symmetry of $A B D E$. Let $C^{\\prime}, F^{\\prime}$ respectively be the points symmetric to $C, F$ with respect to $B E$. The points $G$ and $H$ lie on the circumcircles of $A B C^{\\prime}$ and $D E F^{\\prime}$ respectively (because, for instance, $\\angle A G B=120^{\\circ}=$ $\\left.180^{\\circ}-\\angle A C^{\\prime} B\\right)$; hence from Ptolemy's theorem we have $A G+G B=C^{\\prime} G$ and $D H+H E=H F^{\\prime}$. Therefore  $$ A G+G B+G H+D H+H E=C^{\\prime} G+G H+H F^{\\prime} \\geq C^{\\prime} F^{\\prime}=C F $$  with equality if and only if $G$ and $H$ both lie on $C^{\\prime} F^{\\prime}$. Remark. Since by Ptolemy's inequality $A G+G B \\geq C^{\\prime} G$ and $D H+H E \\geq$ $H F^{\\prime}$, the result holds without the condition $\\angle A G B=\\angle D H E=120^{\\circ}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "12", "problem_type": "Geometry", "exam": "IMO", "problem": "12. G6 (USA) Let $A_{1} A_{2} A_{3} A_{4}$ be a tetrahedron, $G$ its centroid, and $A_{1}^{\\prime}, A_{2}^{\\prime}, A_{3}^{\\prime}$, and $A_{4}^{\\prime}$ the points where the circumsphere of $A_{1} A_{2} A_{3} A_{4}$ intersects $G A_{1}, G A_{2}, G A_{3}$, and $G A_{4}$, respectively. Prove that  $$ G A_{1} \\cdot G A_{2} \\cdot G A_{3} \\cdot G A_{4} \\leq G A_{1}^{\\prime} \\cdot G A_{2}^{\\prime} \\cdot G A_{3}^{\\prime} \\cdot G A_{4}^{\\prime} $$  and  $$ \\frac{1}{G A_{1}^{\\prime}}+\\frac{1}{G A_{2}^{\\prime}}+\\frac{1}{G A_{3}^{\\prime}}+\\frac{1}{G A_{4}^{\\prime}} \\leq \\frac{1}{G A_{1}}+\\frac{1}{G A_{2}}+\\frac{1}{G A_{3}}+\\frac{1}{G A_{4}} $$", "solution": "12. Let $O$ be the circumcenter and $R$ the circumradius of $\\underset{\\longrightarrow}{A_{1} A_{2} A_{3} A_{4}}$. We have $O A_{i}^{2}=\\left(\\overrightarrow{O G}+\\left(\\overrightarrow{O A_{i}}-\\overrightarrow{O G}\\right)\\right)^{2}=O G^{2}+G A_{i}^{2}+2 \\overrightarrow{O G} \\cdot \\overrightarrow{G A_{i}}$. Summing up these equalities for $i=1,2,3,4$ and using that $\\sum_{i=1}^{4} \\overrightarrow{G A_{i}}=\\overrightarrow{0}$, we obtain  $$ \\sum_{i=1}^{4} O A_{i}^{2}=4 O G^{2}+\\sum_{i=1}^{4} G A_{i}^{2} \\Longleftrightarrow \\sum_{i=1}^{4} G A_{i}^{2}=4\\left(R^{2}-O G^{2}\\right) $$  Now we have that the potential of $G$ with respect to the sphere equals $G A_{i} \\cdot G A_{i}^{\\prime}=R^{2}-O G^{2}$. Plugging in these expressions for $G A_{i}^{\\prime}$, we reduce the inequalities we must prove to  $$ \\begin{aligned} G A_{1} \\cdot G A_{2} \\cdot G A_{3} \\cdot G A_{4} & \\leq\\left(R^{2}-O G^{2}\\right)^{2} \\\\ \\text { and } \\quad\\left(R^{2}-O G^{2}\\right) \\sum_{i=1}^{4} \\frac{1}{G A_{i}} & \\geq \\sum_{i=1}^{4} G A_{i} \\end{aligned} $$  Inequality (2) immediately follows from (1) and the quadratic-geometric mean inequality for $G A_{i}$. Since from the Cauchy-Schwarz inequality we have $\\sum_{i=1}^{4} G A_{i}^{4} \\geq \\frac{1}{4}\\left(\\sum_{i=1}^{4} G A_{i}\\right)^{2}$ and $\\left(\\sum_{i=1}^{4} G A_{i}\\right)\\left(\\sum_{i=1}^{4} \\frac{1}{G A_{i}}\\right) \\geq 16$, inequality (3) follows from (1) and from  $$ \\left(\\sum_{i=1}^{4} G A_{i}^{2}\\right)\\left(\\sum_{i=1}^{4} \\frac{1}{G A_{i}}\\right) \\geq \\frac{1}{4}\\left(\\sum_{i=1}^{4} G A_{i}\\right)^{2}\\left(\\sum_{i=1}^{4} \\frac{1}{G A_{i}}\\right) \\geq 4 \\sum_{i=1}^{4} G A_{i} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "13", "problem_type": "Geometry", "exam": "IMO", "problem": "13. G7 (LAT) $O$ is a point inside a convex quadrilateral $A B C D$ of area $S . K, L, M$, and $N$ are interior points of the sides $A B, B C, C D$, and $D A$ respectively. If $O K B L$ and $O M D N$ are parallelograms, prove that $\\sqrt{S} \\geq \\sqrt{S_{1}}+\\sqrt{S_{2}}$, where $S_{1}$ and $S_{2}$ are the areas of $O N A K$ and $O L C M$ respectively.", "solution": "13. If $O$ lies on $A C$, then $A B C D, A K O N$, and $O L C M$ are similar; hence $A C=A O+O C$ implies $\\sqrt{S}=\\sqrt{S_{1}}+\\sqrt{S_{2}}$. Assume that $O$ does not lie on $A C$ and that w.l.o.g. it lies inside triangle $A D C$. Let us denote by $T_{1}, T_{2}$ the areas of parallelograms $K B L O, N O M D$ respectively. Consider a line through $O$ that intersects $A D, D C, C B, B A$ respectively at $X, Y, Z, W$ so that $O W / O X=O Z / O Y$ (such a line exists by a continuity argument: the left side is smaller when $W=X=A$, but greater when $Y=Z=C$ ). The desired inequality is equivalent to $T_{1}+T_{2} \\geq 2 \\sqrt{S_{1} S_{2}}$. Since triangles $W K O, O L Z, W B Z$ are similar and $W O+O Z=W Z$, we have $\\sqrt{S_{W K O}}+\\sqrt{S_{O L Z}}=\\sqrt{S_{W B Z}}=$ $\\sqrt{S_{W K O}+S_{O L Z}+T_{1}}$, which implies $T_{1}=2 \\sqrt{S_{W K O} S_{O L Z}}$. Similarly, $T_{2}=2 \\sqrt{S_{X N O} S_{O M Y}}$. Since $O W / O Z=O X / O Y$, we have $S_{W K O} / S_{X N O}=S_{O L Z} / S_{O M Y}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-603.jpg?height=394&width=489&top_left_y=903&top_left_x=835)  Therefore we obtain  $$ \\begin{aligned} T_{1}+T_{2} & =2 \\sqrt{S_{W K O} S_{O L Z}}+2 \\sqrt{S_{X N O} S_{O M Y}} \\\\ & =2 \\sqrt{\\left(S_{W K O}+S_{X N O}\\right)\\left(S_{O L Z}+S_{O M Y}\\right)} \\geq 2 \\sqrt{S_{1} S_{2}} \\end{aligned} $$  Second solution. By an affine transformation of the plane one can transform any nondegenerate quadrilateral into a cyclic one, thereby preserving parallelness and ratios of areas. Thus we may assume w.l.o.g. that $A B C D$ is cyclic. By a well-known formula, the area of a cyclic quadrilateral with sides $a, b, c, d$ and semiperimeter $p$ is given by  $$ S=\\sqrt{(p-a)(p-b)(p-c)(p-d)} $$  Let us set $A K=a_{1}, K B=b_{1}, B L=a_{2}, L C=b_{2}, C M=a_{3}, M D=b_{3}$, $D N=a_{4}, N A=b_{4}$. Then the sides of quadrilateral $A K O N$ are $a_{i}$, the sides of $C L O M$ are $b_{i}$, and the sides of $A B C D$ are $a_{i}+b_{i}(i=1,2,3,4)$. If $p$ and $q$ are the semiperimeters of $A K O N$ and $C L O M$, and $x_{i}=p-a_{i}$, $y_{i}=q-b_{i}$, then we have $S_{1}=\\sqrt{x_{1} x_{2} x_{3} x_{4}}, S_{2}=\\sqrt{y_{1} y_{2} y_{3} y_{4}}$, and $S=$ $\\sqrt{\\left(x_{1}+y_{1}\\right)\\left(x_{2}+y_{2}\\right)\\left(x_{3}+y_{3}\\right)\\left(x_{4}+y_{4}\\right)}$. Thus we need to show that  $$ \\sqrt[4]{x_{1} x_{2} x_{3} x_{4}}+\\sqrt[4]{y_{1} y_{2} y_{3} y_{4}} \\leq \\sqrt[4]{\\left(x_{1}+y_{1}\\right)\\left(x_{2}+y_{2}\\right)\\left(x_{3}+y_{3}\\right)\\left(x_{4}+y_{4}\\right)} $$  By setting $y_{i}=t_{i} x_{i}$ we reduce this inequality to $1+\\sqrt[4]{t_{1} t_{2} t_{3} t_{4}} \\leq$ $\\sqrt[4]{\\left(1+t_{1}\\right)\\left(1+t_{2}\\right)\\left(1+t_{3}\\right)\\left(1+t_{4}\\right)}$. One way to prove the last inequality is to apply the simple inequality  $$ 1+\\sqrt{u v} \\leq \\sqrt{(1+u)(1+v)} $$  to $\\sqrt{t_{1} t_{2}}, \\sqrt{t_{3} t_{4}}$ and then to $t_{1}, t_{2}$ and $t_{3}, t_{4}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "14", "problem_type": "Geometry", "exam": "IMO", "problem": "14. G8 (COL) Let $A B C$ be a triangle. A circle passing through $B$ and $C$ intersects the sides $A B$ and $A C$ again at $C^{\\prime}$ and $B^{\\prime}$, respectively. Prove that $B B^{\\prime}, C C^{\\prime}$, and $H H^{\\prime}$ are concurrent, where $H$ and $H^{\\prime}$ are the orthocenters of triangles $A B C$ and $A B^{\\prime} C^{\\prime}$ respectively.", "solution": "14. Let $B B^{\\prime}$ cut $C C^{\\prime}$ at $P$. Since $\\angle B^{\\prime} B C^{\\prime}=\\angle B^{\\prime} C C^{\\prime}$, it follows that $\\angle P B H=\\angle P C H$. Let $D$ and $E$ be points such that $B P C D$ and $H P C E$ are parallelograms (consequently, so is $B H E D$ ). Triangles $B A C$ and $C^{\\prime} A B^{\\prime}$ are similar, from which we deduce that $\\triangle B^{\\prime} H^{\\prime} C^{\\prime}$ and $\\triangle B H C$ are similar, as well as $\\triangle B^{\\prime} P C^{\\prime}$ and $\\triangle B D C$. Hence $B^{\\prime} P C^{\\prime} H^{\\prime}$ and $B D C H$ are similar, from which we obtain $\\angle H^{\\prime} P B^{\\prime}=\\angle H D B$. Now $\\angle C D E=\\angle P B H=\\angle P C H=$ $\\angle C H E$ implies that $H C E D$ is a cyclic quadrilateral. Therefore $\\angle B P H=\\angle D C E=\\angle D H E=$ $\\angle H D B=\\angle H^{\\prime} P B^{\\prime}$; hence $H H^{\\prime}$ also passes through $P$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-604.jpg?height=415&width=356&top_left_y=775&top_left_x=906)  Second solution. Observe that $\\triangle H B C \\sim \\triangle H^{\\prime} B^{\\prime} C^{\\prime}, \\angle P B H=\\angle P C H$ and $\\angle P B^{\\prime} H^{\\prime}=\\angle P C^{\\prime} H^{\\prime}$. By Ceva's theorem in trigonometric form applied to $\\triangle B P C$ and the point $H$, we have $\\frac{\\sin \\angle B P H}{\\sin \\angle H P C}=\\frac{\\sin \\angle H B P}{\\sin \\angle H B C} \\cdot \\frac{\\sin \\angle H C B}{\\sin \\angle H C P}=\\frac{\\sin \\angle H C B}{\\sin \\angle H B C}$. Similarly, Ceva's theorem for $\\triangle B^{\\prime} P C^{\\prime}$ and point $H^{\\prime}$ yields $\\frac{\\sin \\angle B^{\\prime} P H^{\\prime}}{\\sin \\angle H^{\\prime} P C^{\\prime}}=\\frac{\\sin \\angle H^{\\prime} C^{\\prime} B^{\\prime}}{\\sin \\angle H^{\\prime} B^{\\prime} C^{\\prime}}$. Thus it follows that  $$ \\frac{\\sin \\angle B^{\\prime} P H^{\\prime}}{\\sin \\angle H^{\\prime} P C^{\\prime}}=\\frac{\\sin \\angle B P H}{\\sin \\angle H P C} $$  which finally implies that $\\angle B P H=\\angle B^{\\prime} P H^{\\prime}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "15", "problem_type": "Number Theory", "exam": "IMO", "problem": "15. N1 (ROM) Let $k$ be a positive integer. Prove that there are infinitely many perfect squares of the form $n 2^{k}-7$, where $n$ is a positive integer.", "solution": "15. We show by induction on $k$ that there exists a positive integer $a_{k}$ for which $a_{k}^{2} \\equiv-7\\left(\\bmod 2^{k}\\right)$. The statement of the problem follows, since every $a_{k}+r 2^{k}(r=0,1, \\ldots)$ also satisfies this condition. Note that for $k=1,2,3$ one can take $a_{k}=1$. Now suppose that $a_{k}^{2} \\equiv-7$ $\\left(\\bmod 2^{k}\\right)$ for some $k>3$. Then either $a_{k}^{2} \\equiv-7\\left(\\bmod 2^{k+1}\\right)$ or $a_{k}^{2} \\equiv 2^{k}-7$ $\\left(\\bmod 2^{k+1}\\right)$. In the former case, take $a_{k+1}=a_{k}$. In the latter case, set $a_{k+1}=a_{k}+2^{k-1}$. Then $a_{k+1}^{2}=a_{k}^{2}+2^{k} a_{k}+2^{2 k-2} \\equiv a_{k}^{2}+2^{k} \\equiv-7(\\bmod$ $2^{k+1}$ ) because $a_{k}$ is odd.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "16", "problem_type": "Number Theory", "exam": "IMO", "problem": "16. N2 (RUS) Let $\\mathbb{Z}$ denote the set of all integers. Prove that for any integers $A$ and $B$, one can find an integer $C$ for which $M_{1}=\\left\\{x^{2}+A x+B: x \\in \\mathbb{Z}\\right\\}$ and $M_{2}=\\left\\{2 x^{2}+2 x+C: x \\in \\mathbb{Z}\\right\\}$ do not intersect.", "solution": "16. If $A$ is odd, then every number in $M_{1}$ is of the form $x(x+A)+B \\equiv B$ $(\\bmod 2)$, while numbers in $M_{2}$ are congruent to $C$ modulo 2 . Thus it is enough to take $C \\equiv B+1(\\bmod 2)$.  If $A$ is even, then all numbers in $M_{1}$ have the form $\\left(X+\\frac{A}{2}\\right)^{2}+B-\\frac{A^{2}}{4}$ and are congruent to $B-\\frac{A^{2}}{4}$ or $B-\\frac{A^{2}}{4}+1$ modulo 4 , while numbers in $M_{2}$ are congruent to $C$ modulo 4 . So one can choose any $C \\equiv B-\\frac{A^{2}}{4}+2$ $(\\bmod 4)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "17", "problem_type": "Number Theory", "exam": "IMO", "problem": "17. N3 $(\\mathbf{C Z E})^{\\mathrm{IMO} 3}$ Determine all integers $n>3$ such that there are $n$ points $A_{1}, A_{2}, \\ldots, A_{n}$ in the plane that satisfy the following two conditions simultaneously: (a) No three lie on the same line. (b) There exist real numbers $p_{1}, p_{2}, \\ldots, p_{n}$ such that the area of $\\triangle A_{i} A_{j} A_{k}$ is equal to $p_{i}+p_{j}+p_{k}$, for $1 \\leq i<j<k \\leq n$.", "solution": "17. For $n=4$, the vertices of a unit square $A_{1} A_{2} A_{3} A_{4}$ and $p_{1}=p_{2}=p_{3}=$ $p_{4}=\\frac{1}{6}$ satisfy the conditions. We claim that there are no solutions for $n=5$ (and thus for any $n \\geq 5$ ). Suppose to the contrary that points $A_{i}$ and $p_{i}, i=1, \\ldots, 5$, satisfy the conditions. Denote the area of $\\triangle A_{i} A_{j} A_{k}$ by $S_{i j k}=p_{i}+p_{j}+p_{k}, 1 \\leq i<$ $j<k \\leq 5$. Observe that all the $p_{i}$ 's must be distinct. Indeed, if $p_{4}=p_{5}$, then $S_{124}=S_{125}$ and $S_{234}=S_{235}$, which implies that $A_{4} A_{5}$ is parallel to $A_{1} A_{2}$ and $A_{2} A_{3}$, so $A_{1}, A_{2}, A_{3}$ are collinear, which is impossible. Also note that if $A_{i} A_{j} A_{k} A_{l}$ is convex, then $S_{i j k}+S_{i k l}=S_{i j l}+S_{j k l}$ gives $p_{i}+p_{k}=p_{j}+p_{l}$. Now consider the convex hull of $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$. There are three cases. (i) The convex hull is the pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$. Then $A_{1} A_{2} A_{3} A_{4}$ and $A_{1} A_{2} A_{3} A_{5}$ are convex, so we have $p_{1}+p_{3}=p_{2}+p_{4}$ and $p_{1}+p_{3}=$ $p_{2}+p_{5}$. Hence $p_{4}=p_{5}$, a contradiction. (ii) The convex hull is w.l.o.g. the quadrilateral $A_{1} A_{2} A_{3} A_{4}$. Assume that $A_{5}$ lies within $A_{1} A_{3} A_{4}$. Then $A_{1} A_{2} A_{3} A_{5}$ is also convex, so as in (1) we get $p_{4}=p_{5}$. (iii) The convex hull is w.l.o.g. the triangle $A_{1} A_{2} A_{3}$. Since $S_{124}+S_{134}+$ $S_{234}=S_{125}+S_{135}+S_{235}$, we conclude that again $p_{4}=p_{5}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "18", "problem_type": "Number Theory", "exam": "IMO", "problem": "18. N4 (BUL) Find all positive integers $x$ and $y$ such that $x+y^{2}+z^{3}=x y z$, where $z$ is the greatest common divisor of $x$ and $y$.", "solution": "18. Let $x=z a$ and $y=z b$, where $a$ and $b$ are relatively prime. The given Diophantine equation becomes $a+z b^{2}+z^{2}=z^{2} a b$, so $a=z c$ for some $c \\in \\mathbb{Z}$. We obtain $c+b^{2}+z=z^{2} c b$, or $c=\\frac{b^{2}+z}{z^{2} b-1}$. (i) If $z=1$, then $c=\\frac{b^{2}+1}{b-1}=b+1+\\frac{2}{b-1}$, so $b=2$ or $b=3$. These values yield two solutions: $(x, y)=(5,2)$ and $(x, y)=(5,3)$. (ii) If $z=2$, then $16 c=\\frac{16 b^{2}+32}{4 b-1}=4 b+1+\\frac{33}{4 b-1}$, so $b=1$ or $b=3$. In this case $(x, y)=(4,2)$ or $(x, y)=(4,6)$. (iii) Let $z \\geq 3$. First, we see that $z^{2} c=\\frac{z^{2} b^{2}+z^{3}}{z^{2} b-1}=b+\\frac{b+z^{3}}{z^{2} b-1}$. Thus $\\frac{b+z^{3}}{z^{2} b-1}$ must be a positive integer, so $b+z^{3} \\geq z^{2} b-1$, which implies $b \\leq$ $\\frac{z^{2}-z+1}{z-1}$. It follows that $b \\leq z$. But then $b^{2}+z \\leq z^{2}+b<z^{2} b-1$, with the last inequality because $\\left(z^{2}-1\\right)(b-1)>2$. Therefore $c=\\frac{b^{2}+z}{z^{2} b-1}<1$, a contradiction. The only solutions for $(x, y)$ are $(4,2),(4,6),(5,2),(5,3)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "19", "problem_type": "Number Theory", "exam": "IMO", "problem": "19. N5 (IRE) At a meeting of $12 k$ people, each person exchanges greetings with exactly $3 k+6$ others. For any two people, the number who exchange greetings with both is the same. How many people are at the meeting?", "solution": "19. For each two people let $n$ be the number of people exchanging greetings with both of them. To determine $n$ in terms of $k$, we shall count in two ways the number of triples $(A, B, C)$ of people such that $A$ exchanged greetings with both $B$ and $C$, but $B$ and $C$ mutually did not. There are $12 k$ possibilities for $A$, and for each $A$ there are $(3 k+6)$ possibilities for $B$. Since there are $n$ people who exchanged greetings with both $A$ and $B$, there are $3 k+5-n$ who did so with $A$ but not with $B$. Thus the number of triples $(A, B, C)$ is $12 k(3 k+6)(3 k+5-n)$. On the other hand, there are $12 k$ possible choices of $B$, and $12 k-1-(3 k+6)=9 k-7$ possible choices of $C$; for every $B, C, A$ can be chosen in $n$ ways, so the number of considered triples equals $12 k n(9 k-7)$. Hence $(3 k+6)(3 k+5-n)=n(9 k-7)$, i.e., $n=\\frac{3(k+2)(3 k+5)}{12 k-1}$. This gives us that $\\frac{4 n}{3}=\\frac{12 k^{2}+44 k+40}{12 k-1}=k+4-\\frac{3 k-44}{12 k-1}$ is an integer too. It is directly verified that only $k=3$ gives an integer value for $n$, namely $n=6$. Remark. The solution is complete under the assumption that such a $k$ exists. We give an example of such a party with 36 persons, $k=3$. Let the people sit in a $6 \\times 6$ array $\\left[P_{i j}\\right]_{i, j=1}^{6}$, and suppose that two persons $P_{i j}, P_{k l}$ exchanged greetings if and only if $i=k$ or $j=l$ or $i-j \\equiv k-l$ (mod 6). Thus each person exchanged greetings with exactly 15 others, and it is easily verified that this party satisfies the conditions.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "2", "problem_type": "Algebra", "exam": "IMO", "problem": "2. A2 (SWE) Let $a$ and $b$ be nonnegative integers such that $a b \\geq c^{2}$, where $c$ is an integer. Prove that there is a number $n$ and integers $x_{1}, x_{2}, \\ldots, x_{n}, y_{1}, y_{2}, \\ldots, y_{n}$ such that  $$ \\sum_{i=1}^{n} x_{i}^{2}=a, \\quad \\sum_{i=1}^{n} y_{i}^{2}=b, \\quad \\text { and } \\quad \\sum_{i=1}^{n} x_{i} y_{i}=c $$", "solution": "2. We may assume $c \\geq 0$ (otherwise, we may simply put $-y_{i}$ in the place of $\\left.y_{i}\\right)$. Also, we may assume $a \\geq b$. If $b \\geq c$, it is enough to take $n=a+b-c$, $x_{1}=\\cdots=x_{a}=1, y_{1}=\\cdots=y_{c}=y_{a+1}=\\cdots=y_{a+b-c}=1$, and the other $x_{i}$ 's and $y_{i}$ 's equal to 0 , so we need only consider the case $a>c>b$. We proceed to prove the statement of the problem by induction on $a+b$. The case $a+b=1$ is trivial. Assume that the statement is true when $a+b \\leq$ $N$, and let $a+b=N+1$. The triple $(a+b-2 c, b, c-b)$ satisfies the condition (since $(a+b-2 c) b-(c-b)^{2}=a b-c^{2}$ ), so by the induction hypothesis there are $n$-tuples $\\left(x_{i}\\right)_{i=1}^{n}$ and $\\left(y_{i}\\right)_{i=1}^{n}$ with the wanted property. It is easy to verify that $\\left(x_{i}+y_{i}\\right)_{i=1}^{n}$ and $\\left(y_{i}\\right)_{i=1}^{n}$ give a solution for $(a, b, c)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "20", "problem_type": "Number Theory", "exam": "IMO", "problem": "20. N6 (POL) ${ }^{\\mathrm{IMO} 06}$ Let $p$ be an odd prime. Find the number of $p$-element subsets $A$ of $\\{1,2, \\ldots, 2 p\\}$ such that the sum of all elements of $A$ is divisible by $p$.", "solution": "20. We shall consider the set $M=\\{0,1, \\ldots, 2 p-1\\}$ instead. Let $M_{1}=$ $\\{0,1, \\ldots, p-1\\}$ and $M_{2}=\\{p, p+1, \\ldots, 2 p-1\\}$. We shall denote by $|A|$ and $\\sigma(A)$ the number of elements and the sum of elements of the set $A$; also, let $C_{p}$ be the family of all $p$-element subsets of $M$. Define the mapping $T: C_{p} \\rightarrow C_{p}$ as $T(A)=\\left\\{x+1 \\mid x \\in A \\cap M_{1}\\right\\} \\cup\\left\\{A \\cap M_{2}\\right\\}$, the addition being modulo $p$. There are exactly two fixed points of $T$ : these are $M_{1}$ and $M_{2}$. Now if $A$ is any subset from $C_{p}$ distinct from $M_{1}, M_{2}$, and $k=\\left|A \\cap M_{1}\\right|$ with $1 \\leq k \\leq p-1$, then for $i=0,1, \\ldots, p-1$, $\\sigma\\left(T^{i}(A)\\right)=\\sigma(A)+i k(\\bmod p)$. Hence subsets $A, T(A), \\ldots, T^{p-1}(A)$ are distinct, and exactly one of them has sum of elements divisible by $p$. Since $\\sigma\\left(M_{1}\\right), \\sigma\\left(M_{2}\\right)$ are divisible by $p$ and $C_{p} \\backslash\\left\\{M_{1}, M_{2}\\right\\}$ decomposes into families of the form $\\left\\{A, T(A), \\ldots, T^{p-1}(A)\\right\\}$, we conclude that the required number is $\\frac{1}{p}\\left(\\left|C_{p}\\right|-2\\right)+2=\\frac{1}{p}\\left(\\binom{2 p}{p}-2\\right)+2$. Second solution. Let $C_{k}$ be the family of all $k$-element subsets of $\\{1,2, \\ldots, 2 p\\}$. Denote by $M_{k}(k=1,2, \\ldots, p)$ the family of $p$-element multisets with $k$ distinct elements from $\\{1,2, \\ldots, 2 p\\}$, exactly one of which appears more than once, that have sum of elements divisible by $p$. It is clear that every subset from $C_{k}, k<p$, can be complemented to a multiset from $M_{k} \\cup M_{k+1}$ in exactly two ways, since the equation $(p-k) a \\equiv 0(\\bmod p)$ has exactly two solutions in $\\{1,2, \\ldots, 2 p\\}$. On the other hand, every multiset from $M_{k}$ can be obtained by completing exactly one subset from $C_{k}$. Additionally, a multiset from $M_{k}$ can be obtained from exactly one subset from $C_{k-1}$ if $k<p$, and from exactly $p$ subsets from $C_{k-1}$ if $k=p$. Therefore $\\left|M_{k}\\right|+\\left|M_{k+1}\\right|=2\\left|C_{k}\\right|=2\\binom{2 p}{k}$ for $k=1,2, \\ldots, p-2$, and $\\left|M_{p-1}\\right|+p\\left|M_{p}\\right|=2\\left|C_{p-1}\\right|=2\\binom{2 p}{p-1}$. Since $M_{1}=2 p$, it is not difficult to show using recursion that $\\left|M_{p}\\right|=\\frac{1}{p}\\left(\\binom{2 p}{p}-2\\right)+2$.  Third solution. Let $\\omega=\\cos \\frac{2 \\pi}{p}+i \\sin \\frac{2 \\pi}{p}$. We have $\\prod_{i=1}^{2 p}\\left(x-\\omega^{i}\\right)=$ $\\left(x^{p}-1\\right)^{2}=x^{2 p}-2 x^{p}+1$; hence comparing the coefficients at $x^{p}$, we obtain $\\sum \\omega^{i_{1}+\\cdots+i_{p}}=\\sum_{i=0}^{p-1} a_{i} \\omega^{i}=2$, where the first sum runs over all $p$-subsets $\\left\\{i_{1}, \\ldots, i_{p}\\right\\}$ of $\\{1, \\ldots, 2 p\\}$, and $a_{i}$ is the number of such subsets for which $i_{1}+\\cdots+i_{p} \\equiv i(\\bmod p)$. Setting $q(x)=-2+\\sum_{i=0}^{p-1} a_{i} x^{i}$, we obtain $q\\left(\\omega^{j}\\right)=0$ for $j=1,2, \\ldots, p-1$. Hence $1+x+\\cdots+x^{p-1} \\mid q(x)$, and since deg $q=p-1$, we have $q(x)=-2+\\sum_{i=0}^{p-1} a_{i} x^{i}=c\\left(1+x+\\cdots+x^{p-1}\\right)$ for some constant $c$. Thus $a_{0}-2=a_{1}=\\cdots=a_{p-1}$, which together with $a_{0}+\\cdots+a_{p-1}=\\binom{2 p}{p}$ yields $a_{0}=\\frac{1}{p}\\left(\\binom{2 p}{p}-2\\right)+2$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "21", "problem_type": "Number Theory", "exam": "IMO", "problem": "21. N7 (BLR) Does there exist an integer $n>1$ that satisfies the following condition? The set of positive integers can be partitioned into $n$ nonempty subsets such that an arbitrary sum of $n-1$ integers, one taken from each of any $n-1$ of the subsets, lies in the remaining subset.", "solution": "21. We shall show that there is no such $n$. Certainly, $n=2$ does not work, so suppose $n \\geq 3$. Let $a, b$ be distinct elements of $A_{1}$, and $c$ any integer greater than $-a$ and $-b$. We claim that $a+c, b+c$ belong to the same subsets. Suppose to the contrary that $a+c \\in A_{1}$ and $b+c \\in A_{2}$, and take arbitrary elements $x_{i} \\in A_{i}, i=3, \\ldots, n$. The number $b+x_{3}+\\cdots+x_{n}$ is in $A_{2}$, so that $s=(a+c)+\\left(b+x_{3}+\\cdots+x_{n}\\right)+x_{4}+\\cdots+x_{n}$ must be in $A_{3}$. On the other hand, $a+x_{3}+\\cdots+x_{n} \\in A_{2}$, so $s=\\left(a+x_{3}+\\cdots+x_{n}\\right)+$ $(b+c)+x_{4}+\\cdots+x_{n}$ is in $A_{1}$, a contradiction. Similarly, if $a+c \\in A_{2}$ and $b+c \\in A_{3}$, then $s=a+(b+c)+x_{4}+\\cdots+x_{n}$ belongs to $A_{2}$, but also $s=b+(a+c)+x_{4}+\\cdots+x_{n} \\in A_{3}$, which is impossible. For $i=1, \\ldots, n$ choose $x_{i} \\in A_{i}$; set $s=x_{1}+\\cdots+x_{n}$ and $y_{i}=s-x_{i}$. Then $y_{i} \\in A_{i}$. By what has been proved above, $2 x_{i}=x_{i}+x_{i}$ belongs to the same subset as $x_{i}+y_{i}=s$ does. It follows that all numbers $2 x_{i}, i=1, \\ldots, n$, are in the same subset. Since we can arbitrarily take $x_{i}$ from each set $A_{i}$, it follows that all even numbers belong to the same set, say $A_{1}$. Similarly, $2 x_{i}+1=\\left(x_{i}+1\\right)+x_{i}$ is in the subset to which $\\left(x_{i}+1\\right)+y_{i}=s+1$ belongs for all $i=1, \\ldots, n$; hence all odd numbers greater than 1 are in the same subset, say $A_{2}$. By the above considerations, $3-2=1 \\in A_{2}$ also. But then nothing remains in $A_{3}, \\ldots, A_{n}$, a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "22", "problem_type": "Number Theory", "exam": "IMO", "problem": "22. N8 (GER) Let $p$ be an odd prime. Determine positive integers $x$ and $y$ for which $x \\leq y$ and $\\sqrt{2 p}-\\sqrt{x}-\\sqrt{y}$ is nonnegative and as small as possible.", "solution": "22. Let $u=\\sqrt{2 p}-\\sqrt{x}-\\sqrt{y}$ and $v=u(2 \\sqrt{2 p}-u)=2 p-(\\sqrt{2 p}-u)^{2}=$ $2 p-x-y-\\sqrt{4 x y}$ for $x, y \\in \\mathbb{N}, x \\leq y$. Obviously $u \\geq 0$ if and only if $v \\geq 0$, and $u, v$ attain minimum positive values simultaneously. Note that $v \\neq 0$. Otherwise $u=0$ too, so $y=(\\sqrt{2 p}-\\sqrt{x})^{2}=2 p-x-2 \\sqrt{2 p x}$, which implies that $2 p x$ is a square, and consequently $x$ is divisible by $2 p$, which is impossible. Now let $z$ be the smallest integer greater than $\\sqrt{4 x y}$. We have $z^{2}-1 \\geq 4 x y$, $z \\leq 2 p-x-y$, and $z \\leq p$ because $\\sqrt{4 x y} \\leq(\\sqrt{x}+\\sqrt{y})^{2}<2 p$. It follows that  $$ v=2 p-x-y-\\sqrt{4 x y} \\geq z-\\sqrt{z^{2}-1}=\\frac{1}{z+\\sqrt{z^{2}-1}} \\geq \\frac{1}{p+\\sqrt{p^{2}-1}} $$  Equality holds if and only if $z=x+y=p$ and $4 x y=p^{2}-1$, which is satisfied only when $x=\\frac{p-1}{2}$ and $y=\\frac{p+1}{2}$. Hence for these values of $x, y$, both $u$ and $v$ attain positive minima.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "23", "problem_type": null, "exam": "IMO", "problem": "23. S1 (UKR) Does there exist a sequence $F(1), F(2), F(3), \\ldots$ of nonnegative integers that simultaneously satisfies the following three conditions? (a) Each of the integers $0,1,2, \\ldots$ occurs in the sequence. (b) Each positive integer occurs in the sequence infinitely often. (c) For any $n \\geq 2$,  $$ F\\left(F\\left(n^{163}\\right)\\right)=F(F(n))+F(F(361)) . $$", "solution": "23. By putting $F(1)=0$ and $F(361)=1$, condition (c) becomes $F\\left(F\\left(n^{163}\\right)\\right)=$ $F(F(n))$ for $n \\geq 2$. For $n=2,3, \\ldots, 360$ let $F(n)=n$, and inductively define $F(n)$ for $n \\geq 362$ as follows:  $$ F(n)= \\begin{cases}F(m), & \\text { if } n=m^{163}, m \\in \\mathbb{N} ; \\\\ \\text { the least number not in }\\{F(k) \\mid k<n\\}, & \\text { otherwise }\\end{cases} $$  Obviously, (a) each nonnegative integer appears in the sequence because there are infinitely many numbers not of the form $m^{163}$, and (b) each positive integer appears infinitely often because $F\\left(m^{163}\\right)=F(m)$. Since $F\\left(n^{163}\\right)=F(n),(c)$ also holds. Second solution. Another example of such a sequence is as follows: If $n=$ $p_{1}^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\cdots p_{k}^{\\alpha_{k}}$, is the factorization of $n$ into primes, we put $F(n)=\\alpha_{1}+$ $\\alpha_{2}+\\cdots+\\alpha_{k}$ and $F(1)=0$. Conditions (a) and (b) are evidently satisfied for this $F$, while (c) follows from $F\\left(F\\left(n^{163}\\right)\\right)=F(163 F(n))=F(F(n))+1$ (because 163 is a prime) and $F(F(361))=F\\left(F\\left(19^{2}\\right)\\right)=F(2)=1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "24", "problem_type": null, "exam": "IMO", "problem": "24. S2 (POL) ${ }^{\\text {IMO4 }}$ The positive real numbers $x_{0}, x_{1}, \\ldots, x_{1995}$ satisfy $x_{0}=$ $x_{1995}$ and  $$ x_{i-1}+\\frac{2}{x_{i-1}}=2 x_{i}+\\frac{1}{x_{i}} $$  for $i=1,2, \\ldots, 1995$. Find the maximum value that $x_{0}$ can have.", "solution": "24. The given condition is equivalent to $\\left(2 x_{i}-x_{i-1}\\right)\\left(x_{i} x_{i-1}-1\\right)=0$, so either $x_{i}=\\frac{1}{2} x_{i-1}$ or $x_{i}=\\frac{1}{x_{i-1}}$. We shall show by induction on $n$ that for any $n \\geq 0, x_{n}=2^{k_{n}} x_{0}^{e_{n}}$ for some integer $k_{n}$, where $\\left|k_{n}\\right| \\leq n$ and $e_{n}=(-1)^{n-k_{n}}$. Indeed, this is true for $n=0$. If it holds for some $n$, then $x_{n+1}=\\frac{1}{2} x_{n}=2^{k_{n}-1} x_{0}^{e_{n}}$ (hence $k_{n+1}=k_{n}-1$ and $e_{n+1}=e_{n}$ ) or $x_{n+1}=\\frac{1}{x_{n}}=2^{-k_{n}} x_{0}^{-e_{n}}$ (hence $k_{n+1}=-k_{n}$ and $e_{n+1}=-e_{n}$ ). Thus $x_{0}=x_{1995}=2^{k_{1995}} x_{0}^{e_{1995}}$. Note that $e_{1995}=1$ is impossible, since in that case $k_{1995}$ would be odd, although it should equal 0 . Therefore $e^{1995}=-1$, which gives $x_{0}^{2}=2^{k_{1995}} \\leq 2^{1994}$, so the maximal value that $x_{0}$ can have is $2^{997}$. This value is attained in the case $x_{i}=2^{997-i}$ for $i=0, \\ldots, 997$ and $x_{i}=2^{i-998}$ for $i=998, \\ldots, 1995$. Second solution. First we show that there is an $n, 0 \\leq n \\leq 1995$, such that $x_{n}=1$. Suppose the contrary. Then each of $x_{n}$ belongs to one of the intervals $I_{-i-1}=\\left[2^{-i-1}, 2^{-i}\\right)$ or $I_{i}=\\left(2^{i}, 2^{i+1}\\right]$, where $i=0,1,2, \\ldots$ Let $x_{n} \\in I_{i_{n}}$. Note that by the formula for $x_{n}, i_{n}$ and $i_{n-1}$ are of different parity. Hence $i_{0}$ and $i_{1995}$ are also of different parity, contradicting $x_{0}=$ $x_{1995}$. It follows that for some $n, x_{n}=1$. Now if $n \\leq 997$, then $x_{0} \\leq 2^{997}$, while if $n \\geq 998$, we also have $x_{0}=x_{1995} \\leq 2^{997}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "25", "problem_type": null, "exam": "IMO", "problem": "25. S3 (POL) For an integer $x \\geq 1$, let $p(x)$ be the least prime that does not divide $x$, and define $q(x)$ to be the product of all primes less than $p(x)$. In particular, $p(1)=2$. For $x$ such that $p(x)=2$, define $q(x)=1$. Consider the sequence $x_{0}, x_{1}, x_{2}, \\ldots$ defined by $x_{0}=1$ and  $$ x_{n+1}=\\frac{x_{n} p\\left(x_{n}\\right)}{q\\left(x_{n}\\right)} $$  for $n \\geq 0$. Find all $n$ such that $x_{n}=1995$.", "solution": "25. By the definition of $q(x)$, it divides $x$ for all integers $x>0$, so $f(x)=$ $x p(x) / q(x)$ is a positive integer too. Let $\\left\\{p_{0}, p_{1}, p_{2}, \\ldots\\right\\}$ be all prime numbers in increasing order. Since it easily follows by induction that all $x_{n}$ 's are square-free, we can assign to each of them a unique code according to which primes divide it: if $p_{m}$ is the largest prime dividing $x_{n}$, the code corresponding to $x_{n}$ will be $\\ldots 0 s_{m} s_{m-1} \\ldots s_{0}$, with $s_{i}=1$ if $p_{i} \\mid x_{n}$ and $s_{i}=0$ otherwise. Let us investigate how $f$ acts on these codes. If the code of $x_{n}$ ends with 0 , then $x_{n}$ is odd, so the code of $f\\left(x_{n}\\right)=x_{n+1}$ is obtained from that of $x_{n}$ by replacing $s_{0}=0$ by $s_{0}=1$. Furthermore, if the code of $x_{n}$ ends with $011 \\ldots 1$, then the code of $x_{n+1}$ ends with $100 \\ldots 0$ instead. Thus if we consider the codes as binary numbers, $f$ acts on them as an addition of 1 . Hence the code of $x_{n}$ is the binary representation of $n$ and thus $x_{n}$ uniquely determines $n$. Specifically, if $x_{n}=1995=3 \\cdot 5 \\cdot 7 \\cdot 19$, then its code is 10001110 and corresponds to $n=142$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "26", "problem_type": null, "exam": "IMO", "problem": "26. S4 (NZL) Suppose that $x_{1}, x_{2}, x_{3}, \\ldots$ are positive real numbers for which  $$ x_{n}^{n}=\\sum_{j=0}^{n-1} x_{n}^{j} $$  for $n=1,2,3, \\ldots$ Prove that for all $n$,  $$ 2-\\frac{1}{2^{n-1}} \\leq x_{n}<2-\\frac{1}{2^{n}} $$", "solution": "26. For $n=1$ the result is trivial, since $x_{1}=1$. Suppose now that $n \\geq 2$ and let $f_{n}(x)=x^{n}-\\sum_{i=0}^{n-1} x^{i}$. Note that $x_{n}$ is the unique positive real root of $f_{n}$, because $\\frac{f_{n}(x)}{x^{n-1}}=x-1-\\frac{1}{x}-\\cdots-\\frac{1}{x^{n-1}}$ is strictly increasing on $\\mathbb{R}^{+}$. Consider $g_{n}(x)=(x-1) f_{n}(x)=(x-2) x^{n}+1$. Obviously $g_{n}(x)$ has no positive roots other than 1 and $x_{n}>1$. Observe that $\\left(1-\\frac{1}{2^{n}}\\right)^{n}>$ $1-\\frac{n}{2^{n}} \\geq \\frac{1}{2}$ for $n \\geq 2$ (by Bernoulli's inequality). Since then  $$ g_{n}\\left(2-\\frac{1}{2^{n}}\\right)=-\\frac{1}{2^{n}}\\left(2-\\frac{1}{2^{n}}\\right)^{n}+1=1-\\left(1-\\frac{1}{2^{n+1}}\\right)^{n}>0 $$  and  $$ g_{n}\\left(2-\\frac{1}{2^{n-1}}\\right)=-\\frac{1}{2^{n-1}}\\left(2-\\frac{1}{2^{n-1}}\\right)^{n}+1=1-2\\left(1-\\frac{1}{2^{n}}\\right)^{n}<0 $$  we conclude that $x_{n}$ is between $2-\\frac{1}{2^{n-1}}$ and $2-\\frac{1}{2^{n}}$, as required. Remark. Moreover, $\\lim _{n \\rightarrow \\infty} 2^{n}\\left(2-x_{n}\\right)=1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "27", "problem_type": null, "exam": "IMO", "problem": "27. S5 (FIN) For positive integers $n$, the numbers $f(n)$ are defined inductively as follows: $f(1)=1$, and for every positive integer $n, f(n+1)$ is the greatest integer $m$ such that there is an arithmetic progression of positive integers $a_{1}<a_{2}<\\cdots<a_{m}=n$ for which  $$ f\\left(a_{1}\\right)=f\\left(a_{2}\\right)=\\cdots=f\\left(a_{m}\\right) $$  Prove that there are positive integers $a$ and $b$ such that $f(a n+b)=n+2$ for every positive integer $n$.", "solution": "27. Computing the first few values of $f(n)$, we observe the following pattern:  $$ \\begin{aligned} f(4 k) & =k, k \\geq 3, & f(8) & =3 ; \\\\ f(4 k+1) & =1, k \\geq 4, & f(5) & =f(13)=2 ; \\\\ f(4 k+2) & =k-3, k \\geq 7, & f(2) & =1, f(6)=f(10)=2, \\\\ & & f(14) & =f(18)=3, f(26)=4 ; \\\\ f(4 k+3) & =2 . & & \\end{aligned} $$  We shall prove these statements simultaneously by induction on $n$, having verified them for $k \\leq 7$. (i) Let $n=4 k$. Since $f(3)=f(7)=\\cdots=f(4 k-1)=2$, we have $f(4 k) \\geq k$. But $f(n) \\leq \\max _{m<n} f(m)+1 \\leq(k-1)+1$, so $f(4 k)=k$. (ii) Let $n=4 k+1, k \\geq 7$. Since $f(4 k)=k$ and $f(m)<k$ for $m<4 k$, we deduce that $f(4 k+1)=1$. (iii) Let $n=4 k+2, k \\geq 7$. Since $f(17)=f(21)=\\cdots=f(4 k+1)=1$, we obtain $f(4 k+2) \\geq k-3$. On the other hand, if $f(4 k+1)=f(4 k+1-$ $d)=1$, then $d \\geq 8$, and $4 k+1-8(k-3)<0$. So $f(4 k+2)=k-3$. (iv) Let $n=4 k+3, k \\geq 7$. We have $f(4 k+2)=k-3$ and $f(m)=k-3$ for exactly one $m<4 k+2$ (namely for $m=4 k-12)$; hence $f(4 k+3)=2$. Therefore, for example, $f(4 n+8)=n+2$ for all $n$; hence we can take $a=4$ and $b=8$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "28", "problem_type": null, "exam": "IMO", "problem": "28. S6 (IND) Let $\\mathbb{N}$ denote the set of all positive integers. Prove that there exists a unique function $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ satisfying  $$ f(m+f(n))=n+f(m+95) $$  for all $m$ and $n$ in $\\mathbb{N}$. What is the value of $\\sum_{k=1}^{19} f(k)$ ?", "solution": "28. Let $F(x)=f(x)-95$ for $x \\geq 1$. Writing $k$ for $m+95$, the given condition becomes  $$ F(k+F(n))=F(k)+n, \\quad k \\geq 96, n \\geq 1 $$  Thus for $x, z \\geq 96$ and an arbitrary $y$ we have $F(x+y)+z=F(x+$ $y+F(z))=F(x+F(F(y)+z))=F(x)+F(y)+z$, and consequently $F(x+y)=F(x)+F(y)$ whenever $x \\geq 96$. Moreover, since then $F(x+$ $y)+F(96)=F(x+y+96)=F(x)+F(y+96)=F(x)+F(y)+F(96)$ for any $x, y$, we obtain  $$ F(x+y)=F(x)+F(y), \\quad x, y \\in \\mathbb{N} . $$  It follows by induction that $F(n)=n c$ for all $n$, where $F(1)=c$. Equation (1) becomes $c k+c^{2} n=c k+n$, and yields $c=1$. Hence $F(n)=n$ and $f(n)=n+95$ for all $n$. Finally, $\\sum_{k=1}^{19} f(k)=96+97+\\cdots+114=1995$. Second solution. First we show that $f(n)>95$ for all $n$. If to the contrary $f(n) \\leq 95$, we have $f(m)=n+f(m+95-f(n))$, so by induction $f(m)=k n+f(m+k(95-f(n))) \\geq k n$ for all $k$, which is impossible. Now for $m>95$ we have $f(m+f(n)-95)=n+f(m)$, and again by induction $f(m+k(f(n)-95))=k n+f(m)$ for all $m, n, k$. It follows that with $n$ fixed,  $$ (\\forall m) \\lim _{k \\rightarrow \\infty} \\frac{f(m+k(f(n)-95))}{m+k(f(n)-95)}=\\frac{n}{f(n)-95} $$  hence  $$ \\lim _{s \\rightarrow \\infty} \\frac{f(s)}{s}=\\frac{n}{f(n)-95} $$  Hence $\\frac{n}{f(n)-95}$ does not depend on $n$, i.e., $f(n) \\equiv c n+95$ for some constant $c$. It is easily checked that only $c=1$ is possible.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "3", "problem_type": "Algebra", "exam": "IMO", "problem": "3. A3 (UKR) Let $n$ be an integer, $n \\geq 3$. Let $a_{1}, a_{2}, \\ldots, a_{n}$ be real numbers such that $2 \\leq a_{i} \\leq 3$ for $i=1,2, \\ldots, n$. If $s=a_{1}+a_{2}+\\cdots+a_{n}$, prove that  $$ \\frac{a_{1}^{2}+a_{2}^{2}-a_{3}^{2}}{a_{1}+a_{2}-a_{3}}+\\frac{a_{2}^{2}+a_{3}^{2}-a_{4}^{2}}{a_{2}+a_{3}-a_{4}}+\\cdots+\\frac{a_{n}^{2}+a_{1}^{2}-a_{2}^{2}}{a_{n}+a_{1}-a_{2}} \\leq 2 s-2 n $$", "solution": "3. Write $A_{i}=\\frac{a_{i}^{2}+a_{i+1}^{2}-a_{i+2}^{2}}{a_{i}+a_{i+1}-a_{i+2}}=a_{i}+a_{i+1}+a_{i+2}-\\frac{2 a_{i} a_{i+1}}{a_{i}+a_{i+1}-a_{i+2}}$. Since $2 a_{i} a_{i+1} \\geq$ $4\\left(a_{i}+a_{i+1}-2\\right)$ (which is equivalent to $\\left.\\left(a_{i}-2\\right)\\left(a_{i+1}-2\\right) \\geq 0\\right)$, it follows that $A_{i} \\leq a_{i}+a_{i+1}+a_{i+2}-4\\left(1+\\frac{a_{i+2}-2}{a_{i}+a_{i+1}-a_{i+2}}\\right) \\leq a_{i}+a_{i+1}+a_{i+2}-$ $4\\left(1+\\frac{a_{i+2}-2}{4}\\right)$, because $1 \\leq a_{i}+a_{i+1}-a_{i+2} \\leq 4$. Therefore $A_{i} \\leq a_{i}+$ $a_{i+1}-2$, so $\\sum_{i=1}^{n} A_{i} \\leq 2 s-2 n$ as required.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "4", "problem_type": "Algebra", "exam": "IMO", "problem": "4. A4 (USA) Let $a, b$, and $c$ be given positive real numbers. Determine all positive real numbers $x, y$, and $z$ such that  $$ x+y+z=a+b+c $$  and  $$ 4 x y z-\\left(a^{2} x+b^{2} y+c^{2} z\\right)=a b c $$", "solution": "4. The second equation is equivalent to $\\frac{a^{2}}{y z}+\\frac{b^{2}}{z x}+\\frac{c^{2}}{x y}+\\frac{a b c}{x y z}=4$. Let $x_{1}=$ $\\frac{a}{\\sqrt{y z}}, y_{1}=\\frac{b}{\\sqrt{z x}}, z_{1}=\\frac{c}{\\sqrt{x y}}$. Then $x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+x_{1} y_{1} z_{1}=4$, where $0<x_{1}, y_{1}, z_{1}<2$. Regarding this as a quadratic equation in $z_{1}$, the discriminant $\\left(4-x_{1}^{2}\\right)\\left(4-y_{1}^{2}\\right)$ suggests that we let $x_{1}=2 \\sin u, y_{1}=2 \\sin v$, $0<u, v<\\pi / 2$. Then it is directly shown that $z_{1}$ will be exactly $2 \\cos (u+v)$ as the only positive solution of the quadratic equation. Thus $a=2 \\sqrt{y z} \\sin u, b=2 \\sqrt{x z} \\sin v, c=2 \\sqrt{x y}(\\cos u \\cos v-\\sin u \\sin v)$, so from $x+y+z-a-b-c=0$ we obtain  $$ (\\sqrt{x} \\cos v-\\sqrt{y} \\cos u)^{2}+(\\sqrt{x} \\sin v+\\sqrt{y} \\sin u-\\sqrt{z})^{2}=0 $$  which implies $\\sqrt{z}=\\sqrt{x} \\sin v+\\sqrt{y} \\sin u=\\frac{1}{2}\\left(y_{1} \\sqrt{x}+x_{1} \\sqrt{y}\\right)=\\frac{1}{2}\\left(\\frac{b}{\\sqrt{z x}} \\sqrt{x}+\\frac{a}{\\sqrt{y z}} \\sqrt{y}\\right)$. Therefore $z=\\frac{a+b}{2}$. Similarly, $x=\\frac{b+c}{2}$ and $y=\\frac{c+a}{2}$. It is clear that the triple $(x, y, z)=\\left(\\frac{b+c}{2}, \\frac{c+a}{2}, \\frac{a+b}{2}\\right)$ is indeed a (unique) solution of the given system of equations. Second solution. Put $x=\\frac{b+c}{2}-u, y=\\frac{c+a}{2}-v, z=\\frac{a+b}{2}-w$, where $u \\leq \\frac{b+c}{2}, v \\leq \\frac{c+a}{2}, w \\leq \\frac{a+b}{2}$ and $u+v+w=0$. The equality $a b c+$ $a^{2} x+b^{2} y+c^{2} z=4 x y z$ becomes $2\\left(a u^{2}+b v^{2}+c w^{2}+2 u v w\\right)=0$. Now $u v w>0$ is clearly impossible. On the other hand, if $u v w \\leq 0$, then two of $u, v, w$ are nonnegative, say $u, v \\geq 0$. Taking into account $w=-u-v$, the above equality reduces to $2\\left[(a+c-2 v) u^{2}+(b+c-2 u) v^{2}+2 c u v\\right]=0$, so $u=v=0$.  Third solution. The fact that we are given two equations and three variables suggests that this is essentially a problem on inequalities. Setting $f(x, y, z)=4 x y z-a^{2} x-b^{2} y-c^{2} z$, we should show that $\\max f(x, y, z)=$ $a b c$, for $0<x, y, z, x+y+z=a+b+c$, and find when this value is attained. Thus we apply Lagrange multipliers to $F(x, y, z)=f(x, y, z)-$ $\\lambda(x+y+z-a-b-c)$, and obtain that $f$ takes a maximum at $(x, y, z)$ such that $4 y z-a^{2}=4 z x-b^{2}=4 x y-c^{2}=\\lambda$ and $x+y+z=a+b+c$. The only solution of this system is $(x, y, z)=\\left(\\frac{b+c}{2}, \\frac{c+a}{2}, \\frac{a+b}{2}\\right)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "5", "problem_type": "Algebra", "exam": "IMO", "problem": "5. A5 (UKR) Let $\\mathbb{R}$ be the set of real numbers. Does there exist a function $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ that simultaneously satisfies the following three conditions? (a) There is a positive number $M$ such that $-M \\leq f(x) \\leq M$ for all $x$. (b) $f(1)=1$. (c) If $x \\neq 0$, then  $$ f\\left(x+\\frac{1}{x^{2}}\\right)=f(x)+\\left[f\\left(\\frac{1}{x}\\right)\\right]^{2} $$", "solution": "5. Suppose that a function $f$ satisfies the condition, and let $c$ be the least upper bound of $\\{f(x) \\mid x \\in \\mathbb{R}\\}$. We have $c \\geq 2$, since $f(2)=f(1+$ $\\left.1 / 1^{2}\\right)=f(1)+f(1)^{2}=2$. Also, since $c$ is the least upper bound, for each $k=1,2, \\ldots$ there is an $x_{k} \\in \\mathbb{R}$ such that $f\\left(x_{k}\\right) \\geq c-1 / k$. Then  $$ c \\geq f\\left(x_{k}+\\frac{1}{x_{k}^{2}}\\right) \\geq c-\\frac{1}{k}+f\\left(\\frac{1}{x_{k}}\\right)^{2} \\Longrightarrow f\\left(\\frac{1}{x_{k}}\\right) \\geq-\\frac{1}{\\sqrt{k}} . $$  On the other hand,  $$ c \\geq f\\left(\\frac{1}{x_{k}}+x_{k}^{2}\\right)=f\\left(\\frac{1}{x_{k}}\\right)+f\\left(x_{k}\\right)^{2} \\geq-\\frac{1}{\\sqrt{k}}+\\left(c-\\frac{1}{k}\\right)^{2} . $$  It follows that  $$ \\frac{1}{\\sqrt{k}}-\\frac{1}{k^{2}} \\geq c\\left(c-1-\\frac{2}{k}\\right) $$  which cannot hold for $k$ sufficiently large. Second solution. Assume that $f$ exists and let $n$ be the least integer such that $f(x) \\leq \\frac{n}{4}$ for all $x$. Since $f(2)=2$, we have $n \\geq 8$. Let $f(x)>\\frac{n-1}{4}$. Then $f(1 / x)=f\\left(x+1 / x^{2}\\right)-f(x)<1 / 4$, so $f(1 / x)>-1 / 2$. On the other hand, this implies $\\left(\\frac{n-1}{4}\\right)^{2}<f(x)^{2}=f\\left(1 / x+x^{2}\\right)-f(1 / x)<\\frac{n}{4}+\\frac{1}{2}$, which is impossible when $n \\geq 8$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "6", "problem_type": "Algebra", "exam": "IMO", "problem": "6. A6 (JAP) Let $n$ be an integer, $n \\geq 3$. Let $x_{1}, x_{2}, \\ldots, x_{n}$ be real numbers such that $x_{i}<x_{i+1}$ for $1 \\leq i \\leq n-1$. Prove that  $$ \\frac{n(n-1)}{2} \\sum_{i<j} x_{i} x_{j}>\\left(\\sum_{i=1}^{n-1}(n-i) x_{i}\\right)\\left(\\sum_{j=2}^{n}(j-1) x_{j}\\right) . $$", "solution": "6. Let $y_{i}=x_{i+1}+\\cdots+x_{n}, Y=\\sum_{j=2}^{n}(j-1) x_{j}$, and $z_{i}=\\frac{n(n-1)}{2} y_{i}-(n-$ $i) Y$. Then $\\frac{n(n-1)}{2} \\sum_{i<j} x_{i} x_{j}-\\left(\\sum_{i=1}^{n-1}(n-i) x_{i}\\right) Y=\\frac{n(n-1)}{2} \\sum_{i=1}^{n-1} x_{i} y_{i}-$ $\\sum_{i=1}^{n-1}(n-i) x_{i} Y=\\sum_{i=1}^{n-1} x_{i} z_{i}$, so it remains to show that $\\sum_{i=1}^{n-1} x_{i} z_{i}>0$. Since $\\sum_{i=1}^{n-1} y_{i}=Y$ and $\\sum_{i=1}^{n-1}(n-i)=\\frac{n(n-1)}{2}$, we have $\\sum z_{i}=0$. Note that $Y<\\sum_{j=2}^{n}(j-1) x_{n}=\\frac{n(n-1)}{2} x_{n}$, and consequently $z_{n-1}=$ $\\frac{n(n-1)}{2} x_{n}-Y>0$. Furthermore, we have  $$ \\frac{z_{i+1}}{n-i-1}-\\frac{z_{i}}{n-i}=\\frac{n(n-1)}{2}\\left(\\frac{y_{i+1}}{n-i-1}-\\frac{y_{i}}{n-i}\\right)>0 $$  which means that $\\frac{z_{1}}{n-1}<\\frac{z_{2}}{n-2}<\\cdots<\\frac{z_{n-1}}{1}$. Therefore there is a $k$ for which $z_{1}, \\ldots, z_{k} \\leq 0$ and $z_{k+1}, \\ldots, z_{n-1}>0$. But then $z_{i}\\left(x_{i}-x_{k}\\right) \\geq 0$, i.e., $x_{i} z_{i} \\geq x_{k} z_{i}$ for all $i$, so $\\sum_{i=1}^{n-1} x_{i} z_{i}>\\sum_{i=1}^{n-1} x_{k} z_{i}=0$ as required.  Second solution. Set $X=\\sum_{j=1}^{n-1}(n-j) x_{j}$ and $Y=\\sum_{j=2}^{n}(j-1) x_{j}$. Since $4 X Y=(X+Y)^{2}-(X-Y)^{2}$, the RHS of the inequality becomes  $$ X Y=\\frac{1}{4}\\left[(n-1)^{2}\\left(\\sum_{i=1}^{n} x_{i}\\right)^{2}-\\left(\\sum_{i=1}^{n}(2 i-1-n) x_{i}\\right)^{2}\\right] $$  The LHS equals $\\frac{1}{4}\\left((n-1)^{2}\\left(\\sum_{i=1}^{n} x_{i}\\right)^{2}-(n-1) \\sum_{i<j}\\left(x_{j}-x_{i}\\right)^{2}\\right)$. Since $\\sum_{i=1}^{n}(2 i-1-n) x_{i}=\\sum_{i<j}\\left(x_{j}-x_{i}\\right)$ also holds, we must prove that  $$ \\left(\\sum_{i<j}\\left(x_{j}-x_{i}\\right)\\right)^{2}>(n-1) \\sum_{i<j}\\left(x_{j}-x_{i}\\right)^{2} $$  Putting $x_{i+1}-x_{i}=d_{i}>0$ (so, $x_{j}-x_{i}=d_{i}+d_{i+1}+\\cdots+d_{j-1}$ ) and expanding the obtained expressions, we reduce this inequality to $\\sum_{k} k^{2}(n-k)^{2} d_{k}^{2}+2 \\sum_{k<l} k l(n-k)(n-l) d_{k} d_{l}>\\sum_{k}(n-1) k(n-k) d_{k}^{2}+$ $2 \\sum_{k<l}(n-1) k(n-l) d_{k} d_{l}$, which is verified immediately by comparing coefficients. Remark. An inequality significantly stronger than (1) in the second solution has appeared later, as IMO 03-5.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "7", "problem_type": "Geometry", "exam": "IMO", "problem": "7. G1 (BUL) ${ }^{\\mathrm{IMO} 1}$ Let $A, B, C$, and $D$ be distinct points on a line, in that order. The circles with diameters $A C$ and $B D$ intersect at $X$ and $Y . O$ is an arbitrary point on the line $X Y$ but not on $A D$. $C O$ intersects the circle with diameter $A C$ again at $M$, and $B O$ intersects the other circle again at $N$. Prove that the lines $A M, D N$, and $X Y$ are concurrent.", "solution": "7. The result is trivial if $O$ coincides with $X$ or $Y$, so let us assume it does not. From $O B \\cdot O N=O C \\cdot O M=O X \\cdot O Y$ we deduce that $B C M N$ is a cyclic quadrilateral. Further, if $O$ lies between $X$ and $Y$, then $\\angle M A D+$ $\\angle M N D=\\angle M A D+\\angle M N B+\\angle B N D=\\angle M A D+\\angle M C A+\\angle A M C=$ $180^{\\circ}$. Similarly, we also have $\\angle M A D+\\angle M N D=180^{\\circ}$ if $O$ is not on the segment $X Y$. Therefore $A D N M$ is cyclic. Now let $A M$ and $D N$ intersect at $Z$ and let the line $Z X$ intersect the two circles at $Y_{1}$ and $Y_{2}$. Then $Z X \\cdot Z Y_{1}=Z M \\cdot Z A=Z N \\cdot Z D=Z X \\cdot Z Y_{2}$. Hence $Y_{1}=Y_{2}=Y$, implying that $Z$ lies on $X Y$.  Second solution. Let $Z_{1}, Z_{2}$ be the points in which $A M, D N$ respectively meet $X Y$, and $P=B C \\cap X Y$. Then, from $\\triangle O P C \\sim \\triangle A P Z_{1}$, we have $P Z_{1}=\\frac{P A \\cdot P C}{P O}=\\frac{P X^{2}}{P O}$ and analogously $P Z_{2}=\\frac{P X^{2}}{P O}$. Hence, we conclude that $Z_{1} \\equiv Z_{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "8", "problem_type": "Geometry", "exam": "IMO", "problem": "8. G2 (GER) Let $A, B$, and $C$ be noncollinear points. Prove that there is a unique point $X$ in the plane of $A B C$ such that $X A^{2}+X B^{2}+A B^{2}=$ $X B^{2}+X C^{2}+B C^{2}=X C^{2}+X A^{2}+C A^{2}$.", "solution": "8. Let $A^{\\prime}, B^{\\prime}, C^{\\prime}$ be the points symmetric to $A, B, C$ with respect to the midpoints of $B C, C A, A B$ respectively. From the condition on $X$ we have $X B^{2}-X C^{2}=A C^{2}-A B^{2}=A^{\\prime} B^{2}-A^{\\prime} C^{2}$, and hence $X$ must lie on the line through $A^{\\prime}$ perpendicular to $B C$. Similarly, $X$ lies on the line through $B^{\\prime}$ perpendicular to $C A$. It follows that there is a unique position for $X$, namely the orthocenter of $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$. It easily follows that this point $X$ satisfies the original equations.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1995", "tier": "T0", "problem_label": "9", "problem_type": "Geometry", "exam": "IMO", "problem": "9. G3 (TUR) The incircle of $A B C$ touches $B C, C A$, and $A B$ at $D, E$, and $F$ respectively. $X$ is a point inside $A B C$ such that the incircle of $X B C$ touches $B C$ at $D$ also, and touches $C X$ and $X B$ at $Y$ and $Z$, respectively. Prove that $E F Z Y$ is a cyclic quadrilateral.", "solution": "9. If $E F$ is parallel to $B C, \\triangle A B C$ must be isosceles and $E, Y$ are symmetric to $F, Z$ with respect to $A D$, so the result follows. Now suppose that $E F$ meets $B C$ at $P$. By Menelaus's theorem, $\\frac{B P}{C P}=\\frac{B F}{F A} \\cdot \\frac{A E}{E C}=\\frac{B D}{D C}$ (since $B D=B F, C D=C E, A E=A F)$. It follows that the point $P$ depends only on $D$ and not on $A$. In particular, the same point is obtained as the intersection of $Z Y$ with $B C$. Therefore $P E \\cdot P F=P D^{2}=P Y \\cdot P Z$, from which it follows that $E F Z Y$ is a cyclic quadrilateral. Second solution. Since $C D=C Y=C E$ and $B D=B Z=B F$, all angles of $E F Z Y$ can be calculated in terms of angles of $A B C$ and $Y Z B C$. In fact, $\\angle F E Y=\\frac{1}{2}(\\angle A+\\angle C+\\angle B C Y)$ and $\\angle F Z Y=\\frac{1}{2}\\left(180^{\\circ}+\\angle B+\\angle B C Y\\right)$, which gives us $\\angle F E Y+\\angle F Z Y=180^{\\circ}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "1", "problem_type": "Algebra", "exam": "IMO", "problem": "1. A1 (SLO) Let $a, b$, and $c$ be positive real numbers such that $a b c=1$. Prove that  $$ \\frac{a b}{a^{5}+b^{5}+a b}+\\frac{b c}{b^{5}+c^{5}+b c}+\\frac{c a}{c^{5}+a^{5}+c a} \\leq 1 $$  When does equality hold?", "solution": "1. We have $a^{5}+b^{5}-a^{2} b^{2}(a+b)=\\left(a^{3}-b^{3}\\right)\\left(a^{2}-b^{2}\\right) \\geq 0$, i.e. $a^{5}+b^{5} \\geq$ $a^{2} b^{2}(a+b)$. Hence  $$ \\frac{a b}{a^{5}+b^{5}+a b} \\leq \\frac{a b}{a^{2} b^{2}(a+b)+a b}=\\frac{a b c^{2}}{a^{2} b^{2} c^{2}(a+b)+a b c^{2}}=\\frac{c}{a+b+c} . $$  Now, the left side of the inequality to be proved does not exceed $\\frac{c}{a+b+c}+$ $\\frac{a}{a+b+c}+\\frac{b}{a+b+c}=1$. Equality holds if and only if $a=b=c$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "10", "problem_type": "Geometry", "exam": "IMO", "problem": "10. G1 (GBR) Let triangle $A B C$ have orthocenter $H$, and let $P$ be a point on its circumcircle, distinct from $A, B, C$. Let $E$ be the foot of the altitude $B H$, let $P A Q B$ and $P A R C$ be parallelograms, and let $A Q$ meet $H R$ in $X$. Prove that $E X$ is parallel to $A P$.", "solution": "10. We first show that $H$ is the common orthocenter of the triangles $A B C$ and $A Q R$.  Let $G, G^{\\prime}, H^{\\prime}$ be respectively the centroid of $\\triangle A B C$, the centroid of $\\triangle P B C$, and the orthocenter of $\\triangle P B C$. Since the triangles $A B C$ and $P B C$ have a common circumcenter, from the properties of the Euler line we get $\\overrightarrow{H H^{\\prime}}=3 \\overrightarrow{G G^{\\prime}}=$ $\\overrightarrow{A P}$. But $\\triangle A Q R$ is exactly the image of $\\triangle P B C$ under translation by $\\overrightarrow{A P}$; hence the orthocenter of $A Q R$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-615.jpg?height=432&width=414&top_left_y=254&top_left_x=882) coincides with H. (Remark: This can be shown by noting that $A H B Q$ is cyclic.) Now we have that $R H \\perp A Q$; hence $\\angle A X H=90^{\\circ}=\\angle A E H$. It follows that $A X E H$ is cyclic; hence  $$ \\angle E X Q=180^{\\circ}-\\angle A H E=180^{\\circ}-\\angle B C A=180^{\\circ}-\\angle B P A=\\angle P A Q $$  (as oriented angles). Hence $E X \\| A P$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "11", "problem_type": "Geometry", "exam": "IMO", "problem": "11. G2 (CAN) ${ }^{\\mathrm{IMO} 2}$ Let $P$ be a point inside $\\triangle A B C$ such that  $$ \\angle A P B-\\angle C=\\angle A P C-\\angle B . $$  Let $D, E$ be the incenters of $\\triangle A P B, \\triangle A P C$ respectively. Show that $A P, B D$ and $C E$ meet in a point.", "solution": "11. Let $X, Y, Z$ respectively be the feet of the perpendiculars from $P$ to $B C$, $C A, A B$. Examining the cyclic quadrilaterals $A Z P Y, B X P Z, C Y P X$, one can easily see that $\\angle X Z Y=\\angle A P B-\\angle C$ and $X Y=P C \\sin \\angle C$. The first relation gives that $X Y Z$ is isosceles with $X Y=X Z$, so from the second relation $P B \\sin \\angle B=P C \\sin \\angle C$. Hence $A B / P B=A C / P C$. This implies that the bisectors $B D$ and $C D$ of $\\angle A B P$ and $\\angle A C P$ divide the segment $A P$ in equal ratios; i.e., they concur with $A P$. Second solution. Take that $X, Y, Z$ are the points of intersection of $A P, B P, C P$ with the circumscribed circle of $A B C$ instead. We similarly obtain $X Y=X Z$. If we write $A P \\cdot P X=B P \\cdot P Y=C P \\cdot P Z=k$, from the similarity of $\\triangle A P C$ and $\\triangle Z P X$ we get  $$ \\frac{A C}{X Z}=\\frac{A P}{P Z}=\\frac{A P \\cdot C P}{k} $$  i.e., $X Z=\\frac{k \\cdot A C \\cdot B P}{A P \\cdot B P \\cdot C P}$. It follows again that $A C / A B=P C / P B$.  Third solution. Apply an inversion with center at $A$ and radius $r$, and denote by $\\bar{Q}$ the image of any point $Q$. Then the given condition becomes $\\angle \\overline{B C P}=\\angle \\overline{C B P}$, i.e., $\\overline{B P}=\\overline{P C}$. But  $$ \\overline{P B}=\\frac{r^{2}}{A P \\cdot A B} P B $$  so $A C / A B=P C / P B$. Remark. Moreover, it follows that the locus of $P$ is an arc of the circle of Apollonius through $C$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "12", "problem_type": "Geometry", "exam": "IMO", "problem": "12. G3 (GBR) Let $A B C$ be an acute-angled triangle with $B C>C A$. Let $O$ be the circumcenter, $H$ its orthocenter, and $F$ the foot of its altitude $C H$. Let the perpendicular to $O F$ at $F$ meet the side $C A$ at $P$. Prove that $\\angle F H P=\\angle B A C$. Possible second part: What happens if $|B C| \\leq|C A|$ (the triangle still being acute-angled)?", "solution": "12. It is easy to see that $P$ lies on the segment $A C$. Let $E$ be the foot of the altitude $B H$ and $Y, Z$ the midpoints of $A C, A B$ respectively. Draw the perpendicular $H R$ to $F P(R \\in F P)$. Since $Y$ is the circumcenter of $\\triangle F C A$, we have $\\angle F Y A=180^{\\circ}-2 \\angle A$. Also, $O F P Y$ is cyclic; hence $\\angle O P F=\\angle O Y F=2 \\angle A-90^{\\circ}$. Next, $\\triangle O Z F$ and $\\triangle H R F$ are similar, so $O Z / O F=H R / H F$. This leads to $H R \\cdot O F=H F \\cdot O Z=\\frac{1}{2} H F$. $H C=\\frac{1}{2} H E \\cdot H B=H E \\cdot O Y \\Longrightarrow$ $H R / H E=O Y / O F$. Moreover, $\\angle E H R=\\angle F O Y$; hence the triangles $E H R$ and $F O Y$ are similar. Consequently $\\angle H P C=\\angle H R E=$ $\\angle O Y F=2 \\angle A-90^{\\circ}$, and finally, $\\angle F H P=\\angle H P C+\\angle H C P=\\angle A$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-616.jpg?height=331&width=387&top_left_y=501&top_left_x=895)  Second solution. As before, $\\angle H F Y=90^{\\circ}-\\angle A$, so it suffices to show that $H P \\perp F Y$. The points $O, F, P, Y$ lie on a circle, say $\\Omega_{1}$ with center at the midpoint $Q$ of $O P$. Furthermore, the points $F, Y$ lie on the nine-point circle $\\Omega$ of $\\triangle A B C$ with center at the midpoint $N$ of $O H$. The segment $F Y$ is the common chord of $\\Omega_{1}$ and $\\Omega$, from which we deduce that $N Q \\perp F Y$. However, $N Q \\| H P$, and the result follows. Third solution. Let $H^{\\prime}$ be the point symmetric to $H$ with respect to $A B$. Then $H^{\\prime}$ lies on the circumcircle of $A B C$. Let the line $F P$ meet the circumcircle at $U, V$ and meet $H^{\\prime} B$ at $P^{\\prime}$. Since $O F \\perp U V, F$ is the midpoint of $U V$. By the butterfly theorem, $F$ is also the midpoint of $P P^{\\prime}$. Therefore $\\triangle H^{\\prime} F P^{\\prime} \\cong F H P$; hence $\\angle F H P=\\angle F H^{\\prime} B=\\angle A$. Remark. It is possible to solve the problem using trigonometry. For example, $\\frac{F Z}{Z O}=\\frac{F K}{K P}=\\frac{\\sin (A-B)}{\\cos C}$, where $K$ is on $C F$ with $P K \\perp C F$. Then $\\frac{C F}{K P}=\\frac{\\sin (A-B)}{\\cos C}+\\tan A$, from which one obtains formulas for $K P$ and $K H$. Finally, we can calculate $\\tan \\angle F H P=\\frac{K P}{K H}=\\cdots=\\tan A$. Second remark. Here is what happens when $B C \\leq C A$. If $\\angle A>45^{\\circ}$, then $\\angle F H P=\\angle A$. If $\\angle A=45^{\\circ}$, the point $P$ escapes to infinity. If $\\angle A<45^{\\circ}$, the point $P$ appears on the extension of $A C$ over $C$, and $\\angle F H P=180^{\\circ}-\\angle A$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "13", "problem_type": "Geometry", "exam": "IMO", "problem": "13. G4 (USA) Let $\\triangle A B C$ be an equilateral triangle and let $P$ be a point in its interior. Let the lines $A P, B P, C P$ meet the sides $B C, C A, A B$ in the points $A_{1}, B_{1}, C_{1}$ respectively. Prove that  $$ A_{1} B_{1} \\cdot B_{1} C_{1} \\cdot C_{1} A_{1} \\geq A_{1} B \\cdot B_{1} C \\cdot C_{1} A $$", "solution": "13. By the law of cosines applied to $\\triangle C A_{1} B_{1}$, we obtain  $$ A_{1} B_{1}^{2}=A_{1} C^{2}+B_{1} C^{2}-A_{1} C \\cdot B_{1} C \\geq A_{1} C \\cdot B_{1} C $$  Analogously, $B_{1} C_{1}^{2} \\geq B_{1} A \\cdot C_{1} A$ and $C_{1} A_{1}^{2} \\geq C_{1} B \\cdot A_{1} B$, so that multiplying these inequalities yields  $$ A_{1} B_{1}^{2} \\cdot B_{1} C_{1}^{2} \\cdot C_{1} A_{1}^{2} \\geq A_{1} B \\cdot A_{1} C \\cdot B_{1} A \\cdot B_{1} C \\cdot C_{1} A \\cdot C_{1} B $$  Now, the lines $A A_{1}, B B_{1}, C C_{1}$ concur, so by Ceva's theorem, $A_{1} B \\cdot B_{1} C$. $C_{1} A=A B_{1} \\cdot B C_{1} \\cdot C A_{1}$, which together with (1) gives the desired inequality. Equality holds if and only if $C A_{1}=C B_{1}$, etc.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "14", "problem_type": "Geometry", "exam": "IMO", "problem": "14. G5 (ARM) ${ }^{\\mathrm{IMO5}}$ Let $A B C D E F$ be a convex hexagon such that $A B$ is parallel to $D E, B C$ is parallel to $E F$, and $C D$ is parallel to $A F$. Let $R_{A}, R_{C}, R_{E}$ be the circumradii of triangles $F A B, B C D, D E F$ respectively, and let $P$ denote the perimeter of the hexagon. Prove that  $$ R_{A}+R_{C}+R_{E} \\geq \\frac{P}{2} $$", "solution": "14. Let $a, b, c, d, e$, and $f$ denote the lengths of the sides $A B, B C, C D, D E$, $E F$, and $F A$ respectively. Note that $\\angle A=\\angle D, \\angle B=\\angle E$, and $\\angle C=\\angle F$. Draw the lines $P Q$ and $R S$ through $A$ and $D$ perpendicular to $B C$ and $E F$ respectively $(P, R \\in B C, Q, S \\in E F)$. Then $B F \\geq P Q=R S$. Therefore $2 B F \\geq$ $P Q+R S$, or ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-617.jpg?height=322&width=389&top_left_y=327&top_left_x=885)  $$ \\begin{array}{ll}  & 2 B F \\geq(a \\sin B+f \\sin C)+(c \\sin C+d \\sin B), \\\\ \\text { and similarly, } & 2 B D \\geq(c \\sin A+b \\sin B)+(e \\sin B+f \\sin A), \\\\ & 2 D F \\geq(e \\sin C+d \\sin A)+(a \\sin A+b \\sin C) \\end{array} $$  Next, we have the following formulas for the considered circumradii:  $$ R_{A}=\\frac{B F}{2 \\sin A}, \\quad R_{C}=\\frac{B D}{2 \\sin C}, \\quad R_{E}=\\frac{D F}{2 \\sin E} $$  It follows from (1) that  $$ \\begin{aligned} R_{A}+R_{C}+R_{E} & \\geq \\frac{1}{4} a\\left(\\frac{\\sin B}{\\sin A}+\\frac{\\sin A}{\\sin B}\\right)+\\frac{1}{4} b\\left(\\frac{\\sin C}{\\sin B}+\\frac{\\sin B}{\\sin C}\\right)+\\cdots \\\\ & \\geq \\frac{1}{2}(a+b+\\cdots)=\\frac{P}{2} \\end{aligned} $$  with equality if and only if $\\angle A=\\angle B=\\angle C=120^{\\circ}$ and $F B \\perp B C$ etc., i.e., if and only if the hexagon is regular.  Second solution. Let us construct points $A^{\\prime \\prime}, C^{\\prime \\prime}, E^{\\prime \\prime}$ such that $A B A^{\\prime \\prime} F$, $C D C^{\\prime \\prime} B$, and $E F E^{\\prime \\prime} D$ are parallelograms. It follows that $A^{\\prime \\prime}, C^{\\prime \\prime}, B$ are collinear and also $C^{\\prime \\prime}, E^{\\prime \\prime}, B$ and $E^{\\prime \\prime}, A^{\\prime \\prime}, F$. Furthermore, let $A^{\\prime}$ be the intersection of the perpendiculars through $F$ and $B$ to $F A^{\\prime \\prime}$ and $B A^{\\prime \\prime}$, respectively, and let $C^{\\prime}$ and $E^{\\prime}$ be analogously defined. Since $A^{\\prime} F A^{\\prime \\prime} B$ is cyclic with the diameter being $A^{\\prime} A^{\\prime \\prime}$ and since $\\triangle F A^{\\prime \\prime} B \\cong$ $\\triangle B A F$, it follows that $2 R_{A}=$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-617.jpg?height=372&width=504&top_left_y=1470&top_left_x=830) $A^{\\prime} A^{\\prime \\prime}=x$. Similarly, $2 R_{C}=C^{\\prime} C^{\\prime \\prime}=y$ and $2 R_{E}=E^{\\prime} E^{\\prime \\prime}=z$. We also have $A B=$ $F A^{\\prime \\prime}=y_{a}, A F=A^{\\prime \\prime} B=z_{a}, C D=C^{\\prime \\prime} B=z_{c}, C B=C^{\\prime \\prime} D=x_{c}$, $E F=E^{\\prime \\prime} D=x_{e}$, and $E D=E^{\\prime \\prime} F=y_{e}$. The original inequality we must prove now becomes  $$ x+y+z \\geq y_{a}+z_{a}+z_{c}+x_{c}+x_{e}+y_{e} . $$  We now follow and generalize the standard proof of the Erdős-Mordell inequality (for the triangle $A^{\\prime} C^{\\prime} E^{\\prime}$ ), which is what (1) is equivalent to when $A^{\\prime \\prime}=C^{\\prime \\prime}=E^{\\prime \\prime}$. We set $C^{\\prime} E^{\\prime}=a, A^{\\prime} E^{\\prime}=c$ and $A^{\\prime} C^{\\prime}=e$. Let $A_{1}$ be the point symmetric to $A^{\\prime \\prime}$ with respect to the bisector of $\\angle E^{\\prime} A^{\\prime} C^{\\prime}$. Let $F_{1}$ and $B_{1}$ be the feet of the perpendiculars from $A_{1}$ to $A^{\\prime} C^{\\prime}$ and $A^{\\prime} E^{\\prime}$, respectively. In that case, $A_{1} F_{1}=A^{\\prime \\prime} F=y_{a}$ and $A_{1} B_{1}=A^{\\prime \\prime} B=z_{a}$. We have  $$ \\begin{aligned} a x=A^{\\prime} A_{1} \\cdot E^{\\prime} C^{\\prime} \\geq 2 S_{A^{\\prime} E^{\\prime} A_{1} C^{\\prime}} & =2 S_{A^{\\prime} E^{\\prime} A_{1}}+2 S_{A^{\\prime} C^{\\prime} A_{1}} \\\\ & =c z_{a}+e y_{a} . \\end{aligned} $$  Similarly, $c y \\geq e x_{c}+a z_{c}$ and $e z \\geq a y_{e}+c x_{e}$. Thus  $$ \\begin{aligned} x+y+z & \\geq \\frac{c}{a} z_{a}+\\frac{a}{c} z_{c}+\\frac{e}{c} x_{c}+\\frac{c}{e} x_{e}+\\frac{a}{e} y_{e}+\\frac{e}{a} y_{a} \\\\ & =\\left(\\frac{c}{a}+\\frac{a}{c}\\right)\\left(\\frac{z_{a}+z_{c}}{2}\\right)+\\left(\\frac{c}{a}-\\frac{a}{c}\\right)\\left(\\frac{z_{a}-z_{c}}{2}\\right)+\\cdots . \\end{aligned} $$  Let us set $a_{1}=\\frac{x_{c}-x_{e}}{2}, c_{1}=\\frac{y_{e}-y_{a}}{2}, e_{1}=\\frac{z_{a}-z_{c}}{2}$. We note that $\\triangle A^{\\prime \\prime} C^{\\prime \\prime} E^{\\prime \\prime} \\sim$ $\\triangle A^{\\prime} C^{\\prime} E^{\\prime}$ and hence $a_{1} / a=c_{1} / c=e_{1} / e=k$. Thus $\\left(\\frac{c}{a}-\\frac{a}{c}\\right) e_{1}+$ $\\left(\\frac{e}{c}-\\frac{c}{e}\\right) a_{1}+\\left(\\frac{a}{e}-\\frac{e}{a}\\right) c_{1}=k\\left(\\frac{c e}{a}-\\frac{a e}{c}+\\frac{e a}{c}-\\frac{c a}{e}+\\frac{a c}{e}-\\frac{e c}{a}\\right)=0$. Equation (2) reduces to  $$ \\begin{aligned} x+y+z \\geq & \\left(\\frac{c}{a}+\\frac{a}{c}\\right)\\left(\\frac{z_{a}+z_{c}}{2}\\right)+\\left(\\frac{e}{c}+\\frac{c}{e}\\right)\\left(\\frac{x_{e}+x_{c}}{2}\\right) \\\\ & +\\left(\\frac{a}{e}+\\frac{e}{a}\\right)\\left(\\frac{y_{a}+y_{e}}{2}\\right) . \\end{aligned} $$  Using $c / a+a / c, e / c+c / e, a / e+e / a \\geq 2$ we finally get $x+y+z \\geq$ $y_{a}+z_{a}+z_{c}+x_{c}+x_{e}+y_{e}$. Equality holds if and only if $a=c=e$ and $A^{\\prime \\prime}=C^{\\prime \\prime}=E^{\\prime \\prime}=$ center of $\\triangle A^{\\prime} C^{\\prime} E^{\\prime}$, i.e., if and only if $A B C D E F$ is regular. Remark. From the second proof it is evident that the Erdős-Mordell inequality is a special case of the problem. if $P_{a}, P_{b}, P_{c}$ are the feet of the perpendiculars from a point $P$ inside $\\triangle A B C$ to the sides $B C, C A, A B$, and $P_{a} P P_{b} P_{c}^{\\prime}, P_{b} P P_{c} P_{a}^{\\prime}, P_{c} P P_{a} P_{b}^{\\prime}$ parallelograms, we can apply the problem to the hexagon $P_{a} P_{c}^{\\prime} P_{b} P_{a}^{\\prime} P_{c} P_{b}^{\\prime}$ to prove the Erdős-Mordell inequality for $\\triangle A B C$ and point $P$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "15", "problem_type": "Geometry", "exam": "IMO", "problem": "15. G6 (ARM) Let the sides of two rectangles be $\\{a, b\\}$ and $\\{c, d\\}$ with $a<c \\leq d<b$ and $a b<c d$. Prove that the first rectangle can be placed within the second one if and only if  $$ \\left(b^{2}-a^{2}\\right)^{2} \\leq(b d-a c)^{2}+(b c-a d)^{2} $$", "solution": "15. Denote by $A B C D$ and $E F G H$ the two rectangles, where $A B=a, B C=$ $b, E F=c$, and $F G=d$. Obviously, the first rectangle can be placed within the second one with the angle $\\alpha$ between $A B$ and $E F$ if and only if  $$ a \\cos \\alpha+b \\sin \\alpha \\leq c, \\quad a \\sin \\alpha+b \\cos \\alpha \\leq d $$  Hence $A B C D$ can be placed within $E F G H$ if and only if there is an $\\alpha \\in[0, \\pi / 2]$ for which (1) holds.  The lines $l_{1}(a x+b y=c)$ and $l_{2}(b x+a y=d)$ and the axes $x$ and $y$ bound a region $\\mathcal{R}$. By (1), the desired placement of the rectangles is possible if and only if $\\mathcal{R}$ contains some point $(\\cos \\alpha, \\sin \\alpha)$ of the unit circle centered at the origin $(0,0)$. This in turn holds if and only if the intersection point $L$ of $l_{1}$ and $l_{2}$ lies outside the unit circle. It is easily computed that $L$ has coordinates $\\left(\\frac{b d-a c}{b^{2}-a^{2}}, \\frac{b c-a d}{b^{2}-a^{2}}\\right)$. Now $L$ being outside the unit circle is exactly equivalent to the inequality we want to prove. Remark. If equality holds, there is exactly one way of placing. This happens, for example, when $(a, b)=(5,20)$ and $(c, d)=(16,19)$. Second remark. This problem is essentially very similar to (SL89-2).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "16", "problem_type": "Geometry", "exam": "IMO", "problem": "16. G7 (GBR) Let $A B C$ be an acute-angled triangle with circumcenter $O$ and circumradius $R$. Let $A O$ meet the circle $B O C$ again in $A^{\\prime}$, let $B O$ meet the circle $C O A$ again in $B^{\\prime}$, and let $C O$ meet the circle $A O B$ again in $C^{\\prime}$. Prove that  $$ O A^{\\prime} \\cdot O B^{\\prime} \\cdot O C^{\\prime} \\geq 8 R^{3} $$  When does equality hold?", "solution": "16. Let $A_{1}$ be the point of intersection of $O A^{\\prime}$ and $B C$; similarly define $B_{1}$ and $C_{1}$. From the similarity of triangles $O B A_{1}$ and $O A^{\\prime} B$ we obtain $O A_{1}$. $O A^{\\prime}=R^{2}$. Now it is enough to show that $8 O A_{1} \\cdot O B^{\\prime} \\cdot O C^{\\prime} \\leq R^{3}$. Thus we must prove that  $$ \\lambda \\mu \\nu \\leq \\frac{1}{8}, \\quad \\text { where } \\quad \\frac{O A_{1}}{O A}=\\lambda, \\quad \\frac{O B_{1}}{O B}=\\mu, \\quad \\frac{O C_{1}}{O C}=\\nu $$  On the other hand, we have  $$ \\frac{\\lambda}{1+\\lambda}+\\frac{\\mu}{1+\\mu}+\\frac{\\nu}{1+\\nu}=\\frac{S_{O B C}}{S_{A B C}}+\\frac{S_{A O C}}{S_{A B C}}+\\frac{S_{A B O}}{S_{A B C}}=1 . $$  Simplifying this relation, we get  $$ 1=\\lambda \\mu+\\mu \\nu+\\nu \\lambda+2 \\lambda \\mu \\nu \\geq 3(\\lambda \\mu \\nu)^{2 / 3}+2 \\lambda \\mu \\nu $$  which cannot hold if $\\lambda \\mu \\nu>\\frac{1}{8}$. Hence $\\lambda \\mu \\nu \\leq \\frac{1}{8}$, with equality if and only if $\\lambda=\\mu=\\nu=\\frac{1}{2}$. This implies that $O$ is the centroid of $A B C$, and consequently, that the triangle is equilateral.  Second solution. In the official solution, the inequality to be proved is transformed into  $$ \\cos (A-B) \\cos (B-C) \\cos (C-A) \\geq 8 \\cos A \\cos B \\cos C $$  Since $\\frac{\\cos (B-C)}{\\cos A}=-\\frac{\\cos (B-C)}{\\cos (B+C)}=\\frac{\\tan B \\tan C+1}{\\tan B \\tan C-1}$, the last inequality becomes $(x y+1)(y z+1)(z x+1) \\geq 8(x y-1)(y z-1)(z x-1)$, where we write $x, y, z$ for $\\tan A, \\tan B, \\tan C$. Using the relation $x+y+z=x y z$, we can reduce this inequality to  $$ (2 x+y+z)(x+2 y+z)(x+y+2 z) \\geq 8(x+y)(y+z)(z+x) $$  This follows from the AM-GM inequality: $2 x+y+z=(x+y)+(x+z) \\geq$ $2 \\sqrt{(x+y)(x+z)}$, etc.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "17", "problem_type": "Geometry", "exam": "IMO", "problem": "17. G8 (RUS) Let $A B C D$ be a convex quadrilateral, and let $R_{A}, R_{B}, R_{C}$, and $R_{D}$ denote the circumradii of the triangles $D A B, A B C, B C D$, and $C D A$ respectively. Prove that $R_{A}+R_{C}>R_{B}+R_{D}$ if and only if  $$ \\angle A+\\angle C>\\angle B+\\angle D . $$", "solution": "17. Let the diagonals $A C$ and $B D$ meet in $X$. Either $\\angle A X B$ or $\\angle A X D$ is geater than or equal to $90^{\\circ}$, so we assume w.l.o.g. that $\\angle A X B \\geq 90^{\\circ}$. Let $\\alpha, \\beta, \\alpha^{\\prime}, \\beta^{\\prime}$ denote $\\angle C A B, \\angle A B D, \\angle B D C, \\angle D C A$. These angles are all acute and satisfy $\\alpha+\\beta=\\alpha^{\\prime}+\\beta^{\\prime}$. Furthermore,  $$ R_{A}=\\frac{A D}{2 \\sin \\beta}, \\quad R_{B}=\\frac{B C}{2 \\sin \\alpha}, \\quad R_{C}=\\frac{B C}{2 \\sin \\alpha^{\\prime}}, \\quad R_{D}=\\frac{A D}{2 \\sin \\beta^{\\prime}} $$  Let $\\angle B+\\angle D=180^{\\circ}$. Then $A, B, C, D$ are concyclic and trivially $R_{A}+$ $R_{C}=R_{B}+R_{D}$. Let $\\angle B+\\angle D>180^{\\circ}$. Then $D$ lies within the circumcircle of $A B C$, which implies that $\\beta>\\beta^{\\prime}$. Similarly $\\alpha<\\alpha^{\\prime}$, so we obtain $R_{A}<R_{D}$ and $R_{C}<R_{B}$. Thus $R_{A}+R_{C}<R_{B}+R_{D}$. Let $\\angle B+\\angle D<180^{\\circ}$. As in the previous case, we deduce that $R_{A}>R_{D}$ and $R_{C}>R_{B}$, so $R_{A}+R_{C}>R_{B}+R_{D}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "18", "problem_type": "Geometry", "exam": "IMO", "problem": "18. G9 (UKR) In the plane are given a point $O$ and a polygon $\\mathcal{F}$ (not necessarily convex). Let $P$ denote the perimeter of $\\mathcal{F}, D$ the sum of the distances from $O$ to the vertices of $\\mathcal{F}$, and $H$ the sum of the distances from $O$ to the lines containing the sides of $\\mathcal{F}$. Prove that  $$ D^{2}-H^{2} \\geq \\frac{P^{2}}{4} $$", "solution": "18. We first prove the result in the simplest case. Given a 2 -gon $A B A$ and a point $O$, let $a, b, c, h$ denote $O A, O B, A B$, and the distance of $O$ from $A B$. Then $D=a+b, P=2 c$, and $H=2 h$, so we should show that  $$ (a+b)^{2} \\geq 4 h^{2}+c^{2} $$  Indeed, let $l$ be the line through $O$ parallel to $A B$, and $D$ the point symmetric to $B$ with respect to $l$. Then $(a+b)^{2}=(O A+O B)^{2}=(O A+$ $O D)^{2} \\geq A D^{2}=c^{2}+4 h^{2}$. Now we pass to the general case. Let $A_{1} A_{2} \\ldots A_{n}$ be the polygon $\\mathcal{F}$ and denote by $d_{i}, p_{i}$, and $h_{i}$ respectively $O A_{i}, A_{i} A_{i+1}$, and the distance of $O$ from $A_{i} A_{i+1}$ (where $A_{n+1}=A_{1}$ ). By the case proved above, we have for each $i, d_{i}+d_{i+1} \\geq \\sqrt{4 h_{i}^{2}+p_{i}^{2}}$. Summing these inequalities for $i=1, \\ldots, n$ and squaring, we obtain  $$ 4 D^{2} \\geq\\left(\\sum_{i=1}^{n} \\sqrt{4 h_{i}^{2}+p_{i}^{2}}\\right)^{2} $$  It remains only to prove that $\\sum_{i=1}^{n} \\sqrt{4 h_{i}^{2}+p_{i}^{2}} \\geq \\sqrt{\\sum_{i=1}^{n}\\left(4 h_{i}^{2}+p_{i}^{2}\\right)}=$ $\\sqrt{4 H^{2}+D^{2}}$. But this follows immediately from the Minkowski inequality. Equality holds if and only if it holds in (1) and in the Minkowski inequality, i.e., if and only if $d_{1}=\\cdots=d_{n}$ and $h_{1} / p_{1}=\\cdots=h_{n} / p_{n}$. This means that $\\mathcal{F}$ is inscribed in a circle with center at $O$ and $p_{1}=\\cdots=p_{n}$, so $\\mathcal{F}$ is a regular polygon and $O$ its center.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "19", "problem_type": "Number Theory", "exam": "IMO", "problem": "19. N1 (UKR) Four integers are marked on a circle. At each step we simultaneously replace each number by the difference between this number and the next number on the circle, in a given direction (that is, the numbers $a, b, c, d$ are replaced by $a-b, b-c, c-d, d-a)$. Is it possible after 1996 such steps to have numbers $a, b, c, d$ such that the numbers $|b c-a d|,|a c-b d|,|a b-c d|$ are primes?", "solution": "19. It is easy to check that after 4 steps we will have all $a, b, c, d$ even. Thus $|a b-c d|,|a c-b d|,|a d-b c|$ remain divisible by 4 , and clearly are not prime. The answer is no. Second solution. After one step we have $a+b+c+d=0$. Then $a c-b d=$ $a c+b(a+b+c)=(a+b)(b+c)$ etc., so  $$ |a b-c d| \\cdot|a c-b d| \\cdot|a d-b c|=(a+b)^{2}(a+c)^{2}(b+c)^{2} $$  However, the product of three primes cannot be a square, hence the answer is $n o$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "2", "problem_type": "Algebra", "exam": "IMO", "problem": "2. A2 (IRE) Let $a_{1} \\geq a_{2} \\geq \\cdots \\geq a_{n}$ be real numbers such that  $$ a_{1}^{k}+a_{2}^{k}+\\cdots+a_{n}^{k} \\geq 0 $$  for all integers $k>0$. Let $p=\\max \\left\\{\\left|a_{1}\\right|, \\ldots,\\left|a_{n}\\right|\\right\\}$. Prove that $p=a_{1}$ and that  $$ \\left(x-a_{1}\\right)\\left(x-a_{2}\\right) \\cdots\\left(x-a_{n}\\right) \\leq x^{n}-a_{1}^{n} $$  for all $x>a_{1}$.", "solution": "2. Clearly $a_{1}>0$, and if $p \\neq a_{1}$, we must have $a_{n}<0,\\left|a_{n}\\right|>\\left|a_{1}\\right|$, and $p=-a_{n}$. But then for sufficiently large odd $k,-a_{n}^{k}=\\left|a_{n}\\right|^{k}>(n-1)\\left|a_{1}\\right|^{k}$, so that $a_{1}^{k}+\\cdots+a_{n}^{k} \\leq(n-1)\\left|a_{1}\\right|^{k}-\\left|a_{n}\\right|^{k}<0$, a contradiction. Hence $p=a_{1}$. Now let $x>a_{1}$. From $a_{1}+\\cdots+a_{n} \\geq 0$ we deduce $\\sum_{j=2}^{n}\\left(x-a_{j}\\right) \\leq$ $(n-1)\\left(x+\\frac{a_{1}}{n-1}\\right)$, so by the AM-GM inequality,  $$ \\left(x-a_{2}\\right) \\cdots\\left(x-a_{n}\\right) \\leq\\left(x+\\frac{a_{1}}{n-1}\\right)^{n-1} \\leq x^{n-1}+x^{n-2} a_{1}+\\cdots+a_{1}^{n-1} $$  The last inequality holds because $\\binom{n-1}{r} \\leq(n-1)^{r}$ for all $r \\geq 0$. Multiplying (1) by $\\left(x-a_{1}\\right)$ yields the desired inequality.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "20", "problem_type": "Number Theory", "exam": "IMO", "problem": "20. N2 (RUS) ${ }^{\\mathrm{IMO} 4}$ The positive integers $a$ and $b$ are such that the numbers $15 a+16 b$ and $16 a-15 b$ are both squares of positive integers. What is the least possible value that can be taken on by the smaller of these two squares?", "solution": "20. Let $15 a+16 b=x^{2}$ and $16 a-15 b=y^{2}$, where $x, y \\in \\mathbb{N}$. Then we obtain $x^{4}+y^{4}=(15 a+16 b)^{2}+(16 a-15 b)^{2}=\\left(15^{2}+16^{2}\\right)\\left(a^{2}+b^{2}\\right)=481\\left(a^{2}+b^{2}\\right)$. In particular, $481=13 \\cdot 37 \\mid x^{4}+y^{4}$. We have the following lemma. Lemma. Suppose that $p \\mid x^{4}+y^{4}$, where $x, y \\in \\mathbb{Z}$ and $p$ is an odd prime, where $p \\not \\equiv 1(\\bmod 8)$. Then $p \\mid x$ and $p \\mid y$. Proof. Since $p \\mid x^{8}-y^{8}$ and by Fermat's theorem $p \\mid x^{p-1}-y^{p-1}$, we deduce that $p \\mid x^{d}-y^{d}$, where $d=(p-1,8)$. But $d \\neq 8$, so $d \\mid 4$. Thus $p \\mid x^{4}-y^{4}$, which implies that $p \\mid 2 y^{4}$, i.e., $p \\mid y$ and $p \\mid x$. In particular, we can conclude that $13 \\mid x, y$ and $37 \\mid x, y$. Hence $x$ and $y$ are divisible by 481 . Thus each of them is at least 481. On the other hand, $x=y=481$ is possible. It is sufficient to take $a=$ $31 \\cdot 481$ and $b=481$.  Second solution. Note that $15 x^{2}+16 y^{2}=481 a^{2}$. It can be directly verified that the divisibility of $15 x^{2}+16 y^{2}$ by 13 and by 37 implies that both $x$ and $y$ are divisible by both primes. Thus $481 \\mid x, y$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "21", "problem_type": "Number Theory", "exam": "IMO", "problem": "21. N3 (BUL) A finite sequence of integers $a_{0}, a_{1}, \\ldots, a_{n}$ is called quadratic if for each $i \\in\\{1,2, \\ldots, n\\}$ we have the equality $\\left|a_{i}-a_{i-1}\\right|=i^{2}$. (a) Prove that for any two integers $b$ and $c$, there exist a natural number $n$ and a quadratic sequence with $a_{0}=b$ and $a_{n}=c$. (b) Find the smallest natural number $n$ for which there exists a quadratic sequence with $a_{0}=0$ and $a_{n}=1996$.", "solution": "21. (a) It clearly suffices to show that for every integer $c$ there exists a quadratic sequence with $a_{0}=0$ and $a_{n}=c$, i.e., that $c$ can be expressed as $\\pm 1^{2} \\pm 2^{2} \\pm \\cdots \\pm n^{2}$. Since  $$ (n+1)^{2}-(n+2)^{2}-(n+3)^{2}+(n+4)^{2}=4 $$  we observe that if our claim is true for $c$, then it is also true for $c \\pm 4$. Thus it remains only to prove the claim for $c=0,1,2,3$. But one immediately finds $1=1^{2}, 2=-1^{2}-2^{2}-3^{2}+4^{2}$, and $3=-1^{2}+2^{2}$, while the case $c=0$ is trivial. (b) We have $a_{0}=0$ and $a_{n}=1996$. Since $a_{n} \\leq 1^{2}+2^{2}+\\cdots+n^{2}=$ $\\frac{1}{6} n(n+1)(2 n+1)$, we get $a_{17} \\leq 1785$, so $n \\geq 18$. On the other hand, $a_{18}$ is of the same parity as $1^{2}+2^{2}+\\cdots+18^{2}=2109$, so it cannot be equal to 1996. Therefore we must have $n \\geq 19$. To construct a required sequence with $n=19$, we note that $1^{2}+2^{2}+\\cdots+19^{2}=$ $2470=1996+2 \\cdot 237$; hence it is enough to write 237 as a sum of distinct squares. Since $237=14^{2}+5^{2}+4^{2}$, we finally obtain  $$ 1996=1^{2}+2^{2}+3^{2}-4^{2}-5^{2}+6^{2}+\\cdots+13^{2}-14^{2}+15^{2}+\\cdots+19^{2} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "22", "problem_type": "Number Theory", "exam": "IMO", "problem": "22. N4 (BUL) Find all positive integers $a$ and $b$ for which  $$ \\left[\\frac{a^{2}}{b}\\right]+\\left[\\frac{b^{2}}{a}\\right]=\\left[\\frac{a^{2}+b^{2}}{a b}\\right]+a b $$  where as usual, $[t]$ refers to greatest integer that is less than or equal to $t$.", "solution": "22. Let $a, b \\in \\mathbb{N}$ satisfy the given equation. It is not possible that $a=b$ (since it leads to $a^{2}+2=2 a$ ), so we assume w.l.o.g. that $a>b$. Next, for $a>b=1$ the equation becomes $a^{2}=2 a$, and one obtains a solution $(a, b)=(2,1)$. Let $b>1$. If $\\left[\\frac{a^{2}}{b}\\right]=\\alpha$ and $\\left[\\frac{b^{2}}{a}\\right]=\\beta$, then we trivially have $a b \\geq$ $\\alpha \\beta$. Since also $\\frac{a^{2}+b^{2}}{a b} \\geq 2$, we obtain $\\alpha+\\beta \\geq \\alpha \\beta+2$, or equivalently $(\\alpha-1)(\\beta-1) \\leq-1$. But $\\alpha \\geq 1$, and therefore $\\beta=0$. It follows that $a>b^{2}$, i.e., $a=b^{2}+c$ for some $c>0$. Now the given equation becomes $b^{3}+2 b c+\\left[\\frac{c^{2}}{b}\\right]=\\left[\\frac{b^{4}+2 b^{2} c+b^{2}+c^{2}}{b^{3}+b c}\\right]+b^{3}+b c$, which reduces to  $$ (c-1) b+\\left[\\frac{c^{2}}{b}\\right]=\\left[\\frac{b^{2}(c+1)+c^{2}}{b^{3}+b c}\\right] $$  If $c=1$, then (1) always holds, since both sides are 0 . We obtain a family of solutions $(a, b)=\\left(n, n^{2}+1\\right)$ or $(a, b)=\\left(n^{2}+1, n\\right)$. Note that the solution $(1,2)$ found earlier is obtained for $n=1$. If $c>1$, then $(1)$ implies that $\\frac{b^{2}(c+1)+c^{2}}{b^{3}+b c} \\geq(c-1) b$. This simplifies to  $$ c^{2}\\left(b^{2}-1\\right)+b^{2}\\left(c\\left(b^{2}-2\\right)-\\left(b^{2}+1\\right)\\right) \\leq 0 $$  Since $c \\geq 2$ and $b^{2}-2 \\geq 0$, the only possibility is $b=2$. But then (2) becomes $3 c^{2}+8 c-20 \\leq 0$, which does not hold for $c \\geq 2$. Hence the only solutions are $\\left(n, n^{2}+1\\right)$ and $\\left(n^{2}+1, n\\right), n \\in \\mathbb{N}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "23", "problem_type": "Number Theory", "exam": "IMO", "problem": "23. N5 (ROM) Let $\\mathbb{N}_{0}$ denote the set of nonnegative integers. Find a bijective function $f$ from $\\mathbb{N}_{0}$ into $\\mathbb{N}_{0}$ such that for all $m, n \\in \\mathbb{N}_{0}$,  $$ f(3 m n+m+n)=4 f(m) f(n)+f(m)+f(n) . $$", "solution": "23. We first observe that the given functional equation is equivalent to  $$ 4 f\\left(\\frac{(3 m+1)(3 n+1)-1}{3}\\right)+1=(4 f(m)+1)(4 f(n)+1) . $$  This gives us the idea of introducing a function $g: 3 \\mathbb{N}_{0}+1 \\rightarrow 4 \\mathbb{N}_{0}+1$ defined as $g(x)=4 f\\left(\\frac{x-1}{3}\\right)+1$. By the above equality, $g$ will be multiplicative, i.e.,  $$ g(x y)=g(x) g(y) \\quad \\text { for all } x, y \\in 3 \\mathbb{N}_{0}+1 $$  Conversely, any multiplicative bijection $g$ from $3 \\mathbb{N}_{0}+1$ onto $4 \\mathbb{N}_{0}+1$ gives us a function $f$ with the required property: $f(x)=\\frac{g(3 x+1)-1}{4}$. It remains to give an example of such a function $g$. Let $P_{1}, P_{2}, Q_{1}, Q_{2}$ be the sets of primes of the forms $3 k+1,3 k+2,4 k+1$, and $4 k+3$, respectively. It is well known that these sets are infinite. Take any bijection $h$ from $P_{1} \\cup P_{2}$ onto $Q_{1} \\cup Q_{2}$ that maps $P_{1}$ bijectively onto $Q_{1}$ and $P_{2}$ bijectively onto $Q_{2}$. Now define $g$ as follows: $g(1)=1$, and for $n=p_{1} p_{2} \\cdots p_{m}\\left(p_{i}\\right.$ 's need not be different) define $g(n)=h\\left(p_{1}\\right) h\\left(p_{2}\\right) \\cdots h\\left(p_{m}\\right)$. Note that $g$ is well-defined. Indeed, among the $p_{i}$ 's an even number are of the form $3 k+2$, and consequently an even number of $h\\left(p_{i}\\right) \\mathrm{s}$ are of the form $4 k+3$. Hence the product of the $h\\left(p_{i}\\right)$ 's is of the form $4 k+1$. Also, it is obvious that $g$ is multiplicative. Thus, the defined $g$ satisfies all the required properties.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "24", "problem_type": "Combinatorics", "exam": "IMO", "problem": "24. C1 (FIN) ${ }^{\\mathrm{IMO} 1}$ We are given a positive integer $r$ and a rectangular board $A B C D$ with dimensions $|A B|=20,|B C|=12$. The rectangle is divided into a grid of $20 \\times 12$ unit squares. The following moves are permitted on the board: One can move from one square to another only if the distance between the centers of the two squares is $\\sqrt{r}$. The task is to find a sequence of moves leading from the square corresponding to vertex $A$ to the square corresponding to vertex $B$. (a) Show that the task cannot be done if $r$ is divisible by 2 or 3 . (b) Prove that the task is possible when $r=73$. (c) Is there a solution when $r=97$ ?", "solution": "24. We shall work on the array of lattice points defined by $\\mathcal{A}=\\left\\{(x, y) \\in \\mathbb{Z}^{2} \\mid\\right.$ $0 \\leq x \\leq 19,0 \\leq y \\leq 11\\}$. Our task is to move from $(0,0)$ to $(19,0)$ via the points of $\\mathcal{A}$ so that each move has the form $(x, y) \\rightarrow(x+a, y+b)$, where $a, b \\in \\mathbb{Z}$ and $a^{2}+b^{2}=r$. (a) If $r$ is even, then $a+b$ is even whenever $a^{2}+b^{2}=r(a, b \\in \\mathbb{Z})$. Thus the parity of $x+y$ does not change after each move, so we cannot reach $(19,0)$ from $(0,0)$. If $3 \\mid r$, then both $a$ and $b$ are divisible by 3 , so if a point $(x, y)$ can be reached from $(0,0)$, we must have $3 \\mid x$. Since $3 \\nmid 19$, we cannot get to $(19,0)$. (b) We have $r=73=8^{2}+3^{2}$, so each move is either $(x, y) \\rightarrow(x \\pm 8, y \\pm 3)$ or $(x, y) \\rightarrow(x \\pm 3, y \\pm 8)$. One possible solution is shown in Fig. 1. (c) We have $97=9^{2}+4^{2}$. Let us partition $\\mathcal{A}$ as $\\mathcal{B} \\cup \\mathcal{C}$, where $\\mathcal{B}=$ $\\{(x, y) \\in \\mathcal{A} \\mid 4 \\leq y \\leq 7\\}$. It is easily seen that moves of the type $(x, y) \\rightarrow(x \\pm 9, y \\pm 4)$ always take us from the set $\\mathcal{B}$ to $\\mathcal{C}$ and vice versa, while the moves $(x, y) \\rightarrow(x \\pm 4, y \\pm 9)$ always take us from $\\mathcal{C}$ to $\\mathcal{C}$. Furthermore, each move of the type $(x, y) \\rightarrow(x \\pm 9, y \\pm 4)$ changes the parity of $x$, so to get from $(0,0)$ to $(19,0)$ we must have an odd number of such moves. On the other hand, with an odd number of such moves, starting from $\\mathcal{C}$ we can end up only in $\\mathcal{B}$, although the point $(19,0)$ is not in $\\mathcal{B}$. Hence, the answer is no. Remark. Part (c) can also be solved by examining all cells that can be reached from $(0,0)$. All these cells are marked in Fig. 2. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-623.jpg?height=259&width=420&top_left_y=1081&top_left_x=289)  Fig. 1 ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-623.jpg?height=256&width=414&top_left_y=1083&top_left_x=886)  Fig. 2", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "25", "problem_type": "Combinatorics", "exam": "IMO", "problem": "25. C2 (UKR) An $(n-1) \\times(n-1)$ square is divided into $(n-1)^{2}$ unit squares in the usual manner. Each of the $n^{2}$ vertices of these squares is to be colored red or blue. Find the number of different colorings such that each unit square has exactly two red vertices. (Two coloring schemes are regarded as different if at least one vertex is colored differently in the two schemes.)", "solution": "25. Let the vertices in the bottom row be assigned an arbitrary coloring, and suppose that some two adjacent vertices receive the same color. The number of such colorings equals $2^{n}-2$. It is easy to see that then the colors of the remaining vertices get fixed uniquely in order to satisfy the requirement. So in this case there are $2^{n}-2$ possible colorings. Next, suppose that the vertices in the bottom row are colored alternately red and blue. There are two such colorings. In this case, the same must hold for every row, and thus we get $2^{n}$ possible colorings. It follows that the total number of considered colorings is $\\left(2^{n}-2\\right)+2^{n}=$ $2^{n+1}-2$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "26", "problem_type": "Combinatorics", "exam": "IMO", "problem": "26. C3 (USA) Let $k, m, n$ be integers such that $1<n \\leq m-1 \\leq k$. Determine the maximum size of a subset $S$ of the set $\\{1,2,3, \\ldots, k\\}$ such that no $n$ distinct elements of $S$ add up to $m$.", "solution": "26. Denote the required maximum size by $M_{k}(m, n)$. If $m<\\frac{n(n+1)}{2}$, then trivially $M=k$, so from now on we assume that $m \\geq \\frac{n(n+1)}{2}$. First we give a lower bound for $M$. Let $r=r_{k}(m, n)$ be the largest integer such that $r+(r+1)+\\cdots+(r+n-1) \\leq m$. This is equivalent to $n r \\leq m-\\frac{n(n-1)}{2} \\leq n(r+1)$, so $r=\\left[\\frac{m}{n}-\\frac{n-1}{2}\\right]$. Clearly no $n$ elements from $\\{r+1, r+2, \\ldots, k\\}$ add up to $m$, so  $$ M \\geq k-r_{k}(m, n)=k-\\left[\\frac{m}{n}-\\frac{n-1}{2}\\right] . $$  We claim that $M$ is actually equal to $k-r_{k}(m, n)$. To show this, we shall prove by induction on $n$ that if no $n$ elements of a set $S \\subseteq\\{1,2, \\ldots, k\\}$ add up to $m$, then $|S| \\leq k-r_{k}(m, n)$. For $n=2$ the claim is true, because then for each $i=1, \\ldots, r_{k}(m, 2)=$ $\\left[\\frac{m-1}{2}\\right]$ at least one of $i$ and $m-i$ must be excluded from $S$. Now let us assume that $n>2$ and that the result holds for $n-1$. Suppose that $S \\subseteq\\{1,2, \\ldots, k\\}$ does not contain $n$ distinct elements with the sum $m$, and let $x$ be the smallest element of $S$. We may assume that $x \\leq r_{k}(m, n)$, because otherwise the statement is clear. Consider the set $S^{\\prime}=\\{y-x \\mid$ $y \\in S, y \\neq x\\}$. Then $S^{\\prime}$ is a subset of $\\{1,2, \\ldots, k-x\\}$ no $n-1$ elements of which have the sum $m-n x$. Also, it is easily checked that $n-1 \\leq$ $m-n x-1 \\leq k-x$, so we may apply the induction hypothesis, which yields that  $$ |S| \\leq 1+k-x-r_{k}(m-n x, n-1)=k-\\left[\\frac{m-x}{n-1}-\\frac{n}{2}\\right] . $$  On the other hand, $\\left(\\frac{m-x}{n-1}-\\frac{n}{2}\\right)-r_{k}(m, n)=\\frac{m-n x-\\frac{n(n-1)}{2}}{n(n-1)} \\geq 0$ because $x \\leq r_{k}(m, n)$; hence (2) implies $|S| \\leq k-r_{k}(m, n)$ as claimed.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "27", "problem_type": "Combinatorics", "exam": "IMO", "problem": "27. C4 (FIN) Determine whether or not there exist two disjoint infinite sets $\\mathcal{A}$ and $\\mathcal{B}$ of points in the plane satisfying the following conditions: (i) No three points in $\\mathcal{A} \\cup \\mathcal{B}$ are collinear, and the distance between any two points in $\\mathcal{A} \\cup \\mathcal{B}$ is at least 1. (ii) There is a point of $\\mathcal{A}$ in any triangle whose vertices are in $\\mathcal{B}$, and there is a point of $\\mathcal{B}$ in any triangle whose vertices are in $\\mathcal{A}$.", "solution": "27. Suppose that such sets of points $\\mathcal{A}, \\mathcal{B}$ exist.  First, we observe that there exist five points $A, B, C, D, E$ in $\\mathcal{A}$ such that their convex hull does not contain any other point of $\\mathcal{A}$. Indeed, take any point $A \\in \\mathcal{A}$. Since any two points of $\\mathcal{A}$ are at distance at least 1 , the number of points $X \\in \\mathcal{A}$ with $X A \\leq r$ is finite for every $r>0$. Thus it is enough to choose four points $B, C, D, E$ of $\\mathcal{A}$ that are closest to $A$. Now consider the convex hull $\\mathcal{C}$ of $A, B, C, D, E$. Suppose that $\\mathcal{C}$ is a pentagon, say $A B C D E$. Then each of the disjoint triangles $A B C, A C D, A D E$ contains a point of $\\mathcal{B}$. Denote these points by $P, Q, R$. Then $\\triangle P Q R$ contains some point $F \\in \\mathcal{A}$, so $F$ is inside $A B C D E$, a contradiction. Suppose that $\\mathcal{C}$ is a quadrilateral, say $A B C D$, with $E$ lying within $A B C D$. Then the triangles $A B E, B C E, C D E, D A E$ contain some points $P, Q, R, S$ of $\\mathcal{B}$ that form two disjoint triangles. It follows that there are two points of $\\mathcal{A}$ inside $A B C D$, which is a contradiction. Finally, suppose that $\\mathcal{C}$ is a triangle with two points of $\\mathcal{A}$ inside. Then $\\mathcal{C}$ is the union of five disjoint triangles with vertices in $\\mathcal{A}$, so there are at least five points of $\\mathcal{B}$ inside $\\mathcal{C}$. These five points make at least three disjoint triangles containing three points of $\\mathcal{A}$. This is again a contradiction. It follows that no such sets $\\mathcal{A}, \\mathcal{B}$ exist.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "28", "problem_type": "Combinatorics", "exam": "IMO", "problem": "28. C5 (FRA) ${ }^{\\mathrm{IMO} 6}$ Let $p, q, n$ be three positive integers with $p+q<n$. Let $\\left(x_{0}, x_{1}, \\ldots, x_{n}\\right)$ be an $(n+1)$-tuple of integers satisfying the following conditions: (i) $x_{0}=x_{n}=0$. (ii) For each $i$ with $1 \\leq i \\leq n$, either $x_{i}-x_{i-1}=p$ or $x_{i}-x_{i-1}=-q$. Show that there exists a pair $(i, j)$ of distinct indices with $(i, j) \\neq(0, n)$ such that $x_{i}=x_{j}$.", "solution": "28. Note that w.l.o.g., we can assume that $p$ and $q$ are coprime. Indeed, otherwise it suffices to consider the problem in which all $x_{i}$ 's and $p, q$ are divided by $\\operatorname{gcd}(p, q)$.  Let $k, l$ be the number of indices $i$ with $x_{i+1}-x_{i}=p$ and the number of those $i$ with $x_{i+1}-x_{i}=-q(0 \\leq i<n)$. From $x_{0}=x_{n}=0$ we get $k p=l q$, so for some integer $t>1, k=q t, l=p t$, and $n=(p+q) t$. Consider the sequence $y_{i}=x_{i+p+q}-x_{i}, i=0, \\ldots, n-p-q$. We claim that at least one of the $y_{i}$ 's equals zero. We begin by noting that each $y_{i}$ is of the form $u p-v q$, where $u+v=p+q$; therefore $y_{i}=(u+v) p-$ $v(p+q)=(p-v)(p+q)$ is always divisible by $p+q$. Moreover, $y_{i+1}-y_{i}=$ $\\left(x_{i+p+q+1}-x_{i+p+q}\\right)-\\left(x_{i+1}-x_{i}\\right)$ is 0 or $\\pm(p+q)$. We conclude that if no $y_{i}$ is 0 then all $y_{i}$ 's are of the same sign. But this is in contradiction with the relation $y_{0}+y_{p+q}+\\cdots+y_{n-p-q}=x_{n}-x_{0}=0$. Consequently some $y_{i}$ is zero, as claimed.  Second solution. As before we assume $(p, q)=1$. Let us define a sequence of points $A_{i}\\left(y_{i}, z_{i}\\right)(i=0,1, \\ldots, n)$ in $\\mathbb{N}_{0}^{2}$ inductively as follows. Set $A_{0}=$ $(0,0)$ and define $\\left(y_{i+1}, z_{i+1}\\right)$ as $\\left(y_{i}, z_{i}+1\\right)$ if $x_{i+1}=x_{i}+p$ and $\\left(y_{i}+1, z_{i}\\right)$ otherwise. The points $A_{i}$ form a trajectory $L$ in $\\mathbb{N}_{0}^{2}$ continuously moving upwards and rightwards by steps of length 1 . Clearly, $x_{i}=p z_{i}-q y_{i}$ for all $i$. Since $x_{n}=0$, it follows that $\\left(z_{n}, y_{n}\\right)=(k q, k p), k \\in \\mathbb{N}$. Since $y_{n}+z_{n}=n>p+q$, it follows that $k>1$. We observe that $x_{i}=x_{j}$ if and only if $A_{i} A_{j} \\| A_{0} A_{n}$. We shall show that such $i, j$ with $i<j$ and $(i, j) \\neq(0, n)$ must exist. If $L$ meets $A_{0} A_{n}$ in an interior point, then our statement trivially holds. From now on we assume the opposite. Let $P_{i j}$ be the rectangle with sides parallel to the coordinate axes and with vertices at $(i p, j q)$ and $((i+$ 1) $p,(j+1) q)$. Let $L_{i j}$ be the part of the trajectory $L$ lying inside $P_{i j}$. We may assume w.l.o.g. that the endpoints of $L_{00}$ lie on the vertical sides of $P_{00}$. Then there obviously exists $d \\in\\{1, \\ldots, k-1\\}$ such that the endpoints of $L_{d d}$ lie on the horizontal sides of $P_{d d}$. Consider the translate $L_{d d}^{\\prime}$ of $L_{d d}$ for the vector $-d(p, q)$. The endpoints of $L_{d d}^{\\prime}$ lie on the vertical sides of $P_{00}$. Hence $L_{00}$ and $L_{d d}^{\\prime}$ have some point $X \\neq A_{0}$ in common. The translate $Y$ of point $X$ for the vector $d(p, q)$ belongs to $L$ and satisfies $X Y \\| A_{0} A_{n}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "29", "problem_type": "Combinatorics", "exam": "IMO", "problem": "29. C6 (CAN) A finite number of beans are placed on an infinite row of squares. A sequence of moves is performed as follows: At each stage a square containing more than one bean is chosen. Two beans are taken from this square; one of them is placed on the square immediately to the left, and the other is placed on the square immediately to the right of the chosen square. The sequence terminates if at some point there is at most one bean on each square. Given some initial configuration, show that any legal sequence of moves will terminate after the same number of steps and with the same final configuration.", "solution": "29. Let the squares be indexed serially by the integers: ..., $-1,0,1,2, \\ldots$. When a bean is moved from $i$ to $i+1$ or from $i+1$ to $i$ for the first time, we may assign the index $i$ to it. Thereafter, whenever some bean is moved in the opposite direction, we shall assume that it is exactly the one marked by $i$, and so on. Thus, each pair of neighboring squares has a bean stuck between it, and since the number of beans is finite, there are only finitely pairs of neighboring squares, and thus finitely many squares on which moves are made. Thus we may assume w.l.o.g. that all moves occur between 0 and $l \\in \\mathbb{N}$ and that all beans exist at all times within $[0, l]$. Defining $b_{i}$ to be the number of beans in the $i$ th cell $(i \\in \\mathbb{Z})$ and $b$ the total number of beans, we define the semi-invariant $S=\\sum_{i \\in \\mathbb{Z}} i^{2} b_{i}$. Since all moves occur above 0 , the semi-invariant $S$ increases by 2 with each move, and since we always have $S<b \\cdot l^{2}$, it follows that the number of moves must be finite. We now prove the uniqueness of the final configuration and the number of moves for some initial configuration $\\left\\{b_{i}\\right\\}$. Let $x_{i} \\geq 0$ be the number of moves made in the $i$ th cell $(i \\in \\mathbb{Z})$ during the game. Since the game is finite, only finitely many of $x_{i}$ 's are nonzero. Also, the number of beans in cell $i$, denoted as $e_{i}$, at the end is  $$ (\\forall i \\in \\mathbb{Z}) e_{i}=b_{i}+x_{i-1}+x_{i+1}-2 x_{i} \\in\\{0,1\\} $$  Thus it is enough to show that given $b_{i} \\geq 0$, the sequence $\\left\\{x_{i}\\right\\}_{i \\in \\mathbb{Z}}$ of nonnegative integers satisfying (1) is unique. Suppose the assertion is false, i.e., that there exists at least one sequence $b_{i} \\geq 0$ for which there exist distinct sequences $\\left\\{x_{i}\\right\\}$ and $\\left\\{x_{i}^{\\prime}\\right\\}$ satisfying (1). We may choose such a $\\left\\{b_{i}\\right\\}$ for which $\\min \\left\\{\\sum_{i \\in \\mathbb{Z}} x_{i}, \\sum_{i \\in \\mathbb{Z}} x_{i}^{\\prime}\\right\\}$ is minimal (since $\\sum_{i \\in \\mathbb{Z}} x_{i}$ is always finite). We choose any index $j$ such that $b_{j}>1$. Such an index $j$ exists, since otherwise the game is over. Then one must make at least one move in the $j$ th cell, which implies that $x_{j}, x_{j}^{\\prime} \\geq 1$. However, then the sequences $\\left\\{x_{i}\\right\\}$ and $\\left\\{x_{i}^{\\prime}\\right\\}$ with $x_{j}$ and $x_{j}^{\\prime}$ decreased by 1 also satisfy (1) for a sequence $\\left\\{b_{i}\\right\\}$ where $b_{j-1}, b_{j}, b_{j+1}$ is replaced with $b_{j-1}+1, b_{j}-2, b_{j+1}+1$. This contradicts the assumption of minimal $\\min \\left\\{\\sum_{i \\in \\mathbb{Z}} x_{i}, \\sum_{i \\in \\mathbb{Z}} x_{i}^{\\prime}\\right\\}$ for the initial $\\left\\{b_{i}\\right\\}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "3", "problem_type": "Algebra", "exam": "IMO", "problem": "3. A3 (GRE) Let $a>2$ be given, and define recursively  $$ a_{0}=1, \\quad a_{1}=a, \\quad a_{n+1}=\\left(\\frac{a_{n}^{2}}{a_{n-1}^{2}}-2\\right) a_{n} $$  Show that for all $k \\in \\mathbb{N}$, we have  $$ \\frac{1}{a_{0}}+\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\cdots+\\frac{1}{a_{k}}<\\frac{1}{2}\\left(2+a-\\sqrt{a^{2}-4}\\right) . $$", "solution": "3. Since $a_{1}>2$, it can be written as $a_{1}=b+b^{-1}$ for some $b>0$. Furthermore, $a_{1}^{2}-2=b^{2}+b^{-2}$ and hence $a_{2}=\\left(b^{2}+b^{-2}\\right)\\left(b+b^{-1}\\right)$. We prove that  $$ a_{n}=\\left(b+b^{-1}\\right)\\left(b^{2}+b^{-2}\\right)\\left(b^{4}+b^{-4}\\right) \\cdots\\left(b^{2^{n-1}}+b^{-2^{n-1}}\\right) $$  by induction. Indeed, $\\frac{a_{n+1}}{a_{n}}=\\left(\\frac{a_{n}}{a_{n-1}}\\right)^{2}-2=\\left(b^{2^{n-1}}+b^{-2^{n-1}}\\right)^{2}-2=$ $b^{2^{n}}+b^{-2^{n}}$. Now we have  $$ \\begin{aligned} \\sum_{i=1}^{n} \\frac{1}{a_{i}}= & 1+\\frac{b}{b^{2}+1}+\\frac{b^{3}}{\\left(b^{2}+1\\right)\\left(b^{4}+1\\right)}+\\cdots \\\\ & \\cdots+\\frac{b^{2^{n}-1}}{\\left(b^{2}+1\\right)\\left(b^{4}+1\\right) \\ldots\\left(b^{2}+1\\right)} \\end{aligned} $$  Note that $\\frac{1}{2}\\left(a+2-\\sqrt{a^{2}-4}\\right)=1+\\frac{1}{b}$; hence we must prove that the right side in (1) is less than $\\frac{1}{b}$. This follows from the fact that  $$ \\begin{aligned} & \\frac{b^{2^{k}}}{\\left(b^{2}+1\\right)\\left(b^{4}+1\\right) \\cdots\\left(b^{2^{k}}+1\\right)} \\\\ & \\quad=\\frac{1}{\\left(b^{2}+1\\right)\\left(b^{4}+1\\right) \\cdots\\left(b^{2^{k-1}}+1\\right)}-\\frac{1}{\\left(b^{2}+1\\right)\\left(b^{4}+1\\right) \\cdots\\left(b^{2^{k}}+1\\right)} \\end{aligned} $$  hence the right side in (1) equals $\\frac{1}{b}\\left(1-\\frac{1}{\\left(b^{2}+1\\right)\\left(b^{4}+1\\right) \\ldots\\left(b^{2^{n}}+1\\right)}\\right)$, and this is clearly less than $1 / b$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "30", "problem_type": "Combinatorics", "exam": "IMO", "problem": "30. C7 (IRE) Let $U$ be a finite set and let $f, g$ be bijective functions from $U$ onto itself. Let $S=\\{w \\in U: f(f(w))=g(g(w))\\}, \\quad T=\\{w \\in U: f(g(w))=g(f(w))\\}$, and suppose that $U=S \\cup T$. Prove that for $w \\in U, f(w) \\in S$ if and only if $g(w) \\in S$.", "solution": "30. For convenience, we shall write $f^{2}, f g, \\ldots$ for the functions $f \\circ f, f \\circ g, \\ldots$ We need two lemmas. Lemma 1. If $f(x) \\in S$ and $g(x) \\in T$, then $x \\in S \\cap T$. Proof. The given condition means that $f^{3}(x)=g^{2} f(x)$ and $g f g(x)=$ $f g^{2}(x)$. Since $x \\in S \\cup T=U$, we have two cases: $x \\in S$. Then $f^{2}(x)=g^{2}(x)$, which also implies $f^{3}(x)=f g^{2}(x)$. Therefore $g f g(x)=f g^{2}(x)=f^{3}(x)=g^{2} f(x)$, and since $g$ is a bijection, we obtain $f g(x)=g f(x)$, i.e., $x \\in T$. $x \\in T$. Then $f g(x)=g f(x)$, so $g^{2} f(x)=g f g(x)$. It follows that $f^{3}(x)=g^{2} f(x)=g f g(x)=f g^{2}(x)$, and since $f$ is a bijection, we obtain $x \\in S$. Hence $x \\in S \\cap T$ in both cases. Similarly, $f(x) \\in T$ and $g(x) \\in S$ again imply $x \\in S \\cap T$. Lemma 2. $f(S \\cap T)=g(S \\cap T)=S \\cap T$. Proof. By symmetry, it is enough to prove $f(S \\cap T)=S \\cap T$, or in other words that $f^{-1}(S \\cap T)=S \\cap T$. Since $S \\cap T$ is finite, this is equivalent to $f(S \\cap T) \\subseteq S \\cap T$. Let $f(x) \\in S \\cap T$. Then if $g(x) \\in S$ (since $f(x) \\in T$ ), Lemma 1 gives $x \\in S \\cap T$; similarly, if $g(x) \\in T$, then by Lemma $1, x \\in S \\cap T$. Now we return to the problem. Assume that $f(x) \\in S$. If $g(x) \\notin S$, then $g(x) \\in T$, so from Lemma 1 we deduce that $x \\in S \\cap T$. Then Lemma 2 claims that $g(x) \\in S \\cap T$ too, a contradiction. Analogously, from $g(x) \\in S$ we are led to $f(x) \\in S$. This finishes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "4", "problem_type": "Algebra", "exam": "IMO", "problem": "4. A4 (KOR) Let $a_{1}, a_{2}, \\ldots, a_{n}$ be nonnegative real numbers, not all zero. (a) Prove that $x^{n}-a_{1} x^{n-1}-\\cdots-a_{n-1} x-a_{n}=0$ has precisely one positive real root. (b) Let $A=\\sum_{j=1}^{n} a_{j}, B=\\sum_{j=1}^{n} j a_{j}$, and let $R$ be the positive real root of the equation in part (a). Prove that  $$ A^{A} \\leq R^{B} $$", "solution": "4. Consider the function  $$ f(x)=\\frac{a_{1}}{x}+\\frac{a_{2}}{x^{2}}+\\cdots+\\frac{a_{n}}{x^{n}} . $$  Since $f$ is strictly decreasing from $+\\infty$ to 0 on the interval $(0,+\\infty)$, there exists exactly one $R>0$ for which $f(R)=1$. This $R$ is also the only positive real root of the given polynomial. Since $\\ln x$ is a concave function on $(0,+\\infty)$, Jensen's inequality gives us  $$ \\sum_{j=1}^{n} \\frac{a_{j}}{A}\\left(\\ln \\frac{A}{R^{j}}\\right) \\leq \\ln \\left(\\sum_{j=1}^{n} \\frac{a_{j}}{A} \\cdot \\frac{A}{R^{j}}\\right)=\\ln f(R)=0 . $$  Therefore $\\sum_{j=1}^{n} a_{j}(\\ln A-j \\ln R) \\leq 0$, which is equivalent to $A \\ln A \\leq$ $B \\ln R$, i.e., $A^{A} \\leq R^{B}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "5", "problem_type": "Algebra", "exam": "IMO", "problem": "5. A5 (ROM) Let $P(x)$ be the real polynomial function $P(x)=a x^{3}+$ $b x^{2}+c x+d$. Prove that if $|P(x)| \\leq 1$ for all $x$ such that $|x| \\leq 1$, then  $$ |a|+|b|+|c|+|d| \\leq 7 $$", "solution": "5. Considering the polynomials $\\pm P( \\pm x)$ we may assume w.l.o.g. that $a, b \\geq$ 0 . We have four cases: (1) $c \\geq 0, d \\geq 0$. Then $|a|+|b|+|c|+|d|=a+b+c+d=P(1) \\leq 1$. (2) $c \\geq 0, d<0$. Then $|a|+|b|+|c|+|d|=a+b+c-d=P(1)-2 P(0) \\leq 3$. (3) $c<0, d \\geq 0$. Then  $$ \\begin{aligned} |a|+|b|+|c|+|d| & =a+b-c+d \\\\ & =\\frac{4}{3} P(1)-\\frac{1}{3} P(-1)-\\frac{8}{3} P(1 / 2)+\\frac{8}{3} P(-1 / 2) \\leq 7 . \\end{aligned} $$  (4) $c<0, d<0$. Then  $$ \\begin{aligned} |a|+|b|+|c|+|d| & =a+b-c-d \\\\ & =\\frac{5}{3} P(1)-4 P(1 / 2)+\\frac{4}{3} P(-1 / 2) \\leq 7 \\end{aligned} $$  Remark. It can be shown that the maximum of 7 is attained only for $P(x)= \\pm\\left(4 x^{3}-3 x\\right)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "6", "problem_type": "Algebra", "exam": "IMO", "problem": "6. A6 (IRE) Let $n$ be an even positive integer. Prove that there exists a positive integer $k$ such that  $$ k=f(x)(x+1)^{n}+g(x)\\left(x^{n}+1\\right) $$  for some polynomials $f(x), g(x)$ having integer coefficients. If $k_{0}$ denotes the least such $k$, determine $k_{0}$ as a function of $n$.  A6 ${ }^{\\prime}$ Let $n$ be an even positive integer. Prove that there exists a positive integer $k$ such that  $$ k=f(x)(x+1)^{n}+g(x)\\left(x^{n}+1\\right) $$  for some polynomials $f(x), g(x)$ having integer coefficients. If $k_{0}$ denotes the least such $k$, show that $k_{0}=2^{q}$, where $q$ is the odd integer determined by $n=q 2^{r}, r \\in \\mathbb{N}$. A6\" Prove that for each positive integer $n$, there exist polynomials $f(x), g(x)$ having integer coefficients such that  $$ f(x)(x+1)^{2^{n}}+g(x)\\left(x^{2^{n}}+1\\right)=2 . $$", "solution": "6. Let $f(x), g(x)$ be polynomials with integer coefficients such that  $$ f(x)(x+1)^{n}+g(x)\\left(x^{n}+1\\right)=k_{0} $$  Write $n=2^{r} m$ for $m$ odd and note that $x^{n}+1=\\left(x^{2^{r}}+1\\right) B(x)$, where $B(x)=x^{2^{r}(m-1)}-x^{2^{r}(m-2)}+\\cdots-x^{2^{r}}+1$. Moreover, $B(-1)=1$; hence $B(x)-1=(x+1) c(x)$ and thus  $$ R(x) B(x)+1=(B(x)-1)^{n}=(x+1)^{n} c(x)^{n} $$  for some polynomials $c(x)$ and $R(x)$. The zeros of the polynomial $x^{2^{r}}+1$ are $\\omega_{j}$, with $\\omega_{1}=\\cos \\frac{\\pi}{2^{r}}+i \\sin \\frac{\\pi}{2^{r}}$, and $\\omega_{j}=\\omega^{2 j-1}$ for $1 \\leq j \\leq 2^{r}$. We have  $$ \\left(\\omega_{1}+1\\right)\\left(\\omega_{2}+1\\right) \\cdots\\left(\\omega_{2^{r+1}}+1\\right)=2 $$  From $(*)$ we also get $f\\left(\\omega_{j}\\right)\\left(\\omega_{j}+1\\right)^{n}=k_{0}$ for $j=1,2, \\ldots, 2^{r}$. Since $A=f\\left(\\omega_{1}\\right) f\\left(\\omega_{2}\\right) \\cdots f\\left(\\omega_{2^{r}}\\right)$ is a symmetric polynomial in $\\omega_{1}, \\ldots, \\omega_{2^{r}}$ with integer coefficients, $A$ is an integer. Consequently, taking the product over $j=1,2, \\ldots, 2^{r}$ and using (2) we deduce that $2^{n} A=k_{0}^{2^{r}}$ is divisible by $2^{n}=2^{2^{r} m}$. Hence $2^{m} \\mid k_{0}$. Furthermore, since $\\omega_{j}+1=\\left(\\omega_{1}+1\\right) p_{j}\\left(\\omega_{1}\\right)$ for some polynomial $p_{j}$ with integer coefficients, (2) gives $\\left(\\omega_{1}+1\\right)^{2^{r}} p\\left(\\omega_{1}\\right)=2$, where $p(x)=$ $p_{2}(x) \\cdots p_{2^{r}}(x)$ has integer coefficients. But then the polynomial $(x+$ $1)^{2^{2}} p(x)-2$ has a zero $x=\\omega_{1}$, so it is divisible by its minimal polynomial $x^{2^{r}}+1$. Therefore  $$ (x+1)^{2^{r}} p(x)=2+\\left(x^{2^{r}}+1\\right) q(x) $$  for some polynomial $q(x)$. Raising (3) to the $m$ th power we get $(x+$ $1)^{n} p(x)^{n}=2^{m}+\\left(x^{2^{r}}+1\\right) Q(x)$ for some polynomial $Q(x)$ with integer coefficients. Now using (1) we obtain  $$ \\begin{aligned} (x+1)^{n} c(x)^{n}\\left(x^{2^{r}}+1\\right) Q(x) & =\\left(x^{2^{r}}+1\\right) Q(x)+\\left(x^{2^{r}}+1\\right) Q(x) B(x) R(x) \\\\ & =(x+1)^{n} p(x)^{n}-2^{m}+\\left(x^{n}+1\\right) Q(X) R(x) . \\end{aligned} $$  Therefore $(x+1)^{n} f(x)+\\left(x^{n}+1\\right) g(x)=2^{m}$ for some polynomials $f(x), g(x)$ with integer coefficients, and $k_{0}=2^{m}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "7", "problem_type": "Algebra", "exam": "IMO", "problem": "7. A7 (ARM) Let $f$ be a function from the set of real numbers $\\mathbb{R}$ into itself such that for all $x \\in \\mathbb{R}$, we have $|f(x)| \\leq 1$ and  $$ f\\left(x+\\frac{13}{42}\\right)+f(x)=f\\left(x+\\frac{1}{6}\\right)+f\\left(x+\\frac{1}{7}\\right) . $$  Prove that $f$ is a periodic function (that is, there exists a nonzero real number $c$ such that $f(x+c)=f(x)$ for all $x \\in \\mathbb{R})$.", "solution": "7. We are given that $f(x+a+b)-f(x+a)=f(x+b)-f(x)$, where $a=1 / 6$ and $b=1 / 7$. Summing up these equations for $x, x+b, \\ldots, x+6 b$ we obtain $f(x+a+1)-f(x+a)=f(x+1)-f(x)$. Summing up the new equations for $x, x+a, \\ldots, x+5 a$ we obtain that  $$ f(x+2)-f(x+1)=f(x+1)-f(x) . $$  It follows by induction that $f(x+n)-f(x)=n[f(x+1)-f(x)]$. If $f(x+1) \\neq f(x)$, then $f(x+n)-f(x)$ will exceed in absolute value an arbitrarily large number for a sufficiently large $n$, contradicting the assumption that $f$ is bounded. Hence $f(x+1)=f(x)$ for all $x$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "8", "problem_type": "Algebra", "exam": "IMO", "problem": "8. A8 (ROM) ${ }^{\\mathrm{IMO} 3}$ Let $\\mathbb{N}_{0}$ denote the set of nonnegative integers. Find all functions $f$ from $\\mathbb{N}_{0}$ into itself such that  $$ f(m+f(n))=f(f(m))+f(n), \\quad \\forall m, n \\in \\mathbb{N}_{0} $$", "solution": "8. Putting $m=n=0$ we obtain $f(0)=0$ and consequently $f(f(n))=f(n)$ for all $n$. Thus the given functional equation is equivalent to  $$ f(m+f(n))=f(m)+f(n), \\quad f(0)=0 $$  Clearly one solution is $(\\forall x) f(x)=0$. Suppose $f$ is not the zero function. We observe that $f$ has nonzero fixed points (for example, any $f(n)$ is a fixed point). Let $a$ be the smallest nonzero fixed point of $f$. By induction, each $k a(k \\in \\mathbb{N})$ is a fixed point too. We claim that all fixed points of $f$ are of this form. Indeed, suppose that $b=k a+i$ is a fixed point, where $i<a$. Then  $$ b=f(b)=f(k a+i)=f(i+f(k a))=f(i)+f(k a)=f(i)+k a $$  hence $f(i)=i$. Hence $i=0$. Since the set of values of $f$ is a set of its fixed points, it follows that for $i=0,1, \\ldots, a-1, f(i)=a n_{i}$ for some integers $n_{i} \\geq 0$ with $n_{0}=0$. Let $n=k a+i$ be any positive integer, $0 \\leq i<a$. As before, the functional equation gives us  $$ f(n)=f(k a+i)=f(i)+k a=\\left(n_{i}+k\\right) a $$  Besides the zero function, this is the general solution of the given functional equation. To verify this, we plug in $m=k a+i, n=l a+j$ and obtain  $$ \\begin{aligned} f(m+f(n)) & =f(k a+i+f(l a+j))=f\\left(\\left(k+l+n_{j}\\right) a+i\\right) \\\\ & =\\left(k+l+n_{j}+n_{i}\\right) a=f(m)+f(n) . \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1996", "tier": "T0", "problem_label": "9", "problem_type": "Algebra", "exam": "IMO", "problem": "9. A9 (POL) Let the sequence $a(n), n=1,2,3, \\ldots$, be generated as follows: $a(1)=0$, and for $n>1$,  $$ a(n)=a([n / 2])+(-1)^{\\frac{n(n+1)}{2}} . \\quad(\\text { Here }[t]=\\text { the greatest integer } \\leq t .) $$  (a) Determine the maximum and minimum value of $a(n)$ over $n \\leq 1996$ and find all $n \\leq 1996$ for which these extreme values are attained. (b) How many terms $a(n), n \\leq 1996$, are equal to 0 ?", "solution": "9. From the definition of $a(n)$ we obtain  $$ a(n)-a([n / 2])=\\left\\{\\begin{array}{r} 1 \\text { if } n \\equiv 0 \\text { or } n \\equiv 3(\\bmod 4) \\\\ -1 \\text { if } n \\equiv 1 \\text { or } n \\equiv 2(\\bmod 4) . \\end{array}\\right. $$  Let $n=\\overline{b_{k} b_{k-1} \\ldots b_{1} b_{0}}$ be the binary representation of $n$, where we assume $b_{k}=1$. If we define $p(n)$ and $q(n)$ to be the number of indices $i=0,1, \\ldots, k-1$ with $b_{i}=b_{i+1}$ and the number of $i=0,1, \\ldots, k-1$ with $b_{i} \\neq b_{i+1}$ respectively, we get  $$ a(n)=p(n)-q(n) $$  (a) The maximum value of $a(n)$ for $n \\leq 1996$ is 9 when $p(n)=9$ and $q(n)=0$, i.e., in the case $n=\\overline{1111111111}_{2}=1023$. The minimum value is -10 and is attained when $p(n)=0$ and $q(n)=$ 10, i.e., only for $n=\\overline{10101010101}_{2}=1365$. (b) From (1) we have that $a(n)=0$ is equivalent to $p(n)=q(n)=k / 2$. Hence $k$ must be even, and the $k / 2$ indices $i$ for which $b_{i}=b_{i+1}$ can be chosen in exactly $\\binom{k}{k / 2}$ ways. Thus the number of positive integers $n<2^{11}=2048$ with $a(n)=0$ is equal to  $$ \\binom{0}{0}+\\binom{2}{1}+\\binom{4}{2}+\\binom{6}{3}+\\binom{8}{4}+\\binom{10}{5}=351 $$  But five of these numbers exceed 1996: these are $2002=\\overline{11111010010}_{2}$, $2004=\\overline{11111010100}_{2}, 2006=\\overline{11111010110}_{2}, 2010=\\overline{11111011010}_{2}$, $2026=\\overline{11111101010}_{2}$. Therefore there are 346 numbers $n \\leq 1996$ for which $a(n)=0$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (BLR) $)^{\\mathrm{IMO}} \\mathrm{An}$ infinite square grid is colored in the chessboard pattern. For any pair of positive integers $m, n$ consider a right-angled triangle whose vertices are grid points and whose legs, of lengths $m$ and $n$, run along the lines of the grid. Let $S_{b}$ be the total area of the black part of the triangle and $S_{w}$ the total area of its white part. Define the function $f(m, n)=\\left|S_{b}-S_{w}\\right|$. (a) Calculate $f(m, n)$ for all $m, n$ that have the same parity. (b) Prove that $f(m, n) \\leq \\frac{1}{2} \\max (m, n)$. (c) Show that $f(m, n)$ is not bounded from above.", "solution": "1. Let $A B C$ be the given triangle, with $\\angle B=90^{\\circ}$ and $A B=m, B C=n$. For an arbitrary polygon $\\mathcal{P}$ we denote by $w(\\mathcal{P})$ and $b(\\mathcal{P})$ respectively the total areas of the white and black parts of $\\mathcal{P}$. (a) Let $D$ be the fourth vertex of the rectangle $A B C D$. When $m$ and $n$ are of the same parity, the coloring of the rectangle $A B C D$ is centrally symmetric with respect to the midpoint of $A C$. It follows that $w(A B C)=\\frac{1}{2} w(A B C D)$ and $b(A B C)=\\frac{1}{2} b(A B C D)$; thus $f(m, n)=\\frac{1}{2}|w(A B C D)-b(A B C D)|$. Hence $f(m, n)$ equals $\\frac{1}{2}$ if $m$ and $n$ are both odd, and 0 otherwise. (b) The result when $m, n$ are of the same parity follows from (a). Suppose that $m>n$, where $m$ and $n$ are of different parity. Choose a point $E$ on $A B$ such that $A E=1$. Since by (a) $|w(E B C)-b(E B C)|=$ $f(m-1, n) \\leq \\frac{1}{2}$, we have $f(m, n) \\leq \\frac{1}{2}+|w(E A C)-b(E A C)| \\leq$ $\\frac{1}{2}+S(E A C)=\\frac{1}{2}+\\frac{n-1}{2}=\\frac{n}{2}$. Therefore $f(m, n) \\leq \\frac{1}{2} \\min (m, n)$. (c) Let us calculate $f(m, n)$ for $m=2 k+1, n=2 k, k \\in \\mathbb{N}$. With $E$ defined as in (b), we have $B E=B C=2 k$. If the square at $B$ is w.l.o.g. white, $C E$ passes only through black squares. The white part of $\\triangle E A C$ then consists of $2 k$ similar triangles with areas $\\frac{1}{2} \\frac{i}{2 k} \\frac{i}{2 k+1}=\\frac{i^{2}}{4 k(2 k+1)}$, where $i=1,2, \\ldots, 2 k$. The total white area of $E A C$ is  $$ \\frac{1}{4 k(2 k+1)}\\left(1^{2}+2^{2}+\\cdots+(2 k)^{2}\\right)=\\frac{4 k+1}{12} . $$  Therefore the black area is $(8 k-1) / 12$, and $f(2 k+1,2 k)=(2 k-1) / 6$, which is not bounded.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. ( $\\mathbf{C Z E}$ ) Find all positive integers $k$ for which the following statement is true: If $F(x)$ is a polynomial with integer coefficients satisfying the condition  $$ 0 \\leq F(c) \\leq k \\quad \\text { for each } c \\in\\{0,1, \\ldots, k+1\\} $$  then $F(0)=F(1)=\\cdots=F(k+1)$.", "solution": "10. Suppose that $k \\geq 4$. Consider any polynomial $F(x)$ with integer coefficients such that $0 \\leq F(x) \\leq k$ for $x=0,1, \\ldots, k+1$. Since $F(k+1)-F(0)$ is divisible by $k+1$, we must have $F(k+1)=F(0)$. Hence  $$ F(x)-F(0)=x(x-k-1) Q(x) $$  for some polynomial $Q(x)$ with integer coefficients. In particular, $F(x)-$ $F(0)$ is divisible by $x(k+1-x)>k+1$ for every $x=2,3, \\ldots, k-1$, so $F(x)=F(0)$ must hold for any $x=2,3, \\ldots, k-1$. It follows that  $$ F(x)-F(0)=x(x-2)(x-3) \\cdots(x-k+1)(x-k-1) R(x) $$  for some polynomial $R(x)$ with integer coefficients. Thus $k \\geq \\mid F(1)-$ $F(0)|=k(k-2)!| R(1) \\mid$, although $k(k-2)!>k$ for $k \\geq 4$. In this case we have $F(1)=F(0)$ and similarly $F(k)=F(0)$. Hence, the statement is true for $k \\geq 4$. It is easy to find counterexamples for $k \\leq 3$. These are, for example,  $$ F(x)= \\begin{cases}x(2-x) & \\text { for } k=1 \\\\ x(3-x) & \\text { for } k=2 \\\\ x(2-x)^{2}(4-x) & \\text { for } k=3\\end{cases} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (NET) Let $P(x)$ be a polynomial with real coefficients such that $P(x)>$ 0 for all $x \\geq 0$. Prove that there exists a positive integer $n$ such that $(1+x)^{n} P(x)$ is a polynomial with nonnegative coefficients.", "solution": "11. All real roots of $P(x)$ (if any) are negative: say $-a_{1},-a_{2}, \\ldots,-a_{k}$. Then $P(x)$ can be factored as  $$ P(x)=C\\left(x+a_{1}\\right) \\cdots\\left(x+a_{k}\\right)\\left(x^{2}-b_{1} x+c_{1}\\right) \\cdots\\left(x^{2}-b_{m} x+c_{m}\\right) $$  where $x^{2}-b_{i} x+c_{i}$ are quadratic polynomials without real roots. Since the product of polynomials with positive coefficients is again a polynomial with positive coefficients, it will be sufficient to prove the result for each of the factors in (1). The case of $x+a_{j}$ is trivial. It remains only to prove the claim for every polynomial $x^{2}-b x+c$ with $b^{2}<4 c$. From the binomial formula, we have for any $n \\in \\mathbb{N}$,  $$ (1+x)^{n}\\left(x^{2}-b x+c\\right)=\\sum_{i=0}^{n+2}\\left[\\binom{n}{i-2}-b\\binom{n}{i-1}+c\\binom{n}{i}\\right] x^{i}=\\sum_{i=0}^{n+2} C_{i} x^{i} $$  where  $$ C_{i}=\\frac{n!\\left((b+c+1) i^{2}-((b+2 c) n+(2 b+3 c+1)) i+c\\left(n^{2}+3 n+2\\right)\\right) x^{i}}{i!(n-i+2)!} $$  The coefficients $C_{i}$ of $x^{i}$ appear in the form of a quadratic polynomial in $i$ depending on $n$. We claim that for large enough $n$ this polynomial has negative discriminant, and is thus positive for every $i$. Indeed, this discriminant equals $D=((b+2 c) n+(2 b+3 c+1))^{2}-4(b+c+1) c\\left(n^{2}+\\right.$ $3 n+2)=\\left(b^{2}-4 c\\right) n^{2}-2 U n+V$, where $U=2 b^{2}+b c+b-4 c$ and $V=(2 b+c+1)^{2}-4 c$, and since $b^{2}-4 c<0$, for large $n$ it clearly holds that $D<0$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (ITA) Let $p$ be a prime number and let $f(x)$ be a polynomial of degree $d$ with integer coefficients such that: (i) $f(0)=0, f(1)=1$; (ii) for every positive integer $n$, the remainder of the division of $f(n)$ by $p$ is either 0 or 1. Prove that $d \\geq p-1$.", "solution": "12. Lemma. For any polynomial $P$ of degree at most $n$, the following equality holds:  $$ \\sum_{i=0}^{n+1}(-1)^{i}\\binom{n+1}{i} P(i)=0 $$  Proof. See (SL81-13). Suppose to the contrary that the degree of $f$ is at most $p-2$. Then it follows from the lemma that  $$ 0=\\sum_{i=0}^{p-1}(-1)^{i}\\binom{p-1}{i} f(i) \\equiv \\sum_{i=0}^{p-1} f(i)(\\bmod p) $$  since $\\binom{p-1}{i}=\\frac{(p-1)(p-2) \\cdots(p-i)}{i!} \\equiv(-1)^{i}(\\bmod p)$. But this is clearly impossible if $f(i)$ equals 0 or 1 modulo $p$ and $f(0)=0, f(1)=1$. Remark. In proving the essential relation $\\sum_{i=0}^{p-1} f(i) \\equiv 0(\\bmod p)$, it is clearly enough to show that $S_{k}=1^{k}+2^{k}+\\cdots+(p-1)^{k}$ is divisible by $p$ for every $k \\leq p-2$. This can be shown in two other ways. (1) By induction. Assume that $S_{0} \\equiv \\cdots \\equiv S_{k-1}(\\bmod p)$. By the binomial formula we have  $$ 0 \\equiv \\sum_{n=0}^{p-1}\\left[(n+1)^{k+1}-n^{k+1}\\right] \\equiv(k+1) S_{k}+\\sum_{i=0}^{k-1}\\binom{k+1}{i} S_{i}(\\bmod p), $$  and the inductive step follows. (2) Using the primitive root $g$ modulo $p$. Then  $$ S_{k} \\equiv 1+g^{k}+\\cdots+g^{k(p-2)}=\\frac{g^{k(p-1)}-1}{g^{k}-1} \\equiv 0(\\bmod p) . $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (IND) In town $A$, there are $n$ girls and $n$ boys, and each girl knows each boy. In town $B$, there are $n$ girls $g_{1}, g_{2}, \\ldots, g_{n}$ and $2 n-1$ boys $b_{1}, b_{2}, \\ldots$, $b_{2 n-1}$. The girl $g_{i}, i=1,2, \\ldots, n$, knows the boys $b_{1}, b_{2}, \\ldots, b_{2 i-1}$, and no others. For all $r=1,2, \\ldots, n$, denote by $A(r), B(r)$ the number of different ways in which $r$ girls from town $A$, respectively town $B$, can dance with $r$ boys from their own town, forming $r$ pairs, each girl with a boy she knows. Prove that $A(r)=B(r)$ for each $r=1,2, \\ldots, n$.", "solution": "13. Denote $A(r)$ and $B(r)$ by $A(n, r)$ and $B(n, r)$ respectively. The numbers $A(n, r)$ can be found directly: one can choose $r$ girls and $r$ boys in $\\binom{n}{r}^{2}$ ways, and pair them in $r$ ! ways. Hence  $$ A(n, r)=\\binom{n}{r}^{2} \\cdot r!=\\frac{n!^{2}}{(n-r)!^{2} r!} $$  Now we establish a recurrence relation between the $B(n, r)$ 's. Let $n \\geq 2$ and $2 \\leq r \\leq n$. There are two cases for a desired selection of $r$ pairs of girls and boys: (i) One of the girls dancing is $g_{n}$. Then the other $r-1$ girls can choose their partners in $B(n-1, r-1)$ ways and $g_{n}$ can choose any of the remaining $2 n-r$ boys. Thus, the total number of choices in this case is $(2 n-r) B(n-1, r-1)$. (ii) $g_{n}$ is not dancing. Then there are exactly $B(n-1, r)$ possible choices. Therefore, for every $n \\geq 2$ it holds that  $$ B(n, r)=(2 n-r) B(n-1, r-1)+B(n-1, r) \\quad \\text { for } r=2, \\ldots, n $$  Here we assume that $B(n, r)=0$ for $r>n$, while $B(n, 1)=1+3+\\cdots+$ $(2 n-1)=n^{2}$ 。 It is directly verified that the numbers $A(n, r)$ satisfy the same initial conditions and recurrence relations, from which it follows that $A(n, r)=$ $B(n, r)$ for all $n$ and $r \\leq n$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. (IND) Let $b, m, n$ be positive integers such that $b>1$ and $m \\neq n$. Prove that if $b^{m}-1$ and $b^{n}-1$ have the same prime divisors, then $b+1$ is a power of 2 .", "solution": "14. We use the following nonstandard notation: ( $1^{\\circ}$ ) for $x, y \\in \\mathbb{N}, x \\sim y$ means that $x$ and $y$ have the same prime divisors; $\\left(2^{\\circ}\\right)$ for a prime $p$ and integers $r \\geq 0$ and $x>0, p^{r} \\| x$ means that $x$ is divisible by $p^{r}$, but not by $p^{r+1}$. First, $b^{m}-1 \\sim b^{n}-1$ is obviously equivalent to $b^{m}-1 \\sim \\operatorname{gcd}\\left(b^{m}-1, b^{n}-\\right.$ $1)=b^{d}-1$, where $d=\\operatorname{gcd}(m, n)$. Setting $b^{d}=a$ and $m=k d$, we reduce the condition of the problem to $a^{k}-1 \\sim a-1$. We are going to show that this implies that $a+1$ is a power of 2 . This will imply that $d$ is odd (for even $d$, $a+1=b^{d}+1$ cannot be divisible by 4 , and consequently $b+1$, as a divisor of $a+1$, is also a power of 2 . But before that, we need the following important lemma (Theorem 2.126). Lemma. Let $a, k$ be positive integers and $p$ an odd prime. If $\\alpha \\geq 1$ and $\\beta \\geq 0$ are such that $p^{\\alpha} \\| a-1$ and $p^{\\beta} \\| k$, then $p^{\\alpha+\\beta} \\| a^{k}-1$. Proof. We use induction on $\\beta$. If $\\beta=0$, then $\\frac{a^{k}-1}{a-1}=a^{k-1}+\\cdots+a+1 \\equiv k$ $(\\bmod p)($ because $a \\equiv 1$ ), and it is not divisible by $p$. Suppose that the lemma is true for some $\\beta \\geq 0$, and let $k=p^{\\beta+1} t$ where $p \\nmid t$. By the induction hypothesis, $a^{k / p}=a^{p^{\\beta} t}=m p^{\\alpha+\\beta}+1$ for some $m$ not divisible by $p$. Furthermore,  $$ a^{k}-1=\\left(m p^{\\alpha+\\beta}+1\\right)^{p}-1=\\left(m p^{\\alpha+\\beta}\\right)^{p}+\\cdots+\\binom{p}{2}\\left(m p^{\\alpha+\\beta}\\right)^{2}+m p^{\\alpha+\\beta+1} $$  Since $p \\left\\lvert\\,\\binom{ p}{2}=\\frac{p(p-1)}{2}\\right.$, all summands except for the last one are divisible by $p^{\\alpha+\\beta+2}$. Hence $p^{\\alpha+\\beta+1} \\| a^{k}-1$, completing the induction. Now let $a^{k}-1 \\sim a-1$ for some $a, k>1$. Suppose that $p$ is an odd prime divisor of $k$, with $p^{\\beta} \\| k$. Then putting $X=a^{p^{\\beta}-1}+\\cdots+a+1$ we also have $(a-1) X=a^{p^{\\beta}}-1 \\sim a-1$; hence each prime divisor $q$ of $X$ must also divide $a-1$. But then $a^{i} \\equiv 1(\\bmod q)$ for each $i \\in \\mathbb{N}_{0}$, which gives us $X \\equiv p^{\\beta}(\\bmod q)$. Therefore $q \\mid p^{\\beta}$, i.e., $q=p$; hence $X$ is a power of $p$. On the other hand, since $p \\mid a-1$, we put $p^{\\alpha} \\| a-1$. From the lemma we obtain $p^{\\alpha+\\beta} \\| a^{p^{\\beta}}-1$, and deduce that $p^{\\beta} \\| X$. But $X$ has no prime divisors other than $p$, so we must have $X=p^{\\beta}$. This is clearly impossible, because $X>p^{\\beta}$ for $a>1$. Thus our assumption that $k$ has an odd prime divisor leads to a contradiction: in other words, $k$ must be a power of 2 . Now $a^{k}-1 \\sim a-1$ implies $a-1 \\sim a^{2}-1=(a-1)(a+1)$, and thus every prime divisor $q$ of $a+1$ must also divide $a-1$. Consequently $q=2$, so it follows that $a+1$ is a power of 2 . As we explained above, this gives that $b+1$ is also a power of 2 . Remark. In fact, one can continue and show that $k$ must be equal to 2 . It is not possible for $a^{4}-1 \\sim a^{2}-1$ to hold. Similarly, we must have $d=1$. Therefore all possible triples $(b, m, n)$ with $m>n$ are $\\left(2^{s}-1,2,1\\right)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (RUS) An infinite arithmetic progression whose terms are positive integers contains the square of an integer and the cube of an integer. Show that it contains the sixth power of an integer.", "solution": "15. Let $a+b t, t=0,1,2, \\ldots$, be a given arithmetic progression that contains a square and a cube $(a, b>0)$. We use induction on the progression step $b$ to prove that the progression contains a sixth power. (i) $b=1$ : this case is trivial. (ii) $b=p^{m}$ for some prime $p$ and $m>0$. The case $p^{m} \\mid a$ trivially reduces to the previous case, so let us have $p^{m} \\nmid a$. Suppose that $\\operatorname{gcd}(a, p)=1$. If $x, y$ are integers such that $x^{2} \\equiv y^{3} \\equiv a$ (here all the congruences will be $\\bmod p^{m}$ ), then $x^{6} \\equiv a^{3}$ and $y^{6} \\equiv a^{2}$. Consider an integer $y_{1}$ such that $y y_{1} \\equiv 1$. It satisfies $a^{2}\\left(x y_{1}\\right)^{6} \\equiv$ $x^{6} y^{6} y_{1}^{6} \\equiv x^{6} \\equiv a^{3}$, and consequently $\\left(x y_{1}\\right)^{6} \\equiv a$. Hence a sixth power exists in the progression. If $\\operatorname{gcd}(a, p)>1$, we can write $a=p^{k} c$, where $k<m$ and $p \\nmid c$. Since the arithmetic progression $x_{t}=a+b t=p^{k}\\left(c+p^{m-k} t\\right)$ contains a square, $k$ must be even; similarly, it contains a cube, so $3 \\mid k$. It follows that $6 \\mid k$. The progression $c+p^{m-k} t$ thus also contains a square and a cube; hence by the previous case it contains a sixth power and thus $x_{t}$ does also. (iii) $b$ is not a power of a prime, and thus can be expressed as $b=b_{1} b_{2}$, where $b_{1}, b_{2}>1$ and $\\operatorname{gcd}\\left(b_{1}, b_{2}\\right)=1$. It is given that progressions $a+b_{1} t$ and $a+b_{2} t$ both contain a square and a cube, and therefore by the inductive hypothesis they both contain sixth powers: say $z_{1}^{6}$ and $z_{2}^{6}$, respectively. By the Chinese remainder theorem, there exists $z \\in \\mathbb{N}$ such that $z \\equiv z_{1}\\left(\\bmod b_{1}\\right)$ and $z \\equiv z_{2}\\left(\\bmod b_{2}\\right)$. But then $z^{6}$ belongs to both of the progressions $a+b_{1} t$ and $a+b_{2} t$. Hence $z^{6}$ is a member of the progression $a+b t$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (BLR) In an acute-angled triangle $A B C$, let $A D, B E$ be altitudes and $A P, B Q$ internal bisectors. Denote by $I$ and $O$ the incenter and the circumcenter of the triangle, respectively. Prove that the points $D, E$, and $I$ are collinear if and only if the points $P, Q$, and $O$ are collinear.", "solution": "16. Let $d_{a}(X), d_{b}(X), d_{c}(X)$ denote the distances of a point $X$ interior to $\\triangle A B C$ from the lines $B C, C A, A B$ respectively. We claim that $X \\in P Q$ if and only if $d_{a}(X)+d_{b}(X)=d_{c}(X)$. Indeed, if $X \\in P Q$ and $P X=$ $k P Q$ then $d_{a}(X)=k d_{a}(Q), d_{b}(X)=(1-k) d_{b}(P)$, and $d_{c}(X)=(1-$ $k) d_{c}(P)+k d_{c}(Q)$, and simple substitution yields $d_{a}(X)+d_{b}(X)=d_{c}(X)$. The converse follows easily. In particular, $O \\in P Q$ if and only if $d_{a}(O)+$ $d_{b}(O)=d_{c}(O)$, i.e., $\\cos \\alpha+\\cos \\beta=\\cos \\gamma$. We shall now show that $I \\in D E$ if and only if $A E+B D=D E$. Let $K$ be the point on the segment $D E$ such that $A E=E K$. Then $\\angle E K A=$ $\\frac{1}{2} \\angle D E C=\\frac{1}{2} \\angle C B A=\\angle I B A$; hence the points $A, B, I, K$ are concyclic. The point $I$ lies on $D E$ if and only if $\\angle B K D=\\angle B A I=\\frac{1}{2} \\angle B A C=$ $\\frac{1}{2} \\angle C D E=\\angle D B K$, which is equivalent to $K D=B D$, i.e., to $A E+B D=$ $D E$. But since $A E=A B \\cos \\alpha, B D=A B \\cos \\beta$, and $D E=A B \\cos \\gamma$, we have that $I \\in D E \\Leftrightarrow \\cos \\alpha+\\cos \\beta=\\cos \\gamma$. The conditions for $O \\in P Q$ and $I \\in D E$ are thus equivalent.  Second solution. We know that three points $X, Y, Z$ are collinear if and only if for some $\\lambda, \\mu \\in \\mathbb{R}$ with sum 1 , we have $\\lambda \\overrightarrow{C X}+\\mu \\overrightarrow{C Y}=\\overrightarrow{C Z}$. Specially, if $\\overrightarrow{C X}=p \\overrightarrow{C A}$ and $\\overrightarrow{C Y}=q \\overrightarrow{C B}$ for some $p, q$, and $\\overrightarrow{C Z}=k \\overrightarrow{C A}+$ $l \\overrightarrow{C B}$, then $Z$ lies on $X Y$ if and only if $k q+l p=p q$. Using known relations in a triangle we directly obtain  $$ \\begin{array}{rlrl} \\overrightarrow{C P} & =\\frac{\\sin \\beta}{\\sin \\beta+\\sin \\gamma} \\overrightarrow{C B}, & \\overrightarrow{C Q}=\\frac{\\sin \\alpha}{\\sin \\alpha+\\sin \\gamma} \\overrightarrow{C A} \\\\ \\overrightarrow{C O} & =\\frac{\\sin 2 \\alpha \\cdot \\overrightarrow{C A}+\\sin 2 \\beta \\cdot \\overrightarrow{C B}}{\\sin 2 \\alpha+\\sin 2 \\beta+\\sin 2 \\gamma} ; & \\overrightarrow{C D}=\\frac{\\tan \\beta}{\\tan \\beta+\\tan \\gamma} \\overrightarrow{C B} \\\\ \\overrightarrow{C E} & =\\frac{\\tan \\beta}{\\tan \\beta+\\tan \\gamma} \\overrightarrow{C A}, & \\overrightarrow{C I} & =\\frac{\\sin \\alpha \\cdot \\overrightarrow{C A}+\\sin \\beta \\cdot \\overrightarrow{C B}}{\\sin \\alpha+\\sin \\beta+\\sin \\gamma} \\end{array} $$  Now by the above considerations we get that the conditions (1) $P, Q, O$ are collinear and (2) $D, E, I$ are collinear are both equivalent to $\\cos \\alpha+\\cos \\beta=$ $\\cos \\gamma$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. $(\\mathbf{C Z E})^{\\mathrm{IMO} 5}$ Find all pairs of integers $x, y \\geq 1$ satisfying the equation $x^{y^{2}}=y^{x}$.", "solution": "17. We note first that $x$ and $y$ must be powers of the same positive integer. Indeed, if $x=p_{1}^{\\alpha_{1}} \\cdots p_{k}^{\\alpha_{k}}$ and $y=p_{1}^{\\beta_{1}} \\cdots p_{k}^{\\beta_{k}}$ (some of $\\alpha_{i}$ and $\\beta_{i}$ may be 0 , but not both for the same index $i$ ), then $x^{y^{2}}=y^{x}$ implies $\\frac{\\alpha_{i}}{\\beta_{i}}=\\frac{x}{y^{2}}=\\frac{p}{q}$ for some $p, q>0$ with $\\operatorname{gcd}(p, q)=1$, so for $a=p_{1}^{\\alpha_{1} / p} \\cdots p_{k}^{\\alpha_{k} / p}$ we can take $x=a^{p}$ and $y=a^{q}$. If $a=1$, then $(x, y)=(1,1)$ is the trivial solution. Let $a>1$. The given equation becomes $a^{p a^{2 q}}=a^{q a^{p}}$, which reduces to $p a^{2 q}=q a^{p}$. Hence $p \\neq q$, so we distinguish two cases: (i) $p>q$. Then from $a^{2 q}<a^{p}$ we deduce $p>2 q$. We can rewrite the equation as $p=a^{p-2 q} q$, and putting $p=2 q+d, d>0$, we obtain $d=q\\left(a^{d}-2\\right)$. By induction, $2^{d}-2>d$ for each $d>2$, so we must have $d \\leq 2$. For $d=1$ we get $q=1$ and $a=p=3$, and therefore $(x, y)=(27,3)$, which is indeed a solution. For $d=2$ we get $q=1$, $a=2$, and $p=4$, so $(x, y)=(16,2)$, which is another solution. (ii) $p<q$. As above, we get $q / p=a^{2 q-p}$, and setting $d=2 q-p>0$, this is transformed to $a^{d}=a^{\\left(2 a^{d}-1\\right) p}$, or equivalently to $d=\\left(2 a^{d}-1\\right) p$. However, this equality cannot hold, because $2 a^{d}-1>d$ for each $a \\geq 2$, $d \\geq 1$. The only solutions are thus $(1,1),(16,2)$, and $(27,3)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (GBR) The altitudes through the vertices $A, B, C$ of an acute-angled triangle $A B C$ meet the opposite sides at $D, E, F$, respectively. The line through $D$ parallel to $E F$ meets the lines $A C$ and $A B$ at $Q$ and $R$, respectively. The line $E F$ meets $B C$ at $P$. Prove that the circumcircle of the triangle $P Q R$ passes through the midpoint of $B C$.", "solution": "18. By symmetry, assume that $A B>A C$. The point $D$ lies between $M$ and $P$ as well as between $Q$ and $R$, and if we show that $D M \\cdot D P=D Q \\cdot D R$, it will imply that $M, P, Q, R$ lie on a circle. Since the triangles $A B C, A E F, A Q R$ are similar, the points $B, C, Q, R$ lie on a circle. Hence $D B \\cdot D C=D Q \\cdot D R$, and it remains to prove that  $$ D B \\cdot D C=D M \\cdot D P $$  However, the points $B, C, E, F$ are concyclic, but so are the points $E, F, D, M$ (they lie on the nine-point circle), and we obtain $P B \\cdot P C=$ $P E \\cdot P F=P D \\cdot P M$. Set $P B=x$ and $P C=y$. We have $P M=\\frac{x+y}{2}$ and hence $P D=\\frac{2 x y}{x+y}$. It follows that $D B=P B-P D=\\frac{x(x-y)}{x+y}$, $D C=\\frac{y(x-y)}{x+y}$, and $D M=\\frac{(x-y)^{2}}{2(x+y)}$, from which we immediately obtain $D B \\cdot D C=D M \\cdot D P=\\frac{x y(x-y)^{2}}{(x+y)^{2}}$, as needed.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (IRE) Let $a_{1} \\geq \\cdots \\geq a_{n} \\geq a_{n+1}=0$ be a sequence of real numbers. Prove that  $$ \\sqrt{\\sum_{k=1}^{n} a_{k}} \\leq \\sum_{k=1}^{n} \\sqrt{k}\\left(\\sqrt{a_{k}}-\\sqrt{a_{k+1}}\\right) $$", "solution": "19. Using that $a_{n+1}=0$ we can transform the desired inequality into  $$ \\begin{aligned} & \\sqrt{a_{1}+}+a_{2}+\\cdots+a_{n+1} \\\\ & \\quad \\leq \\sqrt{1} \\sqrt{a_{1}}+(\\sqrt{2}-\\sqrt{1}) \\sqrt{a_{2}}+\\cdots+(\\sqrt{n+1}-\\sqrt{n}) \\sqrt{a_{n+1}} \\end{aligned} $$  We shall prove by induction on $n$ that (1) holds for any $a_{1} \\geq a_{2} \\geq \\cdots \\geq$ $a_{n+1} \\geq 0$, i.e., not only when $a_{n+1}=0$. For $n=0$ the inequality is obvious. For the inductive step from $n-1$ to $n$, where $n \\geq 1$, we need to prove the inequality  $$ \\sqrt{a_{1}+\\cdots+a_{n+1}}-\\sqrt{a_{1}+\\cdots+a_{n}} \\leq(\\sqrt{n+1}-\\sqrt{n}) \\sqrt{a_{n+1}} $$  Putting $S=a_{1}+a_{2}+\\cdots+a_{n}$, this simplifies to $\\sqrt{S+a_{n+1}}-\\sqrt{S} \\leq$ $\\sqrt{n a_{n+1}+a_{n+1}}-\\sqrt{n a_{n+1}}$. For $a_{n+1}=0$ the inequality is obvious. For $a_{n+1}>0$ we have that the function $f(x)=\\sqrt{x+a_{n+1}}-\\sqrt{x}=$ $\\frac{a_{n+1}}{\\sqrt{x+a_{n+1}}+\\sqrt{x}}$ is strictly decreasing on $\\mathbb{R}^{+}$; hence (2) will follow if we show that $S \\geq n a_{n+1}$. However, this last is true because $a_{1}, \\ldots, a_{n} \\geq a_{n+1}$. Equality holds if and only if $a_{1}=a_{2}=\\cdots=a_{k}$ and $a_{k+1}=\\cdots=a_{n+1}=$ 0 for some $k$.  Second solution. Setting $b_{k}=\\sqrt{a_{k}}-\\sqrt{a_{k+1}}$ for $k=1, \\ldots, n$ we have $a_{i}=\\left(b_{i}+\\cdots+b_{n}\\right)^{2}$, so the desired inequality after squaring becomes  $$ \\sum_{k=1}^{n} k b_{k}^{2}+2 \\sum_{1 \\leq k<l \\leq n} k b_{k} b_{l} \\leq \\sum_{k=1}^{n} k b_{k}^{2}+2 \\sum_{1 \\leq k<l \\leq n} \\sqrt{k l} b_{k} b_{l}, $$  which clearly holds.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (CAN) Let $R_{1}, R_{2}, \\ldots$ be the family of finite sequences of positive integers defined by the following rules: $R_{1}=(1)$, and if $R_{n-1}=\\left(x_{1}, \\ldots, x_{s}\\right)$, then  $$ R_{n}=\\left(1,2, \\ldots, x_{1}, 1,2, \\ldots, x_{2}, \\ldots, 1,2, \\ldots, x_{s}, n\\right) $$  For example, $R_{2}=(1,2), R_{3}=(1,1,2,3), R_{4}=(1,1,1,2,1,2,3,4)$. Prove that if $n>1$, then the $k$ th term from the left in $R_{n}$ is equal to 1 if and only if the $k$ th term from the right in $R_{n}$ is different from 1.", "solution": "2. For any sequence $X=\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)$ let us define  $$ \\bar{X}=\\left(1,2, \\ldots, x_{1}, 1,2, \\ldots, x_{2}, \\ldots, 1,2, \\ldots, x_{n}\\right) $$  Also, for any two sequences $A, B$ we denote their concatenation by $A B$. It clearly holds that $\\overline{A B}=\\bar{A} \\bar{B}$. The sequences $R_{1}, R_{2}, \\ldots$ are given by $R_{1}=(1)$ and $R_{n}=\\overline{R_{n-1}}(n)$ for $n>1$. We consider the family of sequences $Q_{n i}$ for $n, i \\in \\mathbb{N}, i \\leq n$, defined as follows: $Q_{n 1}=(1), \\quad Q_{n n}=(n), \\quad$ and $\\quad Q_{n i}=Q_{n-1, i-1} Q_{n-1, i} \\quad$ if $1<i<n$. These sequences form a Pascal-like triangle, as shown in the picture below:  | $Q_{1 i}:$ |  | 1 |  |  |  |  |  |  | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | $Q_{2 i}$ : |  |  |  | 1 |  | 2 |  |  | | $Q_{3 i}$ : |  |  | 1 |  | 12 |  | 3 |  | | $Q_{4 i}$ : |  | 1 |  | 112 |  | 123 |  | 4 | | $Q_{5 i}$ : | 1 |  | 1112 |  | 2123 |  | 1234 |  |  We claim that $R_{n}$ is in fact exactly $Q_{n 1} Q_{n 2} \\ldots Q_{n n}$. Before proving this, we observe that $Q_{n i}=\\overline{Q_{n-1, i}}$. This follows by induction, because $Q_{n i}=$ $Q_{n-1, i-1} Q_{n-1, i}=\\overline{Q_{n-2, i-1}} \\overline{Q_{n-2, i}}=\\overline{Q_{n-1, i}}$ for $n \\geq 3, i \\geq 2$ (the cases $i=1$ and $n=1,2$ are trivial). Now $R_{1}=Q_{11}$ and  $$ R_{n}=\\overline{R_{n-1}}(n)=\\overline{Q_{n-1,1} \\ldots Q_{n-1, n-1}}(n)=Q_{n, 1} \\ldots Q_{n, n-1} Q_{n, n} $$  for $n \\geq 2$, which justifies our claim by induction. Now we know enough about the sequence $R_{n}$ to return to the question of the problem. We use induction on $n$ once again. The result is obvious for $n=1$ and $n=2$. Given any $n \\geq 3$, consider the $k$ th elements of $R_{n}$ from the left, say $u$, and from the right, say $v$. Assume that $u$ is a member of $Q_{n j}$, and consequently that $v$ is a member of $Q_{n, n+1-j}$. Then $u$ and $v$ come from symmetric positions of $R_{n-1}$ (either from $Q_{n-1, j}, Q_{n-1, n-j}$, or from $\\left.Q_{n-1, j-1}, Q_{n-1, n+1-j}\\right)$, and by the inductive hypothesis exactly one of them is 1 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. (IRE) Let $D$ be an internal point on the side $B C$ of a triangle $A B C$. The line $A D$ meets the circumcircle of $A B C$ again at $X$. Let $P$ and $Q$ be the feet of the perpendiculars from $X$ to $A B$ and $A C$, respectively, and let $\\gamma$ be the circle with diameter $X D$. Prove that the line $P Q$ is tangent to $\\gamma$ if and only if $A B=A C$.", "solution": "20. To avoid dividing into cases regarding the position of the point $X$, we use oriented angles. Let $R$ be the foot of the perpendicular from $X$ to $B C$. It is well known that the points $P, Q, R$ lie on the corresponding Simson line. This line is a tangent to $\\gamma$ (i.e., the circle $X D R$ ) if and only if $\\angle P R D=\\angle R X D$. We have  $$ \\begin{aligned} \\angle P R D & =\\angle P X B=90^{\\circ}-\\angle X B A=90^{\\circ}-\\angle X B C+\\angle A B C \\\\ & =90^{\\circ}-\\angle D A C+\\angle A B C \\end{aligned} $$  and  $$ \\angle R X D=90^{\\circ}-\\angle A D B=90^{\\circ}+\\angle B C A-\\angle D A C $$  hence $\\angle P R D=\\angle R X D$ if and only if $\\angle A B C=\\angle B C A$, i.e, $A B=A C$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. (RUS) ${ }^{\\mathrm{IMO} 3}$ Let $x_{1}, x_{2}, \\ldots, x_{n}$ be real numbers satisfying the conditions  $$ \\left|x_{1}+x_{2}+\\cdots+x_{n}\\right|=1 \\quad \\text { and } \\quad\\left|x_{i}\\right| \\leq \\frac{n+1}{2} \\quad \\text { for } \\quad i=1,2, \\ldots, n $$  Show that there exists a permutation $y_{1}, \\ldots, y_{n}$ of the sequence $x_{1}, \\ldots, x_{n}$ such that  $$ \\left|y_{1}+2 y_{2}+\\cdots+n y_{n}\\right| \\leq \\frac{n+1}{2} $$", "solution": "21. For any permutation $\\pi=\\left(y_{1}, y_{2}, \\ldots, y_{n}\\right)$ of $\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)$, denote by $S(\\pi)$ the sum $y_{1}+2 y_{2}+\\cdots+n y_{n}$. Suppose, contrary to the claim, that $|S(\\pi)|>\\frac{n+1}{2}$ for any $\\pi$. Further, we note that if $\\pi^{\\prime}$ is obtained from $\\pi$ by interchanging two neighboring elements, say $y_{k}$ and $y_{k+1}$, then $S(\\pi)$ and $S\\left(\\pi^{\\prime}\\right)$ differ by $\\left|y_{k}+y_{k+1}\\right| \\leq n+1$, and consequently they must be of the same sign. Now consider the identity permutation $\\pi_{0}=\\left(x_{1}, \\ldots, x_{n}\\right)$ and the reverse permutation $\\overline{\\pi_{0}}=\\left(x_{n}, \\ldots, x_{1}\\right)$. There is a sequence of permutations $\\pi_{0}, \\pi_{1}, \\ldots, \\pi_{m}=\\overline{\\pi_{0}}$ such that for each $i, \\pi_{i+1}$ is obtained from $\\pi_{i}$ by interchanging two neighboring elements. Indeed, by successive interchanges we can put $x_{n}$ in the first place, then $x_{n-1}$ in the second place, etc. Hence all $S\\left(\\pi_{0}\\right), \\ldots, S\\left(\\pi_{m}\\right)$ are of the same sign. However, since $\\left|S\\left(\\pi_{0}\\right)+S\\left(\\pi_{m}\\right)\\right|=(n+1)\\left|x_{1}+\\cdots+x_{n}\\right|=n+1$, this implies that one of $S\\left(\\pi_{0}\\right)$ and $S\\left(\\overline{\\pi_{0}}\\right)$ is smaller than $\\frac{n+1}{2}$ in absolute value, contradicting the initial assumption.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "22", "problem_type": null, "exam": "IMO", "problem": "22. (UKR) (a) Do there exist functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ and $g: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that  $$ f(g(x))=x^{2} \\quad \\text { and } \\quad g(f(x))=x^{3} \\quad \\text { for all } x \\in \\mathbb{R} ? $$  (b) Do there exist functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ and $g: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that  $$ f(g(x))=x^{2} \\quad \\text { and } \\quad g(f(x))=x^{4} \\quad \\text { for all } x \\in \\mathbb{R} ? $$", "solution": "22. (a) Suppose that $f$ and $g$ are such functions. From $g(f(x))=x^{3}$ we have $f\\left(x_{1}\\right) \\neq f\\left(x_{2}\\right)$ whenever $x_{1} \\neq x_{2}$. In particular, $f(-1), f(0)$, and $f(1)$ are three distinct numbers. However, since $f(x)^{2}=f(g(f(x)))=$ $f\\left(x^{3}\\right)$, each of the numbers $f(-1), f(0), f(1)$ is equal to its square, and so must be either 0 or 1 . This contradiction shows that no such $f, g$ exist. (b) The answer is yes. We begin with constructing functions $F, G:(1, \\infty)$ $\\rightarrow(1, \\infty)$ with the property $F(G(x))=x^{2}$ and $G(F(x))=x^{4}$ for $x>$ 1. Define the functions $\\varphi, \\psi$ by $F\\left(2^{2^{t}}\\right)=2^{2^{\\varphi(t)}}$ and $G\\left(2^{2^{t}}\\right)=2^{2^{\\psi(t)}}$. These functions determine $F$ and $G$ on the entire interval $(1, \\infty)$, and satisfy $\\varphi(\\psi(t))=t+1$ and $\\psi(\\varphi(t))=t+2$. It is easy to find examples of $\\varphi$ and $\\psi$ : for example, $\\varphi(t)=\\frac{1}{2} t+1, \\psi(t)=2 t$. Thus we also arrive at an example for $F, G$ :  $$ F(x)=2^{2^{\\frac{1}{2} \\log _{2} \\log _{2} x+1}}=2^{2 \\sqrt{\\log _{2} x}}, \\quad G(x)=2^{2^{2 \\log _{2} \\log _{2} x}}=2^{\\log _{2}^{2} x} $$  It remains only to extend these functions to the whole of $\\mathbb{R}$. This can be done as follows:  $$ \\widetilde{f}(x)= \\begin{cases}F(x) & \\text { for } x>1 \\\\ 1 / F(1 / x) & \\text { for } 0<x<1, \\widetilde{g}(x)=\\left\\{\\begin{array} { c l }  { G ( x ) } & { \\text { for } x > 1 } \\\\ { x } & { \\text { for } x \\in \\{ 0 , 1 \\} ; } \\end{array} \\left\\{\\begin{array}{cl} 1 / G(1 / x) & \\text { for } 0<x<1 \\\\ x & \\text { for } x \\in\\{0,1\\} \\end{array}\\right.\\right.\\end{cases} $$  and then $\\quad f(x)=\\widetilde{f}(|x|), \\quad g(x)=\\widetilde{g}(|x|) \\quad$ for $x \\in \\mathbb{R}$. It is directly verified that these functions have the required property.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "23", "problem_type": null, "exam": "IMO", "problem": "23. (GBR) Let $A B C D$ be a convex quadrilateral and $O$ the intersection of its diagonals $A C$ and $B D$. If  $$ O A \\sin \\angle A+O C \\sin \\angle C=O B \\sin \\angle B+O D \\sin \\angle D, $$  prove that $A B C D$ is cyclic.", "solution": "23. Let $K, L, M$, and $N$ be the projections of $O$ onto the lines $A B, B C, C D$, and $D A$, and let $\\alpha_{1}, \\alpha_{2}, \\alpha_{3}, \\alpha_{4}, \\beta_{1}, \\beta_{2}, \\beta_{3}, \\beta_{4}$ denote the angles $O A B$, $O B C, O C D, O D A, O A D, O B A, O C B, O D C$, respectively. We start with the following observation: Since $N K$ is a chord of the circle with diameter $O A$, we have $O A \\sin \\angle A=N K=O N \\cos \\alpha_{1}+O K \\cos \\beta_{1}$ (because $\\angle O N K=\\alpha_{1}$ and $\\angle O K N=\\beta_{1}$ ). Analogous equalities also hold: $O B \\sin \\angle B=K L=O K \\cos \\alpha_{2}+O L \\cos \\beta_{2}, O C \\sin \\angle C=L M=$ $O L \\cos \\alpha_{3}+O M \\cos \\beta_{3}$ and $O D \\sin \\angle D=M N=O M \\cos \\alpha_{4}+O N \\cos \\beta_{4}$. Now the condition in the problem can be restated as $N K+L M=K L+$ $M N$ (i.e., $K L M N$ is circumscribed), i.e.,  $$ \\begin{aligned} O K\\left(\\cos \\beta_{1}-\\cos \\alpha_{2}\\right)+O L\\left(\\cos \\alpha_{3}-\\cos \\beta_{2}\\right) \\\\ +O M\\left(\\cos \\beta_{3}-\\cos \\alpha_{4}\\right)+O N\\left(\\cos \\alpha_{1}-\\cos \\beta_{4}\\right)=0 \\end{aligned} $$  To prove that $A B C D$ is cyclic, it suffices to show that $\\alpha_{1}=\\beta_{4}$. Assume the contrary, and let w.l.o.g. $\\alpha_{1}>\\beta_{4}$. Then point $A$ lies inside the circle $B C D$, which is further equivalent to $\\beta_{1}>\\alpha_{2}$. On the other hand, from $\\alpha_{1}+\\beta_{2}=\\alpha_{3}+\\beta_{4}$ we deduce $\\alpha_{3}>\\beta_{2}$, and similarly $\\beta_{3}>\\alpha_{4}$. Therefore, since the cosine is strictly decreasing on $(0, \\pi)$, the left side of (1) is strictly negative, yielding a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "24", "problem_type": null, "exam": "IMO", "problem": "24. (LIT) ${ }^{\\mathrm{IMO} 6}$ For a positive integer $n$, let $f(n)$ denote the number of ways to represent $n$ as the sum of powers of 2 with nonnegative integer exponents. Representations that differ only in the ordering in their summands are not considered to be distinct. (For instance, $f(4)=4$ because the number 4 can be represented in the following four ways: $4 ; 2+2 ; 2+1+1 ; 1+1+1+1$.) Prove that the inequality  $$ 2^{n^{2} / 4}<f\\left(2^{n}\\right)<2^{n^{2} / 2} $$  holds for any integer $n \\geq 3$.", "solution": "24. There is a bijective correspondence between representations in the given form of $2 k$ and $2 k+1$ for $k=0,1, \\ldots$, since adding 1 to every representation of $2 k$, we obtain a representation of $2 k+1$, and conversely, every representation of $2 k+1$ contains at least one 1 , which can be removed. Hence, $f(2 k+1)=f(2 k)$. Consider all representations of $2 k$. The number of those that contain at least one 1 equals $f(2 k-1)=f(2 k-2)$, while the number of those not containing a 1 equals $f(k)$ (the correspondence is given by division of summands by 2). Therefore  $$ f(2 k)=f(2 k-2)+f(k) . $$  Summing these equalities over $k=1, \\ldots, n$, we obtain  $$ f(2 n)=f(0)+f(1)+\\cdots+f(n) $$  We first prove the right-hand inequality. Since $f$ is increasing, and $f(0)+$ $f(1)=f(2)$, (2) yields $f(2 n) \\leq n f(n)$ for $n \\geq 2$. Now $f\\left(2^{3}\\right)=f(0)+$ $\\cdots+f(4)=10<2^{3^{2} / 2}$, and one can easily conclude by induction that $f\\left(2^{n+1}\\right) \\leq 2^{n} f\\left(2^{n}\\right)<2^{n} \\cdot 2^{n^{2} / 2}<2^{(n+1)^{2} / 2}$ for each $n \\geq 3$. We now derive the lower estimate. It follows from (1) that $f(x+2)-f(x)$ is increasing. Consequently, for each $m$ and $k<m$ we have $f(2 m+2 k)-$ $f(2 m) \\geq f(2 m+2 k-2)-f(2 m-2) \\geq \\cdots \\geq f(2 m)-f(2 m-2 k)$, so $f(2 m+2 k)+f(2 m-2 k) \\geq 2 f(2 m)$. Adding all these inequalities for $k=1,2, \\ldots, m$, we obtain $f(0)+f(2)+\\cdots+f(4 m) \\geq(2 m+1) f(2 m)$. But since $f(2)=f(3), f(4)=f(5)$ etc., we also have $f(1)+f(3)+\\cdots+$ $f(4 m-1)>(2 m-1) f(2 m)$, which together with the above inequality gives  $$ f(8 m)=f(0)+f(1)+\\cdots+f(4 m)>4 m f(2 m) . $$  Finally, we have that the inequality $f\\left(2^{n}\\right)>2^{n^{2} / 4}$ holds for $n=2$ and $n=3$, while for larger $n$ we have by induction $f\\left(2^{n}\\right)>2^{n-1} f\\left(2^{n-2}\\right)>$ $2^{n-1+(n-2)^{2} / 4}=2^{n^{2} / 4}$. This completes the proof. Remark. Despite the fact that the lower estimate is more difficult, it is much weaker than the upper estimate. It can be shown that $f\\left(2^{n}\\right)$ eventually (for large $n$ ) exceeds $2^{c n^{2}}$ for any $c<\\frac{1}{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "25", "problem_type": null, "exam": "IMO", "problem": "25. (POL) The bisectors of angles $A, B, C$ of a triangle $A B C$ meet its circumcircle again at the points $K, L, M$, respectively. Let $R$ be an internal point on the side $A B$. The points $P$ and $Q$ are defined by the following conditions: $R P$ is parallel to $A K$, and $B P$ is perpendicular to $B L ; R Q$ is parallel to $B L$, and $A Q$ is perpendicular to $A K$. Show that the lines $K P, L Q, M R$ have a point in common.", "solution": "25. Let $M R$ meet the circumcircle of triangle $A B C$ again at a point $X$. We claim that $X$ is the common point of the lines $K P, L Q, M R$. By symmetry, it will be enough to show that $X$ lies on $K P$. It is easy to see that $X$ and $P$ lie on the same side of $A B$ as $K$. Let $I_{a}=A K \\cap B P$ be the excenter of $\\triangle A B C$ corresponding to $A$. It is easy to calculate that $\\angle A I_{a} B=\\gamma / 2$, from which we get $\\angle R P B=\\angle A I_{a} B=\\angle M C B=\\angle R X B$. Therefore $R, B, P, X$ are concyclic. Now if $P$ and $K$ are on distinct sides of $B X$ (the other case is similar), we have $\\angle R X P=180^{\\circ}-\\angle R B P=90^{\\circ}-$ $\\beta / 2=\\angle M A K=180^{\\circ}-\\angle R X K$, from which it follows that $K, X, P$ are collinear, as claimed. Remark. It is not essential for the statement of the problem that $R$ be an internal point of $A B$. Work with cases can be avoided using oriented ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-640.jpg?height=399&width=439&top_left_y=248&top_left_x=858) angles.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "26", "problem_type": null, "exam": "IMO", "problem": "26. (ITA) For every integer $n \\geq 2$ determine the minimum value that the sum $a_{0}+a_{1}+\\cdots+a_{n}$ can take for nonnegative numbers $a_{0}, a_{1}, \\ldots, a_{n}$ satisfying the condition  $$ a_{0}=1, \\quad a_{i} \\leq a_{i+1}+a_{i+2} \\quad \\text { for } i=0, \\ldots, n-2 $$  ### 3.39 The Thirty-Ninth IMO", "solution": "26. Let us first examine the case that all the inequalities in the problem are actually equalities. Then $a_{n-2}=a_{n-1}+a_{n}, a_{n-3}=2 a_{n-1}+a_{n}, \\ldots, a_{0}=$ $F_{n} a_{n-1}+F_{n-1} a_{n}=1$, where $F_{n}$ is the $n$th Fibonacci number. Then it is easy to see (from $F_{1}+F_{2}+\\cdots+F_{k}=F_{k+2}$ ) that $a_{0}+\\cdots+a_{n}=$ $\\left(F_{n+2}-1\\right) a_{n-1}+F_{n+1} a_{n}=\\frac{F_{n+2}-1}{F_{n}}+\\left(F_{n+1}-\\frac{F_{n-1}\\left(F_{n+2}-1\\right)}{F_{n}}\\right) a_{n}$. Since $\\frac{F_{n-1}\\left(F_{n+2}-1\\right)}{F_{n}} \\leq F_{n+1}$, it follows that $a_{0}+a_{1}+\\cdots+a_{n} \\geq \\frac{F_{n+2}-1}{F_{n}}$, with equality holding if and only if $a_{n}=0$ and $a_{n-1}=\\frac{1}{F_{n}}$. We denote by $M_{n}$ the required minimum in the general case. We shall prove by induction that $M_{n}=\\frac{F_{n+2}-1}{F_{n}}$. For $M_{1}=1$ and $M_{2}=2$ it is easy to show that the formula holds; hence the inductive basis is true. Suppose that $n>2$. The sequences $1, \\frac{a_{2}}{a_{1}}, \\ldots, \\frac{a_{n}}{a_{1}}$ and $1, \\frac{a_{3}}{a_{2}}, \\ldots, \\frac{a_{n}}{a_{2}}$ also satisfy the conditions of the problem. Hence we have  $$ a_{0}+\\cdots+a_{n}=a_{0}+a_{1}\\left(1+\\frac{a_{2}}{a_{1}}+\\cdots+\\frac{a_{n}}{a_{1}}\\right) \\geq 1+a_{1} M_{n-1} $$  and  $$ a_{0}+\\cdots+a_{n}=a_{0}+a_{1}+a_{2}\\left(1+\\frac{a_{3}}{a_{2}}+\\cdots+\\frac{a_{n}}{a_{2}}\\right) \\geq 1+a_{1}+a_{2} M_{n-2} $$  Multiplying the first inequality by $M_{n-2}-1$ and the second one by $M_{n-1}$, adding the inequalities and using that $a_{1}+a_{2} \\geq 1$, we obtain ( $M_{n-1}+$ $\\left.M_{n-2}+1\\right)\\left(a_{0}+\\cdots+a_{n}\\right) \\geq M_{n-1} M_{n-2}+M_{n-1}+M_{n-2}+1$, so  $$ M_{n} \\geq \\frac{M_{n-1} M_{n-2}+M_{n-1}+M_{n-2}+1}{M_{n-1}+M_{n-2}+1} $$  Since $M_{n-1}=\\frac{F_{n+1}-1}{F_{n-1}}$ and $M_{n-2}=\\frac{F_{n}-1}{F_{n-2}}$, the above inequality easily yields $M_{n} \\geq \\frac{F_{n+2}-1}{F_{n}}$. However, we have shown above that equality can occur; hence $\\frac{F_{n+2}-1}{F_{n}}$ is indeed the required minimum.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (GER) For each finite set $U$ of nonzero vectors in the plane we define $l(U)$ to be the length of the vector that is the sum of all vectors in $U$. Given a finite set $V$ of nonzero vectors in the plane, a subset $B$ of $V$ is said to be maximal if $l(B)$ is greater than or equal to $l(A)$ for each nonempty subset $A$ of $V$. (a) Construct sets of 4 and 5 vectors that have 8 and 10 maximal subsets respectively. (b) Show that for any set $V$ consisting of $n \\geq 1$ vectors, the number of maximal subsets is less than or equal to $2 n$.", "solution": "3. (a) For $n=4$, consider a convex quadrilateral $A B C D$ in which $A B=$ $B C=A C=B D$ and $A D=D C$, and take the vectors $\\overrightarrow{A B}, \\overrightarrow{B C}$, $\\overrightarrow{C D}, \\overrightarrow{D A}$. For $n=5$, take the vectors $\\overrightarrow{A B}, \\overrightarrow{B C}, \\overrightarrow{C D}, \\overrightarrow{D E}, \\overrightarrow{E A}$ for any regular pentagon $A B C D E$. (b) Let us draw the vectors of $V$ as originated from the same point $O$. Consider any maximal subset $B \\subset V$, and denote by $u$ the sum of all vectors from $B$. If $l$ is the line through $O$ perpendicular to $u$, then $B$ contains exactly those vectors from $V$ that lie on the same side of $l$ as $u$ does, and no others. Indeed, if any $v \\notin B$ lies on the same side of $l$, then $|u+v| \\geq|u|$; similarly, if some $v \\in B$ lies on the other side of $l$, then $|u-v| \\geq|u|$. Therefore every maximal subset is determined by some line $l$ as the set of vectors lying on the same side of $l$. It is obvious that in this way we get at most $2 n$ sets.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (IRN) ${ }^{\\mathrm{IMO} 4} \\mathrm{An} n \\times n$ matrix with entries from $\\{1,2, \\ldots, 2 n-1\\}$ is called a coveralls matrix if for each $i$ the union of the $i$ th row and the $i$ th column contains $2 n-1$ distinct entries. Show that: (a) There exist no coveralls matrices for $n=1997$. (b) Coveralls matrices exist for infinitely many values of $n$.", "solution": "4. (a) Suppose that an $n \\times n$ coveralls matrix $A$ exists for some $n>1$. Let $x \\in\\{1,2, \\ldots, 2 n-1\\}$ be a fixed number that does not appear on the fixed diagonal of $A$. Such an element must exist, since the diagonal can contain at most $n$ different numbers. Let us call the union of the $i$ th row and the $i$ th column the $i$ th cross. There are $n$ crosses, and each of them contains exactly one $x$. On the other hand, each entry $x$ of $A$ is contained in exactly two crosses. Hence $n$ must be even. However, 1997 is an odd number; hence no coveralls matrix exists for $n=1997$. (b) For $n=2, A_{2}=\\left[\\begin{array}{ll}1 & 2 \\\\ 3 & 1\\end{array}\\right]$ is a coveralls matrix. For $n=4$, one such matrix is, for example,  $$ A_{4}=\\left[\\begin{array}{llll} 1 & 2 & 5 & 6 \\\\ 3 & 1 & 7 & 5 \\\\ 4 & 6 & 1 & 2 \\\\ 7 & 4 & 3 & 1 \\end{array}\\right] $$  This construction can be generalized. Suppose that we are given an $n \\times n$ coveralls matrix $A_{n}$. Let $B_{n}$ be the matrix obtained from $A_{n}$ by adding $2 n$ to each entry, and $C_{n}$ the matrix obtained from $B_{n}$ by replacing each diagonal entry (equal to $2 n+1$ by induction) with $2 n$. Then the matrix  $$ A_{2 n}=\\left[\\begin{array}{ll} A_{n} & B_{n} \\\\ C_{n} & A_{n} \\end{array}\\right] $$  is coveralls. To show this, suppose that $i \\leq n$ (the case $i>n$ is similar). The $i$ th cross is composed of the $i$ th cross of $A_{n}$, the $i$ th row of $B_{n}$, and the $i$ th column of $C_{n}$. The $i$ th cross of $A_{i}$ covers $1,2, \\ldots, 2 n-1$. The $i$ th row of $B_{n}$ covers all numbers of the form $2 n+j$, where $j$ is covered by the $i$ th row of $A_{n}$ (including $j=1$ ). Similarly, the $i$ th column of $C_{n}$ covers $2 n$ and all numbers of the form $2 n+k$, where $k>1$ is covered by the $i$ th column of $A_{n}$. Thus we see that all numbers are accounted for in the $i$ th cross of $A_{2 n}$, and hence $A_{2 n}$ is a desired coveralls matrix. It follows that we can find a coveralls matrix whenever $n$ is a power of 2 . Second solution for part $b$. We construct a coveralls matrix explicitly for $n=2^{k}$. We consider the coordinates/cells of the matrix elements modulo $n$ throughout the solution. We define the $i$-diagonal $(0 \\leq i<$ $n$ ) to be the set of cells of the form $(j, j+i)$, for all $j$. We note that each cross contains exactly one cell from the 0 -diagonal (the main diagonal) and two cells from each $i$-diagonal. For two cells within an $i$ diagonal, $x$ and $y$, we define $x$ and $y$ to be related if there exists a cross containing both $x$ and $y$. Evidently, for every cell $x$ not on the 0 -diagonal there are exactly two other cells related to it. The relation thus breaks up each $i$-diagonal $(i>0)$ into cycles of length larger than 1 . Due to the diagonal translational symmetry (modulo $n$ ), all the cycles within a given $i$-diagonal must be of equal length and thus of an even length, since $n=2^{k}$. The construction of a coveralls matrix is now obvious. We select a number, say 1, to place on all the cells of the 0-diagonal. We pair up the remaining numbers and assign each pair to an $i$-diagonal, say $(2 i, 2 i+1)$. Going along each cycle within the $i$-diagonal we alternately assign values of $2 i$ and $2 i+1$. Since the cycle has an even length, a cell will be related only to a cell of a different number, and hence each cross will contain both $2 i$ and $2 i+1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (ROM) Let $A B C D$ be a regular tetrahedron and $M, N$ distinct points in the planes $A B C$ and $A D C$ respectively. Show that the segments $M N, B N, M D$ are the sides of a triangle.", "solution": "5. We shall prove first the 2-dimensional analogue:  Lemma. Given an equilateral triangle $A B C$ and two points $M, N$ on the sides $A B$ and $A C$ respectively, there exists a triangle with sides $C M, B N, M N$. Proof. Consider a regular tetrahedron $A B C D$. Since $C M=D M$ and $B N=D N$, one such triangle is $D M N$.  Now, to solve the problem for a regular tetrahedron $A B C D$, we consider a 4-dimensional polytope $A B C D E$ whose faces $A B C D, A B C E, A B D E$, $A C D E, B C D E$ are regular tetrahedra. We don't know what it looks like, but it yields a desired triangle: for $M \\in A B C$ and $N \\in A D C$, we have $D M=E M$ and $B N=E N$; hence the desired triangle is $E M N$. Remark. A solution that avoids embedding in $\\mathbb{R}^{4}$ is possible, but no longer so short.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (IRE) (a) Let $n$ be a positive integer. Prove that there exist distinct positive integers $x, y, z$ such that  $$ x^{n-1}+y^{n}=z^{n+1} . $$  (b) Let $a, b, c$ be positive integers such that $a$ and $b$ are relatively prime and $c$ is relatively prime either to $a$ or to $b$. Prove that there exist infinitely many triples $(x, y, z)$ of distinct positive integers $x, y, z$ such that  $$ x^{a}+y^{b}=z^{c} . $$  Original formulation: Let $a, b, c, n$ be positive integers such that $n$ is odd and $a c$ is relatively prime to $2 b$. Prove that there exist distinct positive integers $x, y, z$ such that (i) $x^{a}+y^{b}=z^{c}$, and (ii) $x y z$ is relatively prime to $n$.", "solution": "6. (a) One solution is  $$ x=2^{n^{2}} 3^{n+1}, \\quad y=2^{n^{2}-n} 3^{n}, \\quad z=2^{n^{2}-2 n+2} 3^{n-1} . $$  (b) Suppose w.l.o.g. that $\\operatorname{gcd}(c, a)=1$. We look for a solution of the form  $$ x=p^{m}, \\quad y=p^{n}, \\quad z=q p^{r}, \\quad p, q, m, n, r \\in \\mathbb{N} . $$  Then $x^{a}+y^{b}=p^{m a}+p^{n b}$ and $z^{c}=q^{c} p^{r c}$, and we see that it is enough to assume $m a-1=n b=r c$ (there are infinitely many such triples $(m, n, r))$ and $q^{c}=p+1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (RUS) Let $A B C D E F$ be a convex hexagon such that $A B=B C, C D=$ $D E, E F=F A$. Prove that  $$ \\frac{B C}{B E}+\\frac{D E}{D A}+\\frac{F A}{F C} \\geq \\frac{3}{2} $$  When does equality occur?", "solution": "7. Let us set $A C=a, C E=b, E A=c$. Applying Ptolemy's inequality for the quadrilateral $A C E F$ we get  $$ A C \\cdot E F+C E \\cdot A F \\geq A E \\cdot C F $$  Since $E F=A F$, this implies $\\frac{F A}{F C} \\geq \\frac{c}{a+b}$. Similarly $\\frac{B C}{B E} \\geq \\frac{a}{b+c}$ and $\\frac{D E}{D A} \\geq$ $\\frac{b}{c+a}$. Now,  $$ \\frac{B C}{B E}+\\frac{D E}{D A}+\\frac{F A}{F C} \\geq \\frac{a}{b+c}+\\frac{b}{c+a}+\\frac{c}{a+b} $$  Hence it is enough to prove that  $$ \\frac{a}{b+c}+\\frac{b}{c+a}+\\frac{c}{a+b} \\geq \\frac{3}{2} $$  If we now substitute $x=b+c, y=c+a, z=a+b$ and $S=a+b+c$ the inequality (1) becomes equivalent to $S(1 / x+1 / y+1 / y)-3 \\geq 3 / 2$ which follows immediately form $1 / x+1 / y+1 / z \\geq 9 /(x+y+z)=9 /(2 S)$. Equality occurs if it holds in Ptolemy's inequalities and also $a=b=c$. The former happens if and only if the hexagon is cyclic. Hence the only case of equality is when $A B C D E F$ is regular.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (GBR) ${ }^{\\mathrm{IMO} 2}$ Four different points $A, B, C, D$ are chosen on a circle $\\Gamma$ such that the triangle $B C D$ is not right-angled. Prove that: (a) The perpendicular bisectors of $A B$ and $A C$ meet the line $A D$ at certain points $W$ and $V$, respectively, and that the lines $C V$ and $B W$ meet at a certain point $T$. (b) The length of one of the line segments $A D, B T$, and $C T$ is the sum of the lengths of the other two. Original formulation. In triangle $A B C$ the angle at $A$ is the smallest. A line through $A$ meets the circumcircle again at the point $U$ lying on the $\\operatorname{arc} B C$ opposite to $A$. The perpendicular bisectors of $C A$ and $A B$ meet $A U$ at $V$ and $W$, respectively, and the lines $C V, B W$ meet at $T$. Show that $A U=T B+T C$.", "solution": "8. (a) Denote by $b$ and $c$ the perpendicular bisectors of $A B$ and $A C$ respectively. If w.l.o.g. $b$ and $A D$ do not intersect (are parallel), then $\\angle B C D=\\angle B A D=90^{\\circ}$, a contradiction. Hence $V, W$ are well-defined. Now, $\\angle D W B=2 \\angle D A B$ and $\\angle D V C=2 \\angle D A C$ as oriented angles, and therefore $\\angle(W B, V C)=2(\\angle D V C-\\angle D W B)=2 \\angle B A C=$ $2 \\angle B C D$ is not equal to 0 . Consequently $C V$ and $B W$ meet at some $T$ with $\\angle B T C=2 \\angle B A C$. (b) Let $B^{\\prime}$ be the second point of intersection of $B W$ with $\\Gamma$. Clearly $A D=B B^{\\prime}$. But we also have $\\angle B T C=2 \\angle B A C=2 \\angle B B^{\\prime} C$, which implies that $C T=T B^{\\prime}$. It follows that $A D=B B^{\\prime}=\\left|B T \\pm T B^{\\prime}\\right|=$ $|B T \\pm C T|$. Remark. This problem is also solved easily using trigonometry.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1997", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (USA) Let $A_{1} A_{2} A_{3}$ be a nonisosceles triangle with incenter $I$. Let $C_{i}$, $i=1,2,3$, be the smaller circle through $I$ tangent to $A_{i} A_{i+1}$ and $A_{i} A_{i+2}$ (the addition of indices being mod 3 ). Let $B_{i}, i=1,2,3$, be the second point of intersection of $C_{i+1}$ and $C_{i+2}$. Prove that the circumcenters of the triangles $A_{1} B_{1} I, A_{2} B_{2} I, A_{3} B_{3} I$ are collinear.", "solution": "9. For $i=1,2,3$ (all indices in this problem will be modulo 3 ) we denote by $O_{i}$ the center of $C_{i}$ and by $M_{i}$ the midpoint of the $\\operatorname{arc} A_{i+1} A_{i+2}$ that does not contain $A_{i}$. First we have that $O_{i+1} O_{i+2}$ is the perpendicular bisector of $I B_{i}$, and thus it contains the circumcenter $R_{i}$ of $A_{i} B_{i} I$. Additionally, it is easy to show that $T_{i+1} A_{i}=T_{i+1} I$ and $T_{i+2} A_{i}=$ $T_{i+2} I$, which implies that $R_{i}$ lies on the line $T_{i+1} T_{i+2}$. Therefore $R_{i}=$ $O_{i+1} O_{i+2} \\cap T_{i+1} T_{i+2}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-631.jpg?height=360&width=548&top_left_y=579&top_left_x=806)  Now, the lines $T_{1} O_{1}, T_{2} O_{2}, T_{3} O_{3}$ are concurrent at $I$. By Desargues's theorem, the points of intersection of $O_{i+1} O_{i+2}$ and $T_{i+1} T_{i+2}$, i.e., the $R_{i}$ 's, lie on a line for $i=1,2,3$.  Second solution. The centers of three circles passing through the same point $I$ and not touching each other are collinear if and only if they have another common point. Hence it is enough to show that the circles $A_{i} B_{i} I$ have a common point other than $I$. Now apply inversion at center $I$ and with an arbitrary power. We shall denote by $X^{\\prime}$ the image of $X$ under this inversion. In our case, the image of the circle $C_{i}$ is the line $B_{i+1}^{\\prime} B_{i+2}^{\\prime}$ while the image of the line $A_{i+1} A_{i+2}$ is the circle $I A_{i+1}^{\\prime} A_{i+2}^{\\prime}$ that is tangent to $B_{i}^{\\prime} B_{i+2}^{\\prime}$, and $B_{i}^{\\prime} B_{i+2}^{\\prime}$. These three circles have equal radii, so their centers $P_{1}, P_{2}, P_{3}$ form a triangle also homothetic to $\\triangle B_{1}^{\\prime} B_{2}^{\\prime} B_{3}^{\\prime}$. Consequently, points $A_{1}^{\\prime}, A_{2}^{\\prime}, A_{3}^{\\prime}$, that are the reflections of $I$ across the sides of $P_{1} P_{2} P_{3}$, are vertices of a triangle also homothetic to $B_{1}^{\\prime} B_{2}^{\\prime} B_{3}^{\\prime}$. It follows that $A_{1}^{\\prime} B_{1}^{\\prime}, A_{2}^{\\prime} B_{2}^{\\prime}, A_{3}^{\\prime} B_{3}^{\\prime}$ are concurrent at some point $J^{\\prime}$, i.e., that the circles $A_{i} B_{i} I$ all pass through $J$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. (LUX) ${ }^{\\mathrm{IMO}} \\mathrm{A}$ convex quadrilateral $A B C D$ has perpendicular diagonals. The perpendicular bisectors of $A B$ and $C D$ meet at a unique point $P$ inside $A B C D$. Prove that $A B C D$ is cyclic if and only if triangles $A B P$ and $C D P$ have equal areas.", "solution": "1. We begin with the following observation: Suppose that $P$ lies in $\\triangle A E B$, where $E$ is the intersection of $A C$ and $B D$ (the other cases are similar). Let $M, N$ be the feet of the perpendiculars from $P$ to $A C$ and $B D$ respectively. We have $S_{A B P}=S_{A B E}-S_{A E P}-S_{B E P}=\\frac{1}{2}(A E \\cdot B E-A E \\cdot E N-B E$. $E M)=\\frac{1}{2}(A M \\cdot B N-E M \\cdot E N)$. Similarly, $S_{C D P}=\\frac{1}{2}(C M \\cdot D N-E M$. $E N)$. Therefore, we obtain  $$ S_{A B P}-S_{C D P}=\\frac{A M \\cdot B N-C M \\cdot D N}{2} $$  Now suppose that $A B C D$ is cyclic. Then $P$ is the circumcenter of $A B C D$; hence $M$ and $N$ are the midpoints of $A C$ and $B D$. Hence $A M=C M$ and $B N=D N$; thus (1) gives us $S_{A B P}=S_{C D P}$. On the other hand, suppose that ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-641.jpg?height=311&width=410&top_left_y=703&top_left_x=875) $A B C D$ is not cyclic and let w.l.o.g. $P A=P B>P C=P D$. Then we must have $A M>C M$ and $B N>$ $D N$, and consequently by (1), $S_{A B P}>S_{C D P}$. This proves the other implication.  Second solution. Let $F$ and $G$ denote the midpoints of $A B$ and $C D$, and assume that $P$ is on the same side of $F G$ as $B$ and $C$. Since $P F \\perp A B$, $P G \\perp C D$, and $\\angle F E B=\\angle A B E, \\angle G E C=\\angle D C E$, a direct computation yields $\\angle F P G=\\angle F E G=90^{\\circ}+\\angle A B E+\\angle D C E$. Taking into account that $S_{A B P}=\\frac{1}{2} A B \\cdot F P=F E \\cdot F P$, we note that $S_{A B P}=S_{C D P}$ is equivalent to $F E \\cdot F P=G E \\cdot G P$, i.e., to $F E / E G=$ $G P / P F$. But this last is equivalent to triangles $E F G$ and $P G F$ being similar, which holds if and only if $E F P G$ is a parallelogram. This last is equivalent to $\\angle E F P=\\angle E G P$, or $2 \\angle A B E=2 \\angle D C E$. Thus $S_{A B P}=$ $S_{C D P}$ is equivalent to $A B C D$ being cyclic. Remark. The problems also allows an analytic solution, for example putting the $x$ and $y$ axes along the diagonals $A C$ and $B D$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "10", "problem_type": null, "exam": "IMO", "problem": "10. (AUS) Let $r_{1}, r_{2}, \\ldots, r_{n}$ be real numbers greater than or equal to 1 . Prove that  $$ \\frac{1}{r_{1}+1}+\\frac{1}{r_{2}+1}+\\cdots+\\frac{1}{r_{n}+1} \\geq \\frac{n}{\\sqrt[n]{r_{1} r_{2} \\cdots r_{n}}+1} $$", "solution": "10. We shall first prove the inequality for $n$ of the form $2^{k}, k=0,1,2, \\ldots$ The case $k=0$ is clear. For $k=1$, we have  $$ \\frac{1}{r_{1}+1}+\\frac{1}{r_{2}+1}-\\frac{2}{\\sqrt{r_{1} r_{2}}+1}=\\frac{\\left(\\sqrt{r_{1} r_{2}}-1\\right)\\left(\\sqrt{r_{1}}-\\sqrt{r_{2}}\\right)^{2}}{\\left(r_{1}+1\\right)\\left(r_{2}+1\\right)\\left(\\sqrt{r_{1} r_{2}}+1\\right)} \\geq 0 $$  For the inductive step it suffices to show that the claim for $k$ and 2 implies that for $k+1$. Indeed,  $$ \\begin{aligned} \\sum_{i=1}^{2^{k+1}} \\frac{1}{r_{i}+1} & \\geq \\frac{2^{k}}{\\sqrt[2^{k}]{r_{1} r_{2} \\cdots r_{2^{k}}}+1}+\\frac{2^{k}}{\\sqrt[2^{k}]{r_{2^{k}+1^{\\prime} r_{2^{k}}+2^{\\cdots r_{2^{k+1}}}+1}}} \\\\ & \\geq \\frac{2^{k+1}}{\\sqrt[2^{k+1}]{r_{1} r_{2} \\cdots r_{2^{k+1}}}+1} \\end{aligned} $$  and the induction is complete. We now show that if the statement holds for $2^{k}$, then it holds for every $n<2^{k}$ as well. Put $r_{n+1}=r_{n+2}=\\cdots=r_{2^{k}}=\\sqrt[n]{r_{1} r_{2} \\ldots r_{n}}$. Then (1) becomes  $$ \\frac{1}{r_{1}+1}+\\cdots+\\frac{1}{r_{n}+1}+\\frac{2^{k}-n}{\\sqrt[n]{r_{1} \\cdots r_{n}}+1} \\geq \\frac{2^{k}}{\\sqrt[n]{r_{1} \\cdots r_{n}}+1} $$  This proves the claim. Second solution. Define $r_{i}=e^{x_{i}}$, where $x_{i}>0$. The function $f(x)=\\frac{1}{1+e^{x}}$ is convex for $x>0$ : indeed, $f^{\\prime \\prime}(x)=\\frac{e^{x}\\left(e^{x}-1\\right)}{\\left(e^{x}+1\\right)^{3}}>0$. Thus by Jensen's inequality applied to $f\\left(x_{1}\\right), \\ldots, f\\left(x_{n}\\right)$, we get $\\frac{1}{r_{1}+1}+\\cdots+\\frac{1}{r_{n}+1} \\geq \\frac{n}{\\sqrt[n]{r_{1} \\cdots r_{n}}+1}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "11", "problem_type": null, "exam": "IMO", "problem": "11. (RUS) Let $x, y$, and $z$ be positive real numbers such that $x y z=1$. Prove that  $$ \\frac{x^{3}}{(1+y)(1+z)}+\\frac{y^{3}}{(1+z)(1+x)}+\\frac{z^{3}}{(1+x)(1+y)} \\geq \\frac{3}{4} $$", "solution": "11. The given inequality is equivalent to $x^{3}(x+1)+y^{3}(y+1)+z^{3}(z+1) \\geq$ $\\frac{3}{4}(x+1)(y+1)(z+1)$. By the A-G mean inequality, it will be enough to prove a stronger inequality:  $$ x^{4}+x^{3}+y^{4}+y^{3}+z^{4}+z^{3} \\geq \\frac{1}{4}\\left[(x+1)^{3}+(y+1)^{3}+(z+1)^{3}\\right] . $$  If we set $S_{k}=x^{k}+y^{k}+z^{k}$, (1) takes the form $S_{4}+S_{3} \\geq \\frac{1}{4} S_{3}+\\frac{3}{4} S_{2}+\\frac{3}{4} S_{1}+\\frac{3}{4}$. Note that by the A-G mean inequality, $S_{1}=x+y+z \\geq 3$. Thus it suffices to prove the following:  $$ \\text { If } S_{1} \\geq 3 \\text { and } m>n \\text { are positive integers, then } S_{m} \\geq S_{n} \\text {. } $$  This can be shown in many ways. For example, by Hölder's inequality,  $$ \\left(x^{m}+y^{m}+z^{m}\\right)^{n / m}(1+1+1)^{(m-n) / m} \\geq x^{n}+y^{n}+z^{n} . $$  (Another way is using the Chebyshev inequality: if $x \\geq y \\geq z$ then $x^{k-1} \\geq$ $y^{k-1} \\geq z^{k-1}$; hence $S_{k}=x \\cdot x^{k-1}+y \\cdot y^{k-1}+z \\cdot z^{k-1} \\geq \\frac{1}{3} S_{1} S_{k-1}$, and the claim follows by induction.)  Second solution. Assume that $x \\geq y \\geq z$. Then also $\\frac{1}{(y+1)(z+1)} \\geq$ $\\frac{1}{(x+1)(z+1)} \\geq \\frac{1}{(x+1)(y+1)}$. Hence Chebyshev's inequality gives that  $$ \\begin{aligned} & \\frac{x^{3}}{(1+y)(1+z)}+\\frac{y^{3}}{(1+x)(1+z)}+\\frac{z^{3}}{(1+x)(1+y)} \\\\ \\geq & \\frac{1}{3} \\frac{\\left(x^{3}+y^{3}+z^{3}\\right) \\cdot(3+x+y+z)}{(1+x)(1+y)(1+z)} \\end{aligned} $$  Now if we put $x+y+z=3 S$, we have $x^{3}+y^{3}+z^{3} \\geq 3 S$ and $(1+$ $x)(1+y)(1+z) \\leq(1+a)^{3}$ by the A-G mean inequality. Thus the needed inequality reduces to $\\frac{6 S^{3}}{(1+S)^{3}} \\geq \\frac{3}{4}$, which is obviously true because $S \\geq 1$. Remark. Both these solutions use only that $x+y+z \\geq 3$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "12", "problem_type": null, "exam": "IMO", "problem": "12. (POL) Let $n \\geq k \\geq 0$ be integers. The numbers $c(n, k)$ are defined as follows:  $$ \\begin{aligned} c(n, 0) & =c(n, n)=1 & & \\text { for all } n \\geq 0 \\\\ c(n+1, k) & =2^{k} c(n, k)+c(n, k-1) & & \\text { for } n \\geq k \\geq 1 \\end{aligned} $$  Prove that $c(n, k)=c(n, n-k)$ for all $n \\geq k \\geq 0$.", "solution": "12. The assertion is clear for $n=0$. We shall prove the general case by induction on $n$. Suppose that $c(m, i)=c(m, m-i)$ for all $i$ and $m \\leq n$. Then by the induction hypothesis and the recurrence formula we have $c(n+1, k)=2^{k} c(n, k)+c(n, k-1)$ and $c(n+1, n+1-k)=$ $2^{n+1-k} c(n, n+1-k)+c(n, n-k)=2^{n+1-k} c(n, k-1)+c(n, k)$. Thus it remains only to show that  $$ \\left(2^{k}-1\\right) c(n, k)=\\left(2^{n+1-k}-1\\right) c(n, k-1) $$  We prove this also by induction on $n$. By the induction hypothesis,  $$ c(n-1, k)=\\frac{2^{n-k}-1}{2^{k}-1} c(n-1, k-1) $$  and  $$ c(n-1, k-2)=\\frac{2^{k-1}-1}{2^{n+1-k}-1} c(n-1, k-1) $$  Using these formulas and the recurrence formula we obtain $\\left(2^{k}-1\\right) c(n, k)-$ $\\left(2^{n+1-k}-1\\right) c(n, k-1)=\\left(2^{2 k}-2^{k}\\right) c(n-1, k)-\\left(2^{n}-3 \\cdot 2^{k-1}+1\\right) c(n-$ $1, k-1)-\\left(2^{n+1-k}-1\\right) c(n-1, k-2)=\\left(2^{n}-2^{k}\\right) c(n-1, k-1)-\\left(2^{n}-\\right.$ $\\left.3 \\cdot 2^{k-1}+1\\right) c(n-1, k-1)-\\left(2^{k-1}-1\\right) c(n-1, k-1)=0$. This completes the proof. Second solution. The given recurrence formula resembles that of binomial coefficients, so it is natural to search for an explicit formula of the form $c(n, k)=\\frac{F(n)}{F(k) F(n-k)}$, where $F(m)=f(1) f(2) \\cdots f(m)($ with $F(0)=1$ ) and $f$ is a certain function from the natural numbers to the real numbers. If there is such an $f$, then $c(n, k)=c(n, n-k)$ follows immediately. After substitution of the above relation, the recurrence equivalently reduces to $f(n+1)=2^{k} f(n-k+1)+f(k)$. It is easy to see that $f(m)=2^{m}-1$ satisfies this relation. Remark. If we introduce the polynomial $P_{n}(x)=\\sum_{k=0}^{n} c(n, k) x^{k}$, the recurrence relation gives $P_{0}(x)=1$ and $P_{n+1}(x)=x P_{n}(x)+P_{n}(2 x)$. As a consequence of the problem, all polynomials in this sequence are symmetric, i.e., $P_{n}(x)=x^{n} P_{n}\\left(x^{-1}\\right)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "13", "problem_type": null, "exam": "IMO", "problem": "13. (BUL) ${ }^{\\mathrm{IMO} 6}$ Determine the least possible value of $f(1998)$, where $f$ is a function from the set $\\mathbb{N}$ of positive integers into itself such that for all $m, n \\in \\mathbb{N}$,  $$ f\\left(n^{2} f(m)\\right)=m[f(n)]^{2} $$", "solution": "13. Denote by $\\mathcal{F}$ the set of functions considered. Let $f \\in \\mathcal{F}$, and let $f(1)=a$. Putting $n=1$ and $m=1$ we obtain $f(f(z))=a^{2} z$ and $f\\left(a z^{2}\\right)=f(z)^{2}$ for all $z \\in \\mathbb{N}$. These equations, together with the original one, imply $f(x)^{2} f(y)^{2}=f(x)^{2} f\\left(a y^{2}\\right)=f\\left(x^{2} f\\left(f\\left(a y^{2}\\right)\\right)\\right)=f\\left(x^{2} a^{3} y^{2}\\right)=$ $f\\left(a(a x y)^{2}\\right)=f(a x y)^{2}$, or $f(a x y)=f(x) f(y)$ for all $x, y \\in \\mathbb{N}$. Thus $f(a x)=a f(x)$, and we conclude that  $$ a f(x y)=f(x) f(y) \\quad \\text { for all } x, y \\in \\mathbb{N} $$  We now prove that $f(x)$ is divisible by $a$ for each $x \\in \\mathbb{N}$. In fact, we inductively get that $f(x)^{k}=a^{k-1} f\\left(x^{k}\\right)$ is divisible by $a^{k-1}$ for every $k$. If $p^{\\alpha}$ and $p^{\\beta}$ are the exact powers of a prime $p$ that divide $f(x)$ and $a$ respectively, we deduce that $k \\alpha \\geq(k-1) \\beta$ for all $k$, so we must have $\\alpha \\geq \\beta$ for any $p$. Therefore $a \\mid f(x)$. Now we consider the function on natural numbers $g(x)=f(x) / a$. The above relations imply  $$ g(1)=1, \\quad g(x y)=g(x) g(y), \\quad g(g(x))=x \\quad \\text { for all } x, y \\in \\mathbb{N} $$  Since $g \\in \\mathcal{F}$ and $g(x) \\leq f(x)$ for all $x$, we may restrict attention to the functions $g$ only. Clearly $g$ is bijective. We observe that $g$ maps a prime to a prime. Assume to the contrary that $g(p)=u v, u, v>1$. Then $g(u v)=p$, so either $g(u)=1$ and $g(v)=1$. Thus either $g(1)=u$ or $g(1)=v$, which is impossible. We return to the problem of determining the least possible value of $g(1998)$. Since $g(1998)=g\\left(2 \\cdot 3^{3} \\cdot 37\\right)=g(2) \\cdot g(3)^{3} \\cdot g(37)$, and $g(2)$, $g(3), g(37)$ are distinct primes, $g(1998)$ is not smaller than $2^{3} \\cdot 3 \\cdot 5=120$. On the other hand, the value of 120 is attained for any function $g$ satisfying (2) and $g(2)=3, g(3)=2, g(5)=37, g(37)=5$. Hence the answer is 120 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "14", "problem_type": null, "exam": "IMO", "problem": "14. $(\\mathbf{G B R})^{\\mathrm{IMO} 4}$ Determine all pairs $(x, y)$ of positive integers such that $x^{2} y+$ $x+y$ is divisible by $x y^{2}+y+7$.", "solution": "14. If $x^{2} y+x+y$ is divisible by $x y^{2}+y+7$, then so is the number $y\\left(x^{2} y+\\right.$ $x+y)-x\\left(x y^{2}+y+7\\right)=y^{2}-7 x$. If $y^{2}-7 x \\geq 0$, then since $y^{2}-7 x<x y^{2}+y+7$, it follows that $y^{2}-7 x=0$. Hence $(x, y)=\\left(7 t^{2}, 7 t\\right)$ for some $t \\in \\mathbb{N}$. It is easy to check that these pairs really are solutions. If $y^{2}-7 x<0$, then $7 x-y^{2}>0$ is divisible by $x y^{2}+y+7$. But then $x y^{2}+y+7 \\leq 7 x-y^{2}<7 x$, from which we obtain $y \\leq 2$. For $y=1$, we are led to $x+8 \\mid 7 x-1$, and hence $x+8 \\mid 7(x+8)-(7 x-1)=57$. Thus the only possibilities are $x=11$ and $x=49$, and the obtained pairs $(11,1),(49,1)$ are indeed solutions. For $y=2$, we have $4 x+9 \\mid 7 x-4$, so that $7(4 x+9)-4(7 x-4)=79$ is divisible by $4 x+9$. We do not get any new solutions in this case. Therefore all required pairs $(x, y)$ are $\\left(7 t^{2}, 7 t\\right)(t \\in \\mathbb{N}),(11,1)$, and $(49,1)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "15", "problem_type": null, "exam": "IMO", "problem": "15. (AUS) Determine all pairs $(a, b)$ of real numbers such that $a\\lfloor b n\\rfloor=b\\lfloor a n\\rfloor$ for all positive integers $n$. (Note that $\\lfloor x\\rfloor$ denotes the greatest integer less than or equal to $x$.)", "solution": "15. The condition is obviously satisfied if $a=0$ or $b=0$ or $a=b$ or $a, b$ are both integers. We claim that these are the only solutions.  Suppose that $a, b$ belong to none of the above categories. The quotient $a / b=\\lfloor a\\rfloor /\\lfloor b\\rfloor$ is a nonzero rational number: let $a / b=p / q$, where $p$ and $q$ are coprime nonzero integers. Suppose that $p \\notin\\{-1,1\\}$. Then $p$ divides $\\lfloor a n\\rfloor$ for all $n$, so in particular $p$ divides $\\lfloor a\\rfloor$ and thus $a=k p+\\varepsilon$ for some $k \\in \\mathbb{N}$ and $0 \\leq \\varepsilon<1$. Note that $\\varepsilon \\neq 0$, since otherwise $b=k q$ would also be an integer. It follows that there exists an $n \\in \\mathbb{N}$ such that $1 \\leq n \\varepsilon<2$. But then $\\lfloor n a\\rfloor=\\lfloor k n p+n \\varepsilon\\rfloor=k n p+1$ is not divisible by $p$, a contradiction. Similarly, $q \\notin\\{-1,1\\}$ is not possible. Therefore we must have $p, q= \\pm 1$, and since $a \\neq b$, the only possibility is $b=-a$. However, this leads to $\\lfloor-a\\rfloor=-\\lfloor a\\rfloor$, which is not valid if $a$ is not an integer.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "16", "problem_type": null, "exam": "IMO", "problem": "16. (UKR) Determine the smallest integer $n \\geq 4$ for which one can choose four different numbers $a, b, c$, and $d$ from any $n$ distinct integers such that $a+b-c-d$ is divisible by 20 .", "solution": "16. Let $S$ be a set of integers such that for no four distinct elements $a, b, c, d \\in$ $S$, it holds that $20 \\mid a+b-c-d$. It is easily seen that there cannot exist distinct elements $a, b, c, d$ with $a \\equiv b$ and $c \\equiv d(\\bmod 20)$. Consequently, if the elements of $S$ give $k$ different residues modulo 20, then $S$ itself has at most $k+2$ elements. Next, consider these $k$ elements of $S$ with different residues modulo 20. They give $\\frac{k(k-1)}{2}$ different sums of two elements. For $k \\geq 7$ there are at least 21 such sums, and two of them, say $a+b$ and $c+d$, are equal modulo 20 ; it is easy to see that $a, b, c, d$ are discinct. It follows that $k$ cannot exceed 6 , and consequently $S$ has at most 8 elements. An example of a set $S$ with 8 elements is $\\{0,20,40,1,2,4,7,12\\}$. Hence the answer is $n=9$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "17", "problem_type": null, "exam": "IMO", "problem": "17. (GBR) A sequence of integers $a_{1}, a_{2}, a_{3}, \\ldots$ is defined as follows: $a_{1}=1$, and for $n \\geq 1, a_{n+1}$ is the smallest integer greater than $a_{n}$ such that $a_{i}+a_{j} \\neq 3 a_{k}$ for any $i, j, k$ in $\\{1,2, \\ldots, n+1\\}$, not necessarily distinct. Determine $a_{1998}$.", "solution": "17. Initially, we determine that the first few values for $a_{n}$ are $1,3,4,7,10$, $12,13,16,19,21,22,25$. Since these are exactly the numbers of the forms $3 k+1$ and $9 k+3$, we conjecture that this is the general pattern. In fact, it is easy to see that the equation $x+y=3 z$ has no solution in the set $K=\\{3 k+1,9 k+3 \\mid k \\in \\mathbb{N}\\}$. We shall prove that the sequence $\\left\\{a_{n}\\right\\}$ is actually this set ordered increasingly. Suppose $a_{n}>25$ is the first member of the sequence not belonging to $K$. We have several cases: (i) $a_{n}=3 r+2, r \\in \\mathbb{N}$. By the assumption, one of $r+1, r+2, r+3$ is of the form $3 k+1$ (and smaller than $a_{n}$ ), and therefore is a member $a_{i}$ of the sequence. Then $3 a_{i}$ equals $a_{n}+1$, $a_{n}+4$, or $a_{n}+7$, which is a contradiction because $1,4,7$ are in the sequence. (ii) $a_{n}=9 r, r \\in \\mathbb{N}$. Then $a_{n}+a_{2}=3(3 r+1)$, although $3 r+1$ is in the sequence, a contradiction. (iii) $a_{n}=9 r+6, r \\in \\mathbb{N}$. Then one of the numbers $3 r+3,3 r+6,3 r+9$ is a member $a_{j}$ of the sequence, and thus $3 a_{j}$ is equal to $a_{n}+3$, $a_{n}+12$, or $a_{n}+21$, where $3,12,21$ are members of the sequence, again a contradiction. Once we have revealed the structure of the sequence, it is easy to compute $a_{1998}$. We have $1998=4 \\cdot 499+2$, which implies $a_{1998}=9 \\cdot 499+a_{2}=4494$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "18", "problem_type": null, "exam": "IMO", "problem": "18. (BUL) Determine all positive integers $n$ for which there exists an integer $m$ such that $2^{n}-1$ is a divisor of $m^{2}+9$.", "solution": "18. We claim that, if $2^{n}-1$ divides $m^{2}+9$ for some $m \\in \\mathbb{N}$, then $n$ must be a power of 2 . Suppose otherwise that $n$ has an odd divisor $d>1$. Then $2^{d}-1 \\mid 2^{n}-1$ is also a divisor of $m^{2}+9=m^{2}+3^{2}$. However, $2^{d}-1$ has some prime divisor $p$ of the form $4 k-1$, and by a well-known fact, $p$ divides both $m$ and 3 . Hence $p=3$ divides $2^{d}-1$, which is impossible, because for $d$ odd, $2^{d} \\equiv 2(\\bmod 3)$. Hence $n=2^{r}$ for some $r \\in \\mathbb{N}$. Now let $n=2^{r}$. We prove the existence of $m$ by induction on $r$. The case $r=1$ is trivial. Now for any $r>1$ note that $2^{2^{r}}-1=\\left(2^{2^{r-1}}-1\\right)\\left(2^{2^{r-1}}+\\right.$ 1). The induction hypothesis claims that there exists an $m_{1}$ such that $2^{2^{r-1}}-1 \\mid m_{1}^{2}+9$. We also observe that $2^{2^{r-1}}+1 \\mid m_{2}^{2}+9$ for simple $m_{2}=3 \\cdot 2^{2^{r-2}}$. By the Chinese remainder theorem, there is an $m \\in \\mathbb{N}$ that satisfies $m \\equiv m_{1}\\left(\\bmod 2^{2^{r-1}}-1\\right)$ and $m \\equiv m_{2}\\left(\\bmod 2^{2^{r-1}}+1\\right)$. It is easy to see that this $m^{2}+9$ will be divisible by both $2^{2^{r-1}}-1$ and $2^{2^{r-1}}+1$, i.e., that $2^{2^{r}}-1 \\mid m^{2}+9$. This completes the induction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. (BLR) ${ }^{\\mathrm{IMO} 3}$ For any positive integer $n$, let $\\tau(n)$ denote the number of its positive divisors (including 1 and itself). Determine all positive integers $m$ for which there exists a positive integer $n$ such that $\\frac{\\tau\\left(n^{2}\\right)}{\\tau(n)}=m$.", "solution": "19. For $n=p_{1}^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\cdots p_{r}^{\\alpha_{r}}$, where $p_{i}$ are distinct primes and $\\alpha_{i}$ natural numbers, we have $\\tau(n)=\\left(\\alpha_{1}+1\\right) \\cdots\\left(\\alpha_{r}+1\\right)$ and $\\tau\\left(n^{2}\\right)=\\left(2 \\alpha_{1}+1\\right) \\ldots\\left(2 \\alpha_{r}+1\\right)$. Putting $k_{i}=\\alpha_{i}+1$, the problem reduces to determining all natural values of $m$ that can be represented as  $$ m=\\frac{2 k_{1}-1}{k_{1}} \\cdot \\frac{2 k_{2}-1}{k_{2}} \\cdots \\frac{2 k_{r}-1}{k_{r}} $$  Since the numerator $\\tau\\left(n^{2}\\right)$ is odd, $m$ must be odd too. We claim that every odd $m$ has a representation of the form (1). The proof will be done by induction. This is clear for $m=1$. Now for every $m=2 k-1$ with $k$ odd the result follows easily, since $m=\\frac{2 k-1}{k} \\cdot k$, and $k$ can be written as (1). We cannot do the same if $k$ is even; however, in the case $m=4 k-1$ with $k$ odd, we can write it as $m=\\frac{12 k-3}{6 k-1} \\cdot \\frac{6 k-1}{3 k} \\cdot k$, and this works. In general, suppose that $m=2^{t} k-1$, with $k$ odd. Following the same pattern, we can write $m$ as  $$ m=\\frac{2^{t}\\left(2^{t}-1\\right) k-\\left(2^{t}-1\\right)}{2^{t-1}\\left(2^{t}-1\\right) k-\\left(2^{t-1}-1\\right)} \\cdots \\frac{4\\left(2^{t}-1\\right) k-3}{2\\left(2^{t}-1\\right) k-1} \\cdot \\frac{2\\left(2^{t}-1\\right) k-1}{\\left(2^{t}-1\\right) k} \\cdot k $$  The induction is finished. Hence $m$ can be represented as $\\frac{\\tau\\left(n^{2}\\right)}{\\tau(n)}$ if and only if it is odd.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "2. (POL) Let $A B C D$ be a cyclic quadrilateral. Let $E$ and $F$ be variable points on the sides $A B$ and $C D$, respectively, such that $A E: E B=C F$ : $F D$. Let $P$ be the point on the segment $E F$ such that $P E: P F=A B$ : $C D$. Prove that the ratio between the areas of triangles $A P D$ and $B P C$ does not depend on the choice of $E$ and $F$.", "solution": "2. If $A D$ and $B C$ are parallel, then $A B C D$ is an isosceles trapezoid with $A B=C D$, so $P$ is the midpoint of $E F$. Let $M$ and $N$ be the midpoints of $A B$ and $C D$. Then $M N \\| B C$, and the distance $d(E, M N)$ equals the distance $d(F, M N)$ because $B$ and $D$ are the same distance from $M N$ and $E M / B M=F N / D N$. It follows that the midpoint $P$ of $E F$ lies on $M N$, and consequently $S_{A P D}: S_{B P C}=A D: B C$. If $A D$ and $B C$ are not parallel, then they meet at some point $Q$. It is plain that $\\triangle Q A B \\sim \\triangle Q C D$, and since $A E / A B=C F / C D$, we also deduce that $\\triangle Q A E \\sim \\triangle Q C F$. Therefore $\\angle A Q E=\\angle C Q F$. Further, from these similarities we obtain $Q E / Q F=Q A / Q C=A B / C D=P E / P F$, which in turn means that $Q P$ is the internal bisector of $\\angle E Q F$. But since $\\angle A Q E=\\angle C Q F$, this is also the internal bisector of $\\angle A Q B$. Hence $P$ is at equal distances from $A D$ and $B C$, so again $S_{A P D}: S_{B P C}=A D: B C$. Remark. The part $A B \\| C D$ could also be regarded as a limiting case of the other part. Second solution. Denote $\\lambda=\\frac{A E}{A B}, A B=a, B C=b, C D=c, D A=d$, $\\angle D A B=\\alpha, \\angle A B C=\\beta$. Since $d(P, A D)=\\frac{c \\cdot d(E, A D)+a \\cdot d(F, A D)}{a+c}$, we have $S_{A P D}=\\frac{c S_{E A D}+a S_{F A D}}{a+c}=\\frac{\\lambda c S_{A B D}+(1-\\lambda) a S_{A C D}}{a+c}$. Since $S_{A B D}=\\frac{1}{2} a d \\sin \\alpha$ and $S_{A C D}=\\frac{1}{2} c d \\sin \\beta$, we are led to $S_{A P D}=\\frac{a c d}{a+c}[\\lambda \\sin \\alpha+(1-\\lambda) \\sin \\beta]$, and analogously $S_{B P C}=\\frac{a b c}{a+c}[\\lambda \\sin \\alpha+(1-\\lambda) \\sin \\beta]$. Thus we obtain $S_{A P D}: S_{B P C}=d: b$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "20", "problem_type": null, "exam": "IMO", "problem": "20. (ARG) Prove that for each positive integer $n$, there exists a positive integer with the following properties: (i) It has exactly $n$ digits. (ii) None of the digits is 0 . (iii) It is divisible by the sum of its digits.", "solution": "20. We first consider the special case $n=3^{r}$. Then the simplest choice $\\frac{10^{n}-1}{9}=$ $11 \\ldots 1$ ( $n$ digits) works. This can be shown by induction: it is true for $r=$ 1, while the inductive step follows from $10^{3^{r}}-1=\\left(10^{3^{r-1}}-1\\right)\\left(10^{2 \\cdot 3^{r-1}}+\\right.$ $10^{3^{r-1}}+1$ ), because the second factor is divisible by 3 . In the general case, let $k \\geq n / 2$ be a positive integer and $a_{1}, \\ldots, a_{n-k}$ be nonzero digits. We have  $$ \\begin{aligned} A & =\\left(10^{k}-1\\right) \\overline{a_{1} a_{2} \\ldots a_{n-k}} \\\\ & =\\overline{a_{1} a_{2} \\ldots a_{n-k-1} a_{n-k}^{\\prime} \\underbrace{99 \\ldots 99}_{2 k-n}} b_{1} b_{2} \\ldots b_{n-k-1} b_{n-k}^{\\prime} \\end{aligned} $$  where $a_{n-k}^{\\prime}=a_{n-k}-1, b_{i}=9-a_{i}$, and $b_{n-k}^{\\prime}=9-a_{n-k}^{\\prime}$. The sum of digits of $A$ equals $9 k$ independently of the choice of digits $a_{1}, \\ldots, a_{n-k}$. Thus we need only choose $k \\geq \\frac{n}{2}$ and digits $a_{1}, \\ldots, a_{n-k-1} \\notin\\{0,9\\}$ and $a_{n-k} \\in\\{0,1\\}$ in order for the conditions to be fulfilled. Let us choose  $$ k=\\left\\{\\begin{array}{l} 3^{r}, \\quad \\text { if } 3^{r}<n \\leq 2 \\cdot 3^{r} \\text { for some } r \\in \\mathbb{Z} \\\\ 2 \\cdot 3^{r}, \\text { if } 2 \\cdot 3^{r}<n \\leq 3^{r+1} \\text { for some } r \\in \\mathbb{Z} \\end{array}\\right. $$  and $\\overline{a_{1} a_{2} \\ldots a_{n-k}}=\\overline{22 \\ldots 2}$. The number  $$ A=\\overline{\\underbrace{22 \\ldots 2}_{n-k-1} 1 \\underbrace{99 \\ldots 99}_{2 k-n} \\underbrace{77 \\ldots 7}_{n-k-1} 8} $$  thus obtained is divisible by $2 \\cdot\\left(10^{k}-1\\right)$, which is, as explained above, divisible by $18 \\cdot 3^{r}$. Finally, the sum of digits of $A$ is either $9 \\cdot 3^{r}$ or $18 \\cdot 3^{r}$; thus $A$ has the desired properties.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. (CAN) Let $a_{0}, a_{1}, a_{2}, \\ldots$ be an increasing sequence of nonnegative integers such that every nonnegative integer can be expressed uniquely in the form $a_{i}+2 a_{j}+4 a_{k}$, where $i, j, k$ are not necessarily distinct. Determine $a_{1998}$.", "solution": "21. Such a sequence is obviously strictly increasing. We note that it must be unique. Indeed, given $a_{0}, a_{1}, \\ldots, a_{n-1}$, then $a_{n}$ is the least positive integer not of the form $a_{i}+2 a_{j}+4 a_{k}, i, j, k<n$. We easily get that the first few $a_{n}$ 's are $0,1,8,9,64,65,72,73, \\ldots$ Let $\\left\\{c_{n}\\right\\}$ be the increasing sequence of all positive integers that consist of zeros and ones in base 8 , i.e., those of the form $t_{0}+2^{3} t_{1}+\\cdots+2^{3 q} t_{q}$ where $t_{i} \\in\\{0,1\\}$. We claim that $a_{n}=c_{n}$. To prove this, it is enough to show that each $m \\in \\mathbb{N}$ can be uniquely written as $c_{i}+2 c_{j}+4 c_{k}$. If $m=t_{0}+2 t_{1}+\\cdots+2^{r} t_{r}\\left(t_{i} \\in\\{0,1\\}\\right)$, then $m=c_{i}+2 c_{j}+2^{2} c_{k}$ is obviously possible if and only if $c_{i}=t_{0}+2^{3} t_{3}+2^{6} t_{6}+\\cdots, c_{j}=t_{1}+2^{3} t_{4}+\\ldots$, and $c_{k}=t_{2}+2^{3} t_{5}+\\cdots$. Hence for $n=s_{0}+2 s_{1}+\\cdots+2^{r} s_{r}$ we have $a_{n}=s_{0}+8 s_{1}+\\cdots+8^{r} s_{r}$. In particular, $1998=2+2^{2}+2^{3}+2^{6}+2^{7}+2^{8}+2^{9}+2^{10}$, so $a_{1998}=$ $8+8^{2}+8^{3}+8^{6}+8^{7}+8^{8}+8^{9}+8^{10}=1227096648$. Second solution. Define $f(x)=x^{a_{0}}+x^{a_{1}}+\\cdots$. Then the assumed property of $\\left\\{a_{n}\\right\\}$ gives  $$ f(x) f\\left(x^{2}\\right) f\\left(x^{4}\\right)=\\sum_{i, j, k} x^{a_{i}+2 a_{j}+4 a_{k}}=\\sum_{n} x^{n}=\\frac{1}{1-x} $$  We also get as a consequence $f\\left(x^{2}\\right) f\\left(x^{4}\\right) f\\left(x^{8}\\right)=\\frac{1}{1-x^{2}}$, which gives $f(x)=$ $(1+x) f\\left(x^{8}\\right)$. Continuing this, we obtain  $$ f(x)=(1+x)\\left(1+x^{8}\\right)\\left(1+x^{8^{2}}\\right) \\cdots $$  Hence the $a_{n}$ 's are integers that have only 0 's and 1 's in base 8.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "22", "problem_type": null, "exam": "IMO", "problem": "22. (UKR) A rectangular array of numbers is given. In each row and each column, the sum of all numbers is an integer. Prove that each nonintegral number $x$ in the array can be changed into either $\\lceil x\\rceil$ or $\\lfloor x\\rfloor$ so that the row sums and column sums remain unchanged. (Note that $\\lceil x\\rceil$ is the least integer greater than or equal to $x$, while $\\lfloor x\\rfloor$ is the greatest integer less than or equal to $x$.)", "solution": "22. We can obviously change each $x$ into $\\lfloor x\\rfloor$ or $\\lceil x\\rceil$ so that the column sums remain unchanged. However, this does not necessarily match the row sums as well, so let us consider the sum $S$ of the absolute values of the changes in the row sums. It is easily seen that $S$ is even, and we want it to be 0 . A row may have a higher or lower sum than desired. Let us mark a cell by - if its entry $x$ was changed to $\\lfloor x\\rfloor$, and by + if it was changed to $\\lceil x\\rceil$ instead. We call a row $R_{2}$ accessible from a row $R_{1}$ if there is a column $C$ such that $C \\cap R_{1}$ is marked + and $C \\cap R_{2}$ is marked - . Note that a column containing a + must contain a - as well, because column sums are unchanged. Hence from each row with a higher sum we can access another row. Assume that the row sum in $R_{1}$ is higher. If $R_{1}, R_{2}, \\ldots, R_{k}$ is a sequence of rows such that $R_{i+1}$ is accessible from $R_{i}$ via some column $C_{i}$ and such that the row sum in $R_{k}$ is lower, then by changing the signs in $C_{i} \\cap R_{i}$ and $C_{i} \\cap R_{i+1}(i=1,2, \\ldots, k-1)$ we decrease $S$ by 2 , leaving column sums unchanged. We claim that such a sequence of rows always exists. Let $\\mathcal{R}$ be the union of all rows that are accessible from $R_{1}$, directly or indirectly; let $\\overline{\\mathcal{R}}$ be the union of the remaining rows. We show that for any column $C$, the sum in $\\mathcal{R} \\cap C$ is not higher. If $\\mathcal{R} \\cap C$ contains no +'s, then this is clear. If $\\mathcal{R} \\cap C$ contains a + , since the rows of $\\overline{\\mathcal{R}}$ are not accessible, the set $\\overline{\\mathcal{R}} \\cap C$ contains no -'s. It follows that the sum in $\\overline{\\mathcal{R}} \\cap C$ is not lower, and since column sums are unchanged, we again come to the same conclusion. Thus the total sum in $\\mathcal{R}$ is not higher. Therefore, there is a row in $\\mathcal{R}$ with too low a sum, justifying our claim.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "23", "problem_type": null, "exam": "IMO", "problem": "23. (BLR) Let $n$ be an integer greater than 2. A positive integer is said to be attainable if it is 1 or can be obtained from 1 by a sequence of operations with the following properties: (i) The first operation is either addition or multiplication. (ii) Thereafter, additions and multiplications are used alternately. (iii) In each addition one can choose independently whether to add 2 or $n$. (iv) In each multiplication, one can choose independently whether to multiply by 2 or by $n$. A positive integer that cannot be so obtained is said to be unattainable. (a) Prove that if $n \\geq 9$, there are infinitely many unattainable positive integers. (b) Prove that if $n=3$, all positive integers except 7 are attainable.", "solution": "23. (a) If $n$ is even, then every odd integer is unattainable. Assume that $n \\geq 9$ is odd. Let $a$ be obtained by addition from some $b$, and $b$ from $c$ by multiplication. Then $a$ is $2 c+2,2 c+n, n c+2$, or $n c+n$, and is in every case congruent to $2 c+2$ modulo $n-2$. In particular, if $a \\equiv-2$ $(\\bmod n-2)$, then also $b \\equiv-4$ and $c \\equiv-2(\\bmod n-2)$. Now consider any $a=k n(n-2)-2$, where $k$ is odd. If it is attainable, but not divisible by 2 or $n$, it must have been obtained by addition. Thus all predecessors of $a$ are congruent to either -2 or $-4(\\bmod$ $n-2$ ), and none of them equals 1 , a contradiction. (b) Call an attainable number addy if the last operation is addition, and multy if the last operation is multiplication. We prove the following claims by simultaneous induction on $k$ : (1) $n=6 k$ is both addy and multy; (2) $n=6 k+1$ is addy for $k \\geq 2$; (3) $n=6 k+2$ is addy for $k \\geq 1$; (4) $n=6 k+3$ is addy; (5) $n=6 k+4$ is multy for $k \\geq 1$; (6) $n=6 k+5$ is addy.  The cases $k \\leq 1$ are easily verified. For $k \\geq 2$, suppose all six statements hold up to $k-1$.  Since $6 k-3$ is addy, $6 k$ is multy. Next, $6 k-2$ is multy, so both $6 k=(6 k-2)+2$ and $6 k+1=(6 k-2)+3$ are addy. Since $6 k$ is multy, both $6 k+2$ and $6 k+3$ are addy. Number $6 k+4=2 \\cdot(3 k+2)$ is multy, because $3 k+2$ is addy (being either $6 l+2$ or $6 l+5)$. Finally, we have $6 k+5=3 \\cdot(2 k+1)+2$. Since $2 k+1$ is $6 l+1,6 l+3$, or $6 l+5$, it is addy except for 7 . Hence $6 k+5$ is addy except possibly for 23 . But $23=((1 \\cdot 2+2) \\cdot 2+2) \\cdot 2+3$ is also addy. This completes the induction. Now 1 is given and $2=1 \\cdot 2,4=1+3$. It is easily checked that 7 is not attainable, and hence it is the only unattainable number.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "24", "problem_type": null, "exam": "IMO", "problem": "24. (SWE) Cards numbered 1 to 9 are arranged at random in a row. In a move, one may choose any block of consecutive cards whose numbers are in ascending or descending order, and switch the block around. For example, $91 \\underline{6532748}$ may be changed to $91 \\underline{3562748}$. Prove that in at most 12 moves, one can arrange the 9 cards so that their numbers are in ascending or descending order.", "solution": "24. Let $f(n)$ be the minimum number of moves needed to monotonize any permutation of $n$ distinct numbers. Let us be given a permutation $\\pi$ of $\\{1,2, \\ldots, n\\}$, and let $k$ be the first element of $\\pi$. In $f(n-1)$ moves, we can transform $\\pi$ to either $(k, 1,2, \\ldots, k-1, k+1, \\ldots, n)$ or $(k, n, n-1, \\ldots, k+$ $1, k-1, \\ldots, 1)$. Now the former can be changed to $(k, k-1, \\ldots, 2,1, k+$ $1, \\ldots, n)$, which is then monotonized in the next move. Similarly, the latter also can be monotonized in two moves. It follows that $f(n) \\leq f(n-1)+2$. Thus we shall be done if we show that $f(5) \\leq 4$. First we note that $f(3)=1$. Consider a permutation of $\\{1,2,3,4\\}$. If either 1 or 4 is the first or the last element, we need one move to monotonize the other three elements, and at most one more to monotonize the whole permutation. Of the remaining four permutations, $(2,1,4,3)$ and $(3,4,1,2)$ can also be monotonized in two moves. The permutations $(2,4,1,3)$ and $(3,1,4,2)$ require 3 moves, but by this we can choose whether to change them into $(1,2,3,4)$ or $(4,3,2,1)$. We now consider a permutation of $\\{1,2,3,4,5\\}$. If either 1 or 5 is in the first or last position, we can monotonize the rest in 3 moves, but in such a way that the whole permutation can be monotonized in the next move. If this is not the case, then either 1 or 5 is in the second or fourth position. Then we simply switch it to the outside in one move and continue as in the former case. Hence $f(5)=4$, as desired.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "25", "problem_type": null, "exam": "IMO", "problem": "25. (NZL) Let $U=\\{1,2, \\ldots, n\\}$, where $n \\geq 3$. A subset $S$ of $U$ is said to be split by an arrangement of the elements of $U$ if an element not in $S$ occurs in the arrangement somewhere between two elements of $S$. For example, 13542 splits $\\{1,2,3\\}$ but not $\\{3,4,5\\}$. Prove that for any $n-2$ subsets of $U$, each containing at least 2 and at most $n-1$ elements, there is an arrangement of the elements of $U$ that splits all of them.", "solution": "25. We use induction on $n$. For $n=3$, we have a single two-element subset $\\{i, j\\}$ that is split by $(i, k, j)$ (where $k$ is the third element of $U$ ). Assume that the result holds for some $n \\geq 3$, and consider a family $\\mathcal{F}$ of $n-1$ proper subsets of $U=\\{1,2, \\ldots, n+1\\}$, each with at least 2 elements. To continue the induction, we need an element $a \\in U$ that is contained in all $n$-element subsets of $\\mathcal{F}$, but in at most one of the two-element subsets. We claim that such an $a$ exists. Let $\\mathcal{F}$ contain $k n$-element subsets and $m 2$-element subsets $(k+m \\leq n-1)$. The intersection of the $n$-element subsets contains exactly $n+1-k \\geq m+2$ elements. On the other hand, at most $m$ elements belong to more than one 2 -element subset, which justifies our claim.  Now let $A$ be the 2-element subset that contains $a$, if it exists; otherwise, let $A$ be any subset from $\\mathcal{F}$ containing $a$. Excluding $a$ from all the subsets from $\\mathcal{F} \\backslash\\{A\\}$, we get at most $n-2$ subsets of $U \\backslash\\{a\\}$ with at least 2 and at most $n-1$ elements. By the inductive hypothesis, we can arrange $U \\backslash\\{a\\}$ so that we split all the subsets of $\\mathcal{F}$ except $A$. It remains to place $a$, and we shall make a desired arrangement if we put it anywhere away from $A$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "26", "problem_type": null, "exam": "IMO", "problem": "26. (IND) ${ }^{\\mathrm{IMO} 2}$ In a contest, there are $m$ candidates and $n$ judges, where $n \\geq 3$ is an odd integer. Each candidate is evaluated by each judge as either pass or fail. Suppose that each pair of judges agrees on at most $k$ candidates. Prove that $\\frac{k}{m} \\geq \\frac{n-1}{2 n}$.", "solution": "26. Put $n=2 r+1$. Since each of the $\\binom{n}{2}$ pairs of judges agrees on at most two candidates, the total number of agreements is at most $k\\binom{n}{2}$. On the other hand, if the $i$ th candidate is passed by $x_{i}$ judges and failed by $n-x_{i}$ judges, then the number of agreements on this candidate equals  $$ \\binom{x_{i}}{2}+\\binom{n-x_{i}}{2}=\\frac{x_{i}^{2}+\\left(n-x_{i}\\right)^{2}-n}{2} \\geq \\frac{r^{2}+(n-r)^{2}-n}{2}=\\frac{(n-1)^{2}}{4} . $$  Therefore the total number of agreements is at least $\\frac{m(n-1)^{2}}{4}$, which implies that  $$ k\\binom{n}{2} \\geq \\frac{m(n-1)^{2}}{4}, \\quad \\text { hence } \\quad \\frac{k}{m} \\geq \\frac{n-1}{2 n} . $$  Remark. The obtained inequality is sharp. Indeed, if $m=\\binom{2 r+1}{r}$ and each candidate is passed by a different subset of $r$ judges, we get equality. A similar example shows that the result is not valid for even $n$. In that case the weaker estimate $\\frac{k}{m} \\geq \\frac{n-2}{2 n-2}$ holds.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "27", "problem_type": null, "exam": "IMO", "problem": "27. (BLR) Ten points such that no three of them lie on a line are marked in the plane. Each pair of points is connected with a segment. Each of these segments is painted with one of $k$ colors in such a way that for any $k$ of the ten points, there are $k$ segments each joining two of them with no two being painted the same color. Determine all integers $k, 1 \\leq k \\leq 10$, for which this is possible.", "solution": "27. Since this is essentially a graph problem, we call the points and segments vertices and edges of the graph. We first prove that the task is impossible if $k \\leq 4$. Cases $k \\leq 2$ are trivial. If $k=3$, then among the edges from a vertex $A$ there are two of the same color, say $A B$ and $A C$, so we don't have all the three colors among the edges joining $A, B, C$. Now let $k=4$, and assume that there is a desired coloring. Consider the edges incident with a vertex $A$. At least three of them have the same color, say blue. Suppose that four of them, $A B, A C, A D, A E$, are blue. There is a blue edge, say $B C$, among the ones joining $B, C, D, E$. Then four of the edges joining $A, B, C, D$ are blue, and we cannot complete the coloring. So, exactly three edges from $A$ are blue: $A B, A C, A D$. Also, of the edges connecting any three of the 6 vertices other than $A, B, C, D$, one is blue (because the edges joining them with $A$ are not so). By a classical result, there is a blue triangle $E F G$ with vertices among these six. Now one of $E B, E C, E D$ must be blue as well, because none of $B C, B D, C D$ is. Let it be $E B$. Then four of the edges joining $B, E, F, G$ are blue, which is impossible. For $k=5$ the task is possible. Label the vertices $0,1, \\ldots, 9$. For each color, we divide the vertices into four groups and paint in this color every edge joining two from the same group, as shown below. Then among any 5 vertices, 2 must belong to the same group, and the edge connecting them has the considered color.  | yellow: | 011220 | 366993 | 57 | 48 | | :--- | :--- | :--- | :--- | :--- | | red: | 233442 | 588115 | 79 | 60 | | blue: | 455664 | 700337 | 91 | 82 | | green: | 677886 | 922559 | 13 | 04 | | orange: | 899008 | 144771 | 35 | 26. |  A desired coloring can be made for $k \\geq 6$ as well. Paint the edge $i j$ in the $(i+j)$ th color for $i<j \\leq 8$, and in the $2 i$ th color if $j=9$ (the addition being modulo 9 ). We can ignore the edges painted with the extra colors. Then the edges of one color appear as five disjoint segments, so that any complete $k$-graph for $k \\geq 5$ contains one of them.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "28", "problem_type": null, "exam": "IMO", "problem": "28. (IRN) A solitaire game is played on an $m \\times n$ rectangular board, using $m n$ markers that are white on one side and black on the other. Initially, each square of the board contains a marker with its white side up, except for one corner square, which contains a marker with its black side up. In each move, one can take away one marker with its black side up, but must then turn over all markers that are in squares having an edge in common with the square of the removed marker. Determine all pairs $(m, n)$ of positive integers such that all markers can be removed from the board.", "solution": "28. Let $A$ be the number of markers with white side up, and $B$ the number of pairs of markers whose squares share a side. We claim that $A+B$ does not change its parity as the game progresses. Suppose that in some move we remove a marker that has exactly $k$ neighbors, among them $r$ with white side up ( $0 \\leq r \\leq k \\leq 4$ ). Of course, this marker has its black side up. When it is removed, the $r$ white markers get black side up, while the $k-r$ black ones become white. Thus $A$ changes by $k-2 r$. As for $B$, it decreases by $k$. It follows that $A$ decreases by $2 r$ and preserves its parity, as claimed. Initially, $A=m n-1$ and $B=m(n-1)+n(m-1)$; hence $A+B$ equals $3 m n-m-n-1$. If we succeed in removing all the markers, we end up with $A+B=0$. Hence $3 m n-m-n-1=(m-1)(n-1)+2(m n-1)$ must be even, or equivalently at least one of $m$ and $n$ is odd. On the other hand, the game can be finished successfully if $m$ or $n$ is odd. Assume that $m$ is odd. As shown in the picture, we can arrive at the position (1) in $m$ moves; with $\\frac{m+1}{2}$ moves we reduce it to the position $\\left(1 \\frac{1}{2}\\right)$, and with the next $\\frac{m-1}{2}$ moves to the position (2). We continue until we empty all the columns. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-654.jpg?height=336&width=897&top_left_y=1605&top_left_x=342)", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "3. (UKR) ${ }^{\\mathrm{IMO}}$ Let $I$ be the incenter of triangle $A B C$. Let $K, L$, and $M$ be the points of tangency of the incircle of $A B C$ with $A B, B C$, and $C A$, respectively. The line $t$ passes through $B$ and is parallel to $K L$. The lines $M K$ and $M L$ intersect $t$ at the points $R$ and $S$. Prove that $\\angle R I S$ is acute.", "solution": "3. Lemma. If $U, W, V$ are three points on a line $l$ in this order, and $X$ a point in the plane with $X W \\perp U V$, then $\\angle U X V<90^{\\circ}$ if and only if $X W^{2}>U W \\cdot V W$. Proof. Let $X W^{2}>U W \\cdot V W$, and let $X_{0}$ be a point on the segment $X W$ such that $X_{0} W^{2} \\geq U W \\cdot V W$. Then $X_{0} W / U W=V W / X_{0} W$, so that triangles $X_{0} W U$ and $V W X_{0}$ are similar. Thus $\\angle U X_{0} V=\\angle U X_{0} W+$ $\\angle W U X_{0}=90^{\\circ}$, which immediately implies that $\\angle U X V<90^{\\circ}$. Similarly, if $X W^{2} \\leq U W \\cdot V W$, then $\\angle U X V \\geq 90^{\\circ}$. Since $B I \\perp R S$, it will be enough by the lemma to show that $B I^{2}>$ $B R \\cdot B S$. Note that $\\triangle B K R \\sim \\triangle B S L$ : in fact, we have $\\angle K B R=\\angle S B L=$ $90^{\\circ}-\\beta / 2$ and $\\angle B K R=\\angle A K M=\\angle K L M=\\angle B S L=90^{\\circ}-\\alpha / 2$. In particular, we obtain $B R / B K=B L / B S=B K / B S$, so that $B R \\cdot B S=$ $B K^{2}<B I^{2}$ 。 Second solution. Let $E, F$ be the midpoints of $K M$ and $L M$ respectively. The quadrilaterals $R B I E$ and $S B I F$ are inscribed in the circles with diameters $I R$ and $I S$. Now we have $\\angle R I S=\\angle R M S+\\angle I R M+\\angle I S M=$ $90^{\\circ}-\\beta / 2+\\angle I B E+\\angle I B F=90^{\\circ}-\\beta / 2+\\angle E B F$. On the other hand, $B E$ and $B F$ are medians in $\\triangle B K M$ and $\\triangle B L M$ in which $B M>B K$ and $B M>B L$. We conclude that $\\angle M B E<\\frac{1}{2} \\angle M B K$ and $\\angle M B F<\\frac{1}{2} \\angle M B L$. Adding these two inequalities gives $\\angle E B F<$ $\\beta / 2$. Therefore $\\angle R I S<90^{\\circ}$. Remark. It can be shown (using vectors) that the statement remains true for an arbitrary line $t$ passing through $B$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "4. (ARM) Let $M$ and $N$ be points inside triangle $A B C$ such that  $$ \\angle M A B=\\angle N A C \\quad \\text { and } \\quad \\angle M B A=\\angle N B C . $$  Prove that  $$ \\frac{A M \\cdot A N}{A B \\cdot A C}+\\frac{B M \\cdot B N}{B A \\cdot B C}+\\frac{C M \\cdot C N}{C A \\cdot C B}=1 $$", "solution": "4. Let $K$ be the point on the ray $B N$ with $\\angle B C K=\\angle B M A$. Since $\\angle K B C=\\angle A B M$, we get $\\triangle B C K \\sim \\triangle B M A$. It follows that $B C / B M=$ $B K / B A$, which implies that also $\\triangle B A K \\sim \\triangle B M C$. The quadrilateral $A N C K$ is cyclic, because $\\angle B K C=\\angle B A M=\\angle N A C$. Then by Ptolemy's theorem we obtain  $$ A C \\cdot B K=A C \\cdot B N+A N \\cdot C K+C N \\cdot A K $$  On the other hand, from the similarities noted above we get  $$ C K=\\frac{B C \\cdot A M}{B M}, A K=\\frac{A B \\cdot C M}{B M} \\text { and } B K=\\frac{A B \\cdot B C}{B M} . $$  After substitution of these values, the equality (1) becomes  $$ \\frac{A B \\cdot B C \\cdot A C}{B M}=A C \\cdot B N+\\frac{B C \\cdot A M \\cdot A N}{B M}+\\frac{A B \\cdot C M \\cdot C N}{B M}, $$  which is exactly the equality we must prove multiplied by $\\frac{A B \\cdot B C \\cdot C A}{B M}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "5. (FRA) Let $A B C$ be a triangle, $H$ its orthocenter, $O$ its circumcenter, and $R$ its circumradius. Let $D$ be the reflection of $A$ across $B C, E$ that of $B$ across $C A$, and $F$ that of $C$ across $A B$. Prove that $D, E$, and $F$ are collinear if and only if $O H=2 R$.", "solution": "5. Let $G$ be the centroid of $\\triangle A B C$ and $\\mathcal{H}$ the homothety with center $G$ and ratio $-\\frac{1}{2}$. It is well-known that $\\mathcal{H}$ maps $H$ into $O$. For every other point $X$, let us denote by $X^{\\prime}$ its image under $\\mathcal{H}$. Also, let $A_{2} B_{2} C_{2}$ be the triangle in which $A, B, C$ are the midpoints of $B_{2} C_{2}, C_{2} A_{2}$, and $A_{2} B_{2}$, respectively. It is clear that $A^{\\prime}, B^{\\prime}, C^{\\prime}$ are the midpoints of sides $B C, C A, A B$ respectively. Furthermore, $D^{\\prime}$ is the reflection of $A^{\\prime}$ across $B^{\\prime} C^{\\prime}$. Thus $D^{\\prime}$ must lie on $B_{2} C_{2}$ and $A^{\\prime} D^{\\prime} \\perp$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-643.jpg?height=484&width=552&top_left_y=612&top_left_x=806) $B_{2} C_{2}$. However, it also holds that $O A^{\\prime} \\perp B_{2} C_{2}$, so we conclude that $O, D^{\\prime}, A^{\\prime}$ are collinear and $D^{\\prime}$ is the projection of $O$ on $B_{2} C_{2}$. Analogously, $E^{\\prime}, F^{\\prime}$ are the projections of $O$ on $C_{2} A_{2}$ and $A_{2} B_{2}$. Now we apply Simson's theorem. It claims that $D^{\\prime}, E^{\\prime}, F^{\\prime}$ are collinear (which is equivalent to $D, E, F$ being collinear) if and only if $O$ lies on the circumcircle of $A_{2} B_{2} C_{2}$. However, this circumcircle is centered at $H$ with radius $2 R$, so the last condition is equivalent to $H O=2 R$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "6. (POL) Let $A B C D E F$ be a convex hexagon such that $\\angle B+\\angle D+\\angle F=$ $360^{\\circ}$ and  $$ \\frac{A B}{B C} \\cdot \\frac{C D}{D E} \\cdot \\frac{E F}{F A}=1 $$  Prove that  $$ \\frac{B C}{C A} \\cdot \\frac{A E}{E F} \\cdot \\frac{F D}{D B}=1 $$", "solution": "6. Let $P$ be the point such that $\\triangle C D P$ and $\\triangle C B A$ are similar and equally oriented. Since then $\\angle D C P=\\angle B C A$ and $\\frac{B C}{C A}=\\frac{D C}{C P}$, it follows that $\\angle A C P=\\angle B C D$ and $\\frac{A C}{C P}=\\frac{B C}{C D}$, so $\\triangle A C P \\sim \\triangle B C D$. In particular, $\\frac{B C}{C A}=\\frac{D B}{P A}$. Furthermore, by the conditions of the problem we have $\\angle E D P=360^{\\circ}-$ $\\angle B-\\angle D=\\angle F$ and $\\frac{P D}{D E}=\\frac{P D}{C D} \\cdot \\frac{C D}{D E}=\\frac{A B}{B C} \\cdot \\frac{C D}{D E}=\\frac{A F}{F E}$. Therefore $\\triangle E D P \\sim \\triangle E F A$ as well, so that similarly as above we conclude that $\\triangle A E P \\sim \\triangle F E D$ and consequently $\\frac{A E}{E F}=\\frac{P A}{F D}$. Finally, $\\frac{B C}{C A} \\cdot \\frac{A E}{E F} \\cdot \\frac{F D}{D B}=\\frac{D B}{P A} \\cdot \\frac{P A}{F D} \\cdot \\frac{F D}{D B}=1$. Second solution. Let $a, b, c, d, e, f$ be the complex coordinates of $A, B$, $C, D, E, F$, respectively. The condition of the problem implies that $\\frac{a-b}{b-c}$. $\\frac{c-d}{d-e} \\cdot \\frac{e-f}{f-a}=-1$. On the other hand, since $(a-b)(c-d)(e-f)+(b-c)(d-e)(f-a)=$ $(b-c)(a-e)(f-d)+(c-a)(e-f)(d-b)$ holds identically, we immediately deduce that $\\frac{b-c}{c-a} \\cdot \\frac{a-e}{e-f} \\cdot \\frac{f-d}{d-b}=-1$. Taking absolute values gives $\\frac{B C}{C A} \\cdot \\frac{A E}{E F}$. $\\frac{F D}{D B}=1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. (GBR) Let $A B C$ be a triangle such that $\\angle A C B=2 \\angle A B C$. Let $D$ be the point on the side $B C$ such that $C D=2 B D$. The segment $A D$ is extended to $E$ so that $A D=D E$. Prove that  $$ \\angle E C B+180^{\\circ}=2 \\angle E B C . $$", "solution": "7. We shall use the following result.  Lemma. In a triangle $A B C$ with $B C=a, C A=b$, and $A B=c$, i. $\\angle C=2 \\angle B$ if and only if $c^{2}=b^{2}+a b$; ii. $\\angle C+180^{\\circ}=2 \\angle B$ if and only if $c^{2}=b^{2}-a b$.  Proof. i. Take a point $D$ on the extension of $B C$ over $C$ such that $C D=b$. The condition $\\angle C=2 \\angle B$ is equivalent to $\\angle A D C=\\frac{1}{2} \\angle C=\\angle B$, and thus to $A D=A B=c$. This is further equivalent to triangles $C A D$ and $A B D$ being similar, so $C A / A D=A B / B D$, i.e., $c^{2}=$ $b(a+b)$. ii. Take a point $E$ on the ray $C B$ such that $C E=b$. As above, $\\angle C+180^{\\circ}=2 \\angle B$ if and only if $\\triangle C A E \\sim \\triangle A B E$, which is equivalent to $E B / B A=E A / A C$, or $c^{2}=b(b-a)$. Let $F, G$ be points on the ray $C B$ such that $C F=\\frac{1}{3} a$ and $C G=\\frac{4}{3} a$. Set $B C=a, C A=b, A B=c, E C=b_{1}$, and $E B=c_{1}$. By the lemma it follows that $c^{2}=b^{2}+a b$. Also $b_{1}=A G$ and $c_{1}=A F$, so Stewart's theorem gives us $c_{1}^{2}=\\frac{2}{3} b^{2}+\\frac{1}{3} c^{2}-\\frac{2}{9} a^{2}=b^{2}+\\frac{1}{3} a b-\\frac{2}{9} a^{2}$ and $b_{1}^{2}=$ $-\\frac{1}{3} b^{2}+\\frac{4}{3} c^{2}+\\frac{4}{9} a^{2}=b^{2}+\\frac{4}{3} a b+\\frac{4}{9} a^{2}$. It follows that $b_{1}=\\frac{2}{3} a+b$ and $c_{1}^{2}=b_{1}^{2}-\\left(a b+\\frac{2}{3} a^{2}\\right)=b_{1}^{2}-a b_{1}$. The statement of the problem follows immediately by the lemma.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "8", "problem_type": null, "exam": "IMO", "problem": "8. (IND) Let $A B C$ be a triangle such that $\\angle A=90^{\\circ}$ and $\\angle B<\\angle C$. The tangent at $A$ to its circumcircle $\\omega$ meets the line $B C$ at $D$. Let $E$ be the reflection of $A$ across $B C, X$ the foot of the perpendicular from $A$ to $B E$, and $Y$ the midpoint of $A X$. Let the line $B Y$ meet $\\omega$ again at $Z$. Prove that the line $B D$ is tangent to the circumcircle of triangle $A D Z$.", "solution": "8. Let $M$ be the point of intersection of $A E$ and $B C$, and let $N$ be the point on $\\omega$ diametrically opposite $A$. Since $\\angle B<\\angle C$, points $N$ and $B$ are on the same side of $A E$. Furthermore, $\\angle N A E=\\angle B A X=$ $90^{\\circ}-\\angle A B E$; hence the triangles $N A E$ and $B A X$ are similar. Consequently, $\\triangle B A Y$ and $\\triangle N A M$ are also similar, since $M$ is the midpoint ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-644.jpg?height=279&width=555&top_left_y=1171&top_left_x=802) of $A E$. Thus $\\angle A N Z=\\angle A B Z=\\angle A B Y=\\angle A N M$, implying that $N, M, Z$ are collinear. Now we have $\\angle Z M D=90^{\\circ}-\\angle Z M A=\\angle E A Z=$ $\\angle Z E D$ (the last equality because $E D$ is tangent to $\\omega$ ); hence $Z M E D$ is a cyclic quadrilateral. It follows that $\\angle Z D M=\\angle Z E A=\\angle Z A D$, which is enough to conclude that $M D$ is tangent to the circumcircle of $A Z D$. Remark. The statement remains valid if $\\angle B \\geq \\angle C$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1998", "tier": "T0", "problem_label": "9", "problem_type": null, "exam": "IMO", "problem": "9. (MON) Let $a_{1}, a_{2}, \\ldots, a_{n}$ be positive real numbers such that $a_{1}+a_{2}+$ $\\cdots+a_{n}<1$. Prove that  $$ \\frac{a_{1} a_{2} \\cdots a_{n}\\left[1-\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)\\right]}{\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)\\left(1-a_{1}\\right)\\left(1-a_{2}\\right) \\cdots\\left(1-a_{n}\\right)} \\leq \\frac{1}{n^{n+1}} . $$", "solution": "9. Set $a_{n+1}=1-\\left(a_{1}+\\cdots+a_{n}\\right)$. Then $a_{n+1}>0$, and the desired inequality becomes  $$ \\frac{a_{1} a_{2} \\cdots a_{n+1}}{\\left(1-a_{1}\\right)\\left(1-a_{2}\\right) \\cdots\\left(1-a_{n+1}\\right)} \\leq \\frac{1}{n^{n+1}} $$  To prove it, we observe that $1-a_{i}=a_{1}+\\cdots+a_{i-1}+a_{i+1}+\\cdots+a_{n+1} \\geq n \\sqrt[n]{a_{1} \\cdots a_{i-1} a_{i+1} \\cdots a_{n+1}}$. Multiplying these inequalities for $i=1,2, \\ldots, n+1$, we get exactly the inequality we need.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. $\\mathbf{N} \\mathbf{1}(\\mathbf{T W N}){ }^{\\mathrm{IMO}+4}$ Find all pairs of positive integers $(x, p)$ such that $p$ is a prime, $x \\leq 2 p$, and $x^{p-1}$ is a divisor of $(p-1)^{x}+1$.", "solution": "1. Obviously $(1, p)$ (where $p$ is an arbitrary prime) and $(2,2)$ are solutions and the only solutions to the problem for $x<3$ or $p<3$. Let us now assume $x, p \\geq 3$. Since $p$ is odd, $(p-1)^{x}+1$ is odd, and hence $x$ is odd. Let $q$ be the largest prime divisor of $x$, which also must be odd. We have $q|x| x^{p-1} \\mid(p-1)^{x}+1 \\Rightarrow(p-1)^{x} \\equiv-1(\\bmod q)$. Also from Fermat's little theorem $(p-1)^{q-1} \\equiv 1(\\bmod q)$. Since $q-1$ and $x$ are coprime, there exist integers $\\alpha, \\beta$ such that $x \\alpha=(q-1) \\beta+1$. We also note that $\\alpha$ must be odd. We now have $p-1 \\equiv(p-1)^{(q-1) \\beta+1} \\equiv(p-1)^{x \\alpha} \\equiv-1(\\bmod q)$ and hence $q \\mid p \\Rightarrow q=p$. Since $x$ is odd, $p \\mid x$, and $x \\leq 2 p$, it follows $x=p$ for all $x, p \\geq 3$. Thus  $$ p^{p-1} \\left\\lvert\\,(p-1)^{x}+1=p^{2} \\cdot\\left(p^{p-2}-\\binom{p}{1} p^{p-1}+\\cdots-\\binom{p}{p-2}+1\\right) .\\right. $$  Since the expression in parenthesis is not divisible by $p$, it follows that $p^{p-1} \\mid p^{2}$ and hence $p \\leq 3$. One can easily verify that $(3,3)$ is a valid solution. We have shown that the only solutions are $(1, p),(2,2)$, and $(3,3)$, where $p$ is an arbitrary prime.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "10", "problem_type": "Geometry", "exam": "IMO", "problem": "10. G4 (GBR) For a triangle $T=A B C$ we take the point $X$ on the side $(A B)$ such that $A X / X B=4 / 5$, the point $Y$ on the segment $(C X)$ such that $C Y=2 Y X$, and, if possible, the point $Z$ on the ray ( $C A$ such that $\\measuredangle C X Z=180^{\\circ}-\\measuredangle A B C$. We denote by $\\Sigma$ the set of all triangles $T$ for which $\\measuredangle X Y Z=45^{\\circ}$. Prove that all the triangles from $\\Sigma$ are similar and find the measure of their smallest angle.", "solution": "10. We use the following lemma.  Lemma. Let $A B C$ be a triangle and $X \\in A B$ such that $\\overrightarrow{A X}: \\overrightarrow{X B}=m: n$. Then $(m+n) \\cot \\angle C X B=n \\cot A-m \\cot B$ and $m \\cot \\angle A C X=$ $(n+m) \\cot C+n \\cot A$. Proof. Let $C D$ be the altitude from $C$ and $h$ its length. Then using oriented segments we have $A X=A D+D X=h \\cot A-h \\cot \\angle C X B$ and $B X=B D+D X=h \\cot B+h \\cot \\angle C X B$. The first formula in the lemma now follows from $n \\cdot A X=m \\cdot B X$. The second formula immediately follows from the first part applied to the triangle $A C X$ and the point $X^{\\prime} \\in A C$ such that $X X^{\\prime} \\| B C$. Let us set $\\cot A=x, \\cot B=y$, and $\\cot C=z$. Applying the second formula in the lemma to $\\triangle A B C$ and the point $X$, we obtain $4 \\cot \\angle A C X=$ $9 z+5 x$. Applying the first formula in the lemma to $\\triangle C X Z$ and the point $Y$ and using $\\angle X Y Z=45^{\\circ}$ and $\\cot \\angle C X Z=-y$, we obtain $3 \\cot \\angle X Y Z=$ $\\cot \\angle A C X-2 \\cot \\angle C X Z=\\frac{9 z+5 x}{4}+2 y \\Rightarrow 5 x+8 y+9 z=12$. We now use the well-known relation for cotangents of a triangle $x y+y z+$ $x z=1$ to get $9=9(x+y) z+9 x y=(x+y)(12-5 x-8 z)+9 x y=9 \\Rightarrow$ $(4 y+x-3)^{2}+9(x-1)^{2}=0 \\Rightarrow x=1, y=\\frac{1}{2}, z=\\frac{1}{3}$. It follows that $x, y$, and $z$ have fixed values, and hence all triangles $T$ in $\\Sigma$ are similar, with their smallest angle $A$ having cotangent 1 and thus being equal to $\\angle A=45^{\\circ}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "11", "problem_type": "Geometry", "exam": "IMO", "problem": "11. G5 (FRA) Let $A B C$ be a triangle, $\\Omega$ its incircle and $\\Omega_{a}, \\Omega_{b}, \\Omega_{c}$ three circles three circles orthogonal to $\\Omega$ passing through $B$ and $C, A$ and $C$, and $A$ and $B$ respectively. The circles $\\Omega_{a}, \\Omega_{b}$ meet again in $C^{\\prime}$; in the same way we obtain the points $B^{\\prime}$ and $A^{\\prime}$. Prove that the radius of the circumcircle of $A^{\\prime} B^{\\prime} C^{\\prime}$ is half the radius of $\\Omega$.", "solution": "11. Let $\\Omega(I, r)$ be the incircle of $\\triangle A B C$. Let $D, E$, and $F$ denote the points where $\\Omega$ touches $B C, A C$, and $A B$, respectively. Let $P, Q$, and $R$ denote the midpoints of $E F, D F$, and $D E$ respectively. We prove that $\\Omega_{a}$ passes through $Q$ and $R$. Since $\\triangle I Q D \\sim \\triangle I D B$ and $\\triangle I R D \\sim \\triangle I D C$, we obtain $I Q \\cdot I B=I R \\cdot I C=r^{2}$. We conclude that $B, C, Q$, and $R$ lie on a single circle $\\Gamma_{a}$. Moreover, since the power of $I$ with respect to $\\Gamma_{a}$ is $r^{2}$, it follows for a tangent $I X$ from $I$ to $\\Gamma_{a}$ that $X$ lies on $\\Omega$ and hence $\\Omega$ is perpendicular to $\\Gamma_{a}$. From the uniqueness of $\\Omega_{a}$ it follows that $\\Omega_{a}=\\Gamma_{a}$. Thus $\\Omega_{a}$ contains $Q$ and $R$. Similarly $\\Omega_{b}$ contains $P$ and $R$ and $\\Omega_{c}$ contains $P$ and $Q$. Hence, $A^{\\prime}=P, B^{\\prime}=Q$ and $C^{\\prime}=R$. Therefore the radius of the circumcircle of $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ is half the radius of $\\Omega$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "12", "problem_type": "Geometry", "exam": "IMO", "problem": "12. G6 (RUS) ${ }^{\\mathrm{IMO} 5}$ Two circles $\\Omega_{1}$ and $\\Omega_{2}$ touch internally the circle $\\Omega$ in $M$ and $N$, and the center of $\\Omega_{2}$ is on $\\Omega_{1}$. The common chord of the circles $\\Omega_{1}$ and $\\Omega_{2}$ intersects $\\Omega$ in $A$ and $B . M A$ and $M B$ intersect $\\Omega_{1}$ in $C$ and $D$. Prove that $\\Omega_{2}$ is tangent to $C D$.", "solution": "12. We first introduce the following lemmas.  Lemma 1. Let $A B C$ be a triangle, $I$ its inenter and $I_{a}$ the center of the excircle touching $B C$. Let $A^{\\prime}$ be the center of the arc $\\widehat{B C}$ of the circumcircle not containing $A$. Then $A^{\\prime} B=A^{\\prime} C=A^{\\prime} I=A^{\\prime} I_{a}$. Proof. The result follows from a straightforward calculation of the relevant angles. Lemma 2. Let two circles $k_{1}$ and $k_{2}$ meet each other at points $X$ and $Y$ and touch a circle $k$ internally in points $M$ and $N$, respectively. Let $A$ be one of the intersections of the line $X Y$ with $k$. Let $A M$ and $A N$ intersect $k_{1}$ and $k_{2}$ respectively at $C$ and $E$. Then $C E$ is a common tangent of $k_{1}$ and $k_{2}$. Proof. Since $A C \\cdot A M=A X \\cdot A Y=A E \\cdot A N$, the points $M, N, E, C$ lie on a circle. Let $M N$ meet $k_{1}$ again at $Z$. If $M^{\\prime}$ is any point on the common tangent at $M$, then $\\angle M C Z=\\angle M^{\\prime} M Z=\\angle M^{\\prime} M N=\\angle M A N$ (as oriented angles), implying that $C Z \\| A N$. It follows that $\\angle A C E=$ $\\angle A N M=\\angle C Z M$. Hence $C E$ is tangent to $k_{1}$ and analogously to $k_{2}$. In the main problem, let us define $E$ and $F$ respectively as intersections of $N A$ and $N B$ with $\\Omega_{2}$. Then applying Lemma 2 we get that $C E$ and $D F$ are the common tangents of $\\Omega_{1}$ and $\\Omega_{2}$. If the circles have the same radii, the result trivially holds. Otherwise, let $G$ be the intersection of $C E$ and $D F$. Let $O_{1}$ and $O_{2}$ be the centers of $\\Omega_{1}$ and $\\Omega_{2}$. Since $O_{1} D=O_{1} C$ and $\\angle O_{1} D G=\\angle O_{1} C G=90^{\\circ}$, it follows that $O_{1}$ is the midpoint of the shorter arc of the circumcircle of $\\triangle C D G$. The center $O_{2}$ is located on the bisector of $\\angle C G D$, since $\\Omega_{2}$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-660.jpg?height=412&width=529&top_left_y=1000&top_left_x=813) touches both $G C$ and $G D$. However, it also sits on $\\Omega_{1}$, and using Lemma 1 we obtain that $O_{2}$ is either at the incenter or at the excenter of $\\triangle C D G$ opposite $G$. Hence, $\\Omega_{2}$ is either the incircle or the excircle of $C D G$ and thus in both cases touches $C D$. Second solution. Let $O$ be the center of $\\Gamma$, and $r, r_{1}, r_{2}$ the radii of $\\Gamma, \\Gamma_{1}, \\Gamma_{2}$. It suffices to show that the distance $d\\left(O_{2}, C D\\right)$ is equal to $r_{2}$. The homothety with center $M$ and ratio $r / r_{1}$ takes $\\Gamma_{1}, C, D$ into $\\Gamma, A, B$, respectively; hence $C D \\| A B$ and $d(C, A B)=\\frac{r-r_{1}}{r} d(M, A B)$. Let $O_{1} O_{2}$ meet $X Y$ at $R$. Then $d\\left(O_{2}, C D\\right)=O_{2} R+\\frac{r-r_{1}}{r} d(M, A B)$, i.e.,  $$ d\\left(O_{2}, C D\\right)=O_{2} R+\\frac{r-r_{1}}{r}\\left[O_{1} O_{2}-O_{2} R+r_{1} \\cos \\angle O O_{1} O_{2}\\right] $$  since $O, O_{1}$, and $M$ are collinear. We have $O_{1} X=O_{1} O_{2}=r_{1}, O O_{1}=$ $r-r_{1}, O O_{2}=r-r_{2}$, and $O_{2} X=r_{2}$. Using the cosine law in the triangles $O O_{1} O_{2}$ and $X O_{1} O_{2}$, we obtain that $\\cos \\angle O O_{1} O_{2}=\\frac{2 r_{1}^{2}-2 r r_{1}+2 r r_{2}-r_{2}^{2}}{2 r_{1}\\left(r-r_{1}\\right)}$ and $O_{2} R=\\frac{r_{2}^{2}}{2 r_{1}}$. Substituting these values in (1) we get $d\\left(O_{2}, C D\\right)=r_{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "13", "problem_type": "Geometry", "exam": "IMO", "problem": "13. G7 (ARM) The point $M$ inside the convex quadrilateral $A B C D$ is such that $M A=M C, \\angle A M B=\\angle M A D+\\angle M C D, \\angle C M D=\\angle M C B+\\angle M A B$. Prove that $A B \\cdot C M=B C \\cdot M D$ and $B M \\cdot A D=M A \\cdot C D$.", "solution": "13. Let us construct a convex quadrilateral $P Q R S$ and an interior point $T$ such that $\\triangle P T Q \\cong \\triangle A M B, \\triangle Q T R \\sim \\triangle A M D$, and $\\triangle P T S \\sim \\triangle C M D$. We then have $T S=\\frac{M D \\cdot P T}{M C}=M D$ and $\\frac{T R}{T S}=\\frac{T R \\cdot T Q \\cdot T P}{T Q \\cdot T P \\cdot T S}=\\frac{M D \\cdot M B \\cdot M C}{M A \\cdot M A \\cdot M D}=$ $\\frac{M B}{M C}$ (using $M A=M C$ ). We also have $\\angle S T R=\\angle B M C$ and therefore $\\triangle R T S \\sim \\triangle B M C$. Now the relations between angles become  $$ \\angle T P S+\\angle T Q R=\\angle P T Q \\quad \\text { and } \\quad \\angle T P Q+\\angle T S R=\\angle P T S $$  implying that $P Q \\| R S$ and $Q R \\| P S$. Hence $P Q R S$ is a parallelogram and hence $A B=P Q=R S$ and $Q R=P S$. It follows that $\\frac{B C}{M C}=\\frac{R S}{T S}=$ $\\frac{A B}{M D} \\Rightarrow A B \\cdot C M=B C \\cdot M D$ and $\\frac{A D \\cdot B M}{A M}=\\frac{A D \\cdot Q T}{A M}=Q R=P S=$ $\\frac{C D \\cdot T S}{M D}=C D \\Rightarrow B M \\cdot A D=M A \\cdot C D$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "14", "problem_type": "Geometry", "exam": "IMO", "problem": "14. G8 (RUS) Points $A, B, C$ divide the circumcircle $\\Omega$ of the triangle $A B C$ into three arcs. Let $X$ be a variable point on the $\\operatorname{arc} A B$, and let $O_{1}, O_{2}$ be the incenters of the triangles $C A X$ and $C B X$. Prove that the circumcircle of the triangle $X O_{1} O_{2}$ intersects $\\Omega$ in a fixed point.", "solution": "14. We first introduce the same lemma as in problem 12 and state it here without proof. Lemma. Let $A B C$ be a triangle and $I$ the center of its incircle. Let $M$ be the center of the $\\operatorname{arc} \\widehat{B C}$ of the circumcircle not containing $A$. Then $M B=M C=M I$. Let the circle $X O_{1} O_{2}$ intersect the circle $\\Omega$ again at point $T$. Let $M$ and $N$ be respectively the midpoints of $\\operatorname{arcs} \\widehat{B C}$ and $\\widehat{A C}$, and let $P$ be the intersection of $\\Omega$ and the line through $C$ parallel to $M N$. Then the lemma gives $M P=N C=N I=N O_{1}$ and $N P=M C=$ $M I=M O_{2}$. Since $O_{1}$ and $O_{2}$ lie on $X N$ and $X M$ respectively, we have $\\angle N T M=\\angle N X M=\\angle O_{1} X O_{2}=\\angle O_{1} T O_{2}$ and hence $\\angle N T O_{1}=$ $\\angle M T O_{2}$. Moreover, $\\angle T N O_{1}=\\angle T N X=\\angle T M O_{2}$, from which it follows that $\\triangle O_{1} N T \\sim \\triangle O_{2} M T$. Thus $\\frac{N T}{M P}=\\frac{N T}{N O_{1}}=\\frac{M T}{M O_{2}}=\\frac{M T}{N P} \\Rightarrow$ $M P \\cdot M T=N P \\cdot N T \\Rightarrow S_{M P T}=S_{N P T}$. It follows that $T P$ bisects the segment $M N$, and hence it passes through $I$. We conclude that $T$ belongs to the line $P I$ and does not depend on $X$. Remark. An alternative approach is to apply an inversion at point $C$. Points $O_{1}$ and $O_{2}$ become excenters of $\\triangle A X C$ and $\\triangle B X C$, and $T$ becomes the projection of $I_{c}$ onto $A B$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "15", "problem_type": "Algebra", "exam": "IMO", "problem": "15. A1 (POL) ${ }^{\\mathrm{IMO} 2}$ Let $n \\geq 2$ be a fixed integer. Find the least constant $C$ such that the inequality  $$ \\sum_{i<j} x_{i} x_{j}\\left(x_{i}^{2}+x_{j}^{2}\\right) \\leq C\\left(\\sum_{i} x_{i}\\right)^{4} $$  holds for every $x_{1}, \\ldots, x_{n} \\geq 0$ (the sum on the left consists of $\\binom{n}{2}$ summands). For this constant $C$, characterize the instances of equality.", "solution": "15. For all $x_{i}=0$ any $C$ will do, so we may assume the contrary. Since the equation is symmetric and homogeneous, we may assume $\\sum_{i} x_{i}=1$. The equation now becomes $F\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)=\\sum_{i<j} x_{i} x_{j}\\left(x_{i}^{2}+x_{j}^{2}\\right)=$ $\\sum_{i} x_{i}^{2} \\sum_{j \\neq i} x_{j}=\\sum_{i} x_{i}^{3}\\left(1-x_{i}\\right)=\\sum_{i} f\\left(x_{i}\\right) \\leq C$, where we define $f(x)=$ $x^{3}-x^{4}$. We note that for $x, y \\geq 0$ and $x+y \\leq 2 / 3$,  $$ f(x+y)+f(0)-f(x)-f(y)=3 x y(x+y)\\left(\\frac{2}{3}-x-y\\right) \\geq 0 $$  We note that if at least three elements of $\\left\\{x_{1}, x_{2}, \\ldots, x_{n}\\right\\}$ are nonzero the condition of (1) always holds for the two smallest ones. Hence, applying (1) repeatedly, we obtain $F\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right) \\leq F(a, 1-a, 0, \\ldots, 0)=\\frac{1}{2}(2 a(1-$ a)) $(1-2 a(1-a)) \\leq \\frac{1}{8}=F\\left(\\frac{1}{2}, \\frac{1}{2}, 0, \\ldots, 0\\right)$. Thus we have $C=\\frac{1}{8}$ (for all $n$ ), and equality holds only when two $x_{i}$ are equal and the remaining ones are 0 . Second solution. Let $M=x_{1}^{2}+x_{2}^{2}+\\cdots+x_{n}^{2}$. Using $a b \\leq(a+2 b)^{2} / 8$ we have  $$ \\begin{aligned} \\sum_{1 \\leq i<j \\leq n} x_{i} x_{j}\\left(x_{i}^{2}+x_{j}^{2}\\right) & \\leq M \\sum_{i<j} x_{i} x_{j} \\\\ & \\leq \\frac{1}{8}\\left(M+2 \\sum_{i<j} x_{i} x_{j}\\right)^{2}=\\frac{1}{8}\\left(\\sum_{i=1}^{n} x_{i}\\right)^{4} \\end{aligned} $$  Equality holds if and only if $M=2 \\sum_{i<j} x_{i} x_{j}$ and $x_{i} x_{j}\\left(x_{i}^{2}+x_{j}^{2}\\right)=M x_{i} x_{j}$ for all $i<j$, which holds if and only if $n-2$ of the $x_{i}$ are zero and the remaining two are equal. Remark. Problems (SL90-26) and (SL91-27) are very similar.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "16", "problem_type": "Algebra", "exam": "IMO", "problem": "16. A2 (RUS) The numbers from 1 to $n^{2}$ are randomly arranged in the cells of a $n \\times n$ square ( $n \\geq 2$ ). For any pair of numbers situated in the same row or in the same column, the ratio of the greater number to the smaller one is calculated. Let us call the characteristic of the arrangement the smallest of these $n^{2}(n-1)$ fractions. What is the highest possible value of the characteristic?", "solution": "16. Let $C(A)$ denote the characteristic of an arrangement $A$. We shall prove that $\\max C(A)=\\frac{n+1}{n}$. Let us prove first $C(A) \\leq \\frac{n+1}{n}$ for all $A$. Among elements $\\left\\{n^{2}-n, n^{2}-\\right.$ $\\left.n+1, \\ldots, n^{2}\\right\\}$, by the pigeonhole principle, in at least one row and at least one column there exist two elements, and hence one pair in the same row or column that is not $\\left(n^{2}-n, n^{2}\\right)$. Hence  $$ C(A) \\leq \\max \\left\\{\\frac{n^{2}}{n^{2}-n+1}, \\frac{n^{2}-1}{n^{2}-n}\\right\\}=\\frac{n^{2}-1}{n^{2}-n}=\\frac{n+1}{n} . $$  We now consider the following arrangement:  $$ a_{i j}= \\begin{cases}i+n(j-i-1) & \\text { if } i<j \\\\ i+n(n-i+j-1) & \\text { if } i \\geq j\\end{cases} $$  We claim that $C(a)=\\frac{n+1}{n}$. Indeed, in this arrangement no two numbers in the same row or column differ by less than $n-1$, and in addition, $n^{2}$ and $n^{2}-n+1$ are in different rows and columns, and hence  $$ C(A) \\geq \\frac{n^{2}-1}{n^{2}-n}=\\frac{n+1}{n} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "17", "problem_type": "Algebra", "exam": "IMO", "problem": "17. A3 (FIN) A game is played by $n$ girls $(n \\geq 2)$, everybody having a ball. Each of the $\\binom{n}{2}$ pairs of players, in an arbitrary order, exchange the balls they have at that moment. The game is called nice if at the end nobody has her own ball, and it is called tiresome if at the end everybody has her initial ball. Determine the values of $n$ for which there exists a nice game and those for which there exists a tiresome game.", "solution": "17. A game is determined by the ordering $t_{1}, \\ldots, t_{N}$ of the $N=\\binom{n}{2}$ transpositions $(i, j)$ of the set $\\{1,2, \\ldots, n\\}$. The game is nice if the permutation $P=t_{N} t_{N-1} \\ldots t_{1}$ has no fixed point, and tiresome if $P$ is the identity (denoted by $I$ ). Recall that every permutation can be written as a composition of disjoint cycles. We claim that there exists a nice game if and only if $n \\neq 3$. For $n=2, P_{2}=t_{1}=(1,2)$ is obviously nice. For $n=3$ each game has the form $P=(b, c)(a, c)(a, b)=(a, c)$ for an appropriate notation of the players, which cannot be nice. Now for $n \\geq 4$ we define $P_{n}=(1,2)(1,3)(2,3) \\cdots(1, n)(2, n) \\cdots(n-1, n)$. We obtain inductively that $P_{n}=P_{n-1}(1, n, n-1, \\ldots, 2)=(1, n)(2, n-1) \\cdots(i, n+1-i) \\cdots$ is nice for all even $n$. Also, if $n=2 k+1$ is odd, then $Q_{n}=P_{n-1}(1, n)(2, n) \\cdots(k, n)(n-1, n)(n-$ $2, n) \\cdots(k+1, n)$ maps $i$ to $n+1-i$ for $i \\leq k$, to $n-1-i$ for $k+1 \\leq$ $i \\leq 2 k-1$, and to $3 k+1-i$ if $i \\in\\{2 k, 2 k+1\\}$. Hence $Q_{n}$ is nice. This justifies our claim. Now we prove that a tiresome game exists if and only if $n \\equiv 0,1(\\bmod 4)$. Evidently every transposition changes the sign of the permutation. Thus the sign of $P$ is $(-1)^{\\binom{n}{2}}$ and for $P$ to be the identity we must have 2 $\\binom{n}{2} \\Rightarrow n \\equiv 0,1(\\bmod 4)$. Let us now construct tiresome games for the allowed $n$. For $n=4 k$ we divide the girls into groups of 4 . In each group we perform the following game: $(3,4)(1,3)(2,4)(2,3)(1,4)(1,2)=I$. On the other hand, among two different groups (call them $\\{1,2,3,4\\}$ and $\\{5,6,7,8\\}$ ) we perform  $$ \\begin{aligned} & (4,7)(3,7)(4,6)(1,6)(2,8)(3,8)(2,7)(2,6) \\\\ & (4,5)(4,8)(1,7)(1,8)(3,5)(3,6)(2,5)(1,5)=I \\end{aligned} $$  For $n=4 k+1$ we divide into groups of four as before, with one girl remaining. Every time a group (denoted $\\{1,2,3,4\\}$ ) is to play a game the remaining girl (denoted 5 ) joins in, and they play  $$ (3,5)(3,4)(4,5)(1,3)(2,4)(2,3)(1,4)(1,5)(1,2)(2,5)=I \\text {. } $$  This completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "18", "problem_type": "Algebra", "exam": "IMO", "problem": "18. A4 (BLR) Prove that the set of positive integers cannot be partitioned into three nonempty subsets such that for any two integers $x, y$ taken from two different subsets, the number $x^{2}-x y+y^{2}$ belongs to the third subset.", "solution": "18. Define $f(x, y)=x^{2}-x y+y^{2}$. Let us assume that three such sets $A, B$, and $C$ do exist and that w.l.o.g. $1, b$, and $c(c>b)$ are respectively their smallest elements. Lemma 1. Numbers $x, y$, and $x+y$ cannot belong to three different sets. Proof. The number $f(x, x+y)=f(y, x+y)$ must belong to both the set containing $y$ and the set containing $x$, a contradiction. Lemma 2. The subset $C$ contains a multiple of $b$. Moreover, if $k b$ is the smallest such multiple, then $(k-1) b \\in B$ and $(k-1) b+1, k b+1 \\in A$. Proof. Let $r$ be the residue of $c$ modulo $b$. If $r=0$, the first statement automatically holds. Let $0<r<b$. In that case $r \\in A$, and $c-r$ is then not in $B$ according to Lemma 1. Hence $c-r \\in A$ and since $b \\mid c-r$, it follows that $b \\mid f(c-r, b) \\in C$, thus proving the first statement. It follows immediately from Lemma 1 that $(k-1) b \\in B$. Now by Lemma $1,(k-1) b+1=k b-(b-1)$ must be in $A$; similarly, $k b+1=[(k-1) b+1]+b \\in A$ as well. Let us show by induction that $(n k-1) b+1, n k b+1 \\in A$ for all integers $n$. The inductive basis has been shown in Lemma 2. Assuming that [ $n-$ 1) $k-1] b+1 \\in A$ and $(n-1) k b+1 \\in A$, we get that $(n k-1) b+1=$ $((n-1) k b+1)+(k-1) b=[((n-1) k-1) b+1]+k b$ belongs to $A$ and $n k b+1=((n k-1) b+1)+b=((n-1) k b+1)+k b \\Rightarrow n k b+1 \\in A$. This finishes the inductive step. In particular, $f(k b, k b+1)=(k b+1) k b+1 \\in A$. However, since $k b \\in C, k b+1 \\in A$, it follows that $f(k b, k b+1) \\in B$, which is a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "19", "problem_type": "Algebra", "exam": "IMO", "problem": "19. A5 (JAP) ${ }^{\\text {IMO6 }}$ Find all the functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ that satisfy  $$ f(x-f(y))=f(f(y))+x f(y)+f(x)-1 $$  for all $x, y \\in \\mathbb{R}$.", "solution": "19. Let $A=\\{f(x) \\mid x \\in \\mathbb{R}\\}$ and $f(0)=c$. Plugging in $x=y=0$ we get $f(-c)=f(c)+c-1$, hence $c \\neq 0$. If $x \\in A$, then taking $x=f(y)$ in the original functional equation we get $f(x)=\\frac{c+1}{2}-\\frac{x^{2}}{2}$ for all $x \\in A$. We now show that $A-A=\\left\\{x_{1}-x_{2} \\mid x_{1}, x_{2} \\in A\\right\\}=\\mathbb{R}$. Indeed, plugging in $y=0$ into the original equation gives us $f(x-c)-f(x)=c x+f(c)-1$, an expression that evidently spans all the real numbers. Thus, each $x$ can be represented as $x=x_{1}-x_{2}$, where $x_{1}, x_{2} \\in A$. Plugging $x=x_{1}$ and $f(y)=x_{2}$ into the original equation gives us $f(x)=f\\left(x_{1}-x_{2}\\right)=f\\left(x_{1}\\right)+x_{1} x_{2}+f\\left(x_{2}\\right)-1=c-\\frac{x_{1}^{2}+x_{2}^{2}}{2}+x_{1} x_{2}=c-\\frac{x^{2}}{2}$. Hence we must have $c=\\frac{c+1}{2}$, which gives us $c=1$. Thus $f(x)=1-\\frac{x^{2}}{2}$ for all $x \\in \\mathbb{R}$. It is easily checked that this function satisfies the original functional equation.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "2", "problem_type": "Number Theory", "exam": "IMO", "problem": "2. N2 (ARM) Prove that every positive rational number can be represented in the form $\\frac{a^{3}+b^{3}}{c^{3}+d^{3}}$, where $a, b, c, d$ are positive integers.", "solution": "2. We first prove that every rational number in the interval $(1,2)$ can be represented in the form $\\frac{a^{3}+b^{3}}{a^{3}+d^{3}}$. Taking $b, d$ such that $b \\neq d$ and $a=b+d$, we get $a^{2}-a b+b^{2}=a^{2}-a d+d^{2}$ and  $$ \\frac{a^{3}+b^{3}}{a^{3}+d^{3}}=\\frac{(a+b)\\left(a^{2}-a b+b^{2}\\right)}{(a+d)\\left(a^{2}-a d+d^{2}\\right)}=\\frac{a+b}{a+d} . $$  For a given rational number $1<m / n<2$ we can select $a=m+n$ and $b=2 m-n$ such that along with $d=a-b$ we have $\\frac{a+b}{a+d}=\\frac{m}{n}$. This completes the proof of the first statement. For $m / n$ outside of the interval we can easily select a rational number $p / q$ such that $\\sqrt[3]{\\frac{n}{m}}<\\frac{p}{q}<\\sqrt[3]{\\frac{2 n}{m}}$. In other words $1<\\frac{p^{3} m}{q^{3} n}<2$. We now proceed to obtain $a, b$ and $d$ for $\\frac{p^{3} m}{q^{3} n}$ as before, and we finally have  $$ \\frac{p^{3} m}{q^{3} n}=\\frac{a^{3}+b^{3}}{a^{3}+d^{3}} \\Rightarrow \\frac{m}{n}=\\frac{(a q)^{3}+(b q)^{3}}{(a p)^{3}+(d p)^{3}} . $$  Thus we have shown that all positive rational numbers can be expressed in the form $\\frac{a^{3}+b^{3}}{c^{3}+d^{3}}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "20", "problem_type": "Algebra", "exam": "IMO", "problem": "20. A6 (SWE) For $n \\geq 3$ and $a_{1} \\leq a_{2} \\leq \\cdots \\leq a_{n}$ given real numbers we have the following instructions: (1) place the numbers in some order in a circle; (2) delete one of the numbers from the circle; (3) if just two numbers are remaining in the circle, let $S$ be the sum of these two numbers. Otherwise, if there are more than two numbers in the circle, replace $\\left(x_{1}, x_{2}, x_{3}, \\ldots, x_{p-1}, x_{p}\\right)$ with $\\left(x_{1}+x_{2}, x_{2}+\\right.$ $x_{3}, \\ldots, x_{p-1}+x_{p}, x_{p}+x_{1}$ ). Afterwards, start again with step (2). Show that the largest sum $S$ that can result in this way is given by the formula  $$ S_{\\max }=\\sum_{k=2}^{n}\\binom{n-2}{\\left[\\frac{k}{2}\\right]-1} a_{k} $$", "solution": "20. We first introduce some useful notation. An arrangement around the circle will be denoted by $x=\\left\\{x_{1}, x_{2}, \\ldots, x_{n}\\right\\}$, where the elements are arranged clockwise and $x_{1}$ is fixed to be the smallest number. We will call an arrangement balanced if $x_{1} \\leq x_{n} \\leq x_{2} \\leq x_{n-1} \\leq x_{3} \\leq x_{n-2} \\leq \\cdots$ (the string of inequalities continues until all the elements are accounted for). We will denote the permutation of $x=\\left\\{x_{1}, x_{2}, \\ldots, x_{n}\\right\\}$ in ascending order by $x^{\\prime}=\\left\\{x_{1}^{\\prime}, x_{2}^{\\prime}, \\ldots, x_{n}^{\\prime}\\right\\}$. We will let $f_{i}(x)=\\left\\{f_{i}(x)_{1}, f_{i}(x)_{2}, \\ldots, f_{i}(x)_{n-1}\\right\\}$ denote the arrangement after one iteration of the algorithm where $x_{i}$ was the deleted element. Lemma 1. If an arrangement $x$ is balanced, then $f_{1}(x)$ is also balanced. Proof. In one iteration we have $\\left\\{x_{1}, \\ldots, x_{n}\\right\\} \\rightarrow\\left\\{x_{n}+x_{2}, x_{2}+x_{3}, \\ldots\\right.$, $\\left.x_{n-1}+x_{n}\\right\\}$. Since $x_{n} \\leq x_{2} \\leq x_{n-1} \\leq x_{3} \\leq x_{n-2} \\leq \\cdots$, it follows that $x_{n}+x_{2} \\leq x_{n}+x_{n-1} \\leq x_{2}+x_{3} \\leq x_{n-1}+x_{n-2} \\leq \\cdots$, which means that $f_{1}(x)$ is balanced. We will first show by induction that $S_{\\max }$ can be reached by using the balanced initial arrangement $\\left\\{a_{1}, a_{3}, a_{5}, \\ldots, a_{6}, a_{4}, a_{2}\\right\\}$ and repeatedly deleting the smallest member. For $n=3$ we have $S_{3}=a_{2}+a_{3}$, in accordance with the formula. Assuming that the formula holds for a given $n$, we note that for an arrangement $x=\\left\\{a_{1}, a_{3}, a_{5}, \\ldots, a_{6}, a_{4}, a_{2}\\right\\}$ the arrangement $f_{1}(x)$ is also balanced. We now apply the induction hypothesis and use that $\\binom{n-2}{i}+\\binom{n-2}{i-1}=\\binom{n-1}{i}$ :  $$ \\begin{aligned} S(x) & =S\\left(f_{1}(x)\\right) \\\\ & =\\sum_{k=2}^{n-1}\\binom{n-2}{[k / 2]-1}\\left(a_{k}+a_{k+2}\\right)+\\binom{n-2}{[n / 2]-1}\\left(a_{n}+a_{n+1}\\right)=S_{\\max } \\end{aligned} $$  We now prove that every other arrangement yields a smaller value. We shall write $\\left\\{x_{1}, \\ldots, x_{n}\\right\\} \\leq\\left\\{y_{1}, \\ldots, y_{n}\\right\\}$ whenever $x_{n}^{\\prime}+x_{n-1}^{\\prime}+\\cdots+x_{i}^{\\prime} \\leq$ $y_{n}^{\\prime}+y_{n-1}^{\\prime}+\\cdots+y_{i}^{\\prime}$ holds for all $1 \\leq i \\leq n$. Lemma 2. Let $x$ be an arbitrary arrangement and $y$ a balanced arrangement, both of $n$ elements, such that $x \\leq y$. Then it follows that $f_{i}(x) \\leq f_{1}(y)$, for all $i$. Proof. For any $1 \\leq j \\leq n-1$ there exists $k_{j}$ such that $f_{i}(x)_{j}=x_{k_{j}}+x_{k_{j}+1}$ (assuming $k_{j}+1=1$ if $k_{j}=n-1$ ). Then we have  $$ \\begin{aligned} f_{i}(x)_{n-1}+\\cdots+f_{i}(x)_{n-j} & =\\left(x_{k_{1}}+x_{k_{1}+1}\\right)+\\cdots+\\left(x_{k_{j}}+x_{k_{j}+1}\\right) \\\\ & \\leq 2 x_{n}^{\\prime}+\\cdots+2 x_{n-i+1}^{\\prime}+x_{n-i}^{\\prime}+x_{n-i-1}^{\\prime} \\\\ & =f_{1}(y)_{n-1}+\\cdots+f_{1}(y)_{n-j} \\end{aligned} $$  for all $j$, and hence $f_{i}(x) \\leq f_{1}(y)$. An immediate consequence of Lemma 2 is $f^{n-2}(x) \\leq f_{1}^{n-2}(y)$, implying $S=f^{n-2}(x)_{1}+f^{n-2}(x)_{2} \\leq f_{1}^{n-2}(y)_{1}+f_{1}^{n-2}(y)_{2}=S_{\\max }(y)$. Thus the proof is finished.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "21", "problem_type": "Combinatorics", "exam": "IMO", "problem": "21. C1 (IND) Let $n \\geq 1$ be an integer. A path from $(0,0)$ to $(n, n)$ in the $x y$ plane is a chain of consecutive unit moves either to the right (move denoted by $E$ ) or upwards (move denoted by $N$ ), all the moves being made inside the half-plane $x \\geq y$. A step in a path is the occurrence of two consecutive moves of the form $E N$. Show that the number of paths from $(0,0)$ to $(n, n)$ that contain exactly $s$ steps $(n \\geq s \\geq 1)$ is  $$ \\frac{1}{s}\\binom{n-1}{s-1}\\binom{n}{s-1} $$", "solution": "21. Let us call $f(n, s)$ the number of paths from $(0,0)$ to $(n, n)$ that contain exactly $s$ steps. Evidently, for all $n$ we have $f(n, 1)=f(2,2)=1$, in accordance with the formula. Let us thus assume inductively for a given $n>2$ that for all $s$ we have $f(n, s)=\\frac{1}{s}\\binom{n-1}{s-1}\\binom{n}{s-1}$. We shall prove that the given formula holds also for all $f(n+1, s)$, where $s \\geq 2$. We say that an $(n+1, s)$ - or $(n+1, s+1)$-path is related to a given $(n, s)$ path if it is obtained from the given path by inserting a step $E N$ between two moves or at the beginning or the end of the path. We note that by inserting the step between two moves that form a step one obtains an $(n+1, s)$-path; in all other cases one obtains an $(n+1, s+1)$-path. For each $(n, s)$-path there are exactly $2 n+1-s$ related $(n+1, s+1)$-paths, and for each $(n, s+1)$-path there are $s+1$ related $(n+1, s+1)$-paths. Also, each $(n+1, s+1)$-path is related to exactly $s+1$ different $(n, s)$ - or $(n, s+1)$-paths. Thus:  $$ \\begin{aligned} (s+1) f(n+1, s+1) & =(2 n+1-s) f(n, s)+(s+1) f(n, s+1) \\\\ & =\\frac{2 n+1-s}{s}\\binom{n-1}{s-1}\\binom{n}{s-1}+\\binom{n-1}{s}\\binom{n}{s} \\\\ & =\\binom{n}{s}\\binom{n+1}{s}, \\end{aligned} $$  i.e., $f(n+1, s+1)=\\frac{1}{s+1}\\binom{n}{s}\\binom{n+1}{s}$. This completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "22", "problem_type": "Combinatorics", "exam": "IMO", "problem": "22. C2 (CAN) (a) If a $5 \\times n$ rectangle can be tiled using $n$ pieces like those shown in the diagram, prove that $n$ is even. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-314.jpg?height=111&width=365&top_left_y=1607&top_left_x=644) (b) Show that there are more than $2 \\cdot 3^{k-1}$ ways to tile a fixed $5 \\times 2 k$ rectangle ( $k \\geq 3$ ) with $2 k$ pieces. (Symmetric constructions are considered to be different.)", "solution": "22. (a) Color the first, third, and fifth row red, and the remaining squares white. There in total $n$ pieces and $3 n$ red squares. Since each piece can cover at most three red squares, it follows that each piece colors exactly three red squares. Then it follows that the two white squares it covers must be on the same row; otherwise, the piece has to cover at least three. Hence, each white row can be partitioned into pairs of squares belonging to the same piece. Thus it follows that the number of white squares in a row, which is $n$, must be even. (b) Let $a_{k}$ denote the number of different tilings of a $5 \\times 2 k$ rectangle. Let $b_{k}$ be the number of tilings that cannot be partitioned into two smaller tilings along a vertical line (without cutting any pieces). It is easy to see that $a_{1}=b_{1}=2, b_{2}=2, a_{2}=6=2 \\cdot 3, b_{3}=4$, and subsequently, by induction, $b_{3 k} \\geq 4, b_{3 k+1} \\geq 2$, and $b_{3 k+2} \\geq 2$. We also have $a_{k}=b_{k}+\\sum_{i=1}^{k-1} b_{i} a_{k-i}$. For $k \\geq 3$ we now have inductively  $$ a_{k}>2+\\sum_{i=1}^{k-1} 2 a_{k-i} \\geq 2 \\cdot 3^{k-1}+2 a_{k-1} \\geq 2 \\cdot 3^{k} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "23", "problem_type": "Combinatorics", "exam": "IMO", "problem": "23. C3 (GBR) A biologist watches a chameleon. The chameleon catches flies and rests after each catch. The biologist notices that: (i) the first fly is caught after a resting period of one minute; (ii) the resting period before catching the $2 m$ th fly is the same as the resting period before catching the $m$ th fly and one minute shorter than the resting period before catching the $(2 m+1)$ th fly; (iii) when the chameleon stops resting, he catches a fly instantly. (a) How many flies were caught by the chameleon before his first resting period of 9 minutes? (b) After how many minutes will the chameleon catch his 98th fly? (c) How many flies were caught by the chameleon after 1999 minutes passed?", "solution": "23. Let $r(m)$ denote the rest period before the $m$ th catch, $t(m)$ the number of minutes before the $m$ th catch, and $f(n)$ as the number of flies caught in $n$ minutes. We have $r(1)=1, r(2 m)=r(m)$, and $r(2 m+1)=f(m)+1$. We then have by induction that $r(m)$ is the number of ones in the binary representation of $m$. We also have $t(m)=\\sum_{i=1}^{m} r(i)$ and $f(t(m))=m$. From the recursive relations for $r$ we easily derive $t(2 m+1)=2 t(m)+m+1$ and consequently $t(2 m)=2 t(m)+m-r(m)$. We then have, by induction on $p, t\\left(2^{p} m\\right)=2^{p} t(m)+p \\cdot m \\cdot 2^{p-1}-\\left(2^{p}-1\\right) r(m)$. (a) We must find the smallest number $m$ such that $r(m+1)=9$. The smallest number with nine binary digits is $\\overline{111111111}_{2}=511$; hence the required $m$ is 510 . (b) We must calculate $t(98)$. Using the recursive formulas we have $t(98)=$ $2 t(49)+49-r(49), t(49)=2 t(24)+25$, and $t(24)=8 t(3)+36-7 r(3)$. Since we have $t(3)=4, r(3)=2$ and $r(49)=r\\left(\\overline{110001}_{2}\\right)=3$, it follows $t(24)=54 \\Rightarrow t(49)=133 \\Rightarrow t(98)=312$. (c) We must find $m_{c}$ such that $t\\left(m_{c}\\right) \\leq 1999<t\\left(m_{c}+1\\right)$. One can estimate where this occurs using the formula $t\\left(2^{p}\\left(2^{q}-1\\right)\\right)=(p+$ $q) 2^{p+q-1}-p 2^{p-1}-q 2^{p}+q$, provable from the recursive relations. It suffices to note that $t(462)=1993$ and $t(463)=2000$; hence $m_{c}=462$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "24", "problem_type": "Combinatorics", "exam": "IMO", "problem": "24. C4 (GBR) Let $A$ be a set of $N$ residues $\\left(\\bmod N^{2}\\right)$. Prove that there exists a set $B$ of $N$ residues $\\left(\\bmod N^{2}\\right)$ such that the set $A+B=\\{a+b \\mid$ $a \\in A, b \\in B\\}$ contains at least half of all residues $\\left(\\bmod N^{2}\\right)$.", "solution": "24. Let $S=\\left\\{0,1, \\ldots, N^{2}-1\\right\\}$ be the group of residues (with respect to addition modulo $N^{2}$ ) and $A$ an $n$-element subset. We will use $|X|$ to denote the number of elements of a subset $X$ of $S$, and $\\bar{X}$ to refer to the complement of $X$ in $S$. For $i \\in S$ we also define $A_{i}=\\{a+i \\mid a \\in A\\}$. Our task is to select $0 \\leq i_{1}<\\cdots<i_{N} \\leq N^{2}-1$ such that $\\left|\\bigcup_{j=1}^{N} A_{i_{j}}\\right| \\geq \\frac{1}{2}|S|$. Each $x \\in S$ appears in exactly $N$ sets $A_{i}$. We have  $$ \\begin{aligned} \\sum_{i_{1}<\\cdots<i_{N}}\\left|\\bigcap_{j=1}^{N} \\bar{A}_{i_{j}}\\right| & =\\sum_{i_{1}<\\cdots<i_{N}}\\left|\\left\\{x \\in S \\mid x \\notin A_{i_{1}}, \\ldots, A_{i_{N}}\\right\\}\\right| \\\\ & =\\sum_{x \\in S}\\left|\\left\\{i_{1}<\\cdots<i_{N} \\mid x \\notin A_{i_{1}}, \\ldots, A_{i_{N}}\\right\\}\\right| \\\\ & =\\sum_{x \\in S}\\binom{N^{2}-N}{N}=\\binom{N^{2}-N}{N}|S| . \\end{aligned} $$  Hence  $$ \\begin{aligned} \\sum_{i_{1}<\\cdots<i_{N}}\\left|\\bigcup_{j=1}^{N} A_{i_{j}}\\right| & =\\sum_{i_{1}<\\cdots<i_{N}}\\left(|S|-\\left|\\bigcap_{j=1}^{N} \\bar{A}_{i_{j}}\\right|\\right) \\\\ & =\\left(\\binom{N^{2}}{N}-\\binom{N^{2}-N}{N}\\right)|S| . \\end{aligned} $$  Thus, by the pigeonhole principle, one can choose $i_{1}<\\cdots<i_{N}$ such that $\\left|\\bigcup_{j=1}^{N} A_{i_{j}}\\right| \\geq\\left(1-\\binom{N^{2}-N}{N} /\\binom{N^{2}}{N}\\right)|S|$. Since $\\binom{N^{2}}{N} /\\binom{N^{2}-N}{N} \\geq\\left(\\frac{N^{2}}{N^{2}-N}\\right)^{N}$ $=\\left(1+\\frac{1}{N-1}\\right)^{N}>e>2$, it follows that $\\left|\\bigcup_{j=1}^{N} A_{i_{j}}\\right| \\geq \\frac{1}{2}|S|$; hence the chosen $i_{1}<\\cdots<i_{N}$ are indeed the elements of $B$ that satisfy the conditions of the problem.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "25", "problem_type": "Combinatorics", "exam": "IMO", "problem": "25. C5 (BLR) ${ }^{\\mathrm{IMO} 3}$ Let $n$ be an even positive integer. We say that two different cells of an $n \\times n$ board are neighboring if they have a common side. Find the minimal number of cells on the $n \\times n$ board that must be marked so that every cell (marked or not marked) has a marked neighboring cell.", "solution": "25 . Let $n=2 k$. Color the cells neighboring the edge of the board black. Then color the cells neighboring the black cells white. Then in alternation color the still uncolored cells neighboring the white or black cells on the boundary the opposite color ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-667.jpg?height=308&width=311&top_left_y=1188&top_left_x=924) and repeat until all cells are colored. We call the cells colored the same color in each such iteration a \"frame.\" In the color scheme described, each cell (white or black) neighbors exactly two black cells. The number of black cells is $2 k(k+1)$, and hence we need to mark at least $k(k+1)$ cells. On the other hand, going along each black-colored frame, we can alternately mark two consecutive cells and then not mark two consecutive cells. Every cell on the black frame will have one marked neighbor. One can arrange these sequences on two consecutive black frames such that each cell in the white frame in between has exactly one neighbor. Hence, starting from a sequence on the largest frame we obtain a marking that contains exactly half of all the black cells, i.e., $k(k+1)$ and neighbors every cell. It follows that the desired minimal number of markings is $k(k+1)$. Remark. For $n=4 k-1$ and $n=4 k+1$ one can perform similar markings to obtain minimal numbers $4 k^{2}-1$ and $(2 k+1)^{2}$, respectively.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "26", "problem_type": "Combinatorics", "exam": "IMO", "problem": "26. C6 (GBR) Suppose that every integer has been given one of the colors red, blue, green, yellow. Let $x$ and $y$ be odd integers such that $|x| \\neq|y|$. Show that there are two integers of the same color whose difference has one of the following values: $x, y, x+y, x-y$.", "solution": "26. We denote colors by capital initial letters. Let us suppose that there exists a coloring $f: \\mathbb{Z} \\rightarrow\\{R, G, B, Y\\}$ such that for any $a \\in \\mathbb{Z}$ we have $f\\{a, a+$ $x, a+y, a+x+y\\}=\\{R, G, B, Y\\}$. We now define a coloring of an integer lattice $g: \\mathbb{Z} \\times \\mathbb{Z} \\rightarrow\\{R, G, B, Y\\}$ by the rule $g(i, j)=f(x i+y j)$. It follows that every unit square in $g$ must have its vertices colored by four different colors. If there is a row or column with period 2 , then applying the condition to adjacent unit squares, we get (by induction) that all rows or columns, respectively, have period 2 . On the other hand, taking a row to be not of period 2, i.e., containing a sequence of three distinct colors, for example $G R Y$, we get that the next row must contain in these columns $Y B G$, and the following $G R Y$, and so on. It would follow that a column in this case must have period 2. A similar conclusion holds if we start with an aperiodic column. Hence either all rows or all columns must have period 2. Let us assume w.l.o.g. that all rows have a period of 2. Assuming w.l.o.g. $\\{g(0,0), g(1,0)\\}=\\{G, B\\}$, we get that the even rows are painted with $\\{G, B\\}$ and odd with $\\{Y, R\\}$. Since $x$ is odd, it follows that $g(y, 0)$ and $g(0, x)$ are of different color. However, since $g(y, 0)=f(x y)=g(0, x)$, this is a contradiction. Hence the statement of the problem holds.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "27", "problem_type": "Combinatorics", "exam": "IMO", "problem": "27. C7 (IRE) Let $p>3$ be a prime number. For each nonempty subset $T$ of $\\{0,1,2,3, \\ldots, p-1\\}$ let $E(T)$ be the set of all $(p-1)$-tuples $\\left(x_{1}, \\ldots, x_{p-1}\\right)$, where each $x_{i} \\in T$ and $x_{1}+2 x_{2}+\\cdots+(p-1) x_{p-1}$ is divisible by $p$ and let $|E(T)|$ denote the number of elements in $E(T)$. Prove that  $$ |E(\\{0,1,3\\})| \\geq|E(\\{0,1,2\\})|, $$  with equality if and only if $p=5$.", "solution": "27. Denote $A=\\{0,1,2\\}$ and $B=\\{0,1,3\\}$. Let $f_{T}(x)=\\sum_{a \\in T} x^{a}$. Then define $F_{T}(x)=f_{T}(x) f_{T}\\left(x^{2}\\right) \\cdots f_{T}\\left(x^{p-1}\\right)$. We can write $F_{T}(x)=\\sum_{i=0}^{p(p-1)} a_{i} x^{i}$, where $a_{i}$ is the number of ways to select an array $\\left\\{x_{1}, \\ldots, x_{p-1}\\right\\}$ where $x_{i} \\in T$ for all $i$ and $x_{1}+2 x_{2}+\\cdots+(p-1) x_{p-1}=i$. Let $w=\\cos (2 \\pi / p)+$ $i \\sin (2 \\pi / p)$, a $p$ th root of unity. Noting that  $$ 1+w^{j}+w^{2 j}+\\cdots+w^{(p-1) j}=\\left\\{\\begin{array}{c} p, p \\mid j, \\\\ 0, p \\nmid j, \\end{array}\\right. $$  it follows that $F_{T}(1)+F_{T}(w)+\\cdots+F_{T}\\left(w^{p-1}\\right)=p E(T)$. Since $|A|=|B|=3$, it follows that $F_{A}(1)=F_{B}(1)=3^{p-1}$. We also have for $p \\nmid i, j$ that $F_{T}\\left(w^{i}\\right)=F_{T}(w)$. Finally, we have  $$ F_{A}(w)=\\prod_{i=1}^{p-1}\\left(1+w^{i}+w^{2 i}\\right)=\\prod_{i=1}^{p-1} \\frac{1-w^{3 i}}{1-w^{i}}=1 $$  Hence, combining these results, we obtain  $$ E(A)=\\frac{3^{p-1}+p-1}{p} \\text { and } E(B)=\\frac{3^{p-1}+(p-1) F_{B}(w)}{p} $$  It remains to demonstrate that $F_{B}(w) \\geq 1$ for all $p$ and that equality holds only for $p=5$. Since $E(B)$ is an integer, it follows that $F_{B}(w)$ is an integer and $F_{B}(w) \\equiv 1(\\bmod p)$. Since $f_{B}\\left(w^{p-i}\\right)=\\overline{f_{B}\\left(w^{i}\\right)}$, it follows that $F_{B}(w)=\\left|f_{B}(w)\\right|^{2}\\left|f_{B}\\left(w^{2}\\right)\\right|^{2} \\cdots\\left|f_{B}\\left(w^{(p-1) / 2}\\right)\\right|^{2}>0$. Hence $F_{B}(w) \\geq 1$.  It remains to show that $F_{B}(w)=1$ if and only if $p=5$. We have the formula $(x-w)\\left(x-w^{2}\\right) \\cdots\\left(x-w^{p-1}\\right)=x^{p-1}+x^{p-2}+\\cdots+x+1=\\frac{x^{p}-1}{x-1}$. Let $f_{B}(x)=x^{3}+x+1=(x-\\lambda)(x-\\mu)(x-\\nu)$, where $\\lambda$, $\\mu$, and $\\nu$ are the three zeros of the polynomial $f_{B}(x)$. It follows that $F_{B}(w)=\\left(\\frac{\\lambda^{p}-1}{\\lambda-1}\\right)\\left(\\frac{\\mu^{p}-1}{\\mu-1}\\right)\\left(\\frac{\\nu^{p}-1}{\\nu-1}\\right)=-\\frac{1}{3}\\left(\\lambda^{p}-1\\right)\\left(\\mu^{p}-1\\right)\\left(\\nu^{p}-1\\right)$, since $(\\lambda-1)(\\mu-1)(\\nu-1)=-f_{B}(1)=-3$. We also have $\\lambda+\\mu+\\nu=0$, $\\lambda \\mu \\nu=-1, \\lambda \\mu+\\lambda \\nu+\\mu \\nu=1$, and $\\lambda^{2}+\\mu^{2}+\\nu^{2}=(\\lambda+\\mu+\\nu)^{2}-2(\\lambda \\mu+$ $\\lambda \\nu+\\mu \\nu)=-2$. By induction (using that $\\left(\\lambda^{r}+\\mu^{r}+\\nu^{r}\\right)+\\left(\\lambda^{r-2}+\\mu^{r-2}+\\right.$ $\\left.\\nu^{r-2}\\right)+\\left(\\lambda^{r-3}+\\mu^{r-3}+\\nu^{r-3}\\right)=0$ ), it follows that $\\lambda^{r}+\\mu^{r}+\\nu^{r}$ is an integer for all $r \\in \\mathbb{N}$. Let us assume $F_{B}(x)=1$. It follows that $\\left(\\lambda^{p}-1\\right)\\left(\\mu^{p}-1\\right)\\left(\\nu^{p}-1\\right)=-3$. Hence $\\lambda^{p}, \\mu^{p}, \\nu^{p}$ are roots of the polynomial $p(x)=x^{3}-q x^{2}+(1+q) x+1$, where $q=\\lambda^{p}+\\mu^{p}+\\nu^{p}$. Since $f_{B}(x)$ is an increasing function in real numbers, it follows that it has only one real root (w.l.o.g.) $\\lambda$, the other two roots being complex conjugates. From $f_{B}(-1)<0<f_{B}(-1 / 2)$ it follows that $-1<\\lambda<-1 / 2$. It also follows that $\\lambda^{p}$ is the $x$ coordinate of the intersection of functions $y=x^{3}+x+1$ and $y=q\\left(x^{2}-x\\right)$. Since $\\lambda<\\lambda^{p}<0$, it follows that $q>0$; otherwise, $q\\left(x^{2}-x\\right)$ intersects $x^{3}+x+1$ at a value smaller than $\\lambda$. Additionally, as $p$ increases, $\\lambda^{p}$ approaches 0 , and hence $q$ must increase. For $p=5$ we have $1+w+w^{3}=-w^{2}\\left(1+w^{2}\\right)$ and hence $G(w)=\\prod_{i=1}^{p-1}(1+$ $\\left.w^{2 j}\\right)=1$. For a zero of $f_{B}(x)$ we have $x^{5}=-x^{3}-x^{2}=-x^{2}+x+1$ and hence $q=\\lambda^{5}+\\mu^{5}+\\nu^{5}=-\\left(\\lambda^{2}+\\mu^{2}+\\nu^{2}\\right)+(\\lambda+\\mu+\\nu)+3=5$. For $p>5$ we also have $q \\geq 6$. Assuming again $F_{B}(x)=1$ and defining $p(x)$ as before, we have $p(-1)<0, p(0)>0, p(2)<0$, and $p(x)>0$ for a sufficiently large $x>2$. It follows that $p(x)$ must have three distinct real roots. However, since $\\mu^{p}, \\nu^{p} \\in \\mathbb{R} \\Rightarrow \\nu^{p}=\\overline{\\mu^{p}}=\\mu^{p}$, it follows that $p(x)$ has at most two real roots, which is a contradiction. Hence, it follows that $F_{B}(x)>1$ for $p>5$ and thus $E(A) \\leq E(B)$, where equality holds only for $p=5$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "3", "problem_type": "Number Theory", "exam": "IMO", "problem": "3. N3 (RUS) Prove that there exist two strictly increasing sequences $\\left(a_{n}\\right)$ and $\\left(b_{n}\\right)$ such that $a_{n}\\left(a_{n}+1\\right)$ divides $b_{n}^{2}+1$ for every natural number $n$.", "solution": "3. We first prove the following lemma.  Lemma. For $d, c \\in \\mathbb{N}$ and $d^{2} \\mid c^{2}+1$ there exists $b \\in \\mathbb{N}$ such that $d^{2}\\left(d^{2}+1\\right) \\mid b^{2}+1$. Proof. It is enough to set $b=c+d^{2} c-d^{3}=c+d^{2}(c-d)$.  Using the lemma it suffices to find increasing sequences $d_{n}$ and $c_{n}$ such that $c_{n}-d_{n}$ is an increasing sequence and $d_{n}^{2} \\mid c_{n}^{2}+1$. We then obtain the desired sequences $a_{n}$ and $b_{n}$ from $a_{n}=d_{n}^{2}$ and $b_{n}=c_{n}+d_{n}^{2}\\left(c_{n}-d_{n}\\right)$. It is easy to check that $d_{n}=2^{2 n}+1$ and $c_{n}=2^{n d_{n}}$ satisfy the required conditions. Hence we have demonstrated the existence of increasing sequences $a_{n}$ and $b_{n}$ such that $a_{n}\\left(a_{n}+1\\right) \\mid b_{n}^{2}+1$. Remark. There are many solutions to this problem. For example, it is sufficient to prove that the Pell-type equation $5 a_{n}\\left(a_{n}+1\\right)=b_{n}^{2}+1$ has an infinity of solutions in positive integers. Alternatively, one can show that $a_{n}\\left(a_{n}+1\\right)$ can be represented as a sum of two coprime squares for infinitely many $a_{n}$, which implies the existence of $b_{n}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "4", "problem_type": "Number Theory", "exam": "IMO", "problem": "4. N4 (FRA) Denote by $S$ the set of all primes $p$ such that the decimal representation of $1 / p$ has its fundamental period divisible by 3 . For every $p \\in S$ such that $1 / p$ has its fundamental period $3 r$ one may write $1 / p=$ $0 . a_{1} a_{2} \\ldots a_{3 r} a_{1} a_{2} \\ldots a_{3 r} \\ldots$, where $r=r(p)$; for every $p \\in S$ and every integer $k \\geq 1$ define $f(k, p)$ by  $$ f(k, p)=a_{k}+a_{k+r(p)}+a_{k+2 r(p)} $$  (a) Prove that $S$ is infinite. (b) Find the highest value of $f(k, p)$ for $k \\geq 1$ and $p \\in S$.", "solution": "4. (a) The fundamental period of $p$ is the smallest integer $d(p)$ such that $p \\mid 10^{d(p)}-1$. Let $s$ be an arbitrary prime and set $N_{s}=10^{2 s}+10^{s}+1$. In that case $N_{s} \\equiv 3(\\bmod 9)$. Let $p_{s} \\neq 37$ be a prime dividing $N_{s} / 3$. Clearly $p_{s} \\neq 3$. We claim that such a prime exists and that $3 \\mid d\\left(p_{s}\\right)$. The prime $p_{s}$ exists, since otherwise $N_{s}$ could be written in the form $N_{s}=3 \\cdot 37^{k} \\equiv$ $3(\\bmod 4)$, while on the other hand for $s>1$ we have $N_{s} \\equiv 1(\\bmod 4)$. Now we prove $3 \\mid d\\left(p_{s}\\right)$. We have $p_{s}\\left|N_{s}\\right| 10^{3 s}-1$ and hence $d\\left(p_{s}\\right) \\mid 3 s$. We cannot have $d\\left(p_{s}\\right) \\mid s$, for otherwise $p_{s}\\left|10^{s}-1 \\Rightarrow p_{s}\\right|\\left(10^{2 s}+\\right.$ $\\left.10^{s}+1,10^{s}-1\\right)=3$; and we cannot have $d\\left(p_{s}\\right) \\mid 3$, for otherwise $p_{s} \\mid 10^{3}-1=999=3^{3} \\cdot 37$, both of which contradict $p_{s} \\neq 3,37$. It follows that $d\\left(p_{s}\\right)=3 s$. Hence for every prime $s$ there exists a prime $p_{s}$ such that $d\\left(p_{s}\\right)=3 s$. It follows that the cardinality of $S$ is infinite. (b) Let $r=r(s)$ be the fundamental period of $p \\in S$. Then $p \\mid 10^{3 r}-1$, $p \\nmid 10^{r}-1 \\Rightarrow p \\mid 10^{2 r}+10^{r}+1$. Let $x_{j}=\\frac{10^{j-1}}{p}$ and $y_{j}=\\left\\{x_{j}\\right\\}=$ $0 . a_{j} a_{j+1} a_{j+2} \\ldots$. Then $a_{j}<10 y_{j}$, and hence  $$ f(k, p)=a_{k}+a_{k+r}+a_{k+2 r}<10\\left(y_{k}+y_{k+r}+y_{k+2 r}\\right) . $$  We note that $x_{k}+x_{k+s(p)}+x_{k+2 s(p)}=\\frac{10^{k-1} N_{p}}{p}$ is an integer, from which it follows that $y_{k}+y_{k+s(p)}+y_{k+2 s(p)} \\in \\mathbb{N}$. Hence $y_{k}+y_{k+s(p)}+$ $y_{k+2 s(p)} \\leq 2$. It follows that $f(k, p)<20$. We note that $f(2,7)=$ $4+8+7=19$. Hence 19 is the greatest possible value of $f(k, p)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "5", "problem_type": "Number Theory", "exam": "IMO", "problem": "5. N5 (ARM) Let $n, k$ be positive integers such that $n$ is not divisible by 3 and $k \\geq n$. Prove that there exists a positive integer $m$ that is divisible by $n$ and the sum of whose digits in decimal representation is $k$.", "solution": "5. Since one can arbitrarily add zeros at the end of $m$, which increases divisibility by 2 and 5 to an arbitrary exponent, it suffices to assume $2,5 \\nmid n$. If $(n, 10)=1$, there exists an integer $w \\geq 2$ such that $10^{w} \\equiv 1(\\bmod n)$. We also note that $10^{i w} \\equiv 1(\\bmod n)$ and $10^{j w+1} \\equiv 10(\\bmod n)$ for all integers $i$ and $j$. Let us assume that $m$ is of the form $m=\\sum_{i=1}^{u} 10^{i w}+\\sum_{j=1}^{v} 10^{j w+1}$ for integers $u, v \\geq 0$ (where if $u$ or $v$ is 0 , the corresponding sum is 0 ). Obviously, the sum of the digits of $m$ is equal to $u+v$, and also $m \\equiv u+10 v(\\bmod n)$. Hence our problem reduces to finding integers $u, v \\geq 0$ such that $u+v=k$ and $n \\mid u+10 v=k+9 v$. Since $(n, 9)=1$, it follows that there exists some $v_{0}$ such that $0 \\leq v_{0}<n \\leq k$ and $9 v_{0} \\equiv$ $-k(\\bmod n) \\Rightarrow n \\mid k+9 v_{0}$. Taking this $v_{0}$ and setting $u_{0}=k-v_{0}$ we obtain the desired parameters for defining $m$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "6", "problem_type": "Number Theory", "exam": "IMO", "problem": "6. N6 (BLR) Prove that for every real number $M$ there exists an infinite arithmetic progression such that: (i) each term is a positive integer and the common difference is not divisible by 10 ; (ii) the sum of the digits of each term (in decimal representation) exceeds $M$.", "solution": "6. Let $N$ be the smallest integer greater than $M$. We take the difference of the numbers in the progression to be of the form $10^{m}+1, m \\in \\mathbb{N}$. Hence we can take $a_{n}=a_{0}+n\\left(10^{m}+1\\right)=\\overline{b_{s} b_{s-1} \\ldots b_{0}}$ where $a_{0}$ is the initial term in the progression and $\\overline{b_{s} b_{s-1} \\ldots b_{0}}$ is the decimal representation of $a_{n}$. Since $2 m$ is the smallest integer $x$ such that $10^{x} \\equiv 1\\left(\\bmod 10^{m}+1\\right)$, it follows that $10^{k} \\equiv 10^{l}\\left(\\bmod 10^{m}+1\\right) \\Leftrightarrow k \\equiv l(\\bmod 2 m)$. Hence  $$ a_{0} \\equiv a_{n}=\\overline{b_{s} b_{s-1} \\ldots b_{0}} \\equiv \\sum_{i=0}^{2 m-1} c_{i} 10^{i}\\left(\\bmod 10^{m}+1\\right), $$  where $c_{i}=b_{i}+b_{2 m+i}+b_{4 m+i}+\\cdots \\geq 0$ for $i=0,1, \\ldots, 2 m-1$ (these $c_{i}$ also depend on $n$ ). We note that $\\sum_{i=0}^{2 m-1} c_{i} 10^{i}$ is invariant modulo $10^{m}+1$ for all $n$ and that $\\sum_{i=0}^{2 m-1} c_{i}=\\sum_{j=0}^{s} b_{j}$ for a given $n$. Hence we must choose $a_{0}$ and $m$ such that $a_{0}$ is not congruent to any number of the form $\\sum_{i=0}^{2 m-1} c_{i} 10^{i}$, where $c_{0}+c_{1}+\\cdots+c_{2 m-1} \\leq N\\left(c_{0}, c_{1}, \\ldots, c_{2 m-1} \\geq 0\\right)$. The number of ways to select the nonnegative integers $c_{0}, c_{1}, \\ldots, c_{2 m-1}$ such that $c_{0}+c_{1}+\\cdots+c_{2 m-1} \\leq N$ is equal to the number of strictly increasing sequences $0 \\leq c_{0}<c_{0}+c_{1}+1<c_{0}+c_{1}+c_{2}+2+\\cdots<$ $c_{0}+c_{1}+\\cdots+c_{2 m-1}+2 m-1 \\leq N+2 m-1$, which is equal to the number of $2 m$-element subsets of $\\{0,1,2, \\ldots, N+2 m-1\\}$, which is $\\binom{N+2 m}{N}$. For sufficiently large $m$ we have $\\binom{N+2 m}{N}<10^{m}$, and hence in this case one can select $a_{0}$ such that $a_{0}$ is not congruent to $\\sum_{i=0}^{2 m-1} c_{i} 10^{i}$ modulo $10^{m}+1$ for any set of integers $c_{0}, c_{1}, \\ldots, c_{2 m-1}$ such that $c_{0}+c_{1}+\\cdots+c_{2 m-1} \\leq N$. Thus we have found the desired arithmetic progression.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "7", "problem_type": "Geometry", "exam": "IMO", "problem": "7. G1 (ARM) Let $A B C$ be a triangle and $M$ an interior point. Prove that  $$ \\min \\{M A, M B, M C\\}+M A+M B+M C<A B+A C+B C . $$", "solution": "7. We use the following simple lemma.  Lemma. Suppose that $M$ is the interior point of a convex quadrilateral $A B C D$. Then it follows that $M A+M B<A D+D C+C B$. Proof. We repeatedly make use of the triangle inequality. The line $A M$, in addition to $A$, intersects the quadrilateral in a second point $N$. In that case $A M+M B<A N+N B<A D+D C+C B$. We now apply this lemma in the following way. Let $D, E$, and $F$ be median points of $B C, A C$, and $A B$. Any point $M$ in the interior of $\\triangle A B C$ is contained in at least two of the three convex quadrilaterals $A B D E$, $B C E F$, and $C A F D$. Let us assume without loss of generality that $M$ is in the interior of $B C E F$ and $C A F D$. In that case we apply the lemma to obtain $A M+C M<A F+F D+D C$ and $B M+C M<C E+E F+F B$ to obtain  $$ \\begin{aligned} C M+A M+B M+C M & <A F+F D+D C+C E+E F+F B \\\\ & =A B+A C+B C \\end{aligned} $$  from which the required conclusion immediately follows.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "8", "problem_type": "Geometry", "exam": "IMO", "problem": "8. G2 (JAP) A circle is called a separator for a set of five points in a plane if it passes through three of these points, it contains a fourth point in its interior, and the fifth point is outside the circle. Prove that every set of five points such that no three are collinear and no four are concyclic has exactly four separators.", "solution": "8. Let $A, B, C$, and $D$ be inverses of four of the five points, with the fifth point being the pole of the inversion. A separator through the pole transforms into a line containing two of the remaining four points such that the remaining two points are on opposite sides of the line. A separator not containing the pole transforms into a circle through three of the points with the fourth point in its interior. Let $K$ be the convex hull of $A, B, C$, and $D$. We observe two cases: (i) $K$ is a quadrilateral, for example $A B C D$. In that case the four separators are the two diagonals and two circles $A B C$ and $A D C$ if $\\angle A+\\angle C<180^{\\circ}$, or $B A D$ and $B C D$ otherwise. The remaining six viable circles and lines are clearly not separators. (ii) $K$ is a triangle, for example $A B C$ with $D$ in its interior. In that case the separators are lines $D A, D B, D C$ and the circle $A B C$. No other lines and circles qualify. We have thus shown that any set of five points satisfying the stated conditions will have exactly four separators.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "1999", "tier": "T0", "problem_label": "9", "problem_type": "Geometry", "exam": "IMO", "problem": "9. G3 (EST) ${ }^{\\mathrm{IMO} 1}$ A set $S$ of points in space will be called completely symmetric if it has at least three elements and satisfies the following condition: For every two distinct points $A, B$ from $S$ the perpendicular bisector of the segment $A B$ is an axis of symmetry for $S$. Prove that if a completely symmetric set is finite, then it consists of the vertices of either a regular polygon, a regular tetrahedron, or a regular octahedron.", "solution": "9. Let $r_{P Q}$ denote a reflection about the planar bisector of $P Q$ with $P, Q \\in S$. Let $G$ be the centroid of $S$. From $r_{P Q}(S)=S$ it follows that $r_{P Q}(G)=G$. Hence $G$ belongs to the perpendicular bisector of $P Q$ and thus $G P=G Q$. Consequently the whole of $S$ lies on a sphere $\\Sigma$ centered at $G$. We note the following two cases: (a) $S$ is a subset of a plane $\\pi$. In this case $S$ is included in a circle $k, G$ being its center. Hence its $n$ points form a convex polygon $A_{1} A_{2} \\ldots A_{n}$. When applying $r_{A_{i} A_{i+2}}$ for some $0<i<n-1$ the point $A_{i+1}$ transforms into some point of $S$ lying on the same side of $A_{i} A_{i+1}$, which has to be $A_{i+1}$ itself. It thus follows that $A_{i} A_{i+1}=A_{i+1} A_{i+2}$ for all $0<i<n-1$ and hence $A_{1} A_{2} \\ldots A_{n}$ is a regular $n$-gon. (b) The points in $S$ are not coplanar. It follows that $S$ is a polyhedron $P$ inscribed in a sphere $\\Sigma$ centered at $G$. By applying the previous case to the faces of the polyhedron, it follows that all faces are regular $n$-gons. Let us take an arbitrary vertex $V$ and let $V V_{1}, V V_{2}$ and $V V_{3}$ be three consecutive edges stemming from $V\\left(V, V_{1}, V_{2}\\right.$, and $V_{3}$ defining two adjacent faces of $P$ ). We now look at $r_{V_{1} V_{3}}$. Since this transformation leaves the half-planes $\\left[V_{1} V_{3}, V_{2}\\right.$ and $\\left[V_{1} V_{3}, V\\right.$ invariant and since $V_{2}$ and $V$ are the only points of $P$ on the respective half-planes, it follows that $r_{V_{1} V_{3}}$ leaves $V$ and $V_{2}$ invariant. This transform also swaps $V_{1}$ and $V_{3}$. Hence, the face determined by $V V_{1} V_{2}$ is transformed by $r_{V_{1} V_{3}}$ into the face $V V_{3} V_{2}$, and thus the two faces sharing $V V_{2}$ are congruent. We conclude that all faces are congruent and similarly that vertices are endpoints of the same number of edges; hence $P$ is a regular polyhedron. Finally, we have to rule out $S$ being vertices of a cube, a dodecahedron, or an icosahedron. In all of these cases if we select two diametrically opposite points $P$ and $Q$, then $S \\backslash\\{P, Q\\}$ is not symmetric with respect to the bisector of $P Q$, which prevents $r_{P Q}$ from being an invariant transformation of $S$. It thus follows that the only viable finite completely symmetric sets are vertices of regular $n$-gons, the tetrahedron, and the octahedron. It is not explicitly asked for, but it is easy to verify that all of these are indeed completely symmetric. Remark. On the IMO, a simpler version of this problem was adopted, adding the condition that $S$ belongs to a plane and thus eliminating the need for the second case altogether.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "1. $\\mathbf{C 1}$ (HUN) ${ }^{\\mathrm{IMO}}$ A magician has one hundred cards numbered 1 to 100. He puts them into three boxes, a red one, a white one, and a blue one, so that each box contains at least one card. A member of the audience draws two cards from two different boxes and announces the sum of numbers on those cards. Given this information, the magician locates the box from which no card has been drawn. How many ways are there to put the cards in the three boxes so that the trick works?", "solution": "1. In order for the trick to work, whenever $x+y=z+t$ and the cards $x, y$ are placed in different boxes, either $z, t$ are in these boxes as well or they are both in the remaining box. Case 1. The cards $i, i+1, i+2$ are in different boxes for some $i$. Since $i+(i+3)=(i+1)+(i+2)$, the cards $i$ and $i+3$ must be in the same box; moreover, $i-1$ must be in the same box as $i+2$, etc. Hence the cards $1,4,7, \\ldots, 100$ are placed in one box, the cards $2,5, \\ldots, 98$ are in the second, while $3,6, \\ldots, 99$ are in the third box. The number of different arrangements of the cards is 6 in this case. Case 2. No three successive cards are all placed in different boxes. Suppose that 1 is in the blue box, and denote by $w$ and $r$ the smallest numbers on cards lying in the white and red boxes; assume w.l.o.g. that $w<r$. The card $w+1$ is obviously not red, from which it follows that $r>$ $w+1$. Now suppose that $r<100$. Since $w+r=(w-1)+(r+1), r+1$ must be in the blue box. But then $(r+1)+w=r+(w+1)$ implies that $w+1$ must be red, which is a contradiction. Hence the red box contains only the card 100 . Since $99+w=100+(w-1)$, we deduce that the card 99 is in the white box. Moreover, if any of the cards $k$, $2 \\leq k \\leq 99$, were in the blue box, then since $k+99=(k-1)+100$, the card $k-1$ should be in the red box, which is impossible. Hence the blue box contains only the card 1 , whereas the cards $2,3, \\ldots, 99$ are all in the white box. In general, one box contains 1 , another box only 100 , while the remaining contains all the other cards. There are exactly 6 such arrangements, and the trick works in each of them. Therefore the answer is 12.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "10", "problem_type": "Algebra", "exam": "IMO", "problem": "10. A4 (GBR) The function $F$ is defined on the set of nonnegative integers and takes nonnegative integer values satisfying the following conditions: For every $n \\geq 0$, (i) $F(4 n)=F(2 n)+F(n)$; (ii) $F(4 n+2)=F(4 n)+1$; (iii) $F(2 n+1)=F(2 n)+1$.  Prove that for each positive integer $m$, the number of integers $n$ with $0 \\leq n<2^{m}$ and $F(4 n)=F(3 n)$ is $F\\left(2^{m+1}\\right)$.", "solution": "10. Clearly $F(0)=0$ by (i). Moreover, it follows by induction from (i) that $F\\left(2^{n}\\right)=f_{n+1}$ where $f_{n}$ denotes the $n$th Fibonacci's number. In general, if $n=\\epsilon_{k} 2^{k}+\\epsilon_{k-1} 2^{k-1}+\\cdots+\\epsilon_{1} \\cdot 2+\\epsilon_{0}$ (where $\\epsilon_{i} \\in\\{0,1\\}$ ), it is straightforward to verify that  $$ F(n)=\\epsilon_{k} f_{k+1}+\\epsilon_{k-1} f_{k}+\\cdots+\\epsilon_{1} f_{2}+\\epsilon_{0} f_{1} $$  We observe that if the binary representation of $n$ contains no two adjacent ones, then $F(3 n)=F(4 n)$. Indeed, if $n=\\epsilon_{k_{r}} 2^{k_{r}}+\\cdots+\\epsilon_{k_{0}} 2^{k_{0}}$, where $k_{i+1}-k_{i} \\geq 2$ for all $i$, then $3 n=\\epsilon_{k_{r}}\\left(2^{k_{r}+1}+2^{k_{r}}\\right)+\\cdots+\\epsilon_{k_{0}}\\left(2^{k_{0}+1}+2^{k_{0}}\\right)$. According to this, in computing $F(3 n)$ each $f_{i+1}$ in (1) is replaced by $f_{i+1}+f_{i+2}=f_{i+3}$, leading to the value of $F(4 n)$. We shall prove the converse: $F(3 n) \\leq F(4 n)$ holds for all $n \\geq 0$, with equality if and only if the binary representation of $n$ contains no two adjacent ones. We prove by induction on $m \\geq 1$ that this holds for all $n$ satisfying $0 \\leq n<$ $2^{m}$. The verification for the early values of $m$ is direct. Assume it is true for a certain $m$ and let $2^{m} \\leq n \\leq 2^{n+1}$. If $n=2^{m}+p, 0 \\leq p<2^{m}$, then (1) implies $F(4 n)=F\\left(2^{m+2}+4 p\\right)=f_{m+3}+F(4 p)$. Now we distinguish three cases: (i) If $3 p<2^{m}$, then the binary representation of $3 p$ does not carry into that of $3 \\cdot 2^{m}$. Then it follows from (1) and the induction hypothesis that $F(3 n)=F\\left(3 \\cdot 2^{m}\\right)+F(3 p)=f_{m+3}+F(3 p) \\leq f_{m+3}+F(4 p)=F(4 n)$. Equality holds if and only if $F(3 p)=F(4 p)$, i.e. $p$ has no two adjacent binary ones. (ii) If $2^{m} \\leq 3 p<2^{m+1}$, then the binary representation of $3 p$ carries 1 into that of $3 \\cdot 2^{m}$. Thus $F(3 n)=f_{m+3}+\\left(F(3 p)-f_{m+1}\\right)=f_{m+2}+F(3 p)<$ $f_{m+3}+F(4 p)=F(4 n)$. (iii) If $2^{m+1} \\leq p<3 \\cdot 2^{m}$, then the binary representation of $3 p$ caries 10 into that of $3 \\cdot 2^{m}$, which implies  $$ F(3 n)=f_{m+3}+f_{m+1}+\\left(F(3 p)-f_{m+2}\\right)=2 f_{m+1}+F(3 p)<F(4 n) . $$  It remains to compute the number of integers in $\\left[0,2^{m}\\right)$ with no two adjacent binary 1's. Denote their number by $u_{m}$. Among them there are $u_{m-1}$ less than $2^{m-1}$ and $u_{m-2}$ in the segment $\\left[2^{m-1}, 2^{m}\\right)$. Hence $u_{m}=u_{m-1}+u_{m-2}$ for $m \\geq 3$. Since $u_{1}=2=f_{3}, u_{2}=3=f_{4}$, we conclude that $u_{m}=f_{m+2}=F\\left(2^{m+1}\\right)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "11", "problem_type": "Algebra", "exam": "IMO", "problem": "11. A5 (BLR) ${ }^{\\mathrm{IMO} 3}$ Let $n \\geq 2$ be a positive integer and $\\lambda$ a positive real number. Initially there are $n$ fleas on a horizontal line, not all at the same point. We define a move of choosing two fleas at some points $A$ and $B$, with $A$ to the left of $B$, and letting the flea from $A$ jump over the flea from $B$ to the point $C$ such that $B C / A B=\\lambda$. Determine all values of $\\lambda$ such that for any point $M$ on the line and for any initial position of the $n$ fleas, there exists a sequence of moves that will take them all to the position right of $M$.", "solution": "11. We claim that for $\\lambda \\geq \\frac{1}{n-1}$ we can take all fleas as far to the right as we want. In every turn we choose the leftmost flea and let it jump over the rightmost one. Let $d$ and $\\delta$ denote the maximal and the minimal distances between two fleas at some moment. Clearly, $d \\geq(n-1) \\delta$. After the leftmost flea jumps over the rightmost one, the minimal distance does not decrease, because $\\lambda d \\geq \\delta$. However, the position of the leftmost flea moved to the right by at least $\\delta$, and consequently we can move the fleas arbitrarily far to the right after a finite number of moves. Suppose now that $\\lambda<\\frac{1}{n-1}$. Under this assumption we shall prove that there is a number $M$ that cannot be reached by any flea. Let us assign to each flea the coordinate on the real axis in which it is settled. Denote by $s_{k}$ the sum of all the numbers in the $k$ th step, and by $w_{k}$ the coordinate of the rightmost flea. Clearly, $s_{k} \\leq n w_{k}$. We claim that the sequence $w_{k}$ is bounded. In the $(k+1)$ th move let a flea $A$ jump over $B$, landing at $C$, and let $a, b, c$ be their respective coordinates. We have $s_{k+1}-s_{k}=c-a$. Then by the given rule, $\\lambda(b-a)=c-b=s_{k+1}-s_{k}+a-b$, which implies $s_{k+1}-s_{k}=$ $(1+\\lambda)(b-a)=\\frac{1+\\lambda}{\\lambda}(c-b)$. Hence $s_{k+1}-s_{k} \\geq \\frac{1+\\lambda}{\\lambda}\\left(w_{k+1}-w_{k}\\right)$. Summing up these inequalities for $k=0, \\ldots, n-1$ yields $s_{n}-s_{0} \\geq \\frac{1+\\lambda}{\\lambda}\\left(w_{n}-w_{0}\\right)$. Now using $s_{n} \\leq n w_{n}$ we conclude that  $$ \\left(\\frac{1+\\lambda}{\\lambda}-n\\right) w_{n} \\leq \\frac{1+\\lambda}{\\lambda} w_{0}-s_{0} . $$  Since $\\frac{1+\\lambda}{\\lambda}-n>0$, this proves the result.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "12", "problem_type": "Algebra", "exam": "IMO", "problem": "12. A6 (IRE) A nonempty set $A$ of real numbers is called a $B_{3}$-set if the conditions $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6} \\in A$ and $a_{1}+a_{2}+a_{3}=a_{4}+a_{5}+a_{6}$ imply that the sequences $\\left(a_{1}, a_{2}, a_{3}\\right)$ and $\\left(a_{4}, a_{5}, a_{6}\\right)$ are identical up to a permutation. Let $A=\\left\\{a_{0}=0<a_{1}<a_{2}<\\cdots\\right\\}, B=\\left\\{b_{0}=0<b_{1}<b_{2}<\\cdots\\right\\}$ be infinite sequences of real numbers with $D(A)=D(B)$, where, for a set $X$ of real numbers, $D(X)$ denotes the difference set $\\{|x-y| \\mid x, y \\in X\\}$. Prove that if $A$ is a $B_{3}$-set, then $A=B$.", "solution": "12. Since $D(A)=D(B)$, we can define $f(i)>g(i) \\geq 0$ that satisfy $b_{i}-b_{i-1}=$ $a_{f(i)}-a_{g(i)}$ for all $i$. The number $b_{i+1}-b_{i-1} \\in D(B)=D(A)$ can be written in the form $a_{u}-a_{v}, u>v \\geq 0$. Then $b_{i+1}-b_{i-1}=b_{i+1}-b_{i}+b_{i}-b_{i-1}$ implies $a_{f(i+1)}+a_{f(i)}+a_{v}=a_{g(i+1)}+a_{g(i)}+a_{u}$, so the $B_{3}$ property of $A$ implies that $(f(i+1), f(i), v)$ and $(g(i+1), g(i), u)$ coincide up to a permutation. It follows that either $f(i+1)=g(i)$ or $f(i)=g(i+1)$. Hence if we define $R=\\left\\{i \\in \\mathbb{N}_{0} \\mid f(i+1)=g(i)\\right\\}$ and $S=\\left\\{i \\in \\mathbb{N}_{0} \\mid f(i)=g(i+1)\\right\\}$ it holds that $R \\cup S=\\mathbb{N}_{0}$. Lemma. If $i \\in R$, then also $i+1 \\in R$. Proof. Suppose to the contrary that $i \\in R$ and $i+1 \\in S$, i.e., $g(i)=$ $f(i+1)=g(i+2)$. There are integers $x$ and $y$ such that $b_{i+2}-b_{i-1}=$ $a_{x}-a_{y}$. Then $a_{x}-a_{y}=a_{f(i+2)}-a_{g(i+2)}+a_{f(i+1)}-a_{g(i+1)}+a_{f(i)}-$ $a_{g(i)}=a_{f(i+2)}+a_{f(i)}-a_{g(i+1)}-a_{g(i)}$, so by the $B_{3}$ property $(x, g(i+$ $1), g(i))$ and $(y, f(i+2), f(i))$ coincide up to a permutation. But this is impossible, since $f(i+2), f(i)>g(i+2)=g(i)=f(i+1)>g(i+1)$. This proves the lemma. Therefore if $i \\in R \\neq \\emptyset$, then it follows that every $j>i$ belongs to $R$. Consequently $g(i)=f(i+1)>g(i+1)=f(i+2)>g(i+2)=f(i+3)>$ $\\cdots$ is an infinite decreasing sequence of nonnegative integers, which is impossible. Hence $S=\\mathbb{N}_{0}$, i.e.,  $$ b_{i+1}-b_{i}=a_{f(i+1)}-a_{f(i)} \\quad \\text { for all } i \\in \\mathbb{N}_{0} $$  Thus $f(0)=g(1)<f(1)<f(2)<\\cdots$, implying $f(i) \\geq i$. On the other hand, for any $i$ there exist $j, k$ such that $a_{f(i)}-a_{i}=b_{j}-b_{k}=a_{f(j)}-a_{f(k)}$, so by the $B_{3}$ property $i \\in\\{f(i), f(k)\\}$ is a value of $f$. Hence we must have $f(i)=i$ for all $i$, which finally gives $A=B$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "13", "problem_type": "Algebra", "exam": "IMO", "problem": "13. A7 (RUS) For a polynomial $P$ of degree 2000 with distinct real coefficients let $M(P)$ be the set of all polynomials that can be produced from $P$ by permutation of its coefficients. A polynomial $P$ will be called $n$-independent if $P(n)=0$ and we can get from any $Q$ in $M(P)$ a polynomial $Q_{1}$ such that $Q_{1}(n)=0$ by interchanging at most one pair of coefficients of $Q$. Find all integers $n$ for which $n$-independent polynomials exist.", "solution": "13. One can easily find $n$-independent polynomials for $n=0,1$. For example, $P_{0}(x)=2000 x^{2000}+\\cdots+2 x^{2}+x+0$ is 0 -independent (for $Q \\in M\\left(P_{0}\\right)$ it suffices to exchange the coefficient 0 of $Q$ with the last term), and $P_{1}(x)=2000 x^{2000}+\\cdots+2 x^{2}+x-(1+2+\\cdots+2000)$ is 1-independent (since any $Q \\in M\\left(P_{1}\\right)$ vanishes at $\\left.x=1\\right)$. Let us show that no $n$-independent polynomials exist for $n \\notin\\{0,1\\}$. Consider separately the case $n=-1$. For any set $T$ we denote by $S(T)$ the sum of elements of $T$. Suppose that $P(x)=a_{2000} x^{2000}+\\cdots+a_{1} x+a_{0}$ is $-1-$ independent. Since $P(-1)=\\left(a_{0}+a_{2}+\\cdots+a_{2000}\\right)-\\left(a_{1}+a_{3}+\\cdots+a_{1999}\\right)$, this means that for any subset $E$ of the set $C=\\left\\{a_{0}, a_{1}, \\ldots, a_{2000}\\right\\}$ having 1000 or 1001 elements there exist elements $e \\in E$ and $f \\in C \\backslash E$ such that $S(E \\cup\\{f\\} \\backslash\\{e\\})=\\frac{1}{2} S(C)$, or equivalently that $S(E)-\\frac{1}{2} S(C)=e-f$. We may assume w.l.o.g. that $a_{0}<a_{1}<\\cdots<a_{2000}$. Suppose that $E$ is a 1000 -element subset of $C$ containing $b_{0}, b_{1}$ but not $b_{1999}, b_{2000}$. By the -1 -independence of $P$ there exist $e \\in E$ and $f \\in$ $C \\backslash E$ such that $S(E)-\\frac{1}{2} S(C)=e-f$. The same must hold for the set $E^{\\prime}=E \\cup\\left\\{b_{1999}, b_{2000}\\right\\} \\backslash\\left\\{b_{0}, b_{1}\\right\\}$, so for some $e^{\\prime} \\in E^{\\prime}$ and $f^{\\prime} \\in C \\backslash E^{\\prime}$ we have $S\\left(E^{\\prime}\\right)-\\frac{1}{2} S(C)=e^{\\prime}-f^{\\prime}$. It follows that $b_{1999}+b_{2000}-b_{0}-b_{1}=$ $S\\left(E^{\\prime}\\right)-S(E)=e+e^{\\prime}-f-f^{\\prime}$. Therefore the transposition $e \\leftrightarrow f$ must involve at least one of the elements $b_{0}, b_{1}, b_{1999}, b_{2000}$.  There are 7994 possible transpositions involving one of these four elements. On the other hand, by (SL93-12) the subsets $E$ of $C$ containing $b_{0}, b_{1}$ but not $b_{1999}, b_{2000}$ give at least $998 \\cdot 999+1$ distinct sums of elements, far exceeding 7994. This is a contradiction. For the case $|n| \\geq 2$ we need the following lemma. Lemma. Let $n \\geq 2$ be a natural number and $P(x)=a_{m} x^{m}+\\cdots+a_{1} x+a_{0}$ a polynomial with distinct coefficients. Then the set $\\{Q(n) \\mid Q \\in M(P)\\}$ contains at least $2^{m}$ elements. Proof. We shall use induction on $m$. The statement is easily verified for $m=1$. Assume w.l.o.g. that $a_{m}<\\cdots<a_{1}<a_{0}$. Consider two polynomials $Q_{k}$ and $Q_{k+1}$ of the form  $$ \\begin{aligned} Q_{k}(x) & =a_{m} x^{m}+\\cdots+a_{k} x^{k}+a_{0} x^{k-1}+b_{k-1} x^{k-2}+\\cdots+b_{1}, \\\\ Q_{k+1}(x) & =a_{m} x^{m}+\\cdots+a_{k+1} x^{k+1}+a_{0} x^{k}+c_{k} x^{k-1}+\\cdots+c_{1}, \\end{aligned} $$  where $\\left(b_{k-1}, \\ldots, b_{1}\\right)$ and $\\left(c_{k}, \\ldots, c_{1}\\right)$ are permutations of the sets $\\left\\{a_{k-1}, \\ldots, a_{1}\\right\\}$ and $\\left\\{a_{k}, \\ldots, a_{1}\\right\\}$ respectively. We claim that $Q_{k+1}(n) \\geq$ $Q_{k}(n)$. Indeed, since $a_{0}-c_{k} \\leq a_{0}-a_{k}$ and $b_{j}-c_{j}<a_{0}-a_{k}$ for $1 \\leq j \\leq n-1$, we have $Q_{k+1}(n)-Q_{k}(x)=\\left(a_{0}-a_{k}\\right) n^{k}-\\left(a_{0}-c_{k}\\right) n^{k-1}-$ $\\left(b_{k-1}-c_{k-1}\\right) n^{k-2}-\\cdots-\\left(b_{1}-c_{1}\\right) \\geq\\left(a_{0}-a_{k}\\right)\\left(n^{k}-n^{k-1}-\\cdots-n-1\\right)>0$. Furthermore, by the induction hypothesis the polynomials of the form $Q_{k}(x)$ take at least $2^{k-2}$ values at $x=n$. Hence the total number of values of $Q(n)$ for $Q \\in M(P)$ is at least $1+1+2+2^{2}+\\cdots+2^{m-1}=2^{m}$. Now we return to the main result. Suppose that $P(x)=a_{2000} x^{2000}$ $+a_{1999} x^{1999}+a_{0}$ is an $n$-independent polynomial. Since $P_{2}(x)=a_{2000} x^{2000}$ $+a_{1998} x^{1998}+\\cdots+a_{2} x^{2}+a_{0}$ is a polynomial in $t=x^{2}$ of degree 1000 , by the lemma it takes at least $2^{1000}$ distinct values at $x=n$. Hence $\\{Q(n) \\mid Q \\in$ $M(P)\\}$ contains at least $2^{1000}$ elements. On the other hand, interchanging the coefficients $b_{i}$ and $b_{j}$ in a polynomial $Q(x)=b_{2000} x^{2000}+\\cdots+b_{0}$ modifies the value of $Q$ at $x=n$ by $\\left(b_{i}-b_{j}\\right)\\left(n^{i}-n^{j}\\right)=\\left(a_{k}-a_{l}\\right)\\left(n^{i}-n^{j}\\right)$ for some $k, l$. Hence there are fewer than $2001^{4}$ possible modifications of the value at $n$. Since $2001^{4}<2^{1000}$, we have arrived at a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "14", "problem_type": "Number Theory", "exam": "IMO", "problem": "14. N1 (JAP) Determine all positive integers $n \\geq 2$ that satisfy the following condition: For all integers $a, b$ relatively prime to $n$,  $$ a \\equiv b(\\bmod n) \\quad \\text { if and only if } \\quad a b \\equiv 1(\\bmod n) $$", "solution": "14. The given condition is obviously equivalent to $a^{2} \\equiv 1(\\bmod n)$ for all integers $a$ coprime to $n$. Let $n=p_{1}^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\cdots p_{k}^{\\alpha_{k}}$ be the factorization of $n$ onto primes. Since by the Chinese remainder theorem the numbers coprime to $n$ can give any remainder modulo $p_{i}^{\\alpha_{i}}$ except 0 , our condition is equivalent to $a^{2} \\equiv 1\\left(\\bmod p_{i}^{\\alpha_{i}}\\right)$ for all $i$ and integers $a$ coprime to $p_{i}$. Now if $p_{i} \\geq 3$, we have $2^{2} \\equiv 1\\left(\\bmod p_{i}^{\\alpha_{i}}\\right)$, so $p_{i}=3$ and $\\alpha_{i}=2$. If $p_{j}=2$, then $3^{2} \\equiv 1\\left(\\bmod 2^{\\alpha_{j}}\\right)$ implies $\\alpha_{j} \\leq 3$. Hence $n$ is a divisor of $2^{3} \\cdot 3=24$. Conversely, each $n \\mid 24$ has the desired property.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "15", "problem_type": "Number Theory", "exam": "IMO", "problem": "15. N2 (FRA) For a positive integer $n$, let $d(n)$ be the number of all positive divisors of $n$. Find all positive integers $n$ such that $d(n)^{3}=4 n$.", "solution": "15. Let $n=p_{1}^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\cdots p_{k}^{\\alpha_{k}}$ be the factorization of $n$ onto primes $\\left(p_{1}<p_{2}<\\right.$ $\\cdots<p_{k}$ ). Since $4 n$ is a perfect cube, we deduce that $p_{1}=2$ and $\\alpha_{1}=$ $3 \\beta_{1}+1, \\alpha_{2}=3 \\beta_{2}, \\ldots, \\alpha_{k}=3 \\beta_{k}$ for some integers $\\beta_{i} \\geq 0$. Using $d(n)=$ $\\left(\\alpha_{1}+1\\right) \\cdot\\left(\\alpha_{2}+1\\right) \\cdots\\left(\\alpha_{k}+1\\right)$ we can rewrite the equation $d(n)^{3}=4 n$ as  $$ \\left(3 \\beta_{1}+2\\right) \\cdot\\left(3 \\beta_{2}+1\\right) \\cdots\\left(3 \\beta_{k}+1\\right)=2^{\\beta_{1}+1} p_{2}^{\\beta_{2}} \\cdots p_{k}^{\\beta_{k}} $$  Since $d(n)$ is not divisible by 3 , it follows that $p_{i} \\geq 5$ for $i \\geq 2$. Thus the above equation is equivalent to  $$ \\frac{3 \\beta_{1}+2}{2^{\\beta_{1}+1}}=\\frac{p_{2}^{\\beta_{2}}}{3 \\beta_{2}+1} \\cdots \\frac{p_{k}^{\\beta_{k}}}{3 \\beta_{k}+1} $$  For $i \\geq 2$ we have $p_{i}^{\\beta_{i}} \\geq(1+4)^{\\beta_{i}} \\geq 1+4 \\beta_{i}$; hence (1) implies that $\\frac{3 \\beta_{1}+2}{2^{\\beta_{1}+1}} \\geq 1$, which leads to $\\beta_{1} \\leq 2$. For $\\beta_{1}=0$ or $\\beta_{1}=2$ we have that $\\frac{3 \\beta_{1}+2}{2^{\\beta_{1}+1}}=1$, and therefore $\\beta_{2}=\\cdots=$ $\\beta_{k}=0$. This yields the solutions $n=2$ and $n=2^{7}=128$. For $\\beta_{1}=1$ the left-hand side of (1) equals $\\frac{5}{4}$. On the other hand, if $p_{i}>5$ or $\\beta_{i}>1$, then $\\frac{p_{i}^{\\beta_{i}}}{3 \\beta_{i}+1}>\\frac{5}{4}$, which is impossible. We conclude that $p_{2}=5$ and $k=2$, so $n \\stackrel{2000}{=}$. Hence the solutions for $n$ are 2, 128, and 2000.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "16", "problem_type": "Number Theory", "exam": "IMO", "problem": "16. N3 (RUS) ${ }^{\\mathrm{IMO} 5}$ Does there exist a positive integer $n$ such that $n$ has exactly 2000 prime divisors and $2^{n}+1$ is divisible by $n$ ?", "solution": "16. More generally, we will prove by induction on $k$ that for each $k \\in \\mathbb{N}$ there exists $n_{k} \\in \\mathbb{N}$ that has exactly $k$ distinct prime divisors such that $n_{k} \\mid 2^{n_{k}}+1$ and $3 \\mid n_{k}$. For $k=1, n_{1}=3$ satisfies the given conditions. Now assume that $k \\geq 1$ and $n_{k}=3^{\\alpha} m$ where $3 \\nmid m$, so that $m$ has exactly $k-1$ prime divisors. Then the number $3 n_{k}=3^{\\alpha+1} m$ has exactly $k$ prime divisors and $2^{3 n_{k}}+1=$ $\\left(2^{n_{k}}+1\\right)\\left(2^{2 n_{k}}-2^{n_{k}}+1\\right)$ is divisible by $3 n_{k}$, since $3 \\mid 2^{2 n_{k}}-2^{n_{k}}+1$. We shall find a prime $p$ not dividing $n_{k}$ such that $n_{k+1}=3 p n_{k}$. It is enough to find $p$ such that $p \\mid 2^{3 n_{k}}+1$ and $p \\nmid 2^{n_{k}}+1$. Moreover, we shall show that for every integer $a>2$ there exists a prime number $p$ that divides $a^{3}+1=(a+1)\\left(a^{2}-a+1\\right)$ but not $a+1$. To prove this we observe that $\\operatorname{gcd}\\left(a^{2}-a+1, a+1\\right)=\\operatorname{gcd}(3, a+1)$. Now if $3 \\nmid a+1$, we can simply take $p=3$; otherwise, if $a=3 b-1$, then $a^{2}-a+1=9 b^{2}-9 b+3$ is not divisible by $3^{2}$; hence we can take for $p$ any prime divisor of $\\frac{a^{2}-a+1}{3}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "17", "problem_type": "Number Theory", "exam": "IMO", "problem": "17. N4（BRA）Determine all triples of positive integers $(a, m, n)$ such that $a^{m}+1$ divides $(a+1)^{n}$.", "solution": "17. Trivially all triples $(a, 1, n)$ and $(1, m, n)$ are solutions. Assume now that $a>1$ and $m>1$. If $m$ is even, then $a^{m}+1 \\equiv(-1)^{m}+1 \\equiv 2(\\bmod a+1)$, which implies that $a^{m}+1=2^{t}$. In particular, $a$ is odd. But this is impossible, since $2<a^{m}+1=\\left(a^{m / 2}\\right)^{2}+1 \\equiv 2(\\bmod 4)$. Hence $m$ is odd. Let $p$ be an arbitrary prime divisor of $m$ and $m=p m_{1}$. Then $a^{m}+1 \\mid$ $(a+1)^{n} \\mid\\left(a^{m_{1}}+1\\right)^{n}$, so $b^{p}+1 \\mid(b+1)^{n}$ for $b=a^{m_{1}}$. It follows that  $$ \\left.P=\\frac{b^{p}+1}{b+1}=b^{p-1}-b^{p-2}+\\cdots+1 \\right\\rvert\\,(b+1)^{n} $$  Since $P \\equiv p(\\bmod b+1)$, we deduce that $P$ has no prime divisors other than $p$; hence $P$ is a power of $p$ and $p \\mid b+1$. Let $b=k p-1, k \\in \\mathbb{N}$. Then by the binomial formula we have $b^{i}=(k p-1)^{i} \\equiv(-1)^{i+1}(i k p-1)\\left(\\bmod p^{2}\\right)$, and therefore $P \\equiv-k p((p-1)+(p-2)+\\cdots+1)+p \\equiv p\\left(\\bmod p^{2}\\right)$. We conclude that $P \\leq p$. But we also have $P \\geq b^{p-1}-b^{p-2} \\geq b^{p-2}>p$ for $p>3$, so we must have $P=p=3$ and $b=2$. Since $b=a^{m_{1}}$, we obtain $a=2$ and $m=3$. The triple $(2,3, n)$ is indeed a solution if $n \\geq 2$. Hence the set of solutions is $\\{(a, 1, n),(1, m, n) \\mid a, m, n \\in \\mathbb{N}\\} \\cup\\{(2,3, n) \\mid$ $n \\geq 2\\}$. Remark. This problem is very similar to (SL97-14).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "18", "problem_type": "Number Theory", "exam": "IMO", "problem": "18. N5 (BUL) Prove that there exist infinitely many positive integers $n$ such that $p=n r$, where $p$ and $r$ are respectively the semiperimeter and the inradius of a triangle with integer side lengths.", "solution": "18. It is known that the area of the triangle is $S=p r=p^{2} / n$ and $S=$ $\\sqrt{p(p-a)(p-b)(p-c)}$. It follows that $p^{3}=n^{2}(p-a)(p-b)(p-c)$, which by putting $x=p-a, y=p-b$, and $z=p-c$ transforms into  $$ (x+y+z)^{3}=n^{2} x y z . $$  We will be done if we show that (1) has a solution in positive integers for infinitely many natural numbers $n$. Let us assume that $z=k(x+y)$ for an integer $k>0$. Then (1) becomes $(k+1)^{3}(x+y)^{2}=k n^{2} x y$. Further, by setting $n=3(k+1)$ this equation reduces to  $$ (k+1)(x+y)^{2}=9 k x y $$  Set $t=x / y$. Then (2) has solutions in positive integers if and only if $(k+$ 1) $(t+1)^{2}=9 k t$ has a rational solution, i.e., if and only if its discriminant $D=k(5 k-4)$ is a perfect square. Setting $k=u^{2}$, we are led to show that $5 u^{2}-4=v^{2}$ has infinitely many integer solutions. But this is a classic Pell-type equation, whose solution is every Fibonacci number $u=F_{2 i+1}$. This completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "19", "problem_type": "Number Theory", "exam": "IMO", "problem": "19. N6 (ROM) Show that the set of positive integers that cannot be represented as a sum of distinct perfect squares is finite.", "solution": "19. Suppose that a natural number $N$ satisfies $N=a_{1}^{2}+\\cdots+a_{k}^{2}, 2 N=$ $b_{1}^{2}+\\cdots+b_{l}^{2}$, where $a_{i}, b_{j}$ are natural numbers such that none of the ratios $a_{i} / a_{j}, b_{i} / b_{j}, a_{i} / b_{j}, b_{j} / a_{i}$ is a power of 2. We claim that every natural number $n>\\sum_{i=0}^{4 N-2}(2 i N+1)^{2}$ can be represented as a sum of distinct squares. Suppose $n=4 q N+r, 0 \\leq r<4 N$. Then  $$ n=4 N s+\\sum_{i=0}^{r-1}(2 i N+1)^{2} $$  for some positive integer $s$, so it is enough to show that $4 N s$ is a sum of distinct even squares. Let $s=\\sum_{c=1}^{C} 2^{2 u_{c}}+\\sum_{d=1}^{D} 2^{2 v_{d}+1}$ be the binary expansion of $s$. Then  $$ 4 N s=\\sum_{c=1}^{C} \\sum_{i=1}^{k}\\left(2^{u_{c}+1} a_{i}\\right)^{2}+\\sum_{d=1}^{D} \\sum_{j=1}^{l}\\left(2^{u_{d}+1} b_{j}\\right)^{2} $$  where all the summands are distinct by the condition on $a_{i}, b_{j}$.  It remains to choose an appropriate $N$ : for example $N=29$, because $29=5^{2}+2^{2}$ and $58=7^{2}+3^{2}$. Second solution. It can be directly checked that every odd integer $67<$ $n \\leq 211$ can be represented as a sum of distinct squares. For any $n>211$ we can choose an integer $m$ such that $m^{2}>\\frac{n}{2}$ and $n-m^{2}$ is odd and greater than 67 , and therefore by the induction hypothesis can be written as a sum of distinct squares. Hence $n$ is also a sum of distinct squares.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "2", "problem_type": "Combinatorics", "exam": "IMO", "problem": "2. C2 (ITA) A brick staircase with three steps of width 2 is made of twelve unit cubes. Determine all integers $n$ for which it is possible to build a cube of side $n$ using such bricks. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-317.jpg?height=175&width=141&top_left_y=739&top_left_x=725)", "solution": "2. Since the volume of each brick is 12 , the side of any such cube must be divisible by 6 . Suppose that a cube of side $n=6 k$ can be built using $\\frac{n^{3}}{12}=18 k^{3}$ bricks. Set a coordinate system in which the cube is given as $[0, n] \\times[0, n] \\times[0, n]$ and color in black each unit cube $[2 p, 2 p+1] \\times[2 q, 2 q+1] \\times[2 r, 2 r+1]$. There are exactly $\\frac{n^{3}}{9}=27 k^{3}$ black cubes. Each brick covers either one or three black cubes, which is in any case an odd number. It follows that the total number of black cubes must be even, which implies that $k$ is even. Hence $12 \\mid n$. On the other hand, two bricks can be fitted together to give a $2 \\times 3 \\times 4$ box. Using such boxes one can easily build a cube of side 12 , and consequently any cube of side divisible by 12 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "20", "problem_type": "Geometry", "exam": "IMO", "problem": "20. G1 (NET) In the plane we are given two circles intersecting at $X$ and $Y$. Prove that there exist four points $A, B, C, D$ with the following property: For every circle touching the two given circles at $A$ and $B$, and meeting the line $X Y$ at $C$ and $D$, each of the lines $A C, A D, B C, B D$ passes through one of these points.", "solution": "20. Denote by $k_{1}, k_{2}$ the given circles and by $k_{3}$ the circle through $A, B, C, D$. We shall consider the case that $k_{3}$ is inside $k_{1}$ and $k_{2}$, since the other case is analogous. Let $A C$ and $A D$ meet $k_{1}$ at points $P$ and $R$, and $B C$ and $B D$ meet $k_{2}$ at $Q$ and $S$ respectively. We claim that $P Q$ and $R S$ are the common tangents to $k_{1}$ and $k_{2}$, and therefore $P, Q, R, S$ are the desired points. The circles $k_{1}$ and $k_{3}$ are tangent to each other, so we have $D C \\| R P$. Since ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-680.jpg?height=374&width=468&top_left_y=669&top_left_x=848)  $$ A C \\cdot C P=X C \\cdot C Y=B C \\cdot C Q $$  the quadrilateral $A B Q P$ is cyclic, implying that $\\angle A P Q=\\angle A B Q=$ $\\angle A D C=\\angle A R P$. It follows that $P Q$ is tangent to $k_{1}$. Similarly, $P Q$ is tangent to $k_{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "21", "problem_type": "Geometry", "exam": "IMO", "problem": "21. G2 (RUS) ${ }^{\\mathrm{IMO} 1}$ Two circles $G_{1}$ and $G_{2}$ intersect at $M$ and $N$. Let $A B$ be the line tangent to these circles at $A$ and $B$, respectively, such that $M$ lies closer to $A B$ than $N$. Let $C D$ be the line parallel to $A B$ and passing through $M$, with $C$ on $G_{1}$ and $D$ on $G_{2}$. Lines $A C$ and $B D$ meet at $E$; lines $A N$ and $C D$ meet at $P$; lines $B N$ and $C D$ meet at $Q$. Show that $E P=E Q$.", "solution": "21. Let $K$ be the intersection point of the lines $M N$ and $A B$. Since $K A^{2}=K M \\cdot K N=K B^{2}$, it follows that $K$ is the midpoint of the segment $A B$, and consequently $M$ is the midpoint of $A B$. Thus it will be enough to show that $E M \\perp$ $P Q$, or equivalently that $E M \\perp$ $A B$. However, since $A B$ is tangent to the circle $G_{1}$ we have $\\angle B A M=$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-680.jpg?height=356&width=509&top_left_y=1295&top_left_x=812) $\\angle A C M=\\angle E A B$, and similarly $\\angle A B M=\\angle E B A$. This implies that the triangles $E A B$ and $M A B$ are congruent. Hence $E$ and $M$ are symmetric with respect to $A B$; hence $E M \\perp A B$. Remark. The proposer has suggested an alternative version of the problem: to prove that $E N$ bisects the angle $C N D$. This can be proved by noting that $E A N B$ is cyclic.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "22", "problem_type": "Geometry", "exam": "IMO", "problem": "22. G3 (IND) Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $A B C$. Show that there exist points $D, E$, and $F$ on sides $B C$, $C A$, and $A B$ respectively such that $O D+D H=O E+E H=O F+F H$ and the lines $A D, B E$, and $C F$ are concurrent.", "solution": "22. Let $L$ be the point symmetric to $H$ with respect to $B C$. It is well known that $L$ lies on the circumcircle $k$ of $\\triangle A B C$. Let $D$ be the intersection point of $O L$ and $B C$. We similarly define $E$ and $F$. Then  $$ O D+D H=O D+D L=O L=O E+E H=O F+F H $$  We shall prove that $A D, B E$, and $C F$ are concurrent. Let line $A O$ meet $B C$ at $D^{\\prime}$. It is easy to see that $\\angle O D^{\\prime} D=\\angle O D D^{\\prime}$; hence the perpendicular bisector of $B C$ bisects $D D^{\\prime}$ as well. Hence $B D=C D^{\\prime}$. If we define $E^{\\prime}$ and $F^{\\prime}$ analogously, we have $C E=A E^{\\prime}$ and $A F=B F^{\\prime}$. Since the lines $A D^{\\prime}, B E^{\\prime}, C F^{\\prime}$ meet at $O$, it follows that $\\frac{B D}{D C} \\cdot \\frac{C E}{E A} \\cdot \\frac{A F}{F B}=$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-681.jpg?height=399&width=430&top_left_y=252&top_left_x=867) $\\frac{B D^{\\prime}}{D^{\\prime} C} \\cdot \\frac{C E^{\\prime}}{E^{\\prime} A} \\cdot \\frac{A F^{\\prime}}{F^{\\prime} B}=1$. This proves our claim by Ceva's theorem.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "23", "problem_type": "Geometry", "exam": "IMO", "problem": "23. G4 (RUS) Let $A_{1} A_{2} \\ldots A_{n}$ be a convex polygon, $n \\geq 4$. Prove that $A_{1} A_{2} \\ldots A_{n}$ is cyclic if and only if to each vertex $A_{j}$ one can assign a pair $\\left(b_{j}, c_{j}\\right)$ of real numbers, $j=1,2, \\ldots n$, such that  $$ A_{i} A_{j}=b_{j} c_{i}-b_{i} c_{j} \\quad \\text { for all } i, j \\text { with } 1 \\leq i \\leq j \\leq n $$", "solution": "23. First, suppose that there are numbers $\\left(b_{i}, c_{i}\\right)$ assigned to the vertices of the polygon such that  $$ A_{i} A_{j}=b_{j} c_{i}-b_{i} c_{j} \\quad \\text { for all } i, j \\text { with } 1 \\leq i \\leq j \\leq n $$  In order to show that the polygon is cyclic, it is enough to prove that $A_{1}, A_{2}, A_{3}, A_{i}$ lie on a circle for each $i, 4 \\leq i \\leq n$, or equivalently, by Ptolemy's theorem, that $A_{1} A_{2} \\cdot A_{3} A_{i}+A_{2} A_{3} \\cdot A_{i} A_{1}=A_{1} A_{3} \\cdot A_{2} A_{i}$. But this is straightforward with regard to (1). Now suppose that $A_{1} A_{2} \\ldots A_{n}$ is a cyclic quadrilateral. By Ptolemy's theorem we have $A_{i} A_{j}=A_{2} A_{j} \\cdot \\frac{A_{1} A_{i}}{A_{1} A_{2}}-A_{2} A_{i} \\cdot \\frac{A_{1} A_{j}}{A_{1} A_{2}}$ for all $i, j$. This suggests taking $b_{1}=-A_{1} A_{2}, b_{i}=A_{2} A_{i}$ for $i \\geq 2$ and $c_{i}=\\frac{A_{1} A_{i}}{A_{1} A_{2}}$ for all $i$. Indeed, using Ptolemy's theorem, one easily verifies (1).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "24", "problem_type": "Geometry", "exam": "IMO", "problem": "24. G5 (GBR) The tangents at $B$ and $A$ to the circumcircle of an acuteangled triangle $A B C$ meet the tangent at $C$ at $T$ and $U$ respectively. $A T$ meets $B C$ at $P$, and $Q$ is the midpoint of $A P ; B U$ meets $C A$ at $R$, and $S$ is the midpoint of $B R$. Prove that $\\angle A B Q=\\angle B A S$. Determine, in terms of ratios of side lengths, the triangles for which this angle is a maximum.", "solution": "24. Since $\\angle A B T=180^{\\circ}-\\gamma$ and $\\angle A C T=180^{\\circ}-\\beta$, the law of sines gives $\\frac{B P}{P C}=\\frac{S_{A B T}}{S_{A C T}}=\\frac{A B \\cdot B T \\cdot \\sin \\gamma}{A B \\cdot B T \\cdot \\sin \\beta}=\\frac{A B \\sin \\gamma}{A C \\sin \\beta}=\\frac{c^{2}}{b^{2}}$, which implies $B P=\\frac{c^{2} a}{b^{2}+c^{2}}$. Denote by $M$ and $N$ the feet of perpendiculars from $P$ and $Q$ on $A B$. We have $\\cot \\angle A B Q=\\frac{B N}{N Q}=\\frac{2 B N}{P M}=\\frac{B A+B M}{B P \\sin \\beta}=\\frac{c+B P \\cos \\beta}{B P \\sin \\beta}=\\frac{b^{2}+c^{2}+a c \\cos \\beta}{c a \\sin \\beta}=$ $\\frac{2\\left(b^{2}+c^{2}\\right)+a^{2}+c^{2}-b^{2}}{2 c a \\sin \\beta}=\\frac{a^{2}+b^{2}+3 c^{2}}{4 S_{A B C}}=2 \\cot \\alpha+2 \\cot \\beta+\\cot \\gamma$. Similarly, $\\cot \\angle B A S=2 \\cot \\alpha+2 \\cot \\beta+\\cot \\gamma$; hence $\\angle A B Q=\\angle B A S$. Now put $p=\\cot \\alpha$ and $q=\\cot \\beta$. Since $p+q \\geq 0$, the A-G mean inequality gives us $\\cot \\angle A B Q=2 p+2 q+\\frac{1-p q}{p+q} \\geq 2 p+2 q+\\frac{1-(p+q)^{2} / 4}{p+q}=\\frac{7}{4}(p+q)+$ $\\frac{1}{p+q} \\geq 2 \\sqrt{\\frac{7}{4}}=\\sqrt{7}$. Hence $\\angle A B Q \\leq \\arctan \\frac{1}{\\sqrt{7}}$. Equality holds if and only if $\\cot \\alpha=\\cot \\beta=\\frac{1}{\\sqrt{7}}$, i.e., when $a: b: c=1: 1: \\frac{1}{\\sqrt{2}}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "25", "problem_type": "Geometry", "exam": "IMO", "problem": "25. G6 (ARG) Let $A B C D$ be a convex quadrilateral with $A B$ not parallel to $C D$, let $X$ be a point inside $A B C D$ such that $\\measuredangle A D X=\\measuredangle B C X<90^{\\circ}$ and $\\measuredangle D A X=\\measuredangle C B X<90^{\\circ}$. If $Y$ is the point of intersection of the perpendicular bisectors of $A B$ and $C D$, prove that $\\measuredangle A Y B=2 \\measuredangle A D X$.", "solution": "25. By the condition of the problem, $\\triangle A D X$ and $\\triangle B C X$ are similar. Then there exist points $Y^{\\prime}$ and $Z^{\\prime}$ on the perpendicular bisector of $A B$ such that $\\triangle A Y^{\\prime} Z^{\\prime}$ is similar and oriented the same as $\\triangle A D X$, and $\\triangle B Y^{\\prime} Z^{\\prime}$ is (being congruent to $\\triangle A Y^{\\prime} Z^{\\prime}$ ) similar and oriented the same as $\\triangle B C X$. Since then $A D / A Y^{\\prime}=A X / A Z^{\\prime}$ and $\\angle D A Y^{\\prime}=\\angle X A Z^{\\prime}, \\triangle A D Y^{\\prime}$ and $\\triangle A X Z^{\\prime}$ are also similar, implying $\\frac{A D}{A X}=\\frac{D Y^{\\prime}}{X Z^{\\prime}}$. Analogously, $\\frac{B C}{B X}=\\frac{C Y^{\\prime}}{X Z^{\\prime}}$. It follows from $\\frac{A D}{A X}=\\frac{B C}{B X}$ that $C Y^{\\prime}=D Y^{\\prime}$, which means that $Y^{\\prime}$ lies on the perpendicular bisector of $C D$. Hence $Y^{\\prime} \\equiv Y$.  Now $\\angle A Y B=2 \\angle A Y Z^{\\prime}=2 \\angle A D X$, as desired.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "26", "problem_type": "Geometry", "exam": "IMO", "problem": "26. G7 (IRN) Ten gangsters are standing on a flat surface, and the distances between them are all distinct. At twelve o'clock, when the church bells start chiming, each of them fatally shoots the one among the other nine gangsters who is the nearest. At least how many gangsters will be killed?", "solution": "26. The problem can be reformulated in the following way: Given a set $S$ of ten points in the plane such that the distances between them are all distinct, for each point $P \\in S$ we mark the point $Q \\in S \\backslash\\{P\\}$ nearest to $P$. Find the least possible number of marked points. Observe that each point $A \\in S$ is the nearest to at most five other points. Indeed, for any six points $P_{1}, \\ldots, P_{6}$ one of the angles $P_{i} A P_{j}$ is at most $60^{\\circ}$, in which case $P_{i} P_{j}$ is smaller than one of the distances $A P_{i}, A P_{j}$. It follows that at least two points are marked. Now suppose that exactly two points, say $A$ and $B$, are marked. Then $A B$ is the minimal distance of the points from $S$, so by the previous observation the rest of the set $S$ splits into two subsets of four points according to whether the nearest point is $A$ or $B$. Let these subsets be $\\left\\{A_{1}, A_{2}, A_{3}, A_{4}\\right\\}$ and $\\left\\{B_{1}, B_{2}, B_{3}, B_{4}\\right\\}$ respectively. Assume that the points are labelled so that the angles $A_{i} A A_{i+1}$ are successively adjacent as well as the angles $B_{i} B B_{i+1}$, and that $A_{1}, B_{1}$ lie on one side of $A B$, and $A_{4}, B_{4}$ lie on the other side. Since all the angles $A_{i} A A_{i+1}$ and $B_{i} B B_{i+1}$ are greater than $60^{\\circ}$, it follows that  $$ \\angle A_{1} A B+\\angle B A A_{4}+\\angle B_{1} B A+\\angle A B B_{4}<360^{\\circ} . $$  Therefore $\\angle A_{1} A B+\\angle B_{1} B A<180^{\\circ}$ or $\\angle A_{4} A B+\\angle B_{4} B A<180^{\\circ}$. Without loss of generality, let us assume the first inequality. On the other hand, note that the quadrilateral $A B B_{1} A_{1}$ is convex because $A_{1}$ and $B_{1}$ are on different sides of the perpendicular bisector of $A B$. From $A_{1} B_{1}>A_{1} A$ and $B B_{1}>A B$ we obtain $\\angle A_{1} A B_{1}>\\angle A_{1} B_{1} A$ and $\\angle B A B_{1}>\\angle A B_{1} B$. Adding these relations yields $\\angle A_{1} A B>\\angle A_{1} B_{1} B$. Similarly, $\\angle B_{1} B A>\\angle B_{1} A_{1} A$. Adding these two inequalities, we get  $$ 180^{\\circ}>\\angle A_{1} A B+\\angle B_{1} B A>\\angle A_{1} B_{1} B+\\angle B_{1} A_{1} A $$  hence the sum of the angles of the quadrilateral $A B B_{1} A_{1}$ is less than $360^{\\circ}$, which is a contradiction. Thus at least 3 points are marked. An example of a configuration in which exactly 3 gangsters are killed is shown below. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-682.jpg?height=250&width=441&top_left_y=1700&top_left_x=579)", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "27", "problem_type": "Geometry", "exam": "IMO", "problem": "27. G8 (RUS) ${ }^{\\mathrm{IMO}} A_{1} A_{2} A_{3}$ is an acute-angled triangle. The foot of the altitude from $A_{i}$ is $K_{i}$, and the incircle touches the side opposite $A_{i}$ at $L_{i}$. The line $K_{1} K_{2}$ is reflected in the line $L_{1} L_{2}$. Similarly, the line $K_{2} K_{3}$ is reflected in $L_{2} L_{3}$, and $K_{3} K_{1}$ is reflected in $L_{3} L_{1}$. Show that the three new lines form a triangle with vertices on the incircle.", "solution": "27. Denote by $\\alpha_{1}, \\alpha_{2}, \\alpha_{3}$ the angles of $\\triangle A_{1} A_{2} A_{3}$ at vertices $A_{1}, A_{2}, A_{3}$ respectively. Let $T_{1}, T_{2}, T_{3}$ be the points symmetric to $L_{1}, L_{2}, L_{3}$ with respect to $A_{1} I, A_{2} I$, and $A_{3} I$ respectively. We claim that $T_{1} T_{2} T_{3}$ is the desired triangle.  Denote by $S_{1}$ and $R_{1}$ the points symmetric to $K_{1}$ and $K_{3}$ with respect to $L_{1} L_{3}$. It is enough to show that $T_{1}$ and $T_{3}$ lie on the line $R_{1} S_{1}$. To prove this, we shall prove that $\\angle K_{1} S_{1} T_{1}=\\angle K^{\\prime} K_{1} S_{1}$ for a point $K^{\\prime}$ on the line $K_{1} K_{3}$ such that $K_{3}$ and $K^{\\prime}$ lie on different sides of $K_{1}$. We show first that $S_{1} \\in A_{1} I$. Let $X$ be the point of intersection of lines $A_{1} I$ and $L_{1} L_{3}$. We see from the triangle $A_{1} L_{3} X$ that $\\angle L_{1} X I=$ $\\alpha_{3} / 2=\\angle L_{1} A_{3} I$, which implies that ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-683.jpg?height=489&width=559&top_left_y=289&top_left_x=805) $L_{1} X A_{3} I$ is cyclic. We now have $\\angle A_{1} X A_{3}=90^{\\circ}=\\angle A_{1} K_{1} A_{3}$; hence $A_{1} K_{1} X A_{3}$ is also cyclic. It follows that $\\angle K_{1} X I=\\angle K_{1} A_{3} A_{1}=\\alpha_{3}=2 \\angle L_{1} X I$; hence $X_{1} L_{1}$ bisects the angle $K_{1} X_{1} I$. Hence $S_{1} \\in X I$ as claimed. Now we have $\\angle K_{1} S_{1} T_{1}=\\angle K_{1} S_{1} L_{1}+2 \\angle L_{1} S_{1} X=\\angle S_{1} K_{1} L_{1}+2 \\angle L_{1} K_{1} X$. It remains to prove that $K_{1} X$ bisects $\\angle A_{3} K_{1} K^{\\prime}$. From the cyclic quadrilateral $A_{1} K_{1} X A_{3}$ we see that $\\angle X K_{1} A_{3}=\\alpha_{1} / 2$. Since $A_{1} K_{3} K_{1} A_{3}$ is cyclic, we also have $\\angle K^{\\prime} K_{1} A_{3}=\\alpha_{1}=2 \\angle X K_{1} A_{3}$, which proves the claim.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "3", "problem_type": "Combinatorics", "exam": "IMO", "problem": "3. C3 (COL) Let $n \\geq 4$ be a fixed positive integer. Given a set $S=$ $\\left\\{P_{1}, P_{2}, \\ldots, P_{n}\\right\\}$ of points in the plane such that no three are collinear and no four concyclic, let $a_{t}, 1 \\leq t \\leq n$, be the number of circles $P_{i} P_{j} P_{k}$ that contain $P_{t}$ in their interior, and let  $$ m(S)=a_{1}+a_{2}+\\cdots+a_{n} $$  Prove that there exists a positive integer $f(n)$, depending only on $n$, such that the points of $S$ are the vertices of a convex polygon if and only if $m(S)=f(n)$.", "solution": "3. Clearly $m(S)$ is the number of pairs of point and triangle $\\left(P_{t}, P_{i} P_{j} P_{k}\\right)$ such that $P_{t}$ lies inside the circle $P_{i} P_{j} P_{k}$. Consider any four-element set $S_{i j k l}=\\left\\{P_{i}, P_{j}, P_{k}, P_{l}\\right\\}$. If the convex hull of $S_{i j k l}$ is the triangle $P_{i} P_{j} P_{k}$, then we have $a_{i}=a_{j}=a_{k}=0, a_{l}=1$. Suppose that the convex hull is the quadrilateral $P_{i} P_{j} P_{k} P_{l}$. Since this quadrilateral is not cyclic, we may suppose that $\\angle P_{i}+\\angle P_{k}<180^{\\circ}<\\angle P_{j}+\\angle P_{l}$. In this case $a_{i}=a_{k}=0$ and $a_{j}=a_{l}=1$. Therefore $m\\left(S_{i j k l}\\right)$ is 2 if $P_{i}, P_{j}, P_{k}, P_{l}$ are vertices of a convex quadrilateral, and 1 otherwise. There are $\\binom{n}{4}$ four-element subsets $S_{i j k l}$. If $a(S)$ is the number of such subsets whose points determine a convex quadrilateral, we have $m(S)=$ $2 a(S)+\\left(\\binom{n}{4}-a(S)\\right)=\\binom{n}{4}+a(S) \\leq 2\\binom{n}{4}$. Equality holds if and only if every four distinct points of $S$ determine a convex quadrilateral, i.e. if and only if the points of $S$ determine a convex polygon. Hence $f(n)=2\\binom{n}{4}$ has the desired property.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "4", "problem_type": "Combinatorics", "exam": "IMO", "problem": "4. C4 (CZE) Let $n$ and $k$ be positive integers such that $n / 2<k \\leq 2 n / 3$. Find the least number $m$ for which it is possible to place $m$ pawns on $m$ squares of an $n \\times n$ chessboard so that no column or row contains a block of $k$ adjacent unoccupied squares.", "solution": "4. By a good placement of pawns we mean the placement in which there is no block of $k$ adjacent unoccupied squares in a row or column. We can make a good placement as follows: Label the rows and columns with $0,1, \\ldots, n-1$ and place a pawn on a square $(i, j)$ if and only if $k$ divides $i+j+1$. This is obviously a good placement in which the pawns are placed on three lines with $k, 2 n-2 k$, and $2 n-3 k$ squares, which adds up to $4 n-4 k$ pawns in total. Now we shall prove that a good placement must contain at least $4 n-4 k$ pawns. Suppose we have a good placement of $m$ pawns. Partition the board into nine rectangular regions as shown in the picture. Let $a, b, \\ldots, h$ be the numbers of pawns in the rectangles $A, B, \\ldots, H$ respectively. Note that each row that ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-671.jpg?height=394&width=450&top_left_y=962&top_left_x=850) passes through $A, B$, and $C$ either contains a pawn inside $B$, or contains a pawn in both $A$ and $C$. It follows that $a+c+2 b \\geq 2(n-k)$. We similarly obtain that $c+e+2 d, e+g+2 f$, and $g+a+2 h$ are all at least $2(n-k)$. Adding and dividing by 2 yields $a+b+\\cdots+h \\geq 4(n-k)$, which proves the statement.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "5", "problem_type": "Combinatorics", "exam": "IMO", "problem": "5. C5 (RUS) In the plane we have $n$ rectangles with parallel sides. The sides of distinct rectangles lie on distinct lines. The boundaries of the rectangles cut the plane into connected regions. A region is nice if it has at least one of the vertices of the $n$ rectangles on its boundary. Prove that the sum of the numbers of the vertices of all nice regions is less than $40 n$. (There can be nonconvex regions as well as regions with more than one boundary curve.)", "solution": "5. We say that a vertex of a nice region is convex if the angle of the region at that vertex equals $90^{\\circ}$; otherwise (if the angle is $270^{\\circ}$ ), we say that a vertex is concave. For a simple broken line $C$ contained in the boundary of a nice region $R$ we call the pair $(R, C)$ a boundary pair. Such a pair is called outer if the region $R$ is inside the broken line $C$, and inner otherwise. Let $\\mathcal{B}_{i}, \\mathcal{B}_{o}$ be the sets of inner and outer boundary pairs of nice regions respectively, and let $\\mathcal{B}=\\mathcal{B}_{i} \\cup \\mathcal{B}_{o}$. For a boundary pair $b=(R, C)$ denote by $c_{b}$ and $v_{b}$ respectively the number of convex and concave vertices of $R$ that belong to $C$. We have the following facts: (1) Each vertex of a rectangle corresponds to one concave angle of a nice region and vice versa. This correspondence is bijective, so $\\sum_{b \\in \\mathcal{B}} v_{b}=$ $4 n$. (2) For a boundary pair $b=(R, C)$ the sum of angles of $R$ that are on $C$ equals $\\left(c_{b}+v_{b}-2\\right) 180^{\\circ}$ if $b$ is outer, and $\\left(c_{b}+v_{b}+2\\right) 180^{\\circ}$ if $b$ is inner. On the other hand the sum of angles is obviously equal to $c_{b} \\cdot 90^{\\circ}+v_{b} \\cdot 270^{\\circ}$. It immediately follows that $c_{b}-v_{b}=\\left\\{\\begin{array}{r}4 \\text { if } b \\in \\mathcal{B}_{o}, \\\\ -4 \\text { if } b \\in \\mathcal{B}_{i} .\\end{array}\\right.$ (3) Since every vertex of a rectangle appears in exactly two boundary pairs and each boundary pair contains at least one vertex of a rectangle, the number $K$ of boundary pairs is less than or equal to $8 n$. (4) The set $\\mathcal{B}_{i}$ is nonempty, because every boundary of the infinite region is inner. Consequently, the sum of the numbers of the vertices of all nice regions is equal to  $$ \\sum_{b \\in \\mathcal{B}}\\left(c_{b}+v_{b}\\right)=\\sum_{b \\in \\mathcal{B}}\\left(2 v_{b}+\\left(c_{b}-v_{b}\\right)\\right) \\leq 2 \\cdot 4 n+4(K-1)-4 \\leq 40 n-8 $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "6", "problem_type": "Combinatorics", "exam": "IMO", "problem": "6. C6 (FRA) Let $p$ and $q$ be relatively prime positive integers. A subset $S$ of $\\{0,1,2, \\ldots\\}$ is called ideal if $0 \\in S$ and for each element $n \\in S$, the integers $n+p$ and $n+q$ belong to $S$. Determine the number of ideal subsets of $\\{0,1,2 \\ldots\\}$.", "solution": "6. Every integer $z$ has a unique representation $z=p x+q y$, where $x, y \\in \\mathbb{Z}$, $0 \\leq x \\leq q-1$. Consider the region $T$ in the $x y$-plane defined by the last inequality and $p x+q y \\geq 0$. There is a bijective correspondence between lattice points of this region and nonnegative integers given by $(x, y) \\mapsto$ $z=p x+q y$. Let us mark all lattice points of $T$ whose corresponding integers belong to $S$ and color in black the unit squares whose left-bottom vertices are at marked points. Due to the condition for $S$, this coloring has the property that all points lying on the right or above a colored point are colored as well. In particular, since the point $(0,0)$ is colored, all points above or on the line $y=0$ are colored. What we need is the number of such colorings of $T$. The border of the colored subregion $C$ of $T$ determines a path from $(0,0)$ to $(q,-p)$ consisting of consecutive unit moves either to the right or downwards. There are $\\binom{p+q}{p}$ such paths in total. We must find the number of such paths not going below the line $l: p x+q y=0$. Consider any path $\\gamma=A_{0} A_{1} \\ldots A_{p+q}$ from $A_{0}=(0,0)$ to $A_{p+q}=(q,-p)$. We shall see the path $\\gamma$ as a sequence $G_{1} G_{2} \\ldots G_{p+q}$ of moves to the right $(R)$ or downwards ( $D$ ) with exactly $p D$ 's and $q R$ 's. Two paths are said to be equivalent if one is obtained from the other by a circular shift of the corresponding sequence $G_{1} G_{2} \\ldots G_{p+q}$. We note that all the $p+q$ circular shifts of a path are distinct. Indeed, $G_{1} \\ldots G_{p+q} \\equiv$ $G_{i+1} \\ldots G_{i+p+q}$ would imply $G_{1}=G_{i+1}=G_{2 i+1}=\\cdots$ (where $G_{j+p+q}=$ $G_{j}$ ), so $G_{1}=\\cdots=G_{p+q}$, which is impossible. Hence each equivalence class contains exactly $p+q$ paths. Let $l_{i}, 0 \\leq i<p+q$, be the line through $A_{i}$ that is parallel to the line $l$. Since $\\operatorname{gcd}(p, q)=1$, all these lines are distinct. Let $l_{m}$ be the unique lowest line among the $l_{i}$ 's. Then the path $G_{m+1} G_{m+2} \\ldots G_{m+p+q}$ is above the line $l$. Every other cyclic shift gives rise to a path having at least one vertex below the line $l$. Thus each equiv- alence class contains exactly one path above the line $l$, so the number of such paths is equal to $\\frac{1}{p+q}\\binom{p+q}{p}$. Therefore the answer is $\\frac{1}{p+q}\\binom{p+q}{p}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "7", "problem_type": "Algebra", "exam": "IMO", "problem": "7. A1 (USA) ${ }^{\\mathrm{IMO} 2}$ Let $a, b, c$ be positive real numbers with product 1. Prove that  $$ \\left(a-1+\\frac{1}{b}\\right)\\left(b-1+\\frac{1}{c}\\right)\\left(c-1+\\frac{1}{a}\\right) \\leq 1 . $$", "solution": "7. Elementary computation gives $\\left(a-1+\\frac{1}{b}\\right)\\left(b-1+\\frac{1}{c}\\right)=a b-a+\\frac{a}{c}-b+$ $1-\\frac{1}{c}+1-\\frac{1}{b}+\\frac{1}{b c}$. Using $a b=\\frac{1}{c}$ and $\\frac{1}{b c}=a$ we obtain  $$ \\left(a-1+\\frac{1}{b}\\right)\\left(b-1+\\frac{1}{c}\\right)=\\frac{a}{c}-b-\\frac{1}{b}+2 \\leq \\frac{a}{c} $$  since $b+\\frac{1}{b} \\geq 2$. Similarly we obtain  $$ \\left(b-1+\\frac{1}{c}\\right)\\left(c-1+\\frac{1}{a}\\right) \\leq \\frac{b}{a} \\text { and }\\left(c-1+\\frac{1}{a}\\right)\\left(a-1+\\frac{1}{b}\\right) \\leq \\frac{c}{b} . $$  The desired inequality follows from the previous three inequalities. Equality holds if and only if $a=b=c=1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "8", "problem_type": "Algebra", "exam": "IMO", "problem": "8. A2 (GBR) Let $a, b, c$ be positive integers satisfying the conditions $b>2 a$ and $c>2 b$. Show that there exists a real number $t$ with the property that all the three numbers $t a, t b, t c$ have their fractional parts lying in the interval (1/3,2/3].", "solution": "8. We note that $\\{t a\\}$ lies in $\\left(\\frac{1}{3}, \\frac{2}{3}\\right]$ if and only if there is an integer $k$ such that $k+\\frac{1}{3}<t a \\leq k+\\frac{2}{3}$, i.e., if and only if $t \\in I_{k}=\\left(\\frac{k+1 / 3}{a}, \\frac{k+2 / 3}{a}\\right]$ for some $k$. Similarly, $t$ should belong to the sets $J_{m}=\\left(\\frac{m+1 / 3}{b}, \\frac{m+2 / 3}{b}\\right]$ and $K_{n}=\\left(\\frac{n+1 / 3}{c}, \\frac{n+2 / 3}{c}\\right]$ for some $m, n$. We have to show that $I_{k} \\cap J_{m} \\cap K_{n}$ is nonempty for some integers $k, m, n$. The intervals $K_{n}$ are separated by a distance $\\frac{2}{3 c}$, and since $\\frac{2}{3 c}<\\frac{1}{3 b}$, each of the intervals $J_{m}$ intersects at least one of the $K_{n}$ 's. Hence it is enough to prove that $J_{m} \\subset I_{k}$ for some $k, m$. Let $u_{m}$ and $v_{m}$ be the left and right endpoints of $J_{m}$. Since $a v_{m}=a u_{m}+$ $\\frac{a}{3 b}<a u_{m}+\\frac{1}{6}$, it will suffice to show that there is an integer $m$ such that the fractional part of $a u_{m}$ lies in $\\left[\\frac{1}{3}, \\frac{1}{2}\\right]$. Let $a=d \\alpha, b=d \\beta, \\operatorname{gcd}(\\alpha, \\beta)=1$. Setting $m=d \\mu$ we obtain that $a u_{m}=a \\frac{m+1 / 3}{b}=\\frac{\\alpha m}{d \\beta}+\\frac{\\alpha}{3 \\beta}=\\frac{\\alpha \\mu}{\\beta}+\\frac{\\alpha}{3 \\beta}$. Since $\\alpha \\mu$ gives all possible residues modulo $\\beta$, every term of the arithmetic progression $\\frac{j}{\\beta}+\\frac{\\alpha}{3 \\beta}(j \\in \\mathbb{Z})$ has its fractional part equal to the fractional part of some $a u_{m}$. Now for $\\beta \\geq 6$ the progression step is $\\frac{1}{\\beta} \\leq \\frac{1}{6}$, so at least one of the $a u_{m}$ has its fractional part in $[1 / 3,1 / 2]$. If otherwise $\\beta \\leq 5$, the only irreducible fractions $\\frac{\\alpha}{\\beta}$ that satisfy $2 \\alpha<\\beta$ are $\\frac{1}{3}, \\frac{1}{4}, \\frac{1}{5}, \\frac{2}{5}$; hence one can take $m$ to be $1,1,2,3$ respectively. This justifies our claim.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2000", "tier": "T0", "problem_label": "9", "problem_type": "Algebra", "exam": "IMO", "problem": "9. A3 (BLR) Find all pairs of functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}, g: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that  $$ f(x+g(y))=x f(y)-y f(x)+g(x) \\quad \\text { for all } x, y \\in R $$", "solution": "9. Let us first solve the problem under the assumption that $g(\\alpha)=0$ for some $\\alpha$. Setting $y=\\alpha$ in the given equation yields $g(x)=(\\alpha+1) f(x)-x f(\\alpha)$. Then the given equation becomes $f(x+g(y))=(\\alpha+1-y) f(x)+(f(y)-f(\\alpha)) x$, so setting $y=\\alpha+1$ we get $f(x+n)=m x$, where $n=g(\\alpha+1)$ and $m=f(\\alpha+1)-f(\\alpha)$. Hence $f$ is a linear function, and consequently $g$ is also linear. If we now substitute $f(x)=a x+b$ and $g(x)=c x+d$ in the given equation and compare the coefficients, we easily find that  $$ f(x)=\\frac{c x-c^{2}}{1+c} \\quad \\text { and } \\quad g(x)=c x-c^{2}, \\quad c \\in \\mathbb{R} \\backslash\\{-1\\} $$  Now we prove the existence of $\\alpha$ such that $g(\\alpha)=0$. If $f(0)=0$ then putting $y=0$ in the given equation we obtain $f(x+g(0))=g(x)$, so we can take $\\alpha=-g(0)$. Now assume that $f(0)=b \\neq 0$. By replacing $x$ by $g(x)$ in the given equation we obtain $f(g(x)+g(y))=g(x) f(y)-y f(g(x))+g(g(x))$ and, analogously, $f(g(x)+g(y))=g(y) f(x)-x f(g(y))+g(g(y))$. The given functional equation for $x=0$ gives $f(g(y))=a-b y$, where $a=g(0)$. In particular, $g$ is injective and $f$ is surjective, so there exists $c \\in \\mathbb{R}$ such that $f(c)=0$. Now the above two relations yield  $$ g(x) f(y)-a y+g(g(x))=g(y) f(x)-a x+g(g(y)) $$  Plugging $y=c$ in (1) we get $g(g(x))=g(c) f(x)-a x+g(g(c))+a c=$ $k f(x)-a x+d$. Now (1) becomes $g(x) f(y)+k f(x)=g(y) f(x)+k f(y)$. For $y=0$ we have $g(x) b+k f(x)=a f(x)+k b$, whence  $$ g(x)=\\frac{a-k}{b} f(x)+k . $$  Note that $g(0)=a \\neq k=g(c)$, since $g$ is injective. From the surjectivity of $f$ it follows that $g$ is surjective as well, so it takes the value 0 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "1", "problem_type": "Algebra", "exam": "IMO", "problem": "1. A1 (IND) Let $T$ denote the set of all ordered triples $(p, q, r)$ of nonnegative integers. Find all functions $f: T \\rightarrow \\mathbb{R}$ such that  $$ f(p, q, r)=\\left\\{\\begin{array}{rrr} 0 & & \\text { if } p q r=0 \\\\ 1+\\frac{1}{6} & (f(p+1, q-1, r)+f(p-1, q+1, r) & \\\\ & +f(p-1, q, r+1)+f(p+1, q, r-1) & \\\\ & +f(p, q+1, r-1)+f(p, q-1, r+1)) & \\\\ & \\text { otherwise. } \\end{array}\\right. $$", "solution": "1. First, let us show that such a function is at most unique. Suppose that $f_{1}$ and $f_{2}$ are two such functions, and consider $g=f_{1}-f_{2}$. Then $g$ is zero on the boundary and satisfies  $$ g(p, q, r)=\\frac{1}{6}[g(p+1, q-1, r)+\\cdots+g(p, q-1, r+1)] $$  i.e., $g(p, q, r)$ is equal to the average of the values of $g$ at six points $(p+$ $1, q-1, r), \\ldots$ that lie in the plane $\\pi$ given by $x+y+z=p+q+r$. Suppose that $(p, q, r)$ is the point at which $g$ attains its maximum in absolute value on $\\pi \\cap T$. The averaging property of $g$ implies that the values of $g$ at $(p+1, q-1, r)$ etc. are all equal to $g(p, q, r)$. Repeating this argument we obtain that $g$ is constant on the whole of $\\pi \\cap T$, and hence it equals 0 everywhere. Therefore $f_{1} \\equiv f_{2}$. It remains to guess $f$. It is natural to try $\\bar{f}(p, q, r)=p q r$ first: it satisfies $\\bar{f}(p, q, r)=\\frac{1}{6}[\\bar{f}(p+1, q-1, r)+\\cdots+\\bar{f}(p, q-1, r+1)]+\\frac{p+q+r}{3}$. Thus we simply take  $$ f(p, q, r)=\\frac{3}{p+q+r} \\bar{f}(p, q, r)=\\frac{3 p q r}{p+q+r} $$  and directly check that it satisfies the required property. Hence this is the unique solution.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "10", "problem_type": "Combinatorics", "exam": "IMO", "problem": "10. C4 (NZL) A set of three nonnegative integers $\\{x, y, z\\}$ with $x<y<z$ is called historic if $\\{z-y, y-x\\}=\\{1776,2001\\}$. Show that the set of all nonnegative integers can be written as the union of disjoint historic sets.", "solution": "10. For convenience let us write $a=1776, b=2001,0<a<b$. There are two types of historic sets:  $$ \\text { (1) }\\{x, x+a, x+a+b\\} \\quad \\text { and } \\quad \\text { (2) }\\{x, x+b, x+a+b\\} \\text {. } $$  We construct a sequence of historic sets $H_{1}, H_{2}, H_{3}, \\ldots$ inductively as follows: (i) $H_{1}=\\{0, a, a+b\\}$, and (ii) Let $y_{n}$ be the least nonnegative integer not occurring in $U_{n}=H_{1} \\cap$ $\\cdots \\cap H_{n}$. We take $H_{n+1}$ to be $\\left\\{y_{n}, y_{n}+a, y_{n}+a+b\\right\\}$ if $y_{n}+a \\notin U_{n}$, and $\\left\\{y_{n}, y_{n}+b, y_{n}+a+b\\right\\}$ otherwise. It remains to show that this construction never fails. Suppose that it failed at the construction of $H_{n+1}$. The element $y_{n}+a+b$ is not contained in $U_{n}$, since by the construction the smallest elements of $H_{1}, \\ldots, H_{n}$ are all less than $y_{n}$. Hence the reason for the failure must be the fact that both $y_{n}+a$ and $y_{n}+b$ are covered by $U_{n}$. Further, $y_{n}+b$ must have been the largest element of its set $H_{k}$, so the smallest element of $H_{k}$ equals $y_{n}-a$. But since $y_{n}$ is not covered, we conclude that $H_{k}$ is of type (2). This is a contradiction, because $y_{n}$ was free, so by the algorithm we had to choose for $H_{k}$ the set of type (1) (that is, $\\left\\{y_{n}-a, y_{n}, y_{n}+b\\right\\}$ ) first.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "11", "problem_type": "Combinatorics", "exam": "IMO", "problem": "11. C5 (FIN) Find all finite sequences $\\left(x_{0}, x_{1}, \\ldots, x_{n}\\right)$ such that for every $j, 0 \\leq j \\leq n, x_{j}$ equals the number of times $j$ appears in the sequence.", "solution": "11. Let $\\left(x_{0}, x_{1}, \\ldots, x_{n}\\right)$ be any such sequence: its terms are clearly nonnegative integers. Also, $x_{0}=0$ yields a contradiction, so $x_{0}>0$. Let $m$ be the number of positive terms among $x_{1}, \\ldots, x_{n}$. Since $x_{i}$ counts the terms equal to $i$, the sum $x_{1}+\\cdots+x_{n}$ counts the total number of positive terms in the sequence, which is known to be $m+1$. Therefore among $x_{1}, \\ldots, x_{n}$ exactly $m-1$ terms are equal to 1 , one is equal to 2 , and the others are 0 . Only $x_{0}$ can exceed 2 , and consequently at most one of $x_{3}, x_{4}, \\ldots$ can be positive. It follows that $m \\leq 3$. (i) $m=1$ : Then $x_{2}=2$ (since $x_{1}=2$ is impossible), so $x_{0}=2$. The resulting sequence is $(2,0,2,0)$. (ii) $m=2$ : Either $x_{1}=2$ or $x_{2}=2$. These cases yield $(1,2,1,0)$ and $(2,1,2,0,0)$ respectively. (iii) $m=3$ : This means that $x_{k}>0$ for some $k>2$. Hence $x_{0}=k$ and $x_{k}=1$. Further, $x_{1}=1$ is impossible, so $x_{1}=2$ and $x_{2}=1$; there are no more positive terms in the sequence. The resulting sequence is $(p, 2,1, \\underbrace{0, \\ldots, 0}_{p-3}, 1,0,0,0)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "12", "problem_type": "Combinatorics", "exam": "IMO", "problem": "12. C6 (CAN) For a positive integer $n$ define a sequence of zeros and ones to be balanced if it contains $n$ zeros and $n$ ones. Two balanced sequences $a$ and $b$ are neighbors if you can move one of the $2 n$ symbols of $a$ to another position to form $b$. For instance, when $n=4$, the balanced sequences 01101001 and 00110101 are neighbors because the third (or fourth) zero in the first sequence can be moved to the first or second position to form the second sequence. Prove that there is a set $S$ of at most $\\frac{1}{n+1}\\binom{2 n}{n}$ balanced sequences such that every balanced sequence is equal to or is a neighbor of at least one sequence in $S$.", "solution": "12. For each balanced sequence $a=\\left(a_{1}, a_{2}, \\ldots, a_{2 n}\\right)$ denote by $f(a)$ the sum of $j$ 's for which $a_{j}=1$ (for example, $f(100101)=1+4+6=11$ ). Partition the $\\binom{2 n}{n}$ balanced sequences into $n+1$ classes according to the residue of $f$ modulo $n+1$. Now take $S$ to be a class of minimum size: obviously $|S| \\leq \\frac{1}{n+1}\\binom{2 n}{n}$. We claim that every balanced sequence $a$ is either a member of $S$ or a neighbor of a member of $S$. We consider two cases. (i) Let $a_{1}$ be 1 . It is easy to see that moving this 1 just to the right of the $k$ th 0 , we obtain a neighboring balanced sequence $b$ with $f(b)=$ $f(a)+k$. Thus if $a \\notin S$, taking a suitable $k \\in\\{1,2, \\ldots, n\\}$ we can achieve that $b \\in S$. (ii) Let $a_{1}$ be 0 . Taking this 0 just to the right of the $k$ th 1 gives a neighbor $b$ with $f(b)=f(a)-k$, and the conclusion is similar to that of (i). This justifies our claim.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "13", "problem_type": "Combinatorics", "exam": "IMO", "problem": "13. C7 (FRA) A pile of $n$ pebbles is placed in a vertical column. This configuration is modified according to the following rules. A pebble can be moved if it is at the top of a column that contains at least two more pebbles than the column immediately to its right. (If there are no pebbles to the right, think of this as a column with 0 pebbles.) At each stage, choose a pebble from among those that can be moved (if there are any) and place it at the top of the column to its right. If no pebbles can be moved, the configuration is called a final configuration. For each $n$, show that no matter what choices are made at each stage, the final configuration is unique. Describe that configuration in terms of $n$.", "solution": "13. At any moment, let $p_{i}$ be the number of pebbles in the $i$ th column, $i=$ $1,2, \\ldots$ The final configuration has obvious properties $p_{1} \\geq p_{2} \\geq \\cdots$ and $p_{i+1} \\in\\left\\{p_{i}, p_{i}-1\\right\\}$. We claim that $p_{i+1}=p_{i}>0$ is possible for at most one $i$. Assume the opposite. Then the final configuration has the property that for some $r$ and $s>r$ we have $p_{r+1}=p_{r}, p_{s+1}=p_{s}>0$ and $p_{r+k}=$ $p_{r+1}-k+1$ for all $k=1, \\ldots, s-r$. Consider the earliest configuration, say $C$, with this property. What was the last move before $C$ ? The only possibilities are moving a pebble either from the $r$ th or from the $s$ th column; however, in both cases the configuration preceding this last move had the same property, contradicting the assumption that $C$ is the earliest. Therefore the final configuration looks as follows: $p_{1}=a \\in \\mathbb{N}$, and for some $r, p_{i}$ equals $a-(i-1)$ if $i \\leq r$, and $a-(i-2)$ otherwise. It is easy to determine $a, r$ : since $n=p_{1}+p_{2}+\\cdots=\\frac{(a+1)(a+2)}{2}-r$, we get $\\frac{a(a+1)}{2} \\leq n<\\frac{(a+1)(a+2)}{2}$, from which we uniquely find $a$ and then $r$ as well.  The final configuration for $n=13$ : ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-690.jpg?height=162&width=155&top_left_y=922&top_left_x=1027)", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "14", "problem_type": "Combinatorics", "exam": "IMO", "problem": "14. C8 (GER) ${ }^{\\mathrm{IMO} 3}$ Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that (i) each contestant solved at most six problems, and (ii) for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy. Show that there is a problem that was solved by at least three girls and at least three boys.", "solution": "14. We say that a problem is difficult for boys if at most two boys solved it, and difficult for girls if at most two girls solved it. Let us estimate the number of pairs boy-girl both of whom solved some problem difficult for boys. Consider any girl. By the condition (ii), among the six problems she solved, at least one was solved by at least 3 boys, and hence at most 5 were difficult for boys. Since each of these problems was solved by at most 2 boys and there are 21 girls, the considered number of pairs does not exceed $5 \\cdot 2 \\cdot 21=210$. Similarly, there are at most 210 pairs boy-girl both of whom solved some problem difficult for girls. On the other hand, there are $21^{2}>2 \\cdot 210$ pairs boy-girl, and each of them solved one problem in common. Thus some problems were difficult neither for girls nor for boys, as claimed. Remark. The statement can be generalized: if $2(m-1)(n-1)+1$ boys and as many girls participated, and nobody solved more than $m$ problems, then some problem was solved by at least $n$ boys and $n$ girls.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "15", "problem_type": "Geometry", "exam": "IMO", "problem": "15. G1 (UKR) Let $A_{1}$ be the center of the square inscribed in acute triangle $A B C$ with two vertices of the square on side $B C$. Thus one of the two remaining vertices of the square is on side $A B$ and the other is on $A C$. Points $B_{1}, C_{1}$ are defined in a similar way for inscribed squares with two vertices on sides $A C$ and $A B$, respectively. Prove that lines $A A_{1}, B B_{1}, C C_{1}$ are concurrent.", "solution": "15. Let $M N P Q$ be the square inscribed in $\\triangle A B C$ with $M \\in A B, N \\in A C$, $P, Q \\in B C$, and let $A A_{1}$ meet $M N, P Q$ at $K, X$ respectively. Put $M K=$ $P X=m, N K=Q X=n$, and $M N=d$. Then  $$ \\frac{B X}{X C}=\\frac{m}{n}=\\frac{B X+m}{X C+n}=\\frac{B P}{C Q}=\\frac{d \\cot \\beta+d}{d \\cot \\gamma+d}=\\frac{\\cot \\beta+1}{\\cot \\gamma+1} . $$  Similarly, if $B B_{1}$ and $C C_{1}$ meet $A C$ and $B C$ at $Y, Z$ respectively then $\\frac{C Y}{Y A}=\\frac{\\cot \\gamma+1}{\\cot \\alpha+1}$ and $\\frac{A Z}{Z B}=\\frac{\\cot \\alpha+1}{\\cot \\beta+1}$. Therefore $\\frac{B X}{X C} \\frac{C Y}{Y A} \\frac{A Z}{Z B}=1$, so by Ceva's theorem, $A X, B Y, C Z$ have a common point.  Second solution. Let $A_{2}$ be the center of the square constructed over $B C$ outside $\\triangle A B C$. Since this square and the inscribed square corresponding to the side $B C$ are homothetic, $A, A_{1}$, and $A_{2}$ are collinear. Points $B_{2}, C_{2}$ are analogously defined. Denote the angles $B A A_{2}, A_{2} A C, C B B_{2}$, $B_{2} B A, A C C_{2}, C_{2} C B$ by $\\alpha_{1}, \\alpha_{2}, \\beta_{1}, \\beta_{2}, \\gamma_{1}, \\gamma_{2}$. By the law of sines we have  $$ \\frac{\\sin \\alpha_{1}}{\\sin \\alpha_{2}}=\\frac{\\sin \\left(\\beta+45^{\\circ}\\right)}{\\sin \\left(\\gamma+45^{\\circ}\\right)}, \\frac{\\sin \\beta_{1}}{\\sin \\beta_{2}}=\\frac{\\sin \\left(\\gamma+45^{\\circ}\\right)}{\\sin \\left(\\alpha+45^{\\circ}\\right)}, \\frac{\\sin \\gamma_{1}}{\\sin \\gamma_{2}}=\\frac{\\sin \\left(\\alpha+45^{\\circ}\\right)}{\\sin \\left(\\beta+45^{\\circ}\\right)} $$  Since the product of these ratios is 1 , by the trigonometric Ceva's theorem $A A_{2}, B B_{2}, C C_{2}$ are concurrent.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "16", "problem_type": "Geometry", "exam": "IMO", "problem": "16. G2 (KOR) ${ }^{\\mathrm{IMO} 1}$ In acute triangle $A B C$ with circumcenter $O$ and altitude $A P, \\measuredangle C \\geq \\measuredangle B+30^{\\circ}$. Prove that $\\measuredangle A+\\measuredangle C O P<90^{\\circ}$.", "solution": "16. Since $\\angle O C P=90^{\\circ}-\\angle A$, we are led to showing that $\\angle O C P>\\angle C O P$, i.e., $O P>C P$. By the triangle inequality it suffices to prove $C P<\\frac{1}{2} C O$. Let $C O=R$. The law of sines yields $C P=A C \\cos \\gamma=2 R \\sin \\beta \\cos \\gamma<$ $2 R \\sin \\beta \\cos \\left(\\beta+30^{\\circ}\\right)$. Finally, we have  $$ 2 \\sin \\beta \\cos \\left(\\beta+30^{\\circ}\\right)=\\sin \\left(2 \\beta+30^{\\circ}\\right)-\\sin 30^{\\circ} \\leq \\frac{1}{2} $$  which completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "17", "problem_type": "Geometry", "exam": "IMO", "problem": "17. G3 (GBR) Let $A B C$ be a triangle with centroid $G$. Determine, with proof, the position of the point $P$ in the plane of $A B C$ such that  $$ A P \\cdot A G+B P \\cdot B G+C P \\cdot C G $$  is a minimum, and express this minimum value in terms of the side lengths of $A B C$.", "solution": "17. Let us investigate a more general problem, in which $G$ is any point of the plane such that $A G, B G, C G$ are sides of a triangle. Let $F$ be the point in the plane such that $B C: C F: F B=A G: B G: C G$ and $F, A$ lie on different sides of $B C$. Then by Ptolemy's inequality, on $B P C F$ we have $A G \\cdot A P+B G \\cdot B P+C G \\cdot C P=A G \\cdot A P+\\frac{A G}{B C}(C F$. $B P+B F \\cdot C P) \\geq A G \\cdot A P+\\frac{A G}{B C} B C \\cdot P F$. Hence  $$ A G \\cdot A P+B G \\cdot B P+C G \\cdot C P \\geq A G \\cdot A F $$  where equality holds if and only if $P$ lies on the segment $A F$ and on the circle $B C F$. Now we return to the case of $G$ the centroid of $\\triangle A B C$. We claim that $F$ is then the point $\\widehat{G}$ in which the line $A G$ meets again the circumcircle of $\\triangle B G C$. Indeed, if $M$ is the midpoint of $A B$, by the law of sines we have $\\frac{B C}{C \\widehat{G}}=$ $\\frac{\\sin \\angle B \\widehat{G} C}{\\sin \\angle C B \\widehat{G}}=\\frac{\\sin \\angle B G M}{\\sin \\angle A G M}=\\frac{A G}{B G}$, and similarly $\\frac{B C}{B \\widehat{G}}=\\frac{A G}{C G}$. Thus (1) implies ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-691.jpg?height=342&width=507&top_left_y=1514&top_left_x=831)  $$ A G \\cdot A P+B G \\cdot B P+C G \\cdot C P \\geq A G \\cdot A \\widehat{G} $$  It is easily seen from the above considerations that equality holds if and only if $P \\equiv G$, and then the (minimum) value of $A G \\cdot A P+B G \\cdot B P+$ $C G \\cdot C P$ equals  $$ A G^{2}+B G^{2}+C G^{2}=\\frac{a^{2}+b^{2}+c^{2}}{3} $$  Second solution. Notice that $A G \\cdot A P \\geq \\overrightarrow{A G} \\cdot \\overrightarrow{A P}=\\overrightarrow{A G} \\cdot(\\overrightarrow{A G}+\\overrightarrow{P G})$. Summing this inequality with analogous inequalities for $B G \\cdot B P$ and $C G \\cdot C P$ gives us $A G \\cdot A P+B G \\cdot B P+C G \\cdot C P \\geq A G^{2}+B G^{2}+C G^{2}+$ $(\\overrightarrow{A G}+\\overrightarrow{B G}+\\overrightarrow{C G}) \\cdot \\overrightarrow{P G}=A G^{2}+B G^{2}+C G^{2}=\\frac{a^{2}+b^{2}+c^{2}}{3}$. Equality holds if and only if $P \\equiv Q$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "18", "problem_type": "Geometry", "exam": "IMO", "problem": "18. G4 (FRA) Let $M$ be a point in the interior of triangle $A B C$. Let $A^{\\prime}$ lie on $B C$ with $M A^{\\prime}$ perpendicular to $B C$. Define $B^{\\prime}$ on $C A$ and $C^{\\prime}$ on $A B$ similarly. Define  $$ p(M)=\\frac{M A^{\\prime} \\cdot M B^{\\prime} \\cdot M C^{\\prime}}{M A \\cdot M B \\cdot M C} $$  Determine, with proof, the location of $M$ such that $p(M)$ is maximal. Let $\\mu(A B C)$ denote the maximum value. For which triangles $A B C$ is the value of $\\mu(A B C)$ maximal?", "solution": "18. Let $\\alpha_{1}, \\beta_{1}, \\gamma_{1}, \\alpha_{2}, \\beta_{2}, \\gamma_{2}$ denote the angles $\\angle M A B, \\angle M B C, \\angle M C A$, $\\angle M A C, \\angle M B A, \\angle M C B$ respectively. Then $\\frac{M B^{\\prime} \\cdot M C^{\\prime}}{M A^{2}}=\\sin \\alpha_{1} \\sin \\alpha_{2}$, $\\frac{M C^{\\prime} \\cdot M A^{\\prime}}{M B^{2}}=\\sin \\beta_{1} \\sin \\beta_{2}, \\frac{M A^{\\prime} \\cdot M B^{\\prime}}{M C^{2}}=\\sin \\gamma_{1} \\sin \\gamma_{2}$; hence  $$ p(M)^{2}=\\sin \\alpha_{1} \\sin \\alpha_{2} \\sin \\beta_{1} \\sin \\beta_{2} \\sin \\gamma_{1} \\sin \\gamma_{2} $$  Since  $$ \\sin \\alpha_{1} \\sin \\alpha_{2}=\\frac{1}{2}\\left(\\cos \\left(\\alpha_{1}-\\alpha_{2}\\right)-\\cos \\left(\\alpha_{1}+\\alpha_{2}\\right) \\leq \\frac{1}{2}(1-\\cos \\alpha)=\\sin ^{2} \\frac{\\alpha}{2}\\right. $$  we conclude that  $$ p(M) \\leq \\sin \\frac{\\alpha}{2} \\sin \\frac{\\beta}{2} \\sin \\frac{\\gamma}{2} $$  Equality occurs when $\\alpha_{1}=\\alpha_{2}, \\beta_{1}=\\beta_{2}$, and $\\gamma_{1}=\\gamma_{2}$, that is, when $M$ is the incenter of $\\triangle A B C$. It is well known that $\\mu(A B C)=\\sin \\frac{\\alpha}{2} \\sin \\frac{\\beta}{2} \\sin \\frac{\\gamma}{2}$ is maximal when $\\triangle A B C$ is equilateral (it follows, for example, from Jensen's inequality applied to $\\ln \\sin x)$. Hence $\\max \\mu(A B C)=\\frac{1}{8}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "19", "problem_type": "Geometry", "exam": "IMO", "problem": "19. G5 (GRE) Let $A B C$ be an acute triangle. Let $D A C, E A B$, and $F B C$ be isosceles triangles exterior to $A B C$, with $D A=D C, E A=E B$, and $F B=F C$ such that  $$ \\angle A D C=2 \\angle B A C, \\quad \\angle B E A=2 \\angle A B C, \\quad \\angle C F B=2 \\angle A C B . $$  Let $D^{\\prime}$ be the intersection of lines $D B$ and $E F$, let $E^{\\prime}$ be the intersection of $E C$ and $D F$, and let $F^{\\prime}$ be the intersection of $F A$ and $D E$. Find, with proof, the value of the sum  $$ \\frac{D B}{D D^{\\prime}}+\\frac{E C}{E E^{\\prime}}+\\frac{F A}{F F^{\\prime}} $$", "solution": "19. It is easy to see that the hexagon $A E B F C D$ is convex and $\\angle A E B+$ $\\angle B F C+\\angle C D A=360^{\\circ}$. Using this relation we obtain that the circles $\\omega_{1}, \\omega_{2}, \\omega_{3}$ with centers at $D, E, F$ and radii $D A, E B, F C$ respectively all pass through a common point $O$. Indeed, if $\\omega_{1} \\cap \\omega_{2}=\\{O\\}$, then $\\angle A O B=180^{\\circ}-\\angle A E B / 2$ and $\\angle B O C=180^{\\circ}-\\angle B F C / 2$; hence $\\angle C O A=180^{\\circ}-\\angle C D A / 2$ as well, i.e., $O \\in \\omega_{3}$. The point $O$ is the re- ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-692.jpg?height=491&width=419&top_left_y=1318&top_left_x=870) flection of $A$ with respect to $D E$. Similarly, it is also the reflection of $B$ with respect to $E F$, and that of $C$ with respect to $F D$. Hence  $$ \\frac{D B}{D D^{\\prime}}=1+\\frac{D^{\\prime} B}{D D^{\\prime}}=1+\\frac{S_{E B F}}{S_{E D F}}=1+\\frac{S_{O E F}}{S_{D E F}} $$  Analogously $\\frac{E C}{E E^{\\prime}}=1+\\frac{S_{O D F}}{S_{D E F}}$ and $\\frac{F A}{F F^{\\prime}}=1+\\frac{S_{O D E}}{S_{D E F}}$. Adding these relations gives us  $$ \\frac{D B}{D D^{\\prime}}+\\frac{E C}{E E^{\\prime}}+\\frac{F A}{F F^{\\prime}}=3+\\frac{S_{O E F}+S_{O D F}+S_{O D E}}{S_{D E F}}=4 . $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "2", "problem_type": "Algebra", "exam": "IMO", "problem": "2. A2 (POL) Let $a_{0}, a_{1}, a_{2}, \\ldots$ be an arbitrary infinite sequence of positive numbers. Show that the inequality $1+a_{n}>a_{n-1} \\sqrt[n]{2}$ holds for infinitely many positive integers $n$.", "solution": "2. It follows from Bernoulli's inequality that for each $n \\in \\mathbb{N},\\left(1+\\frac{1}{n}\\right)^{n} \\geq 2$, or $\\sqrt[n]{2} \\leq 1+\\frac{1}{n}$. Consequently, it will be enough to show that $1+a_{n}>$ $\\left(1+\\frac{1}{n}\\right) a_{n-1}$. Assume the opposite. Then there exists $N$ such that for each $n \\geq N$,  $$ 1+a_{n} \\leq\\left(1+\\frac{1}{n}\\right) a_{n-1}, \\quad \\text { i.e., } \\quad \\frac{1}{n+1}+\\frac{a_{n}}{n+1} \\leq \\frac{a_{n-1}}{n} $$  Summing for $n=N, \\ldots, m$ yields $\\frac{a_{m}}{m+1} \\leq \\frac{a_{N-1}}{N}-\\left(\\frac{1}{N+1}+\\cdots+\\frac{1}{m+1}\\right)$. However, it is well known that the sum $\\frac{1}{N+1}+\\cdots+\\frac{1}{m+1}$ can be arbitrarily large for $m$ large enough, so that $\\frac{a_{m}^{N+1}}{m+1}$ is eventually negative. This contradiction yields the result. Second solution. Suppose that $1+a_{n} \\leq \\sqrt[n]{2} a_{n-1}$ for all $n \\geq N$. Set $b_{n}=2^{-(1+1 / 2+\\cdots+1 / n)}$ and multiply both sides of the above inequality to obtain $b_{n}+b_{n} a_{n} \\leq b_{n-1} a_{n-1}$. Thus  $$ b_{N} a_{N}>b_{N} a_{N}-b_{n} a_{n} \\geq b_{N}+b_{N+1}+\\cdots+b_{n} $$  However, it can be shown that $\\sum_{n>N} b_{N}$ diverges: in fact, since $1+\\frac{1}{2}+$ $\\cdots+\\frac{1}{n}<1+\\ln n$, we have $b_{n}>2^{-1-\\ln n}=\\frac{1}{2} n^{-\\ln 2}>\\frac{1}{2 n}$, and we already know that $\\sum_{n>N} \\frac{1}{2 n}$ diverges. Remark. As can be seen from both solutions, the value 2 in the problem can be increased to $e$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "20", "problem_type": "Geometry", "exam": "IMO", "problem": "20. G6 (IND) Let $A B C$ be a triangle and $P$ an exterior point in the plane of the triangle. Suppose $A P, B P, C P$ meet the sides $B C, C A, A B$ (or extensions thereof) in $D, E, F$, respectively. Suppose further that the areas of triangles $P B D, P C E, P A F$ are all equal. Prove that each of these areas is equal to the area of triangle $A B C$ itself.", "solution": "20. By Ceva's theorem, we can choose real numbers $x, y, z$ such that  $$ \\frac{\\overrightarrow{B D}}{\\overrightarrow{D C}}=\\frac{z}{y}, \\frac{\\overrightarrow{C E}}{\\overrightarrow{E A}}=\\frac{x}{z}, \\text { and } \\frac{\\overrightarrow{A F}}{\\overrightarrow{F B}}=\\frac{y}{x} $$  The point $P$ lies outside the triangle $A B C$ if and only if $x, y, z$ are not all of the same sign. In what follows, $S_{X}$ will denote the signed area of a figure $X$. Let us assume that the area $S_{A B C}$ of $\\triangle A B C$ is 1 . Since $S_{P B C}: S_{P C A}$ : $S_{P A B}=x: y: z$ and $S_{P B D}: S_{P D C}=z: y$, it follows that $S_{P B D}=\\frac{z}{y+z} \\frac{x}{x+y+z}$. Hence $S_{P B D}=\\frac{1}{y(y+z)} \\frac{x y z}{x+y+z}, S_{P C E}=\\frac{1}{z(z+x)} \\frac{x y z}{x+y+z}$, $S_{P A F}=\\frac{1}{x(x+y)} \\frac{x y z}{x+y+z}$. By the condition of the problem we have $\\left|S_{P B D}\\right|=$ $\\left|S_{P C E}\\right|=\\left|S_{P A F}\\right|$, or  $$ |x(x+y)|=|y(y+z)|=|z(z+x)| . $$  Obviously $x, y, z$ are nonzero, so that we can put w.l.o.g. $z=1$. At least two of the numbers $x(x+y), y(y+1), 1(1+x)$ are equal, so we can assume that $x(x+y)=y(y+1)$. We distinguish two cases: (i) $x(x+y)=y(y+1)=1+x$. Then $x=y^{2}+y-1$, from which we obtain $\\left(y^{2}+y-1\\right)\\left(y^{2}+2 y-1\\right)=y(y+1)$. Simplification gives $y^{4}+3 y^{3}-y^{2}-4 y+1=0$, or  $$ (y-1)\\left(y^{3}+4 y^{2}+3 y-1\\right)=0 $$  If $y=1$, then also $z=x=1$, so $P$ is the centroid of $\\triangle A B C$, which is not an exterior point. Hence $y^{3}+4 y^{2}+3 y-1=0$. Now the signed area of each of the triangles $P B D, P C E, P A F$ equals  $$ \\begin{aligned} S_{P A F} & =\\frac{y z}{(x+y)(x+y+z)} \\\\ & =\\frac{y}{\\left(y^{2}+2 y-1\\right)\\left(y^{2}+2 y\\right)}=\\frac{1}{y^{3}+4 y^{2}+3 y-2}=-1 . \\end{aligned} $$  It is easy to check that not both of $x, y$ are positive, implying that $P$ is indeed outside $\\triangle A B C$. This is the desired result. (ii) $x(x+y)=y(y+1)=-1-x$. In this case we are led to  $$ f(y)=y^{4}+3 y^{3}+y^{2}-2 y+1=0 . $$  We claim that this equation has no real solutions. In fact, assume that $y_{0}$ is a real root of $f(y)$. We must have $y_{0}<0$, and hence $u=-y_{0}>0$ satisfies $3 u^{3}-u^{4}=(u+1)^{2}$. On the other hand,  $$ \\begin{aligned} 3 u^{3}-u^{4} & =u^{3}(3-u)=4 u\\left(\\frac{u}{2}\\right)\\left(\\frac{u}{2}\\right)(3-u) \\\\ & \\leq 4 u\\left(\\frac{u / 2+u / 2+3-u}{3}\\right)^{3}=4 u \\\\ & \\leq(u+1)^{2} \\end{aligned} $$  where at least one of the inequalities is strict, a contradiction. Remark. The official solution was incomplete, missing the case (ii).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "21", "problem_type": "Geometry", "exam": "IMO", "problem": "21. G7 (BUL) Let $O$ be an interior point of acute triangle $A B C$. Let $A_{1}$ lie on $B C$ with $O A_{1}$ perpendicular to $B C$. Define $B_{1}$ on $C A$ and $C_{1}$ on $A B$ similarly. Prove that $O$ is the circumcenter of $A B C$ if and only if the perimeter of $A_{1} B_{1} C_{1}$ is not less than any one of the perimeters of $A B_{1} C_{1}, B C_{1} A_{1}$, and $C A_{1} B_{1}$.", "solution": "21. We denote by $p(X Y Z)$ the perimeter of a triangle $X Y Z$.  If $O$ is the circumcenter of $\\triangle A B C$, then $A_{1}, B_{1}, C_{1}$ are the midpoints of the corresponding sides of the triangle, and hence $p\\left(A_{1} B_{1} C_{1}\\right)=$ $p\\left(A B_{1} C_{1}\\right)=p\\left(A_{1} B C_{1}\\right)=p\\left(A_{1} B_{1} C\\right)$. Conversely, suppose that $p\\left(A_{1} B_{1} C_{1}\\right) \\geq p\\left(A B_{1} C_{1}\\right), p\\left(A_{1} B C_{1}\\right), p\\left(A_{1} B_{1} C\\right)$. Let $\\alpha_{1}, \\alpha_{2}, \\beta_{1}, \\beta_{2}, \\gamma_{1}, \\gamma_{2}$ denote $\\angle B_{1} A_{1} C, \\angle C_{1} A_{1} B, \\angle C_{1} B_{1} A, \\angle A_{1} B_{1} C$, $\\angle A_{1} C_{1} B, \\angle B_{1} C_{1} A$. Suppose that $\\gamma_{1}, \\beta_{2} \\geq \\alpha$. If $A_{2}$ is the fourth vertex of the parallelogram $B_{1} A C_{1} A_{2}$, then these conditions imply that $A_{1}$ is in the interior or on the border of $\\triangle B_{1} C_{1} A_{2}$, and therefore $p\\left(A_{1} B_{1} C_{1}\\right) \\leq p\\left(A_{2} B_{1} C_{1}\\right)=$ $p\\left(A B_{1} C_{1}\\right)$. Moreover, if one of the inequalities $\\gamma_{1} \\geq \\alpha, \\beta_{2} \\geq \\alpha$ is strict, ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-694.jpg?height=346&width=543&top_left_y=902&top_left_x=804) then $p\\left(A_{1} B_{1} C_{1}\\right)$ is strictly less than $p\\left(A B_{1} C_{1}\\right)$, contrary to the assumption. Hence  $$ \\begin{aligned} & \\beta_{2} \\geq \\alpha \\Longrightarrow \\gamma_{1} \\leq \\alpha \\\\ & \\gamma_{2} \\geq \\beta \\Longrightarrow \\alpha_{1} \\leq \\beta \\\\ & \\alpha_{2} \\geq \\gamma \\Longrightarrow \\beta_{1} \\leq \\gamma \\end{aligned} $$  the last two inequalities being obtained analogously to the first one. Because of the symmetry, there is no loss of generality in assuming that $\\gamma_{1} \\leq \\alpha$. Then since $\\gamma_{1}+\\alpha_{2}=180^{\\circ}-\\beta=\\alpha+\\gamma$, it follows that $\\alpha_{2} \\geq \\gamma$. From (1) we deduce $\\beta_{1} \\leq \\gamma$, which further implies $\\gamma_{2} \\geq \\beta$. Similarly, this leads to $\\alpha_{1} \\leq \\beta$ and $\\beta_{2} \\geq \\alpha$. To sum up,  $$ \\gamma_{1} \\leq \\alpha \\leq \\beta_{2}, \\quad \\alpha_{1} \\leq \\beta \\leq \\gamma_{2}, \\quad \\beta_{1} \\leq \\gamma \\leq \\alpha_{2} $$  Since $O A_{1} B C_{1}$ and $O B_{1} C A_{1}$ are cyclic, we have $\\angle A_{1} O B=\\gamma_{1}$ and $\\angle A_{1} O C=\\beta_{2}$. Hence $B O: C O=\\cos \\beta_{2}: \\cos \\gamma_{1}$, hence $B O \\leq C O$. Analogously, $C O \\leq A O$ and $A O \\leq B O$. Therefore $A O=B O=C O$, i.e., $O$ is the circumcenter of $A B C$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "22", "problem_type": "Geometry", "exam": "IMO", "problem": "22. G8 (ISR) ${ }^{\\mathrm{IMO}}$ Let $A B C$ be a triangle with $\\measuredangle B A C=60^{\\circ}$. Let $A P$ bisect $\\angle B A C$ and let $B Q$ bisect $\\angle A B C$, with $P$ on $B C$ and $Q$ on $A C$. If $A B+$ $B P=A Q+Q B$, what are the angles of the triangle?", "solution": "22. Let $S$ and $T$ respectively be the points on the extensions of $A B$ and $A Q$ over $B$ and $Q$ such that $B S=B P$ and $Q T=Q B$. It is given that $A S=$ $A B+B P=A Q+Q B=A T$. Since $\\angle P A S=\\angle P A T$, the triangles $A P S$ and $A P T$ are congruent, from which we deduce that $\\angle A T P=\\angle A S P=$ $\\beta / 2=\\angle Q B P$. Hence $\\angle Q T P=\\angle Q B P$. If $P$ does not lie on $B T$, then the last equality implies that $\\triangle Q B P$ and $\\triangle Q T P$ are congruent, so $P$ lies on the internal bisector of $\\angle B Q T$. But $P$ also lies on the internal bisector of $\\angle Q A B$; consequently, $P$ is an excenter of $\\triangle Q A B$, thus lying on the internal bisector of $\\angle Q B S$ as well. It follows that $\\angle P B Q=\\beta / 2=\\angle P B S=180^{\\circ}-\\beta$, so $\\beta=120^{\\circ}$, which is impossible. Therefore $P \\in B T$, which means that $T \\equiv C$. Now from $Q C=Q B$ we conclude that $120^{\\circ}-\\beta=\\gamma=\\beta / 2$, i.e., $\\beta=80^{\\circ}$ and $\\gamma=40^{\\circ}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "23", "problem_type": "Number Theory", "exam": "IMO", "problem": "23. N1 (AUS) Prove that there is no positive integer $n$ such that for $k=$ $1,2, \\ldots, 9$, the leftmost digit (in decimal notation) of $(n+k)$ ! equals $k$.", "solution": "23. For each positive integer $x$, define $\\alpha(x)=x / 10^{r}$ if $r$ is the positive integer satisfying $10^{r} \\leq x<10^{r+1}$. Observe that if $\\alpha(x) \\alpha(y)<10$ for some $x, y \\in \\mathbb{N}$, then $\\alpha(x y)=\\alpha(x) \\alpha(y)$. If, as usual, $[t]$ means the integer part of $t$, then $[\\alpha(x)]$ is actually the leftmost digit of $x$. Now suppose that $n$ is a positive integer such that $k \\leq \\alpha((n+k)!)<k+1$ for $k=1,2, \\ldots, 9$. We have  $$ 1<\\alpha(n+k)=\\frac{\\alpha((n+k)!)}{\\alpha((n+k-1)!)}<\\frac{k+1}{k-1} \\leq 3 \\quad \\text { for } 2 \\leq k \\leq 9 $$  from which we obtain $\\alpha(n+k+1)>\\alpha(n+k)$ (the opposite can hold only if $\\alpha(n+k) \\geq 9)$. Therefore  $$ 1<\\alpha(n+2)<\\cdots<\\alpha(n+9) \\leq \\frac{5}{4} . $$  On the other hand, this implies that $\\alpha((n+4)!)=\\alpha((n+1)!) \\alpha(n+2) \\alpha(n+$ 3) $\\alpha(n+4)<(5 / 4)^{3} \\alpha((n+1)$ ! $)<4$, contradicting the assumption that the leftmost digit of $(n+4)$ ! is 4 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "24", "problem_type": "Number Theory", "exam": "IMO", "problem": "24. N2 (COL) Consider the system  $$ \\begin{aligned} x+y & =z+u \\\\ 2 x y & =z u . \\end{aligned} $$  Find the greatest value of the real constant $m$ such that $m \\leq x / y$ for every positive integer solution $x, y, z, u$ of the system with $x \\geq y$.", "solution": "24. We shall find the general solution to the system. Squaring both sides of the first equation and subtracting twice the second equation we obtain $(x-y)^{2}=z^{2}+u^{2}$. Thus $(z, u, x-y)$ is a Pythagorean triple. Then it is well known that there are positive integers $t, a, b$ such that $z=t\\left(a^{2}-b^{2}\\right)$, $u=2 t a b$ (or vice versa), and $x-y=t\\left(a^{2}+b^{2}\\right)$. Using that $x+y=z+u$ we come to the general solution:  $$ x=t\\left(a^{2}+a b\\right), \\quad y=t\\left(a b-b^{2}\\right) ; \\quad z=t\\left(a^{2}-b^{2}\\right), \\quad u=2 t a b . $$  Putting $a / b=k$ we obtain  $$ \\frac{x}{y}=\\frac{k^{2}+k}{k-1}=3+(k-1)+\\frac{2}{k-1} \\geq 3+2 \\sqrt{2} $$  with equality for $k-1=\\sqrt{2}$. On the other hand, $k$ can be arbitrarily close to $1+\\sqrt{2}$, and so $x / y$ can be arbitrarily close to $3+2 \\sqrt{2}$. Hence $m=3+2 \\sqrt{2}$. Remark. There are several other techniques for solving the given system. The exact lower bound of $m$ itself can be obtained as follows: by the $\\operatorname{system}\\left(\\frac{x}{y}\\right)^{2}-6 \\frac{x}{y}+1=\\left(\\frac{z-u}{y}\\right)^{2} \\geq 0$, so $x / y \\geq 3+2 \\sqrt{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "25", "problem_type": "Number Theory", "exam": "IMO", "problem": "25. N3 (GBR) Let $a_{1}=11^{11}, a_{2}=12^{12}, a_{3}=13^{13}$, and  $$ a_{n}=\\left|a_{n-1}-a_{n-2}\\right|+\\left|a_{n-2}-a_{n-3}\\right|, \\quad n \\geq 4 $$  Determine $a_{14^{14}}$.", "solution": "25. Define $b_{n}=\\left|a_{n+1}-a_{n}\\right|$ for $n \\geq 1$. From the equalities $a_{n+1}=b_{n-1}+b_{n-2}$, from $a_{n}=b_{n-2}+b_{n-3}$ we obtain $b_{n}=\\left|b_{n-1}-b_{n-3}\\right|$. From this relation we deduce that $b_{m} \\leq \\max \\left(b_{n}, b_{n+1}, b_{n+2}\\right)$ for all $m \\geq n$, and consequently $b_{n}$ is bounded. Lemma. If $\\max \\left(b_{n}, b_{n+1}, b_{n+2}\\right)=M \\geq 2$, then $\\max \\left(b_{n+6}, b_{n+7}, b_{n+8}\\right) \\leq$ $M-1$. Proof. Assume the opposite. Suppose that $b_{j}=M, j \\in\\{n, n+1, n+2\\}$, and let $b_{j+1}=x$ and $b_{j+2}=y$. Thus $b_{j+3}=M-y$. If $x, y, M-y$ are all less than $M$, then the contradiction is immediate. The remaining cases are these: (i) $x=M$. Then the sequence has the form $M, M, y, M-y, y, \\ldots$, and since $\\max (y, M-y, y)=M$, we must have $y=0$ or $y=M$. (ii) $y=M$. Then the sequence has the form $M, x, M, 0, x, M-x, \\ldots$, and since $\\max (0, x, M-x)=M$, we must have $x=0$ or $x=M$. (iii) $y=0$. Then the sequence is $M, x, 0, M, M-x, M-x, x, \\ldots$, and since $\\max (M-x, x, x)=M$, we have $x=0$ or $x=M$. In every case $M$ divides both $x$ and $y$. From the recurrence formula $M$ also divides $b_{i}$ for every $i<j$. However, $b_{2}=12^{12}-11^{11}$ and $b_{4}=11^{11}$ are relatively prime, a contradiction. From $\\max \\left(b_{1}, b_{2}, b_{3}\\right) \\leq 13^{13}$ and the lemma we deduce inductively that $b_{n} \\leq 1$ for all $n \\geq 6 \\cdot 13^{13}-5$. Hence $a_{n}=b_{n-2}+b_{n-3}$ takes only the values $0,1,2$ for $n \\geq 6 \\cdot 13^{13}-2$. In particular, $a_{14^{14}}$ is 0,1 , or 2 . On the other hand, the sequence $a_{n}$ modulo 2 is as follows: $1,0,1,0,0,1,1 ; 1,0,1,0, \\ldots$; and therefore it is periodic with period 7 . Finally, $14^{14} \\equiv 0$ modulo 7 , from which we obtain $a_{14^{14}} \\equiv a_{7} \\equiv 1(\\bmod 2)$. Therefore $a_{14^{14}}=1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "26", "problem_type": "Number Theory", "exam": "IMO", "problem": "26. N4 (VIE) Let $p \\geq 5$ be a prime number. Prove that there exists an integer $a$ with $1 \\leq a \\leq p-2$ such that neither $a^{p-1}-1$ nor $(a+1)^{p-1}-1$ is divisible by $p^{2}$.", "solution": "26. Let $C$ be the set of those $a \\in\\{1,2, \\ldots, p-1\\}$ for which $a^{p-1} \\equiv 1\\left(\\bmod p^{2}\\right)$. At first, we observe that $a, p-a$ do not both belong to $C$, regardless of the value of $a$. Indeed, by the binomial formula,  $$ (p-a)^{p-1}-a^{p-1} \\equiv-(p-1) p a^{p-2} \\not \\equiv 0 \\quad\\left(\\bmod p^{2}\\right) $$  As a consequence we deduce that $|C| \\leq \\frac{p-1}{2}$. Further, we observe that $p-k \\in C \\Leftrightarrow k \\equiv k(p-k)^{p-1}\\left(\\bmod p^{2}\\right)$, i.e.,  $$ p-k \\in C \\Leftrightarrow k \\equiv k\\left(k^{p-1}-(p-1) p k^{p-2}\\right) \\equiv k^{p}+p\\left(\\bmod p^{2}\\right) $$  Now assume the contrary to the claim, that for every $a=1, \\ldots, p-2$ one of $a, a+1$ is in $C$. In this case it is not possible that $a, a+1$ are both in $C$, for then $p-a, p-a-1 \\notin C$. Thus, since $1 \\in C$, we inductively obtain that $2,4, \\ldots, p-1 \\notin C$ and $1,3,5, \\ldots, p-2 \\in C$. In particular, $p-2, p-4 \\in C$, which is by $(1)$ equivalent to $2 \\equiv 2^{p}+p$ and $4 \\equiv 4^{p}+p\\left(\\bmod p^{2}\\right)$. However, squaring the former equality and subtracting the latter, we obtain $2^{p+1} p \\equiv p\\left(\\bmod p^{2}\\right)$, or $4 \\equiv 1(\\bmod p)$, which is a contradiction unless $p=3$. This finishes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "27", "problem_type": "Number Theory", "exam": "IMO", "problem": "27. N5 (BUL) ${ }^{\\mathrm{IMO} 6}$ Let $a>b>c>d$ be positive integers and suppose  $$ a c+b d=(b+d+a-c)(b+d-a+c) . $$  Prove that $a b+c d$ is not prime.", "solution": "27. The given equality is equivalent to $a^{2}-a c+c^{2}=b^{2}+b d+d^{2}$. Hence $(a b+c d)(a d+b c)=a c\\left(b^{2}+b d+d^{2}\\right)+b d\\left(a^{2}-a c+c^{2}\\right)$, or equivalently,  $$ (a b+c d)(a d+b c)=(a c+b d)\\left(a^{2}-a c+c^{2}\\right) $$  Now suppose that $a b+c d$ is prime. It follows from $a>b>c>d$ that  $$ a b+c d>a c+b d>a d+b c $$  hence $a c+b d$ is relatively prime with $a b+c d$. But then (1) implies that $a c+b d$ divides $a d+b c$, which is impossible by (2). Remark. Alternatively, (1) could be obtained by applying the law of cosines and Ptolemy's theorem on a quadrilateral $X Y Z T$ with $X Y=a$, $Y Z=c, Z T=b, T X=d$ and $\\angle Y=60^{\\circ}, \\angle T=120^{\\circ}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "28", "problem_type": "Number Theory", "exam": "IMO", "problem": "28. N6 (RUS) Is it possible to find 100 positive integers not exceeding 25,000 such that all pairwise sums of them are different?", "solution": "28. Yes. The desired result is an immediate consequence of the following fact applied on $p=101$. Lemma. For any odd prime number $p$, there exist $p$ nonnegative integers less than $2 p^{2}$ with all pairwise sums mutually distinct. Proof. We claim that the numbers $a_{n}=2 n p+\\left(n^{2}\\right)$ have the desired property, where $(x)$ denotes the remainder of $x$ upon division by $p$. Suppose that $a_{k}+a_{l}=a_{m}+a_{n}$. By the construction of $a_{i}$, we have $2 p(k+l) \\leq a_{k}+a_{l}<2 p(k+l+1)$. Hence we must have $k+l=m+n$, and therefore also $\\left(k^{2}\\right)+\\left(l^{2}\\right)=\\left(m^{2}\\right)+\\left(n^{2}\\right)$. Thus  $$ k+l \\equiv m+n \\quad \\text { and } \\quad k^{2}+l^{2} \\equiv m^{2}+n^{2} \\quad(\\bmod p) . $$  But then it holds that $(k-l)^{2}=2\\left(k^{2}+l^{2}\\right)-(k+l)^{2} \\equiv(m-n)^{2}(\\bmod$ $p)$, so $k-l \\equiv \\pm(m-n)$, which leads to $(k, l)=(m, n)$. This proves the lemma.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "3", "problem_type": "Algebra", "exam": "IMO", "problem": "3. A3 (ROM) Let $x_{1}, x_{2}, \\ldots, x_{n}$ be arbitrary real numbers. Prove the inequality  $$ \\frac{x_{1}}{1+x_{1}^{2}}+\\frac{x_{2}}{1+x_{1}^{2}+x_{2}^{2}}+\\cdots+\\frac{x_{n}}{1+x_{1}^{2}+\\cdots+x_{n}^{2}}<\\sqrt{n} $$", "solution": "3. By the arithmetic-quadratic mean inequality, it suffices to prove that  $$ \\frac{x_{1}^{2}}{\\left(1+x_{1}^{2}\\right)^{2}}+\\frac{x_{2}^{2}}{\\left(1+x_{1}^{2}+x_{2}^{2}\\right)^{2}}+\\cdots+\\frac{x_{n}^{2}}{\\left(1+x_{1}^{2}+\\cdots+x_{n}^{2}\\right)^{2}}<1 . $$  Observe that for $k \\geq 2$ the following holds:  $$ \\begin{aligned} \\frac{x_{k}^{2}}{\\left(1+x_{1}^{2}+\\cdots+x_{k}^{2}\\right)^{2}} & \\leq \\frac{x_{k}^{2}}{\\left(1+\\cdots+x_{k-1}^{2}\\right)\\left(1+\\cdots+x_{k}^{2}\\right)} \\\\ & =\\frac{1}{1+x_{1}^{2}+\\cdots+x_{k-1}^{2}}-\\frac{1}{1+x_{1}^{2}+\\cdots+x_{k}^{2}} \\end{aligned} $$  For $k=1$ we have $\\frac{x_{1}^{2}}{\\left(1+x_{1}\\right)^{2}} \\leq 1-\\frac{1}{1+x_{1}^{2}}$. Summing these inequalities, we obtain  $$ \\frac{x_{1}^{2}}{\\left(1+x_{1}^{2}\\right)^{2}}+\\cdots+\\frac{x_{n}^{2}}{\\left(1+x_{1}^{2}+\\cdots+x_{n}^{2}\\right)^{2}} \\leq 1-\\frac{1}{1+x_{1}^{2}+\\cdots+x_{n}^{2}}<1 $$  Second solution. Let $a_{n}(k)=\\sup \\left(\\frac{x_{1}}{k^{2}+x_{1}^{2}}+\\cdots+\\frac{x_{n}}{k^{2}+x_{1}^{2}+\\cdots+x_{n}^{2}}\\right)$ and $a_{n}=$ $a_{n}(1)$. We must show that $a_{n}<\\sqrt{n}$. Replacing $x_{i}$ by $k x_{i}$ shows that $a_{n}(k)=a_{n} / k$. Hence  $$ a_{n}=\\sup _{x_{1}}\\left(\\frac{x_{1}}{1+x_{1}^{2}}+\\frac{a_{n-1}}{\\sqrt{1+x_{1}^{2}}}\\right)=\\sup _{\\theta}\\left(\\sin \\theta \\cos \\theta+a_{n-1} \\cos \\theta\\right), $$  where $\\tan \\theta=x_{1}$. The above supremum can be computed explicitly:  $$ a_{n}=\\frac{1}{8 \\sqrt{2}}\\left(3 a_{n-1}+\\sqrt{a_{n-1}^{2}+8}\\right) \\sqrt{4-a_{n-1}^{2}+a_{n-1} \\sqrt{a_{n-1}^{2}+8}} . $$  However, the required inequality is weaker and can be proved more easily: if $a_{n-1}<\\sqrt{n-1}$, then by (1) $a_{n}<\\sin \\theta+\\sqrt{n-1} \\cos \\theta=\\sqrt{n} \\sin (\\theta+\\alpha) \\leq$ $\\sqrt{n}$, for $\\alpha \\in(0, \\pi / 2)$ with $\\tan \\alpha=\\sqrt{n}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "4", "problem_type": "Algebra", "exam": "IMO", "problem": "4. A4 (LIT) Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfying  $$ f(x y)(f(x)-f(y))=(x-y) f(x) f(y) $$  for all $x, y$.", "solution": "4. Let $(*)$ denote the given functional equation. Substituting $y=1$ we get $f(x)^{2}=x f(x) f(1)$. If $f(1)=0$, then $f(x)=0$ for all $x$, which is the trivial solution. Suppose $f(1)=C \\neq 0$. Let $G=\\{y \\in \\mathbb{R} \\mid f(y) \\neq 0\\}$. Then  $$ f(x)=\\left\\{\\begin{array}{cl} C x & \\text { if } x \\in G \\\\ 0 & \\text { otherwise } \\end{array}\\right. $$  We must determine the structure of $G$ so that the function defined by (1) satisfies (*). (1) Clearly $1 \\in G$, because $f(1) \\neq 0$. (2) If $x \\in G, y \\notin G$, then by $(*)$ it holds $f(x y) f(x)=0$, so $x y \\notin G$. (3) If $x, y \\in G$, then $x / y \\in G$ (otherwise by $2^{\\circ}, y(x / y)=x \\notin G$ ). (4) If $x, y \\in G$, then by $2^{\\circ}$ we have $x^{-1} \\in G$, so $x y=y / x^{-1} \\in G$.  Hence $G$ is a set that contains 1 , does not contain 0 , and is closed under multiplication and division. Conversely, it is easy to verify that every such $G$ in (1) gives a function satisfying (*).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "5", "problem_type": "Algebra", "exam": "IMO", "problem": "5. A5 (BUL) Find all positive integers $a_{1}, a_{2}, \\ldots, a_{n}$ such that  $$ \\frac{99}{100}=\\frac{a_{0}}{a_{1}}+\\frac{a_{1}}{a_{2}}+\\cdots+\\frac{a_{n-1}}{a_{n}} $$  where $a_{0}=1$ and $\\left(a_{k+1}-1\\right) a_{k-1} \\geq a_{k}^{2}\\left(a_{k}-1\\right)$ for $k=1,2, \\ldots, n-1$.", "solution": "5. Let $a_{1}, a_{2}, \\ldots, a_{n}$ satisfy the conditions of the problem. Then $a_{k}>a_{k-1}$, and hence $a_{k} \\geq 2$ for $k=1, \\ldots, n$. The inequality $\\left(a_{k+1}-1\\right) a_{k-1} \\geq$ $a_{k}^{2}\\left(a_{k}-1\\right)$ can be rewritten as  $$ \\frac{a_{k-1}}{a_{k}}+\\frac{a_{k}}{a_{k+1}-1} \\leq \\frac{a_{k-1}}{a_{k}-1} . $$  Summing these inequalities for $k=i+1, \\ldots, n-1$ and using the obvious inequality $\\frac{a_{n-1}}{a_{n}}<\\frac{a_{n-1}}{a_{n}-1}$, we obtain $\\frac{a_{i}}{a_{i+1}}+\\cdots+\\frac{a_{n-1}}{a_{n}}<\\frac{a_{i}}{a_{i+1}-1}$. Therefore  $$ \\frac{a_{i}}{a_{i+1}} \\leq \\frac{99}{100}-\\frac{a_{0}}{a_{1}}-\\cdots-\\frac{a_{i-1}}{a_{i}}<\\frac{a_{i}}{a_{i+1}-1} \\quad \\text { for } i=1,2, \\ldots, n-1 $$  Consequently, given $a_{0}, a_{1}, \\ldots, a_{i}$, there is at most one possibility for $a_{i+1}$. In our case, (1) yields $a_{1}=2, a_{2}=5, a_{3}=56, a_{4}=280^{2}=78400$. These values satisfy the conditions of the problem, so that this is a unique solution.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "6", "problem_type": "Algebra", "exam": "IMO", "problem": "6. A6 (KOR) ${ }^{\\mathrm{IMO} 2}$ Prove that for all positive real numbers $a, b, c$,  $$ \\frac{a}{\\sqrt{a^{2}+8 b c}}+\\frac{a}{\\sqrt{b^{2}+8 c a}}+\\frac{c}{\\sqrt{c^{2}+8 a b}} \\geq 1 $$", "solution": "6. We shall determine a constant $k>0$ such that  $$ \\frac{a}{\\sqrt{a^{2}+8 b c}} \\geq \\frac{a^{k}}{a^{k}+b^{k}+c^{k}} \\quad \\text { for all } a, b, c>0 $$  This inequality is equivalent to $\\left(a^{k}+b^{k}+c^{k}\\right)^{2} \\geq a^{2 k-2}\\left(a^{2}+8 b c\\right)$, which further reduces to  $$ \\left(a^{k}+b^{k}+c^{k}\\right)^{2}-a^{2 k} \\geq 8 a^{2 k-2} b c $$  On the other hand, the AM-GM inequality yields  $$ \\left(a^{k}+b^{k}+c^{k}\\right)^{2}-a^{2 k}=\\left(b^{k}+c^{k}\\right)\\left(2 a^{k}+b^{k}+c^{k}\\right) \\geq 8 a^{k / 2} b^{3 k / 4} c^{3 k / 4} $$  and therefore $k=4 / 3$ is a good choice. Now we have  $$ \\begin{aligned} & \\frac{a}{\\sqrt{a^{2}+8 b c}}+\\frac{b}{\\sqrt{b^{2}+8 c a}}+\\frac{c}{\\sqrt{c^{2}+8 a b}} \\\\ & \\geq \\frac{a^{4 / 3}}{a^{4 / 3}+b^{4 / 3}+c^{4 / 3}}+\\frac{b^{4 / 3}}{a^{4 / 3}+b^{4 / 3}+c^{4 / 3}}+\\frac{c^{4 / 3}}{a^{4 / 3}+b^{4 / 3}+c^{4 / 3}}=1 \\end{aligned} $$  Second solution. The numbers $x=\\frac{a}{\\sqrt{a^{2}+8 b c}}, y=\\frac{b}{\\sqrt{b^{2}+8 c a}}$ and $z=\\frac{c}{\\sqrt{c^{2}+8 a b}}$ satisfy  $$ f(x, y, z)=\\left(\\frac{1}{x^{2}}-1\\right)\\left(\\frac{1}{y^{2}}-1\\right)\\left(\\frac{1}{z^{2}}-1\\right)=8^{3} . $$  Our task is to prove $x+y+z \\geq 1$. Since $f$ is decreasing on each of the variables $x, y, z$, this is the same as proving that $x, y, z>0, x+y+z=1$ implies $f(x, y, z) \\geq 8^{3}$. However, since $\\frac{1}{x^{2}}-1=\\frac{(x+y+z)^{2}-x^{2}}{x^{2}}=\\frac{(2 x+y+z)(y+z)}{x^{2}}$, the inequality $f(x, y, z) \\geq 8^{3}$ becomes  $$ \\frac{(2 x+y+z)(x+2 y+z)(x+y+2 z)(y+z)(z+x)(x+y)}{x^{2} y^{2} z^{2}} \\geq 8^{3} $$  which follows immediately by the AM-GM inequality. Third solution. We shall prove a more general fact: the inequality $\\frac{a}{\\sqrt{a^{2}+k b c}}+\\frac{b}{\\sqrt{b^{2}+k c a}}+\\frac{c}{\\sqrt{c^{2}+k a b}} \\geq \\frac{3}{\\sqrt{1+k}}$ is true for all $a, b, c>0$ if and only if $k \\geq 8$. Firstly suppose that $k \\geq 8$. Setting $x=b c / a^{2}, y=c a / b^{2}, z=a b / c^{2}$, we reduce the desired inequality to  $$ F(x, y, z)=f(x)+f(y)+f(z) \\geq \\frac{3}{\\sqrt{1+k}}, \\quad \\text { where } f(t)=\\frac{1}{\\sqrt{1+k t}} $$  for $x, y, z>0$ such that $x y z=1$. We shall prove (2) using the method of Lagrange multipliers. The boundary of the set $D=\\left\\{(x, y, z) \\in \\mathbb{R}_{+}^{3} \\mid x y z=1\\right\\}$ consists of points $(x, y, z)$ with one of $x, y, z$ being 0 and another one being $+\\infty$. If w.l.o.g. $x=0$, then $F(x, y, z) \\geq f(x)=1 \\geq 3 / \\sqrt{1+k}$. Suppose now that $(x, y, z)$ is a point of local minimum of $F$ on $D$. There exists $\\lambda \\in \\mathbb{R}$ such that $(x, y, z)$ is stationary point of the function $F(x, y, z)+\\lambda x y z$. Then $(x, y, z, \\lambda)$ is a solution to the system $f^{\\prime}(x)+\\lambda y z=$ $f^{\\prime}(y)+\\lambda x z=f^{\\prime}(z)+\\lambda x y=0, x y z=1$. Eliminating $\\lambda$ gives us  $$ x f^{\\prime}(x)=y f^{\\prime}(y)=z f^{\\prime}(z), \\quad x y z=1 $$  The function $t f^{\\prime}(t)=\\frac{-k t}{2(1+k t)^{3 / 2}}$ decreases on the interval $(0,2 / k]$ and increases on $[2 / k,+\\infty)$ because $\\left(t f^{\\prime}(t)\\right)^{\\prime}=\\frac{k(k t-2)}{4(1+k t)^{5 / 2}}$. It follows that two of the numbers $x, y, z$ are equal. If $x=y=z$, then $(1,1,1)$ is the only solution to (3). Suppose that $x=y \\neq z$. Since $\\left(y f^{\\prime}(y)\\right)^{2}-\\left(z f^{\\prime}(z)\\right)^{2}=$ $\\frac{k^{2}(z-y)\\left(k^{3} y^{2} z^{2}-3 k y z-y-z\\right)}{4(1+k y)^{3}(1+k z)^{3}},(3)$ gives us $y^{2} z=1$ and $k^{3} y^{2} z^{2}-3 k y z-y-z=$ 0 . Eliminating $z$ we obtain an equation in $y, k^{3} / y^{2}-3 k / y-y-1 / y^{2}=0$, whose only real solution is $y=k-1$. Thus $\\left(k-1, k-1,1 /(k-1)^{2}\\right)$ and the cyclic permutations are the only solutions to (3) with $x, y, z$ being not all equal. Since $F\\left(k-1, k-1,1 /(k-1)^{2}\\right)=(k+1) / \\sqrt{k^{2}-k+1}>$ $F(1,1,1)=1$, the inequality (2) follows. For $0<k<8$ we have that $\\frac{a}{\\sqrt{a^{2}+k b c}}+\\frac{b}{\\sqrt{b^{2}+k c a}}+\\frac{c}{\\sqrt{c^{2}+k a b}}>\\frac{a}{\\sqrt{a^{2}+8 b c}}+$ $\\frac{b}{\\sqrt{b^{2}+8 c a}}+\\frac{c}{\\sqrt{c^{2}+8 a b}} \\geq 1$. If we fix $c$ and let $a, b$ tend to 0 , the first two summands will tend to 0 while the third will tend to 1 . Hence the inequality cannot be improved.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "7", "problem_type": null, "exam": "IMO", "problem": "7. $\\mathbf{C 1}$ (COL) Let $A=\\left(a_{1}, a_{2}, \\ldots, a_{2001}\\right)$ be a sequence of positive integers. Let $m$ be the number of 3 -element subsequences $\\left(a_{i}, a_{j}, a_{k}\\right)$ with $1 \\leq i<$ $j<k \\leq 2001$ such that $a_{j}=a_{i}+1$ and $a_{k}=a_{j}+1$. Considering all such sequences $A$, find the greatest value of $m$.", "solution": "7. It is evident that arranging of $A$ in increasing order does not diminish $m$. Thus we can assume that $A$ is nondecreasing. Assume w.l.o.g. that $a_{1}=1$, and let $b_{i}$ be the number of elements of $A$ that are equal to $i$ $\\left(1 \\leq i \\leq n=a_{2001}\\right)$. Then we have $b_{1}+b_{2}+\\cdots+b_{n}=2001$ and  $$ m=b_{1} b_{2} b_{3}+b_{2} b_{3} b_{4}+\\cdots+b_{n-2} b_{n-1} b_{n} $$  Now if $b_{i}, b_{j}(i<j)$ are two largest $b$ 's, we deduce from (1) and the AMGM inequality that $m \\leq b_{i} b_{j}\\left(b_{1}+\\cdots+b_{i-1}+b_{i+1}+\\cdots+b_{j-1}+b_{j+1}+b_{n}\\right) \\leq$ $\\left(\\frac{2001}{3}\\right)^{3}=667^{3} \\quad\\left(b_{1} b_{2} b_{3} \\leq b_{1} b_{i} b_{j}\\right.$, etc.). The value $667^{3}$ is attained for $b_{1}=b_{2}=b_{3}=667$ (i.e., $a_{1}=\\cdots=a_{667}=1, a_{668}=\\cdots=a_{1334}=2$, $a_{1335}=\\cdots=a_{2001}=3$ ). Hence the maximum of $m$ is $667^{3}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "8", "problem_type": "Combinatorics", "exam": "IMO", "problem": "8. C2 (CAN) ${ }^{\\mathrm{IMO} 4}$ Let $n$ be an odd integer greater than 1 and let $c_{1}, c_{2}, \\ldots$, $c_{n}$ be integers. For each permutation $a=\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ of $\\{1,2, \\ldots, n\\}$, define $S(a)=\\sum_{i=1}^{n} c_{i} a_{i}$. Prove that there exist permutations $a \\neq b$ of $\\{1,2, \\ldots, n\\}$ such that $n$ ! is a divisor of $S(a)-S(b)$.", "solution": "8. Suppose to the contrary that all the $S(a)$ 's are different modulo $n$ !. Then the sum of $S(a)$ 's over all permutations $a$ satisfies $\\sum_{a} S(a) \\equiv 0+1+\\cdots+$ $(n!-1)=\\frac{(n!-1) n!}{2} \\equiv \\frac{n!}{2}(\\bmod n!)$. On the other hand, the coefficient of $c_{i}$ in $\\sum_{a} S(a)$ is equal to $(n-1)!(1+2+\\cdots+n)=\\frac{n+1}{2} n!$ for all $i$, from which we obtain  $$ \\sum_{a} S(a) \\equiv \\frac{n+1}{2}\\left(c_{1}+\\cdots+c_{n}\\right) n!\\equiv 0(\\bmod n!) $$  for odd $n$. This is a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2001", "tier": "T0", "problem_label": "9", "problem_type": "Combinatorics", "exam": "IMO", "problem": "9. C3 (RUS) Define a $k$-clique to be a set of $k$ people such that every pair of them are acquainted with each other. At a certain party, every pair of 3 -cliques has at least one person in common, and there are no 5 -cliques. Prove that there are two or fewer people at the party whose departure leaves no 3-clique remaining.", "solution": "9. Consider one such party. The result is trivially true if there is only one 3 -clique, so suppose there exist at least two 3 -cliques $C_{1}$ and $C_{2}$. We distinguish two cases: (i) $C_{1}=\\{a, b, c\\}$ and $C_{2}=\\{a, d, e\\}$ for some distinct people $a, b, c, d, e$. If the departure of $a$ destroys all 3-cliques, then we are done. Otherwise, there is a third 3 -clique $C_{3}$, which has a person in common with each of $C_{1}, C_{2}$ and does not include $a$ : say, $C_{3}=\\{b, d, f\\}$ for some $f$. We thus obtain another 3 -clique $C_{4}=\\{a, b, d\\}$, which has two persons in common with $C_{3}$, and the case (ii) is applied. (ii) $C_{1}=\\{a, b, c\\}$ and $C_{2}=\\{a, b, d\\}$ for distinct people $a, b, c, d$. If the departure of $a, b$ leaves no 3-clique, then we are done. Otherwise, for some $e$ there is a clique $\\{c, d, e\\}$. We claim that then the departure of $c, d$ breaks all 3 -cliques. Suppose the opposite, that a 3 -clique $C$ remains. Since $C$ shares a person with each of the 3 -cliques $\\{c, d, a\\},\\{c, d, b\\},\\{c, d, e\\}$, it must be $C=\\{a, b, e\\}$. However, then $\\{a, b, c, d, e\\}$ is a 5 -clique, which is assumed to be impossible.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "1", "problem_type": "Number Theory", "exam": "IMO", "problem": "1. N1 (UZB) What is the smallest positive integer $t$ such that there exist integers $x_{1}, x_{2}, \\ldots, x_{t}$ with  $$ x_{1}^{3}+x_{2}^{3}+\\cdots+x_{t}^{3}=2002^{2002} ? $$", "solution": "1. Consider the given equation modulo 9 . Since each cube is congruent to either $-1,0$ or 1 , whereas $2002^{2002} \\equiv 4^{2002}=4 \\cdot 64^{667} \\equiv 4(\\bmod 9)$, we conclude that $t \\geq 4$. On the other hand, $2002^{2002}=2002 \\cdot\\left(2002^{667}\\right)^{3}=\\left(10^{3}+10^{3}+1^{3}+\\right.$ $\\left.1^{3}\\right)\\left(2002^{667}\\right)^{3}$ is a solution with $t=4$. Hence the answer is 4.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "10", "problem_type": "Geometry", "exam": "IMO", "problem": "10. G4 (RUS) Circles $S_{1}$ and $S_{2}$ intersect at points $P$ and $Q$. Distinct points $A_{1}$ and $B_{1}$ (not at $P$ or $Q$ ) are selected on $S_{1}$. The lines $A_{1} P$ and $B_{1} P$ meet $S_{2}$ again at $A_{2}$ and $B_{2}$ respectively, and the lines $A_{1} B_{1}$ and $A_{2} B_{2}$ meet at $C$. Prove that as $A_{1}$ and $B_{1}$ vary, the circumcenters of triangles $A_{1} A_{2} C$ all lie on one fixed circle.", "solution": "10. Let $O$ be the circumcenter of $A_{1} A_{2} C$, and $O_{1}, O_{2}$ the centers of $S_{1}, S_{2}$ respectively. First, from $\\angle A_{1} Q A_{2}=180^{\\circ}-\\angle P A_{1} Q-\\angle Q A_{2} P=\\frac{1}{2}\\left(360^{\\circ}-\\angle P O_{1} Q-\\right.$ $\\left.\\angle Q O_{2} P\\right)=\\angle O_{1} Q O_{2}$ we obtain $\\angle A_{1} Q A_{2}=\\angle B_{1} Q B_{2}=\\angle O_{1} Q O_{2}$. Therefore $\\angle A_{1} Q A_{2}=\\angle B_{1} Q P+$ $\\angle P Q B_{2}=\\angle C A_{1} P+\\angle C A_{2} P=$ $180^{\\circ}-\\angle A_{1} C A_{2}$, from which we conclude that $Q$ lies on the circumcircle of $\\triangle A_{1} A_{2} C$. Hence $O A_{1}=$ $O Q$. However, we also have $O_{1} A_{2}=$ $O_{1} Q$. Consequently, $O, O_{1}$ both lie on the perpendicular bisector of $A_{1} Q$, so $O O_{1} \\perp A_{1} Q$. Similarly, $O O_{2} \\perp A_{2} Q$, leading to $\\angle O_{2} O O_{1}=$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-701.jpg?height=412&width=518&top_left_y=598&top_left_x=825) $180^{\\circ}-\\angle A_{1} Q A_{2}=180^{\\circ}-\\angle O_{1} Q O_{2}$. Hence, $O$ lies on the circle through $O_{1}, O_{2}, Q$, which is fixed.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "11", "problem_type": "Geometry", "exam": "IMO", "problem": "11. G5 (AUS) For any set $S$ of five points in the plane, no three of which are collinear, let $M(S)$ and $m(S)$ denote the greatest and smallest areas, respectively, of triangles determined by three points from $S$. What is the minimum possible value of $M(S) / m(S)$ ?", "solution": "11. When $S$ is the set of vertices of a regular pentagon, then it is easily verified that $\\frac{M(S)}{m(S)}=\\frac{1+\\sqrt{5}}{2}=\\alpha$. We claim that this is the best possible. Let $A, B, C, D, E$ be five arbitrary points, and assume that $\\triangle A B C$ has the area $M(S)$. We claim that some triangle has area less than or equal to $M(S) / \\alpha$. Construct a larger triangle $A^{\\prime} B^{\\prime} C^{\\prime}$ with $C \\in A^{\\prime} B^{\\prime}\\left\\|A B, A \\in B^{\\prime} C^{\\prime}\\right\\| B C$, $B \\in C^{\\prime} A^{\\prime} \\| C A$. The point $D$, as well as $E$, must lie on the same side of $B^{\\prime} C^{\\prime}$ as $B C$, for otherwise $\\triangle D B C$ would have greater area than $\\triangle A B C$. A similar result holds for the other edges, and therefore $D, E$ lie inside the triangle $A^{\\prime} B^{\\prime} C^{\\prime}$ or on its boundary. Moreover, at least one of the triangles $A^{\\prime} B C, A B^{\\prime} C, A B C^{\\prime}$, say $A B C^{\\prime}$, contains neither $D$ nor $E$. Hence we can assume that $D, E$ are contained inside the quadrilateral $A^{\\prime} B^{\\prime} A B$. An affine linear transformation does not change the ratios between areas. Thus if we apply such an affine transformation mapping $A, B, C$ into the vertices $A B M C N$ of a regular pentagon, we won't change $M(S) / m(S)$. If now $D$ or $E$ lies inside $A B M C N$, then we are done. Suppose that both $D$ and $E$ are inside the triangles $C M A^{\\prime}, C N B^{\\prime}$. Then $C D, C E \\leq C M$ (because $C M=C N=C A^{\\prime}=C B^{\\prime}$ ) and $\\angle D C E$ is either less than or equal to $36^{\\circ}$ or greater than or equal to $108^{\\circ}$, from which we obtain that the area of $\\triangle C D E$ cannot exceed the area of $\\triangle C M N=M(S) / \\alpha$. This completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "12", "problem_type": "Geometry", "exam": "IMO", "problem": "12. G6 (UKR) ${ }^{\\mathrm{IMO} 6}$ Let $n \\geq 3$ be a positive integer. Let $C_{1}, C_{2}, C_{3}, \\ldots, C_{n}$ be unit circles in the plane, with centers $O_{1}, O_{2}, O_{3}, \\ldots, O_{n}$ respectively. If no line meets more than two of the circles, prove that  $$ \\sum_{1 \\leq i<j \\leq n} \\frac{1}{O_{i} O_{j}} \\leq \\frac{(n-1) \\pi}{4} $$", "solution": "12. Let $l(M N)$ denote the length of the shorter $\\operatorname{arc} M N$ of a given circle.  Lemma. Let $P R, Q S$ be two chords of a circle $k$ of radius $r$ that meet each other at a point $X$, and let $\\angle P X Q=\\angle R X S=2 \\alpha$. Then $l(P Q)+$ $l(R S)=4 \\alpha r$. Proof. Let $O$ be the center of the circle. Then $l(P Q)+l(R S)=\\angle P O Q$. $r+\\angle R O S \\cdot r=2(\\angle Q S P+\\angle R P S) r=2 \\angle Q X P \\cdot r=4 \\alpha r$. Consider a circle $k$, sufficiently large, whose interior contains all the given circles. For any two circles $C_{i}, C_{j}$, let their exterior common tangents $P R, Q S(P, Q, R, S \\in k)$ form an angle $2 \\alpha$. Then $O_{i} O_{j}=\\frac{2}{\\sin \\alpha}$, so $\\alpha>$ $\\sin \\alpha=\\frac{2}{O_{i} O_{j}}$. By the lemma we have $l(P Q)+l(R S)=4 \\alpha r \\geq \\frac{8 r}{O_{i} O_{j}}$, and hence  $$ \\frac{1}{O_{i} O_{j}} \\leq \\frac{l(P Q)+l(R S)}{8 r} $$  Now sum all these inequalities for $i<j$. The result will follow if we show that every point of the circle $k$ belongs to at most $n-1$ arcs such as $P Q, R S$. Indeed, that will imply that the sum of all the arcs is at most $2(n-1) \\pi r$, hence from (1) we conclude that $\\sum \\frac{1}{O_{i} O_{j}} \\leq \\frac{(n-1) \\pi}{4}$. Consider an arbitrary point $T$ of $k$. We prove by induction (the basis $n=1$ is trivial) that the number of pairs of circles that are simultaneously intercepted by a ray from $T$ is at most $n-1$. Let $T u$ be a ray touching $k$ at $T$. If we let this ray rotate around $T$, it will at some moment intercept a pair of circles for the first time, say $C_{1}, C_{2}$. At some further moment the interception with one of these circles, say $C_{1}$, is lost and never established again. Thus the pair $\\left(C_{1}, C_{2}\\right)$ is the only pair containing $C_{1}$ that is intercepted by some ray from $T$. On the other hand, by the inductive hypothesis the number of such pairs not containing $C_{1}$ does not exceed $n-2$, justifying our claim.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "13", "problem_type": "Geometry", "exam": "IMO", "problem": "13. G7 (BUL) The incircle $\\Omega$ of the acute-angled triangle $A B C$ is tangent to $B C$ at $K$. Let $A D$ be an altitude of triangle $A B C$ and let $M$ be the midpoint of $A D$. If $N$ is the other common point of $\\Omega$ and $K M$, prove that $\\Omega$ and the circumcircle of triangle $B C N$ are tangent at $N$.", "solution": "13. Let $k$ be the circle through $B, C$ that is tangent to the circle $\\Omega$ at point $N^{\\prime}$. We must prove that $K, M, N^{\\prime}$ are collinear. Since the statement is trivial for $A B=A C$, we may assume that $A C>A B$. As usual, $R, r, \\alpha, \\beta, \\gamma$ denote the circumradius and the inradius and the angles of $\\triangle A B C$, respectively. We have $\\tan \\angle B K M=D M / D K$. Straightforward calculation gives $D M=\\frac{1}{2} A D=R \\sin \\beta \\sin \\gamma$ and $D K=\\frac{D C-D B}{2}-\\frac{K C-K B}{2}=R \\sin (\\beta-$ $\\gamma)-R(\\sin \\beta-\\sin \\gamma)=4 R \\sin \\frac{\\beta-\\gamma}{2} \\sin \\frac{\\beta}{2} \\sin \\frac{\\gamma}{2}$, so we obtain  $$ \\tan \\angle B K M=\\frac{\\sin \\beta \\sin \\gamma}{4 \\sin \\frac{\\beta-\\gamma}{2} \\sin \\frac{\\beta}{2} \\sin \\frac{\\gamma}{2}}=\\frac{\\cos \\frac{\\beta}{2} \\cos \\frac{\\gamma}{2}}{\\sin \\frac{\\beta-\\gamma}{2}} $$  To calculate the angle $B K N^{\\prime}$, we apply the inversion $\\psi$ with center at $K$ and power $B K \\cdot C K$. For each object $X$, we denote by $\\widehat{X}$ its image under $\\psi$. The incircle $\\Omega$ maps to a ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-702.jpg?height=215&width=545&top_left_y=1844&top_left_x=807) line $\\widehat{\\Omega}$ parallel to $\\widehat{B} \\widehat{C}$, at distance $\\frac{B K \\cdot C K}{2 r}$ from $\\widehat{B} \\widehat{C}$. Thus the point $\\widehat{N^{\\prime}}$ is the projection of the midpoint $\\widehat{U}$ of $\\widehat{B} \\widehat{C}$ onto $\\widehat{\\Omega}$. Hence  $$ \\tan \\angle B K N^{\\prime}=\\tan \\angle \\widehat{B} K \\widehat{N^{\\prime}}=\\frac{\\widehat{U} \\widehat{N^{\\prime}}}{\\widehat{U} K}=\\frac{B K \\cdot C K}{r(C K-B K)} . $$  Again, one easily checks that $K B \\cdot K C=b c \\sin ^{2} \\frac{\\alpha}{2}$ and $r=4 R \\sin \\frac{\\alpha}{2}$. $\\sin \\frac{\\beta}{2} \\cdot \\sin \\frac{\\gamma}{2}$, which implies  $$ \\begin{aligned} \\tan \\angle B K N^{\\prime} & =\\frac{b c \\sin ^{2} \\frac{\\alpha}{2}}{r(b-c)} \\\\ & =\\frac{4 R^{2} \\sin \\beta \\sin \\gamma \\sin ^{2} \\frac{\\alpha}{2}}{4 R \\sin \\frac{\\alpha}{2} \\sin \\frac{\\beta}{2} \\sin \\frac{\\gamma}{2} \\cdot 2 R(\\sin \\beta-\\sin \\gamma)}=\\frac{\\cos \\frac{\\beta}{2} \\cos \\frac{\\gamma}{2}}{\\sin \\frac{\\beta-\\gamma}{2}} . \\end{aligned} $$  Hence $\\angle B K M=\\angle B K N^{\\prime}$, which implies that $K, M, N^{\\prime}$ are indeed collinear; thus $N^{\\prime} \\equiv N$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "14", "problem_type": "Geometry", "exam": "IMO", "problem": "14. G8 (ARM) Let $S_{1}$ and $S_{2}$ be circles meeting at the points $A$ and $B$. A line through $A$ meets $S_{1}$ at $C$ and $S_{2}$ at $D$. Points $M, N, K$ lie on the line segments $C D, B C, B D$ respectively, with $M N$ parallel to $B D$ and $M K$ parallel to $B C$. Let $E$ and $F$ be points on those $\\operatorname{arcs} B C$ of $S_{1}$ and $B D$ of $S_{2}$ respectively that do not contain $A$. Given that $E N$ is perpendicular to $B C$ and $F K$ is perpendicular to $B D$, prove that $\\measuredangle E M F=90^{\\circ}$.", "solution": "14. Let $G$ be the other point of intersection of the line $F K$ with the arc $B A D$. Since $B N / N C=D K / K B$ and $\\angle C E B=\\angle B G D$ the triangles $C E B$ and $B G D$ are similar. Thus $B N / N E=D K / K G=F K / K B$. From $B N=M K$ and $B K=$ $M N$ it follows that $M N / N E=$ $F K / K M$. But we also have that $\\angle M N E=90^{\\circ}+\\angle M N B=90^{\\circ}+$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-703.jpg?height=327&width=539&top_left_y=1011&top_left_x=817) $\\angle M K B=\\angle F K M$, and hence $\\triangle M N E \\sim \\triangle F K M$. Now $\\angle E M F=\\angle N M K-\\angle N M E-\\angle K M F=\\angle N M K-\\angle N M E-$ $\\angle N E M=\\angle N M K-90^{\\circ}+\\angle B N M=90^{\\circ}$ as claimed.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "15", "problem_type": "Algebra", "exam": "IMO", "problem": "15. A1 (CZE) Find all functions $f$ from the reals to the reals such that  $$ f(f(x)+y)=2 x+f(f(y)-x) $$  for all real $x, y$.", "solution": "15. We observe first that $f$ is surjective. Indeed, setting $y=-f(x)$ gives $f(f(-f(x))-x)=f(0)-2 x$, where the right-hand expression can take any real value. In particular, there exists $x_{0}$ for which $f\\left(x_{0}\\right)=0$. Now setting $x=x_{0}$ in the functional equation yields $f(y)=2 x_{0}+f\\left(f(y)-x_{0}\\right)$, so we obtain  $$ f(z)=z-x_{0} \\quad \\text { for } z=f(y)-x_{0} $$  Since $f$ is surjective, $z$ takes all real values. Hence for all $z, f(z)=z+c$ for some constant $c$, and this is indeed a solution.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "16", "problem_type": "Algebra", "exam": "IMO", "problem": "16. A2 (YUG) Let $a_{1}, a_{2}, \\ldots$ be an infinite sequence of real numbers for which there exists a real number $c$ with $0 \\leq a_{i} \\leq c$ for all $i$ such that  $$ \\left|a_{i}-a_{j}\\right| \\geq \\frac{1}{i+j} \\quad \\text { for all } i, j \\text { with } i \\neq j $$  Prove that $c \\geq 1$.", "solution": "16. For $n \\geq 2$, let $\\left(k_{1}, k_{2}, \\ldots, k_{n}\\right)$ be the permutation of $\\{1,2, \\ldots, n\\}$ with $a_{k_{1}} \\leq a_{k_{2}} \\leq \\cdots \\leq a_{k_{n}}$. Then from the condition of the problem, using the Cauchy-Schwarz inequality, we obtain  $$ \\begin{aligned} c & \\geq a_{k_{n}}-a_{k_{1}}=\\left|a_{k_{n}}-a_{k_{n-1}}\\right|+\\cdots+\\left|a_{k_{3}}-a_{k_{2}}\\right|+\\left|a_{k_{2}}-a_{k_{1}}\\right| \\\\ & \\geq \\frac{1}{k_{1}+k_{2}}+\\frac{1}{k_{2}+k_{3}}+\\cdots+\\frac{1}{k_{n-1}+k_{n}} \\\\ & \\geq \\frac{(n-1)^{2}}{\\left(k_{1}+k_{2}\\right)+\\left(k_{2}+k_{3}\\right)+\\cdots+\\left(k_{n-1}+k_{n}\\right)} \\\\ & =\\frac{(n-1)^{2}}{2\\left(k_{1}+k_{2}+\\cdots+k_{n}\\right)-k_{1}-k_{n}} \\geq \\frac{(n-1)^{2}}{n^{2}+n-3} \\geq \\frac{n-1}{n+2} . \\end{aligned} $$  Therefore $c \\geq 1-\\frac{3}{n+2}$ for every positive integer $n$. But if $c<1$, this inequality is obviously false for all $n>\\frac{3}{1-c}-2$. We conclude that $c \\geq 1$. Remark. The least value of $c$ is not greater than $2 \\ln 2$. An example of a sequence $\\left\\{a_{n}\\right\\}$ with $0 \\leq a_{n} \\leq 2 \\ln 2$ can be constructed inductively as follows: Given $a_{1}, a_{2}, \\ldots, a_{n-1}$, then $a_{n}$ can be any number from $[0,2 \\ln 2]$ that does not belong to any of the intervals $\\left(a_{i}-\\frac{1}{i+n}, a_{i}+\\frac{1}{i+n}\\right)(i=$ $1,2, \\ldots, n-1)$, and the total length of these intervals is always less than or equal to  $$ \\frac{2}{n+1}+\\frac{2}{n+2}+\\cdots+\\frac{2}{2 n-1}<2 \\ln 2 $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "17", "problem_type": "Algebra", "exam": "IMO", "problem": "17. A3 (POL) Let $P$ be a cubic polynomial given by $P(x)=a x^{3}+b x^{2}+c x+$ $d$, where $a, b, c, d$ are integers and $a \\neq 0$. Suppose that $x P(x)=y P(y)$ for infinitely many pairs $x, y$ of integers with $x \\neq y$. Prove that the equation $P(x)=0$ has an integer root.", "solution": "17. Let $x, y$ be distinct integers satisfying $x P(x)=y P(y)$; this is equivalent to $a\\left(x^{4}-y^{4}\\right)+b\\left(x^{3}-y^{3}\\right)+c\\left(x^{2}-y^{2}\\right)+d(x-y)=0$. Dividing by $x-y$ we obtain  $$ a\\left(x^{3}+x^{2} y+x y^{2}+y^{3}\\right)+b\\left(x^{2}+x y+y^{2}\\right)+c(x+y)+d=0 . $$  Putting $x+y=p, x^{2}+y^{2}=q$ leads to $x^{2}+x y+y^{2}=\\frac{p^{2}+q}{2}$, so the above equality becomes  $$ a p q+\\frac{b}{2}\\left(p^{2}+q\\right)+c p+d=0, \\quad \\text { i.e. } \\quad(2 a p+b) q=-\\left(b p^{2}+2 c p+2 d\\right) $$  Since $q \\geq p^{2} / 2$, it follows that $p^{2}|2 a p+b| \\leq 2\\left|b p^{2}+2 c p+2 d\\right|$, which is possible only for finitely many values of $p$, although there are infinitely many pairs $(x, y)$ with $x P(x)=y P(y)$. Hence there exists $p$ such that $x P(x)=(p-x) P(p-x)$ for infinitely many $x$, and therefore for all $x$. If $p \\neq 0$, then $p$ is a root of $P(x)$. If $p=0$, the above relation gives $P(x)=-P(-x)$. This forces $b=d=0$, so $P(x)=x\\left(a x^{2}+c\\right)$. Thus 0 is a root of $P(x)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "18", "problem_type": "Algebra", "exam": "IMO", "problem": "18. A4 (IND) ${ }^{\\text {IMO5 }}$ Find all functions $f$ from the reals to the reals such that  $$ (f(x)+f(z))(f(y)+f(t))=f(x y-z t)+f(x t+y z) $$  for all real $x, y, z, t$.", "solution": "18. Putting $x=z=0$ and $t=y$ into the given equation gives $4 f(0) f(y)=$ $2 f(0)$ for all $y$. If $f(0) \\neq 0$, then we deduce $f(y)=\\frac{1}{2}$, i.e., $f$ is identically equal to $\\frac{1}{2}$. Now we suppose that $f(0)=0$. Setting $z=t=0$ we obtain  $$ f(x y)=f(x) f(y) \\quad \\text { for all } x, y \\in \\mathbb{R} $$  Thus if $f(y)=0$ for some $y \\neq 0$, then $f$ is identically zero. So, assume $f(y) \\neq 0$ whenever $y \\neq 0$.  Next, we observe that $f$ is strictly increasing on the set of positive reals. Actually, it follows from (1) that $f(x)=f(\\sqrt{x})^{2} \\geq 0$ for all $x \\geq 0$, so that the given equation for $t=x$ and $z=y$ yields $f\\left(x^{2}+y^{2}\\right)=(f(x)+f(y))^{2} \\geq$ $f\\left(x^{2}\\right)$ for all $x, y \\geq 0$. Using (1) it is easy to get $f(1)=1$. Now plugging $t=y$ into the given equation, we are led to  $$ 2[f(x)+f(z)]=f(x-z)+f(x+z) \\quad \\text { for all } x, z $$  In particular, $f(z)=f(-z)$. Further, it is easy to get by induction from (2) that $f(n x)=n^{2} f(x)$ for all integers $n$ (and consequently for all rational numbers as well). Therefore $f(q)=f(-q)=q^{2}$ for all $q \\in \\mathbb{Q}$. But $f$ is increasing for $x>0$, so we must have $f(x)=x^{2}$ for all $x$. It is easy to verify that $f(x)=0, f(x)=\\frac{1}{2}$ and $f(x)=x^{2}$ are indeed solutions.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "19", "problem_type": "Algebra", "exam": "IMO", "problem": "19. A5 (IND) Let $n$ be a positive integer that is not a perfect cube. Define real numbers $a, b, c$ by  $$ a=\\sqrt[3]{n}, \\quad b=\\frac{1}{a-[a]}, \\quad c=\\frac{1}{b-[b]}, $$  where $[x]$ denotes the integer part of $x$. Prove that there are infinitely many such integers $n$ with the property that there exist integers $r, s, t$, not all zero, such that $r a+s b+t c=0$.", "solution": "19. Write $m=[\\sqrt[3]{n}]$. To simplify the calculation, we shall assume that $[b]=1$. Then $a=\\sqrt[3]{n}, b=\\frac{1}{\\sqrt[3]{n}-m}=\\frac{1}{n-m^{3}}\\left(m^{2}+m \\sqrt[3]{n}+\\sqrt[3]{n^{2}}\\right), c=\\frac{1}{b-1}=$ $u+v \\sqrt[3]{n}+w \\sqrt[3]{n^{2}}$ for certain rational numbers $u, v, w$. Obviously, integers $r, s, t$ with $r a+s b+t c=0$ exist if (and only if) $u=m^{2} w$, i.e., if ( $b-$ 1) $\\left(m^{2} w+v \\sqrt[3]{n}+w \\sqrt[3]{n^{2}}\\right)=1$ for some rational $v, w$.  When the last equality is expanded and simplified, comparing the coefficients at $1, \\sqrt[3]{n}, \\sqrt[3]{n^{2}}$ one obtains  $$ \\begin{array}{rlrl} 1: & v+\\left(\\left(m^{2}+m^{3}-n\\right) m^{2}+m\\right) w & =n-m^{3}, \\\\ \\sqrt[3]{n}: & \\left(m^{2}+m^{3}-n\\right) v+ & \\left(m^{3}+n\\right) w & =0, \\\\ \\sqrt[3]{n^{2}}: & m v+ & \\left(2 m^{2}+m^{3}-n\\right) w & =0 . \\end{array} $$  In order for the system (1) to have a solution $v, w$, we must have $\\left(2 m^{2}+\\right.$ $\\left.m^{3}-n\\right)\\left(m^{2}+m^{3}-n\\right)=m\\left(m^{3}+n\\right)$. This quadratic equation has solutions $n=m^{3}$ and $n=m^{3}+3 m^{2}+m$. The former is not possible, but the latter gives $a-[a]>\\frac{1}{2}$, so $[b]=1$, and the system (1) in $v, w$ is solvable. Hence every number $n=m^{3}+3 m^{2}+m, m \\in \\mathbb{N}$, satisfies the condition of the problem.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "2", "problem_type": "Number Theory", "exam": "IMO", "problem": "2. N2 (ROM) ${ }^{\\mathrm{IMO} 4}$ Let $n \\geq 2$ be a positive integer, with divisors $1=d_{1}<$ $d_{2}<\\cdots<d_{k}=n$. Prove that $d_{1} d_{2}+d_{2} d_{3}+\\cdots+d_{k-1} d_{k}$ is always less than $n^{2}$, and determine when it is a divisor of $n^{2}$.", "solution": "2. Set $S=d_{1} d_{2}+\\cdots+d_{k-1} d_{k}$. Since $d_{i} / n=1 / d_{k+1-i}$, we have $\\frac{S}{n^{2}}=$ $\\frac{1}{d_{k} d_{k-1}}+\\cdots+\\frac{1}{d_{2} d_{1}}$. Hence  $$ \\frac{1}{d_{2} d_{1}} \\leq \\frac{S}{n^{2}} \\leq\\left(\\frac{1}{d_{k-1}}-\\frac{1}{d_{k}}\\right)+\\cdots+\\left(\\frac{1}{d_{1}}-\\frac{1}{d_{2}}\\right)=1-\\frac{1}{d_{k}}<1 $$  or $\\left(\\right.$ since $\\left.d_{1}=1\\right) 1<\\frac{n^{2}}{S} \\leq d_{2}$. This shows that $S<n^{2}$. Also, if $S$ is a divisor of $n^{2}$, then $n^{2} / S$ is a nontrivial divisor of $n^{2}$ not exceeding $d_{2}$. But $d_{2}$ is obviously the least prime divisor of $n$ (and also of $n^{2}$ ), so we must have $n^{2} / S=d_{2}$, which holds if and only if $n$ is prime.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "20", "problem_type": "Algebra", "exam": "IMO", "problem": "20. A6 (IRN) Let $A$ be a nonempty set of positive integers. Suppose that there are positive integers $b_{1}, \\ldots, b_{n}$ and $c_{1}, \\ldots, c_{n}$ such that (i) for each $i$ the set $b_{i} A+c_{i}=\\left\\{b_{i} a+c_{i} \\mid a \\in A\\right\\}$ is a subset of $A$, and (ii) the sets $b_{i} A+c_{i}$ and $b_{j} A+c_{j}$ are disjoint whenever $i \\neq j$.  Prove that  $$ \\frac{1}{b_{1}}+\\cdots+\\frac{1}{b_{n}} \\leq 1 $$", "solution": "20. Assume to the contrary that $\\frac{1}{b_{1}}+\\cdots+\\frac{1}{b_{n}}>1$. Certainly $n \\geq 2$ and $A$ is infinite. Define $f_{i}: A \\rightarrow A$ as $f_{i}(x)=b_{i} x+c_{i}$ for each $i$. By condition (ii), $f_{i}(x)=f_{j}(y)$ implies $i=j$ and $x=y$; iterating this argument, we deduce that $f_{i_{1}}\\left(\\ldots f_{i_{m}}(x) \\ldots\\right)=f_{j_{1}}\\left(\\ldots f_{j_{m}}(x) \\ldots\\right)$ implies $i_{1}=j_{1}, \\ldots, i_{m}=j_{m}$ and $x=y$. As an illustration, we shall consider the case $b_{1}=b_{2}=b_{3}=2$ first. If $a$ is large enough, then for any $i_{1}, \\ldots, i_{m} \\in\\{1,2,3\\}$ we have $f_{i_{1}} \\circ \\cdots \\circ f_{i_{m}}(a) \\leq$ $2.1^{m} a$. However, we obtain $3^{m}$ values in this way, so they cannot be all distinct if $m$ is sufficiently large, a contradiction. In the general case, let real numbers $d_{i}>b_{i}, i=1,2 \\ldots, n$, be chosen such that $\\frac{1}{d_{1}}+\\cdots+\\frac{1}{d_{n}}>1$ : for $a$ large enough, $f_{i}(x)<d_{i} a$ for each $x \\geq a$.  Also, let $k_{i}>0$ be arbitrary rational numbers with sum 1 ; denote by $N_{0}$ the least common multiple of their denominators. Let $N$ be a fixed multiple of $N_{0}$, so that each $k_{j} N$ is an integer. Consider all combinations $f_{i_{1}} \\circ \\cdots \\circ f_{i_{N}}$ of $N$ functions, among which each $f_{i}$ appears exactly $k_{i} N$ times. There are $F_{N}=\\frac{N!}{\\left(k_{1} N\\right)!\\cdots\\left(k_{n} N\\right)!}$ such combinations, so they give $F_{N}$ distinct values when applied to $a$. On the other hand, $f_{i_{1}} \\circ \\cdots \\circ f_{i_{N}}(a) \\leq\\left(d_{1}^{k_{1}} \\cdots d_{n}^{k_{n}}\\right)^{N} a$. Therefore  $$ \\left(d_{1}^{k_{1}} \\cdots d_{n}^{k_{n}}\\right)^{N} a \\geq F_{N} \\quad \\text { for all } N, N_{0} \\mid N $$  It remains to find a lower estimate for $F_{N}$. In fact, it is straightforward to verify that $F_{N+N_{0}} / F_{N}$ tends to $Q^{N_{0}}$, where $Q=1 /\\left(k_{1}^{k_{1}} \\cdots k_{n}^{k_{n}}\\right)$. Consequently, for every $q<Q$ there exists $p>0$ such that $F_{N}>p q^{N}$. Then (1) implies that  $$ \\left(\\frac{d_{1}^{k_{1}} \\cdots d_{n}^{k_{n}}}{q}\\right)^{N}>\\frac{p}{a} \\text { for every multiple } N \\text { of } N_{0} $$  and hence $d_{1}^{k_{1}} \\cdots d_{n}^{k_{n}} / q \\geq 1$. This must hold for every $q<Q$, and so we have $d_{1}^{k_{1}} \\cdots d_{n}^{k_{n}} \\geq Q$, i.e.,  $$ \\left(k_{1} d_{1}\\right)^{k_{1}} \\cdots\\left(k_{n} d_{n}\\right)^{k_{n}} \\geq 1 $$  However, if we choose $k_{1}, \\ldots, k_{n}$ such that $k_{1} d_{1}=\\cdots=k_{n} d_{n}=u$, then we must have $u \\geq 1$. Therefore $\\frac{1}{d_{1}}+\\cdots+\\frac{1}{d_{n}} \\leq k_{1}+\\cdots+k_{n}=1$, a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "21", "problem_type": null, "exam": "IMO", "problem": "21. $\\mathbf{C 1}(\\mathbf{C O L})^{\\mathrm{IMO} 1}$ Let $n$ be a positive integer. Each point $(x, y)$ in the plane, where $x$ and $y$ are nonnegative integers with $x+y \\leq n$, is colored red or blue, subject to the following condition: If a point $(x, y)$ is red, then so are all points $\\left(x^{\\prime}, y^{\\prime}\\right)$ with $x^{\\prime} \\leq x$ and $y^{\\prime} \\leq y$. Let $A$ be the number of ways to choose $n$ blue points with distinct $x$-coordinates, and let $B$ be the number of ways to choose $n$ blue points with distinct $y$-coordinates. Prove that $A=B$.", "solution": "21. Let $a_{i}$ be the number of blue points with $x$-coordinate $i$, and $b_{i}$ the number of blue points with $y$-coordinate $i$. Our task is to show that $a_{0} a_{1} \\cdots a_{n-1}=$ $b_{0} b_{1} \\cdots b_{n-1}$. Moreover, we claim that $a_{0}, \\ldots, a_{n-1}$ is a permutation of $b_{0}, \\ldots, b_{n-1}$, and to show this we use induction on the number of red points. The result is trivial if all the points are blue. So, choose a red point $(x, y)$ with $x+y$ maximal: clearly $a_{x}=b_{y}=n-x-y-1$. If we change this point to blue, $a_{x}$ and $b_{y}$ will decrease by 1 . Then by the induction hypothesis, $a_{0}, \\ldots, a_{n-1}$ with $a_{x}$ decreased by 1 is a permutation of $b_{0}, \\ldots, b_{n-1}$ with $b_{y}$ decreased by 1 . However, $a_{x}=b_{y}$, and the claim follows. Remark. One can also use induction on $n$ : it is not more difficult.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "22", "problem_type": "Combinatorics", "exam": "IMO", "problem": "22. C2 (ARM) For $n$ an odd positive integer, the unit squares of an $n \\times n$ chessboard are colored alternately black and white, with the four corners colored black. A tromino is an $L$-shape formed by three connected unit squares. For which values of $n$ is it possible to cover all the black squares with nonoverlapping trominos? When it is possible, what is the minimum number of trominos needed?", "solution": "22. Write $n=2 k+1$. Consider the black squares at an odd height: there are $(k+1)^{2}$ of them in total and no two can be covered by one trimino. Thus, we always need at least $(k+1)^{2}$ triminoes, which cover $3(k+1)^{2}$ squares in total. However, $3(k+1)^{2}$ is greater than $n^{2}$ for $n=1,3,5$, so we must have $n \\geq 7$. The case $n=7$ admits such a covering, as shown in Figure 1. For $n>7$ this is possible as well: it follows by induction from Figure 2. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-707.jpg?height=186&width=195&top_left_y=302&top_left_x=402)  Fig. 1 ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-707.jpg?height=239&width=249&top_left_y=251&top_left_x=903)  Fig. 2", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "23", "problem_type": "Combinatorics", "exam": "IMO", "problem": "23. C3 (COL) Let $n$ be a positive integer. A sequence of $n$ positive integers (not necessarily distinct) is called full if it satisfies the following condition: For each positive integer $k \\geq 2$, if the number $k$ appears in the sequence, then so does the number $k-1$, and moreover, the first occurrence of $k-1$ comes before the last occurrence of $k$. For each $n$, how many full sequences are there?", "solution": "23. We claim that there are $n$ ! full sequences. To show this, we construct a bijection with the set of permutations of $\\{1,2, \\ldots, n\\}$. Consider a full sequence $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$, and let $m$ be the greatest of the numbers $a_{1}, \\ldots, a_{n}$. Let $S_{k}, 1 \\leq k \\leq m$, be the set of those indices $i$ for which $a_{i}=k$. Then $S_{1}, \\ldots S_{m}$ are nonempty and form a partition of the set $\\{1,2, \\ldots, n\\}$. Now we write down the elements of $S_{1}$ in descending order, then the elements of $S_{2}$ in descending order and so on. This maps the full sequence to a permutation of $\\{1,2, \\ldots, n\\}$. Moreover, this map is reversible, since each permutation uniquely breaks apart into decreasing sequences $S_{1}^{\\prime}, S_{2}^{\\prime}, \\ldots, S_{m}^{\\prime}$, so that $\\max S_{i}^{\\prime}>\\min S_{i-1}^{\\prime}$. Therefore the full sequences are in bijection with the permutations of $\\{1,2, \\ldots, n\\}$. Second solution. Let there be given a full sequence of length $n$. Removing from it the first occurrence of the highest number, we obtain a full sequence of length $n-1$. On the other hand, each full sequence of length $n-1$ can be obtained from exactly $n$ full sequences of length $n$. Therefore, if $x_{n}$ is the number of full sequences of length $n$, we deduce $x_{n}=n x_{n-1}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "24", "problem_type": "Combinatorics", "exam": "IMO", "problem": "24. C4 (BUL) Let $T$ be the set of ordered triples $(x, y, z)$, where $x, y, z$ are integers with $0 \\leq x, y, z \\leq 9$. Players $A$ and $B$ play the following guessing game: Player $A$ chooses a triple $(x, y, z)$ in $T$, and Player $B$ has to discover A's triple in as few moves as possible. A move consists of the following: $B$ gives $A$ a triple $(a, b, c)$ in $T$, and $A$ replies by giving $B$ the number $|x+y-a-b|+|y+z-b-c|+|z+x-c-a|$. Find the minimum number of moves that $B$ needs to be sure of determining $A$ 's triple.", "solution": "24. Two moves are not sufficient. Indeed, the answer to each move is an even number between 0 and 54 , so the answer takes at most 28 distinct values. Consequently, two moves give at most $28^{2}=784$ distinct outcomes, which is less than $10^{3}=1000$. We now show that three moves are sufficient. With the first move $(0,0,0)$, we get the reply $2(x+y+z)$, so we now know the value of $s=x+y+z$. Now there are several cases: (i) $s \\leq 9$. Then we ask $(9,0,0)$ as the second move and get $(9-x-y)+$ $(9-x-z)+(y+z)=18-2 x$, so we come to know $x$. Asking $(0,9,0)$ we obtain $y$, which is enough, since $z=s-x-y$. (ii) $10 \\leq s \\leq 17$. In this case the second move is $(9, s-9,0)$. The answer is $z+(9-x)+|x+z-9|=2 k$, where $k=z$ if $x+z \\geq 9$, or $k=9-x$ if $x+z<9$. In both cases we have $z \\leq k \\leq y+z \\leq s$. Let $s-k \\leq 9$. Then in the third move we ask $(s-k, 0, k)$ and obtain $|z-k|+|k-y-z|+y$, which is actually $(k-z)+(y+z-k)+y=2 y$. Thus we also find out $x+z$, and thus deduce whether $k$ is $z$ or $9-x$. Consequently we determine both $x$ and $z$. Let $s-k>9$. In this case, the third move is $(9, s-k-9, k)$. The answer is $|s-k-x-y|+|s-9-y-z|+|k+9-z-x|=$ $(k-z)+(9-x)+(9-x+k-z)=18+2 k-2(x+z)$, from which we find out again whether $k$ is $z$ or $9-x$. Now we are easily done. (iii) $18 \\leq s \\leq 27$. Then as in the first case, asking $(0,9,9)$ and $(9,0,9)$ we obtain $x$ and $y$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "25", "problem_type": "Combinatorics", "exam": "IMO", "problem": "25. C5 (BRA) Let $r \\geq 2$ be a fixed positive integer, and let $\\mathcal{F}$ be an infinite family of sets, each of size $r$, no two of which are disjoint. Prove that there exists a set of size $r-1$ that meets each set in $\\mathcal{F}$.", "solution": "25. Assume to the contrary that no set of size less than $r$ meets all sets in $\\mathcal{F}$. Consider any set $A$ of size less than $r$ that is contained in infinitely many sets of $\\mathcal{F}$. By the assumption, $A$ is disjoint from some set $B \\in \\mathcal{F}$. Then of the infinitely many sets that contain $A$, each must meet $B$, so some element $b$ of $B$ belongs to infinitely many of them. But then the set $A \\cup\\{b\\}$ is contained in infinitely many sets of $\\mathcal{F}$ as well. Such a set $A$ exists: for example, the empty set. Now taking for $A$ the largest such set we come to a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "26", "problem_type": "Combinatorics", "exam": "IMO", "problem": "26. C6 (POL) Let $n$ be an even positive integer. Show that there is a permutation $x_{1}, x_{2}, \\ldots, x_{n}$ of $1,2, \\ldots, n$ such that for every $1 \\leq i \\leq n$ the number $x_{i+1}$ is one of $2 x_{i}, 2 x_{i}-1,2 x_{i}-n, 2 x_{i}-n-1$ (where we take $x_{n+1}=x_{1}$ ).", "solution": "26. Write $n=2 m$. We shall define a directed graph $G$ with vertices $1, \\ldots, m$ and edges labelled $1,2, \\ldots, 2 m$ in such a way that the edges issuing from $i$ are labelled $2 i-1$ and $2 i$, and those entering it are labelled $i$ and $i+m$. What we need is an Euler circuit in $G$, namely a closed path that passes each edge exactly once. Indeed, if $x_{i}$ is the $i$ th edge in such a circuit, then $x_{i}$ enters some vertex $j$ and $x_{i+1}$ leaves it, so $x_{i} \\equiv j(\\bmod m)$ and $x_{i+1}=2 j-1$ or $2 j$. Hence $2 x_{i} \\equiv 2 j$ and $x_{i+1} \\equiv 2 x_{i}$ or $2 x_{i}-1(\\bmod n)$, as required. The graph $G$ is connected: by induction on $k$ there is a path from 1 to $k$, since 1 is connected to $j$ with $2 j=k$ or $2 j-1=k$, and there is an edge from $j$ to $k$. Also, the in-degree and out-degree of each vertex of $G$ are equal (to 2), and thus by a known result, $G$ contains an Euler circuit.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "27", "problem_type": "Combinatorics", "exam": "IMO", "problem": "27. C7 (NZL) Among a group of 120 people, some pairs are friends. A weak quartet is a set of four people containing exactly one pair of friends. What is the maximum possible number of weak quartets?", "solution": "27. For a graph $G$ on 120 vertices (i.e., people at the party), write $q(G)$ for the number of weak quartets in $G$. Our solution will consist of three parts. First, we prove that some graph $G$ with maximal $q(G)$ breaks up as a disjoint union of complete graphs. This will follow if we show that any two adjacent vertices $x, y$ have the same neighbors (apart from themselves). Let $G_{x}$ be the graph obtained from $G$ by \"copying\" $x$ to $y$ (i.e., for each $z \\neq x, y$, we add the edge $z y$ if $z x$ is an edge, and delete $z y$ if $z x$ is not an edge). Similarly $G_{y}$ is the graph obtained from $G$ by copying $y$ to $x$. We claim that $2 q(G) \\leq q\\left(G_{x}\\right)+q\\left(G_{y}\\right)$. Indeed, the number of weak quartets containing neither $x$ nor $y$ is the same in $G, G_{x}$, and $G_{y}$, while the number of those containing both $x$ and $y$ is not less in $G_{x}$ and $G_{y}$ than in $G$. Also, the number containing exactly one of $x$ and $y$ in $G_{x}$ is at least twice the number in $G$ containing $x$ but not $y$, while the number containing exactly one of $x$ and $y$ in $G_{y}$ is at least twice the number in $G$ containing $y$ but not $x$. This justifies our claim by adding. It follows that for an extremal graph $G$ we must have $q(G)=q\\left(G_{x}\\right)=q\\left(G_{y}\\right)$. Repeating the copying operation pair by pair ( $y$ to $x$, then their common neighbor $z$ to both $x, y$, etc.) we eventually obtain an extremal graph consisting of disjoint complete graphs. Second, suppose the complete graphs in $G$ have sizes $a_{1}, a_{2}, \\ldots, a_{n}$. Then  $$ q(G)=\\sum_{i=1}^{n}\\binom{a_{i}}{2} \\sum_{\\substack{j<k \\\\ j, k \\neq i}} a_{j} a_{k} . $$  If we fix all the $a_{i}$ except two, say $p, q$, then $p+q=s$ is fixed, and for some constants $C_{i}, q(G)=C_{1}+C_{2} p q+C_{3}\\left(\\binom{p}{2}+\\binom{q}{2}\\right)+C_{4}\\left(q\\binom{p}{2}+p\\binom{q}{2}\\right)=$ $A+B p q$, where $A$ and $B$ depend only on $s$. Hence the maximum of $q(G)$ is attained if $|p-q| \\leq 1$ or $p q=0$. Thus if $q(G)$ is maximal, any two nonzero $a_{i}$ 's differ by at most 1 . Finally, if $G$ consists of $n$ disjoint complete graphs, then $q(G)$ cannot exceed the value obtained if $a_{1}=\\cdots=a_{n}$ (not necessarily integral), which equals  $$ Q_{n}=\\frac{120^{2}}{n}\\binom{120 / n}{2}\\binom{n-1}{2}=30 \\cdot 120^{2} \\frac{(n-1)(n-2)(120-n)}{n^{3}} . $$  It is easy to check that $Q_{n}$ takes its maximum when $n=5$ and $a_{1}=\\cdots=$ $a_{5}=24$, and that this maximum equals $15 \\cdot 23 \\cdot 24^{3}=4769280$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "3", "problem_type": "Number Theory", "exam": "IMO", "problem": "3. N3 (MON) Let $p_{1}, p_{2}, \\ldots, p_{n}$ be distinct primes greater than 3 . Show that $2^{p_{1} p_{2} \\cdots p_{n}}+1$ has at least $4^{n}$ divisors.", "solution": "3. We observe that if $a, b$ are coprime odd numbers, then $\\operatorname{gcd}\\left(2^{a}+1,2^{b}+1\\right)=$ 3. In fact, this g.c.d. divides $\\operatorname{gcd}\\left(2^{2 a}-1,2^{2 b}-1\\right)=2^{\\operatorname{gcd}(2 a, 2 b)}-1=2^{2}-1=$ 3 , while 3 obviously divides both $2^{a}+1$ and $2^{b}+1$. In particular, if $3 \\nmid b$, then $3^{2} \\nmid 2^{b}+1$, so $2^{a}+1$ and $\\left(2^{b}+1\\right) / 3$ are coprime; consequently $2^{a b}+1$ (being divisible by $\\left.2^{a}+1,2^{b}+1\\right)$ is divisible by $\\frac{\\left(2^{a}+1\\right)\\left(2^{b}+1\\right)}{3}$. Now we prove the desired result by induction on $n$. For $n=1,2^{p_{1}}+1$ is divisible by 3 and exceeds $3^{2}$, so it has at least 4 divisors. Assume that $2^{a}+1=2^{p_{1} \\cdots p_{n-1}}+1$ has at least $4^{n-1}$ divisors and consider $N=2^{a b}+1=$ $2^{p_{1} \\cdots p_{n}}+1\\left(\\right.$ where $\\left.b=p_{n}\\right)$. As above, $2^{a}+1$ and $\\frac{2^{b}+1}{3}$ are coprime, and thus $Q=\\left(2^{a}+1\\right)\\left(2^{b}+1\\right) / 3$ has at least $2 \\cdot 4^{n-1}$ divisors. Moreover, $N$ is divisible by $Q$ and is greater than $Q^{2}$ (indeed, $N>2^{a b}>2^{2 a} 2^{2 b}>Q^{2}$ if $a, b \\geq 5$ ). Then $N$ has at least twice as many divisors as $Q$ (because for every $d \\mid Q$ both $d$ and $N / d$ are divisors of $N$ ), which counts up to $4^{n}$ divisors, as required. Remark. With some knowledge of cyclotomic polynomials, one can show that $2^{p_{1} \\cdots p_{n}}+1$ has at least $2^{2^{n-1}}$ divisors, far exceeding $4^{n}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "4", "problem_type": "Number Theory", "exam": "IMO", "problem": "4. N4 (GER) Is there a positive integer $m$ such that the equation  $$ \\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{a b c}=\\frac{m}{a+b+c} $$  has infinitely many solutions in positive integers $a, b, c$ ?", "solution": "4. For $a=b=c=1$ we obtain $m=12$. We claim that the given equation has infinitely many solutions in positive integers $a, b, c$ for this value of $m$. After multiplication by $a b c(a+b+c)$ the equation $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{a b c}-\\frac{12}{a+b+c}=0$ becomes  $$ a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)+a+b+c-9 a b c=0 $$  We must show that this equation has infinitely many solutions in positive integers. Suppose that $(a, b, c)$ is one such solution with $a<b<c$. Regarding (1) as a quadratic equation in $a$, we see by Vieta's formula that ( $b, c, \\frac{b c+1}{a}$ ) also satisfies (1).  Define $\\left(a_{n}\\right)_{n=0}^{\\infty}$ by $a_{0}=a_{1}=a_{2}=1$ and $a_{n+1}=\\frac{a_{n} a_{n-1}+1}{a_{n-2}}$ for each $n>1$. We show that all $a_{n}$ 's are integers. This procedure is fairly standard. The above relation for $n$ and $n-1$ gives $a_{n+1} a_{n-2}=a_{n} a_{n-1}+1$ and $a_{n-1} a_{n-2}+1=a_{n} a_{n-3}$, so that adding yields $a_{n-2}\\left(a_{n-1}+a_{n+1}\\right)=$ $a_{n}\\left(a_{n-1}+a_{n-3}\\right)$. Therefore $\\frac{a_{n+1}+a_{n-1}}{a_{n}}=\\frac{a_{n-1}+a_{n-3}}{a_{n-2}}=\\cdots$, from which it follows that  $$ \\frac{a_{n+1}+a_{n-1}}{a_{n}}=\\left\\{\\begin{array}{l} \\frac{a_{2}+a_{0}}{a_{1}}=2 \\text { for } n \\text { odd; } \\\\ \\frac{a_{3}+a_{1}}{a_{2}}=3 \\text { for } n \\text { even. } \\end{array}\\right. $$  It is now an immediate consequence that every $a_{n}$ is integral. Also, the above consideration implies that $\\left(a_{n-1}, a_{n}, a_{n+1}\\right)$ is a solution of (1) for each $n \\geq 1$. Since $a_{n}$ is strictly increasing, this gives an infinity of solutions in integers. Remark. There are infinitely many values of $m \\in \\mathbb{N}$ for which the given equation has at least one solution in integers, and each of those values admits an infinity of solutions.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "5", "problem_type": "Number Theory", "exam": "IMO", "problem": "5. N5 (IRN) Let $m, n \\geq 2$ be positive integers, and let $a_{1}, a_{2}, \\ldots, a_{n}$ be integers, none of which is a multiple of $m^{n-1}$. Show that there exist integers $e_{1}, e_{2}, \\ldots, e_{n}$, not all zero, with $\\left|e_{i}\\right|<m$ for all $i$, such that $e_{1} a_{1}+e_{2} a_{2}+\\cdots+e_{n} a_{n}$ is a multiple of $m^{n}$.", "solution": "5. Consider all possible sums $c_{1} a_{1}+c_{2} a_{2}+\\cdots+c_{n} a_{n}$, where each $c_{i}$ is an integer with $0 \\leq c_{i}<m$. There are $m^{n}$ such sums, and if any two of them give the same remainder modulo $m^{n}$, say $\\sum c_{i} a_{i} \\equiv \\sum d_{i} a_{i}\\left(\\bmod m^{n}\\right)$, then $\\sum\\left(c_{i}-d_{i}\\right) a_{i}$ is divisible by $m^{n}$, and since $\\left|c_{i}-d_{i}\\right|<m$, we are done. We claim that two such sums must exist. Suppose to the contrary that the sums $\\sum_{i} c_{i} a_{i}\\left(0 \\leq c_{i}<m\\right)$ give all the different remainders modulo $m^{n}$. Consider the polynomial  $$ P(x)=\\sum x^{c_{1} a_{1}+\\cdots+c_{n} a_{n}}, $$  where the sum is taken over all $\\left(c_{1}, \\ldots, c_{n}\\right)$ with $0 \\leq c_{i}<m$. If $\\xi$ is a primitive $m^{n}$ th root of unity, then by the assumption we have  $$ P(\\xi)=1+\\xi+\\cdots+\\xi^{m^{n}-1}=0 $$  On the other hand, $P(x)$ can be factored as  $$ P(x)=\\prod_{i=1}^{n}\\left(1+x^{a_{i}}+\\cdots+x^{(m-1) a_{i}}\\right)=\\prod_{i=1}^{n} \\frac{1-x^{m a_{i}}}{1-x^{a_{i}}}, $$  so that none of its factors is zero at $x=\\xi$ because $m a_{i}$ is not divisible by $m^{n}$. This is obviously a contradiction. Remark. The example $a_{i}=m^{i-1}$ for $i=1, \\ldots, n$ shows that the condition that no $a_{i}$ is a multiple of $m^{n-1}$ cannot be removed.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "6", "problem_type": "Number Theory", "exam": "IMO", "problem": "6. N6 (ROM) ${ }^{\\mathrm{IMO} 3}$ Find all pairs of positive integers $m, n \\geq 3$ for which there exist infinitely many positive integers $a$ such that  $$ \\frac{a^{m}+a-1}{a^{n}+a^{2}-1} $$  is itself an integer.", "solution": "6. Suppose that $(m, n)$ is such a pair. Assume that division of the polynomial $F(x)=x^{m}+x-1$ by $G(x)=x^{n}+x^{2}-1$ gives the quotient $Q(x)$ and remainder $R(x)$. Since $\\operatorname{deg} R(x)<\\operatorname{deg} G(x)$, for $x$ large enough $|R(x)|<$ $|G(x)|$; however, $R(x)$ is divisible by $G(x)$ for infinitely many integers $x$, so it is equal to zero infinitely often. Hence $R \\equiv 0$, and thus $F(x)$ is exactly divisible by $G(x)$. The polynomial $G(x)$ has a root $\\alpha$ in the interval $(0,1)$, because $G(0)=-1$ and $G(1)=1$. Then also $F(\\alpha)=0$, so that  $$ \\alpha^{m}+\\alpha=\\alpha^{n}+\\alpha^{2}=1 $$  If $m \\geq 2 n$, then $1-\\alpha=\\alpha^{m} \\leq\\left(\\alpha^{n}\\right)^{2}=\\left(1-\\alpha^{2}\\right)^{2}$, which is equivalent to $\\alpha(\\alpha-1)\\left(\\alpha^{2}+\\alpha-1\\right) \\geq 0$. But this last is not possible, because $\\alpha^{2}+\\alpha-1>$ $\\alpha^{m}+\\alpha-1=0$; hence $m<2 n$. Now we have $F(x) / G(x)=x^{m-n}-\\left(x^{m-n+2}-x^{m-n}-x+1\\right) / G(x)$, so $H(x)=x^{m-n+2}-x^{m-n}-x+1$ is also divisible by $G(x)$; but $\\operatorname{deg} H(x)=$ $m-n+2 \\leq n+1=\\operatorname{deg} G(x)+1$, from which we deduce that either $H(x)=G(x)$ or $H(x)=(x-a) G(x)$ for some $a \\in \\mathbb{Z}$. The former case is impossible. In the latter case we must have $m=2 n-1$, and thus $H(x)=x^{n+1}-x^{n-1}-x+1$; on the other hand, putting $x=1$ gives $a=1$, so $H(x)=(x-1)\\left(x^{n}+x^{2}-1\\right)=x^{n+1}-x^{n}+x^{3}-x^{2}-x+1$. This is possible only if $n=3$ and $m=5$. Remark. It is an old (though difficult) result that the polynomial $x^{n} \\pm$ $x^{k} \\pm 1$ is either irreducible or equals $x^{2} \\pm x+1$ times an irreducible factor.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "7", "problem_type": "Geometry", "exam": "IMO", "problem": "7. G1 (FRA) Let $B$ be a point on a circle $S_{1}$, and let $A$ be a point distinct from $B$ on the tangent at $B$ to $S_{1}$. Let $C$ be a point not on $S_{1}$ such that the line segment $A C$ meets $S_{1}$ at two distinct points. Let $S_{2}$ be the circle touching $A C$ at $C$ and touching $S_{1}$ at a point $D$ on the opposite side of $A C$ from $B$. Prove that the circumcenter of triangle $B C D$ lies on the circumcircle of triangle $A B C$.", "solution": "7. To avoid working with cases, we use oriented angles modulo $180^{\\circ}$. Let $K$ be the circumcenter of $\\triangle B C D$, and $X$ any point on the common tangent to the circles at $D$. Since the tangents at the ends of a chord are equally inclined to the chord, we have $\\angle B A C=\\angle A B D+\\angle B D C+\\angle D C A=$ $\\angle B D X+\\angle B D C+\\angle X D C=2 \\angle B D C=\\angle B K C$. It follows that $B, C, A, K$ are concyclic, as required.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "8", "problem_type": "Geometry", "exam": "IMO", "problem": "8. G2 (KOR) Let $A B C$ be a triangle for which there exists an interior point $F$ such that $\\angle A F B=\\angle B F C=\\angle C F A$. Let the lines $B F$ and $C F$ meet the sides $A C$ and $A B$ at $D$ and $E$ respectively. Prove that  $$ A B+A C \\geq 4 D E $$", "solution": "8. Construct equilateral triangles $A C P$ and $A B Q$ outside the triangle $A B C$. Since $\\angle A P C+\\angle A F C=60^{\\circ}+120^{\\circ}=180^{\\circ}$, the points $A, C, F, P$ lie on a circle; hence $\\angle A F P=\\angle A C P=60^{\\circ}=\\angle A F D$, so $D$ lies on the segment $F P$; similarly, $E$ lies on $F Q$. Further, note that  $$ \\frac{F P}{F D}=1+\\frac{D P}{F D}=1+\\frac{S_{A P C}}{S_{A F C}} \\geq 4 $$  with equality if $F$ is the midpoint of the smaller $\\operatorname{arc} A C$ : hence $F D \\leq \\frac{1}{4} F P$ and $F E \\leq \\frac{1}{4} F Q$. Then by the law of cosines,  $$ \\begin{aligned} D E & =\\sqrt{F D^{2}+F E^{2}+F D \\cdot F E} \\\\ & \\leq \\frac{1}{4} \\sqrt{F P^{2}+F Q^{2}+F P \\cdot F Q}=\\frac{1}{4} P Q \\leq A P+A Q=A B+A C . \\end{aligned} $$  Equality holds if and only if $\\triangle A B C$ is equilateral.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2002", "tier": "T0", "problem_label": "9", "problem_type": "Geometry", "exam": "IMO", "problem": "9. G3 (KOR) ${ }^{\\mathrm{IMO} 2}$ The circle $S$ has center $O$, and $B C$ is a diameter of $S$. Let $A$ be a point of $S$ such that $\\measuredangle A O B<120^{\\circ}$. Let $D$ be the midpoint of the arc $A B$ that does not contain $C$. The line through $O$ parallel to $D A$ meets the line $A C$ at $I$. The perpendicular bisector of $O A$ meets $S$ at $E$ and at $F$. Prove that $I$ is the incenter of the triangle $C E F$.", "solution": "9. Since $\\angle B C A=\\frac{1}{2} \\angle B O A=\\angle B O D$, the lines $C A$ and $O D$ are parallel, so that $O D A I$ is a parallelogram. It follows that $A I=O D=O E=A E=$ $A F$. Hence $\\angle I F E=\\angle I F A-\\angle E F A=\\angle A I F-\\angle E C A=\\angle A I F-\\angle A C F=\\angle C F I$. Also, from $A E=A F$ we get that $C I$ bisects $\\angle E C F$. Therefore $I$ is the incenter of $\\triangle C E F$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "1", "problem_type": "Algebra", "exam": "IMO", "problem": "1. A1 (USA) Let $a_{i j}, i=1,2,3, j=1,2,3$, be real numbers such that $a_{i j}$ is positive for $i=j$ and negative for $i \\neq j$. Prove that there exist positive real numbers $c_{1}, c_{2}, c_{3}$ such that the numbers  $$ a_{11} c_{1}+a_{12} c_{2}+a_{13} c_{3}, \\quad a_{21} c_{1}+a_{22} c_{2}+a_{23} c_{3}, \\quad a_{31} c_{1}+a_{32} c_{2}+a_{33} c_{3} $$  are all negative, all positive, or all zero.", "solution": "1. Consider the points $O(0,0,0), P\\left(a_{11}, a_{21}, a_{31}\\right), Q\\left(a_{12}, a_{22}, a_{32}\\right), R\\left(a_{13}, a_{23}\\right.$, $\\left.a_{33}\\right)$ in three-dimensional Euclidean space. It is enough to find a point $U\\left(u_{1}, u_{2}, u_{3}\\right)$ in the interior of the triangle $P Q R$ whose coordinates are all positive, all negative, or all zero (indeed, then we have $\\overrightarrow{O U}=c_{1} \\overrightarrow{O P}+$ $c_{2} \\overrightarrow{O Q}+c_{3} \\overrightarrow{O R}$ for some $c_{1}, c_{2}, c_{3}>0$ with $\\left.c_{1}+c_{2}+c_{3}=1\\right)$. Let $P^{\\prime}\\left(a_{11}, a_{21}, 0\\right), Q^{\\prime}\\left(a_{12}, a_{22}, 0\\right)$, and $R^{\\prime}\\left(a_{13}, a_{23}, 0\\right)$ be the projections of $P, Q$, and $R$ onto the $O x y$ plane. We see that $P^{\\prime}, Q^{\\prime}, R^{\\prime}$ lie in the fourth, second, and third quadrants, respectively. We have the following two cases: (i) $O$ is in the exterior of $\\triangle P^{\\prime} Q^{\\prime} R^{\\prime}$. Set $S^{\\prime}=O R^{\\prime} \\cap P^{\\prime} Q^{\\prime}$ and let $S$ be the point of the segment $P Q$ that projects to $S^{\\prime}$. The point $S$ has its $z$ coordinate negative (because the $z$ coordinates of $P$ and $Q$ are negative). Thus any point ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-710.jpg?height=290&width=446&top_left_y=702&top_left_x=875) of the segment $S R$ sufficiently close to $S$ has all coordinates negative. (ii) $O$ is in the interior or on the boundary of $\\triangle P^{\\prime} Q^{\\prime} R^{\\prime}$.  Let $T$ be the point in the plane $P Q R$ whose projection is $O$. If $T=O$, then all coordinates of $T$ are zero, and we are done. Otherwise $O$ is interior to $\\triangle P^{\\prime} Q^{\\prime} R^{\\prime}$. Suppose that the $z$ coordinate of $T$ is positive (negative). Since $x$ and $y$ coordinates of $T$ are equal to 0 , there is a point $U$ inside $P Q R$ close to $T$ with both $x$ and $y$ coordinates positive (respectively negative), and this point $U$ has all coordinates of the same sign.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "10", "problem_type": "Combinatorics", "exam": "IMO", "problem": "10. C4 (IRN) Let $x_{1}, \\ldots, x_{n}$ and $y_{1}, \\ldots, y_{n}$ be real numbers. Let $A=$ $\\left(a_{i j}\\right)_{1 \\leq i, j \\leq n}$ be the matrix with entries  $$ a_{i j}= \\begin{cases}1, & \\text { if } x_{i}+y_{j} \\geq 0 \\\\ 0, & \\text { if } x_{i}+y_{j}<0\\end{cases} $$  Suppose that $B$ is an $n \\times n$ matrix whose entries are 0,1 such that the sum of the elements in each row and each column of $B$ is equal to the corresponding sum for the matrix $A$. Prove that $A=B$.", "solution": "10. Denote by $b_{i j}$ the entries of the matrix $B$. Suppose the contrary, i.e., that there is a pair $\\left(i_{0}, j_{0}\\right)$ such that $a_{i_{0}, j_{0}} \\neq b_{i_{0}, j_{0}}$. We may assume without loss of generality that $a_{i_{0}, j_{0}}=0$ and $b_{i_{0}, j_{0}}=1$. Since the sums of elements in the $i_{0}$ th rows of the matrices $A$ and $B$ are equal, there is some $j_{1}$ for which $a_{i_{0}, j_{1}}=1$ and $b_{i_{0}, j_{1}}=0$. Similarly, from the fact that the sums in the $j_{1}$ th columns of the matrices $A$ and $B$ are equal, we conclude that there exists $i_{1}$ such that $a_{i_{1}, j_{1}}=0$ and $b_{i_{1}, j_{1}}=1$. Continuing this procedure, we construct two sequences $i_{k}, j_{k}$ such that $a_{i_{k}, j_{k}}=0, b_{i_{k}, j_{k}}=1, a_{i_{k}, j_{k+1}}=1, b_{i_{k}, j_{k+1}}=0$. Since the set of the pairs $\\left(i_{k}, j_{k}\\right)$ is finite, there are two different numbers $t, s$ such that $\\left(i_{t}, j_{t}\\right)=\\left(i_{s}, j_{s}\\right)$. From the given condition we have that $x_{i_{k}}+y_{i_{k}}<0$ and $x_{i_{k+1}}+y_{i_{k+1}} \\geq 0$. But $j_{t}=j_{s}$, and hence $0 \\leq \\sum_{k=s}^{t-1}\\left(x_{i_{k}}+y_{j_{k+1}}\\right)=$ $\\sum_{k=s}^{t-1}\\left(x_{i_{k}}+y_{j_{k}}\\right)<0$, a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "11", "problem_type": "Combinatorics", "exam": "IMO", "problem": "11. C5 (ROM) Every point with integer coordinates in the plane is the center of a disk with radius $1 / 1000$. (a) Prove that there exists an equilateral triangle whose vertices lie in different disks. (b) Prove that every equilateral triangle with vertices in different disks has side length greater than 96.", "solution": "11. (a) By the pigeonhole principle there are two different integers $x_{1}, x_{2}$, $x_{1}>x_{2}$, such that $\\left|\\left\\{x_{1} \\sqrt{3}\\right\\}-\\left\\{x_{2} \\sqrt{3}\\right\\}\\right|<0.001$. Set $a=x_{1}-x_{2}$. Consider the equilateral triangle with vertices $(0,0),(2 a, 0),(a, a \\sqrt{3})$. The points $(0,0)$ and $(2 a, 0)$ are lattice points, and we claim that the point $(a, a \\sqrt{3})$ is at distance less than 0.001 from a lattice point. Indeed, since $0.001>\\left|\\left\\{x_{1} \\sqrt{3}\\right\\}-\\left\\{x_{2} \\sqrt{3}\\right\\}\\right|=\\left|a \\sqrt{3}-\\left(\\left[x_{1} \\sqrt{3}\\right]-\\left[x_{2} \\sqrt{3}\\right]\\right)\\right|$, we see that the distance between the points $(a, a \\sqrt{3})$ and $\\left(a,\\left[x_{1} \\sqrt{3}\\right]-\\right.$ $\\left.\\left[x_{2} \\sqrt{3}\\right]\\right)$ is less than 0.001 , and the point $\\left(a,\\left[x_{1} \\sqrt{3}\\right]-\\left[x_{2} \\sqrt{3}\\right]\\right)$ is with integer coefficients. (b) Suppose that $P^{\\prime} Q^{\\prime} R^{\\prime}$ is an equilateral triangle with side length $l \\leq 96$ such that each of its vertices $P^{\\prime}, Q^{\\prime}, R^{\\prime}$ lies in a disk of radius 0.001 centered at a lattice point. Denote by $P, Q, R$ the centers of these disks. Then we have $l-0.002 \\leq P Q, Q R, R P \\leq l+0.002$. Since $P Q R$ is not an equilateral triangle, two of its sides are different, say $P Q \\neq Q R$. On the other hand, $P Q^{2}, Q R^{2}$ are integers, so we have $1 \\leq\\left|P Q^{2}-Q R^{2}\\right|=(P Q+Q R)|P Q-Q R| \\leq 0.004(P Q+Q R) \\leq$ $(2 l+0.004) \\cdot 0.004 \\leq 2 \\cdot 96.002 \\cdot 0.004<1$, which is a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "12", "problem_type": "Combinatorics", "exam": "IMO", "problem": "12. C6 (SAF) Let $f(k)$ be the number of integers $n$ that satisfy the following conditions: (i) $0 \\leq n<10^{k}$, so $n$ has exactly $k$ digits (in decimal notation), with leading zeros allowed; (ii) the digits of $n$ can be permuted in such a way that they yield an integer divisible by 11 . Prove that $f(2 m)=10 f(2 m-1)$ for every positive integer $m$.", "solution": "12. Denote by $\\overline{a_{k-1} a_{k-2} \\ldots a_{0}}$ the decimal representation of a number whose digits are $a_{k-1}, \\ldots, a_{0}$. We will use the following well-known fact:  $$ \\overline{a_{k-1} a_{k-2} \\ldots a_{0}} \\equiv i(\\bmod 11) \\Longleftrightarrow \\sum_{l=0}^{k-1}(-1)^{l} a_{l} \\equiv i(\\bmod 11) . $$  Let $m$ be a positive integer. Define $A$ as the set of integers $n(0 \\leq n<$ $10^{2 m}$ ) whose right $2 m-1$ digits can be so permuted to yield an integer divisible by 11 , and $B$ as the set of integers $n\\left(0 \\leq n<10^{2 m-1}\\right)$ whose digits can be permuted resulting in an integer divisible by 11. Suppose that $a=\\overline{a_{2 m-1} \\ldots a_{0}} \\in A$. Then there that satisfies  $$ \\sum_{l=0}^{2 m-1}(-1)^{l} b_{l} \\equiv 0(\\bmod 11) $$  The $2 m$-tuple $\\left(b_{2 m-1}, \\ldots, b_{0}\\right)$ satisfies (1) if and only if the $2 m$-tuple $\\left(k b_{2 m-1}+l, \\ldots, k b_{0}+l\\right)$ satisfies ( 1 ), where $k, l \\in \\mathbb{Z}, 11 \\nmid k$. Since $a_{0}+1 \\not \\equiv 0(\\bmod 11)$, we can choose $k$ from the set $\\{1, \\ldots, 10\\}$ such that $\\left(a_{0}+1\\right) k \\equiv 1(\\bmod 11)$. Thus there is a permutation of the $2 m$-tuple $\\left(\\left(a_{2 m-1}+1\\right) k-1, \\ldots,\\left(a_{1}+1\\right) k-1,0\\right)$ satisfying (1). Interchanging odd and even positions if necessary, we may assume that this permutation keeps the 0 at the last position. Since $\\left(a_{i}+1\\right) k$ is not divisible by 11 for any $i$, there is a unique $b_{i} \\in\\{0,1, \\ldots, 9\\}$ such that $b_{i} \\equiv\\left(a_{i}+1\\right) k-1(\\bmod 11)$. Hence the number $\\overline{b_{2 m-1} \\ldots b_{1}}$ belongs to $B$. Thus for fixed $a_{0} \\in\\{0,1,2, \\ldots, 9\\}$, to each $a \\in A$ such that the last digit of $a$ is $a_{0}$ we associate a unique $b \\in B$. Conversely, having $a_{0} \\in$ $\\{0,1,2, \\ldots, 9\\}$ fixed, from any number $\\overline{b_{2 m-1} \\ldots b_{1}} \\in B$ we can reconstruct $\\overline{a_{2 m-1} \\ldots a_{1} a_{0}} \\in A$. Hence $|A|=10|B|$, i.e., $f(2 m)=10 f(2 m-1)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "13", "problem_type": "Geometry", "exam": "IMO", "problem": "13. G1 (FIN) ${ }^{\\mathrm{IMO} 4}$ Let $A B C D$ be a cyclic quadrilateral. Let $P, Q, R$ be the feet of the perpendiculars from $D$ to the lines $B C, C A, A B$, respectively. Show that $P Q=Q R$ if and only if the bisectors of $\\angle A B C$ and $\\angle A D C$ are concurrent with $A C$.", "solution": "13. Denote by $K$ and $L$ the intersections of the bisectors of $\\angle A B C$ and $\\angle A D C$ with the line $A C$, respectively. Since $A B: B C=A K: K C$ and $A D: D C=A L: L C$, we have to prove that  $$ P Q=Q R \\Leftrightarrow \\frac{A B}{B C}=\\frac{A D}{D C} . $$  Since the quadrilaterals $A Q D R$ and $Q P C D$ are cyclic, we see that ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-715.jpg?height=403&width=399&top_left_y=1567&top_left_x=862) $\\angle R D Q=\\angle B A C$ and $\\angle Q D P=\\angle A C B$. By the law of sines it follows that $\\frac{A B}{B C}=\\frac{\\sin (\\angle A C B)}{\\sin (\\angle B A C)}$ and that $Q R=A D \\sin (\\angle R D Q), Q P=$ $C D \\sin (\\angle Q D P)$. Now we have  $$ \\frac{A B}{B C}=\\frac{\\sin (\\angle A C B)}{\\sin (\\angle B A C)}=\\frac{\\sin (\\angle Q D P)}{\\sin (\\angle R D Q)}=\\frac{A D \\cdot Q P}{Q R \\cdot C D} $$  The statement (1) follows directly.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "14", "problem_type": "Geometry", "exam": "IMO", "problem": "14. G2 (GRE) Three distinct points $A, B, C$ are fixed on a line in this order. Let $\\Gamma$ be a circle passing through $A$ and $C$ whose center does not lie on the line $A C$. Denote by $P$ the intersection of the tangents to $\\Gamma$ at $A$ and $C$. Suppose $\\Gamma$ meets the segment $P B$ at $Q$. Prove that the intersection of the bisector of $\\angle A Q C$ and the line $A C$ does not depend on the choice of $\\Gamma$.", "solution": "14. Denote by $R$ the intersection point of the bisector of $\\angle A Q C$ and the line $A C$. From $\\triangle A C Q$ we get  $$ \\frac{A R}{R C}=\\frac{A Q}{Q C}=\\frac{\\sin \\angle Q C A}{\\sin \\angle Q A C} $$  By the sine version of Ceva's theorem we have $\\frac{\\sin \\angle A P B}{\\sin \\angle B P C} \\cdot \\frac{\\sin \\angle Q A C}{\\sin \\angle P A Q}$. $\\frac{\\sin \\angle Q C P}{\\sin \\angle Q C A}=1$, which is equivalent to  $$ \\frac{\\sin \\angle A P B}{\\sin \\angle B P C}=\\left(\\frac{\\sin \\angle Q C A}{\\sin \\angle Q A C}\\right)^{2} $$  because $\\angle Q C A=\\angle P A Q$ and $\\angle Q A C=\\angle Q C P$. Denote by $S(X Y Z)$ the area of a triangle $X Y Z$. Then  $$ \\frac{\\sin \\angle A P B}{\\sin \\angle B P C}=\\frac{A P \\cdot B P \\cdot \\sin \\angle A P B}{B P \\cdot C P \\cdot \\sin \\angle B P C}=\\frac{S(\\Delta A B P)}{S(\\Delta B C P)}=\\frac{A B}{B C}, $$  which implies that $\\left(\\frac{A R}{R C}\\right)^{2}=\\frac{A B}{B C}$. Hence $R$ does not depend on $\\Gamma$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "15", "problem_type": "Geometry", "exam": "IMO", "problem": "15. G3 (IND) Let $A B C$ be a triangle and let $P$ be a point in its interior. Denote by $D, E, F$ the feet of the perpendiculars from $P$ to the lines $B C$, $C A$, and $A B$, respectively. Suppose that  $$ A P^{2}+P D^{2}=B P^{2}+P E^{2}=C P^{2}+P F^{2} $$  Denote by $I_{A}, I_{B}, I_{C}$ the excenters of the triangle $A B C$. Prove that $P$ is the circumcenter of the triangle $I_{A} I_{B} I_{C}$.", "solution": "15. From the given equality we see that $0=\\left(B P^{2}+P E^{2}\\right)-\\left(C P^{2}+P F^{2}\\right)=$ $B F^{2}-C E^{2}$, so $B F=C E=x$ for some $x$. Similarly, there are $y$ and $z$ such that $C D=A F=y$ and $B D=A E=z$. It is easy to verify that $D$, $E$, and $F$ must lie on the segments $B C, C A, A B$. Denote by $a, b, c$ the length of the segments $B C, C A, A B$. It follows that $a=z+y, b=z+x, c=x+y$, so $D, E, F$ are the points where the excircles touch the sides of $\\triangle A B C$. Hence $P, D$, and $I_{A}$ are collinear and  $$ \\angle P I_{A} C=\\angle D I_{A} C=90^{\\circ}-\\frac{180^{\\circ}-\\angle A C B}{2}=\\frac{\\angle A C B}{2} $$  In the same way we obtain that $\\angle P I_{B} C=\\frac{\\angle A C B}{2}$ and $P I_{B}=P I_{A}$. Analogously, we get $P I_{C}=P I_{B}$, which implies that $P$ is the circumcenter of the triangle $I_{A} I_{B} I_{C}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "16", "problem_type": "Geometry", "exam": "IMO", "problem": "16. G4 (ARM) Let $\\Gamma_{1}, \\Gamma_{2}, \\Gamma_{3}, \\Gamma_{4}$ be distinct circles such that $\\Gamma_{1}, \\Gamma_{3}$ are externally tangent at $P$, and $\\Gamma_{2}, \\Gamma_{4}$ are externally tangent at the same point $P$. Suppose that $\\Gamma_{1}$ and $\\Gamma_{2} ; \\Gamma_{2}$ and $\\Gamma_{3} ; \\Gamma_{3}$ and $\\Gamma_{4} ; \\Gamma_{4}$ and $\\Gamma_{1}$ meet at $A, B, C, D$, respectively, and that all these points are different from $P$. Prove that  $$ \\frac{A B \\cdot B C}{A D \\cdot D C}=\\frac{P B^{2}}{P D^{2}} $$", "solution": "16. Apply an inversion with center at $P$ and radius $r$; let $\\widehat{X}$ denote the image of $X$. The circles $\\Gamma_{1}, \\Gamma_{2}, \\Gamma_{3}, \\Gamma_{4}$ are transformed into lines $\\widehat{\\Gamma_{1}}, \\widehat{\\Gamma_{2}}, \\widehat{\\Gamma_{3}}, \\widehat{\\Gamma}_{4}$, where $\\widehat{\\Gamma_{1}} \\| \\widehat{\\Gamma_{3}}$ and $\\widehat{\\Gamma_{2}} \\| \\widehat{\\Gamma_{4}}$, and therefore $\\widehat{A} \\widehat{B} \\widehat{C} \\widehat{D}$ is a parallelogram. Further, we have $A B=\\frac{r^{2}}{P \\widehat{A} \\cdot P \\widehat{B}} \\widehat{A} \\widehat{B}, B C=\\frac{r^{2}}{P \\widehat{B} \\cdot P \\widehat{C}} \\widehat{B} \\widehat{C}, C D=\\frac{r^{2}}{P \\widehat{C} \\cdot P \\widehat{D}} \\widehat{C} \\widehat{D}$, $D A=\\frac{r^{2}}{P \\widehat{D} \\cdot P \\widehat{A}} \\widehat{D} \\widehat{A}$ and $P B=\\frac{r^{2}}{P \\widehat{B}}, P D=\\frac{r^{2}}{P \\widehat{D}}$. The equality to be proven becomes  $$ \\frac{P \\widehat{D}^{2}}{P \\widehat{B}^{2}} \\cdot \\frac{\\widehat{A} \\widehat{B} \\cdot \\widehat{B} \\widehat{C}}{\\widehat{A} \\cdot \\widehat{D} \\widehat{C}}=\\frac{P \\widehat{D}^{2}}{P \\widehat{B}^{2}} $$  which holds because $\\widehat{A} \\widehat{B}=\\widehat{C} \\widehat{D}$ and $\\widehat{B} \\widehat{C}=\\widehat{D} \\widehat{A}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "17", "problem_type": "Geometry", "exam": "IMO", "problem": "17. G5 (KOR) Let $A B C$ be an isosceles triangle with $A C=B C$, whose incenter is $I$. Let $P$ be a point on the circumcircle of the triangle $A I B$ lying inside the triangle $A B C$. The lines through $P$ parallel to $C A$ and $C B$ meet $A B$ at $D$ and $E$, respectively. The line through $P$ parallel to $A B$ meets $C A$ and $C B$ at $F$ and $G$, respectively. Prove that the lines $D F$ and $E G$ intersect on the circumcircle of the triangle $A B C$.", "solution": "17. The triangles $P D E$ and $C F G$ are homothetic; hence lines $F D, G E$, and $C P$ intersect at one point. Let $Q$ be the intersection point of the line $C P$ and the circumcircle of $\\triangle A B C$. The required statement will follow if we show that $Q$ lies on the lines $G E$ and $F D$. Since $\\angle C F G=\\angle C B A=\\angle C Q A$, the quadrilateral $A Q P F$ is cyclic. Analogously, $B Q P G$ is cyclic. However, the isosceles trapezoid $B D P G$ is also cyclic; it follows that $B, Q, D, P, G$ lie on a circle. Therefore we get  $$ \\angle P Q F=\\angle P A C, \\angle P Q D=\\angle P B A . $$  Since $I$ is the incenter of $\\triangle A B C$, we have $\\angle C A I=\\frac{1}{2} \\angle C A B=$ $\\frac{1}{2} \\angle C B A=\\angle I B A$; hence $C A$ is the tangent at $A$ to the circumcircle of $\\triangle A B I$. This implies that $\\angle P A C=$ $\\angle P B A$, and it follows from (1) that $\\angle P Q F=\\angle P Q D$, i.e., that $F, D, Q$ are also collinear. Similarly, $G, E, Q$ are collinear and the claim is thus proved. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-717.jpg?height=412&width=358&top_left_y=632&top_left_x=903)", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "18", "problem_type": "Geometry", "exam": "IMO", "problem": "18. G6 (POL) ${ }^{\\mathrm{IMO} 3}$ Each pair of opposite sides of a convex hexagon has the following property: The distance between their midpoints is equal to $\\sqrt{3} / 2$ times the sum of their lengths. Prove that all the angles of the hexagon are equal.", "solution": "18. Let $A B C D E F$ be the given hexagon. We shall use the following lemma. Lemma. If $\\angle X Z Y \\geq 60^{\\circ}$ and if $M$ is the midpoint of $X Y$, then $M Z \\leq$ $\\frac{\\sqrt{3}}{2} X Y$, with equality if and only if $\\triangle X Y Z$ is equilateral. Proof. Let $Z^{\\prime}$ be the point such that $\\triangle X Y Z^{\\prime}$ is equilateral. Then $Z$ is inside the circle circumscribed about $\\triangle X Y Z^{\\prime}$. Consequently $M Z \\leq$ $M Z^{\\prime}=\\frac{\\sqrt{3}}{2} X Y$, with equality if and only if $Z=Z^{\\prime}$. Set $A D \\cap B E=P, B E \\cap C F=Q$, and $C F \\cap A D=R$. Suppose $\\angle A P B=$ $\\angle D P E>60^{\\circ}$, and let $K, L$ be the midpoints of the segments $A B$ and $D E$ respectively. Then by the lemma,  $$ \\frac{\\sqrt{3}}{2}(A B+D E)=K L \\leq P K+P L<\\frac{\\sqrt{3}}{2}(A B+D E), $$  which is impossible. Therefore $\\angle A P B \\leq 60^{\\circ}$ and similarly $\\angle B Q C \\leq 60^{\\circ}$, $\\angle C R D \\leq 60^{\\circ}$. But the sum of the angles $A P B, B Q C, C R D$ is $180^{\\circ}$, from which we conclude that these angles are all equal to $60^{\\circ}$, and moreover that the triangles $A P B, B Q C, C R D$ are equilateral. Thus $\\angle A B C=\\angle A B P+$ $\\angle Q B C=120^{\\circ}$, and in the same way all angles of the hexagon are equal to $120^{\\circ}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "19", "problem_type": "Geometry", "exam": "IMO", "problem": "19. G7 (SAF) Let $A B C$ be a triangle with semiperimeter $s$ and inradius $r$. The semicircles with diameters $B C, C A, A B$ are drawn outside of the triangle $A B C$. The circle tangent to all three semicircles has radius $t$. Prove that  $$ \\frac{s}{2}<t \\leq \\frac{s}{2}+\\left(1-\\frac{\\sqrt{3}}{2}\\right) r $$", "solution": "19. Let $D, E, F$ be the midpoints of $B C, C A, A B$, respectively. We construct smaller semicircles $\\Gamma_{d}, \\Gamma_{e}, \\Gamma_{f}$ inside $\\triangle A B C$ with centers $D, E, F$ and radii $d=\\frac{s-a}{2}, e=\\frac{s-b}{2}, f=\\frac{s-c}{2}$ respectively. Since $D E=d+e, D F=d+f$, and $E F=e+f$, we deduce that $\\Gamma_{d}, \\Gamma_{e}$, and $\\Gamma_{f}$ touch each other at the points $D_{1}, E_{1}, F_{1}$ of tangency of the incircle $\\gamma$ of $\\triangle D E F$ with its sides ( $D_{1} \\in E F$, etc.). Consider the circle $\\Gamma_{g}$ with center $O$ and radius $g$ that lies inside $\\triangle D E F$ and tangents $\\Gamma_{d}, \\Gamma_{e}, \\Gamma_{f}$.  Now let $O D, O E, O F$ meet the semicircles $\\Gamma_{d}, \\Gamma_{e}, \\Gamma_{f}$ at $D^{\\prime}, E^{\\prime}, F^{\\prime}$ respectively. We have $O D^{\\prime}=O D+$ $D D^{\\prime}=g+d+\\frac{a}{2}=g+\\frac{s}{2}$ and similarly $O E^{\\prime}=O F^{\\prime}=g+\\frac{s}{2}$. It follows that the circle with center $O$ and radius $g+\\frac{s}{2}$ touches all three semicircles, and consequently $t=$ $g+\\frac{s}{2}>\\frac{s}{2}$. Now set the coordinate system such that we have the points $D_{1}(0,0), E(-e, 0), F(f, 0)$ and such that the $y$ coordinate of $D$ is positive. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-718.jpg?height=528&width=529&top_left_y=233&top_left_x=831)  Apply the inversion with center $D_{1}$ and unit radius. This inversion maps the circles $\\Gamma_{e}$ and $\\Gamma_{f}$ to the lines $\\widehat{\\Gamma_{e}}\\left[x=-\\frac{1}{2 e}\\right]$ and $\\widehat{\\Gamma_{e}}\\left[x=\\frac{1}{2 f}\\right]$ respectively, and the circle $\\gamma$ goes to the line $\\widehat{\\gamma}\\left[y=\\frac{1}{r}\\right]$. The images $\\widehat{\\Gamma_{d}}$ and $\\widehat{\\Gamma_{g}}$ of $\\Gamma_{d}, \\Gamma_{g}$ are the circles that touch the lines $\\widehat{\\Gamma_{e}}$ and $\\widehat{\\Gamma_{f}}$. Since $\\widehat{\\Gamma_{d}}, \\widehat{\\Gamma_{g}}$ are perpendicular to $\\gamma$, they have radii equal to $R=\\frac{1}{4 e}+\\frac{1}{4 f}$ and centers at $\\left(-\\frac{1}{4 e}+\\frac{1}{4 f}, \\frac{1}{r}\\right)$ and $\\left(-\\frac{1}{4 e}+\\frac{1}{4 f}, \\frac{1}{r}+2 R\\right)$ respectively. Let $p$ and $P$ be the distances from $D_{1}(0,0)$ to the centers of $\\Gamma_{g}$ and $\\widehat{\\Gamma_{g}}$ respectively. We have that $P^{2}=\\left(\\frac{1}{4 e}-\\frac{1}{4 f}\\right)^{2}+\\left(\\frac{1}{r}+2 R\\right)^{2}$, and that the circles $\\Gamma_{g}$ and $\\widehat{\\Gamma_{g}}$ are homothetic with center of homothety $D_{1}$; hence $p / P=g / R$. On the other hand, $\\widehat{\\Gamma_{g}}$ is the image of $\\Gamma_{g}$ under inversion; hence the product of the tangents from $D_{1}$ to these two circles is equal to 1 . In other words, we obtain $\\sqrt{p^{2}-g^{2}} \\cdot \\sqrt{P^{2}-R^{2}}=1$. Using the relation $p / P=g / R$ we get $g=\\frac{R}{P^{2}-R^{2}}$. The inequality we have to prove is equivalent to $(4+2 \\sqrt{3}) g \\leq r$. This can be proved as follows:  $$ \\begin{aligned} r-(4+2 \\sqrt{3}) g & =\\frac{r\\left(P^{2}-R^{2}-(4+2 \\sqrt{3}) R / r\\right)}{P^{2}-R^{2}} \\\\ & =\\frac{r\\left(\\left(\\frac{1}{r}+2 R\\right)^{2}+\\left(\\frac{1}{4 e}-\\frac{1}{4 f}\\right)^{2}-R^{2}-(4+2 \\sqrt{3}) \\frac{R}{r}\\right)}{P^{2}-R^{2}} \\\\ & =\\frac{r}{P^{2}-R^{2}}\\left(\\left(R \\sqrt{3}-\\frac{1}{r}\\right)^{2}+\\left(\\frac{1}{4 e}-\\frac{1}{4 f}\\right)^{2}\\right) \\geq 0 \\end{aligned} $$  Remark. One can obtain a symmetric formula for $g$ :  $$ \\frac{1}{2 g}=\\frac{1}{s-a}+\\frac{1}{s-b}+\\frac{1}{s-c}+\\frac{2}{r} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "2", "problem_type": "Algebra", "exam": "IMO", "problem": "2. A2 (AUS) Find all nondecreasing functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that (i) $f(0)=0, f(1)=1$; (ii) $f(a)+f(b)=f(a) f(b)+f(a+b-a b)$ for all real numbers $a, b$ such that $a<1<b$.", "solution": "2. We can rewrite (ii) as $-(f(a)-1)(f(b)-1)=f(-(a-1)(b-1)+1)-1$. So putting $g(x)=f(x+1)-1$, this equation becomes $-g(a-1) g(b-1)=$ $g(-(a-1)(b-1))$ for $a<1<b$. Hence  $$ -g(x) g(y)=g(-x y) \\text { for } x<0<y $$  and $g$ is nondecreasing with $g(-1)=-1, g(0)=0$. Conversely, if $g$ satisfies (1), than $f$ is a solution of our problem. Setting $y=1$ in (1) gives $-g(-x) g(1)=g(x)$ for each $x>0$, and therefore (1) reduces to $g(1) g(y z)=g(y) g(z)$ for all $y, z>0$. We have two cases: (i) $g(1)=0$. By (1) we have $g(z)=0$ for all $z>0$. Then any nondecreasing function $g: \\mathbb{R} \\rightarrow \\mathbb{R}$ with $g(-1)=-1$ and $g(z)=0$ for $z \\geq 0$ satisfies (1) and gives a solution: $f$ is nondecreasing, $f(0)=0$ and $f(x)=1$ for every $x \\geq 1$ (ii) $g(1) \\neq 0$. Then the function $h(x)=\\frac{g(x)}{g(1)}$ is nondecreasing and satisfies $h(0)=0, h(1)=1$, and $h(x y)=h(x) h(y)$. Fix $a>0$, and let $h(a)=$ $b=a^{k}$ for some $k \\in \\mathbb{R}$. It follows by induction that $h\\left(a^{q}\\right)=h(a)^{q}=$ $\\left(a^{q}\\right)^{k}$ for every rational number $q$. But $h$ is nondecreasing, so $k \\geq 0$, and since the set $\\left\\{a^{q} \\mid q \\in \\mathbb{Q}\\right\\}$ is dense in $\\mathbb{R}^{+}$, we conclude that $h(x)=x^{k}$ for every $x>0$. Finally, putting $g(1)=c$, we obtain $g(x)=c x^{k}$ for all $x>0$. Then $g(-x)=-x^{k}$ for all $x>0$. This $g$ obviously satisfies (1). Hence  $$ f(x)=\\left\\{\\begin{array}{ll} c(x-1)^{k}, & \\text { if } x>1 ; \\\\ 1, & \\text { if } x=1 ; \\\\ 1-(1-x)^{k}, & \\text { if } x<1 \\end{array} \\quad \\text { where } c>0 \\text { and } k \\geq 0\\right. $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "20", "problem_type": "Number Theory", "exam": "IMO", "problem": "20. N1 (POL) Let $m$ be a fixed integer greater than 1 . The sequence $x_{0}, x_{1}, x_{2}, \\ldots$ is defined as follows:  $$ x_{i}= \\begin{cases}2^{i}, & \\text { if } 0 \\leq i \\leq m-1 \\\\ \\sum_{j=1}^{m} x_{i-j}, & \\text { if } i \\geq m\\end{cases} $$  Find the greatest $k$ for which the sequence contains $k$ consecutive terms divisible by $m$.", "solution": "20. Let $r_{i}$ be the remainder when $x_{i}$ is divided by $m$. Since there are at most $m^{m}$ types of $m$-consecutive blocks in the sequence $\\left(r_{i}\\right)$, some type will repeat at least twice. Then since the entire sequence is determined by one $m$-consecutive block, the entire sequence will be periodic. The formula works both forward and backward; hence using the rule $x_{i}=$ $x_{i+m}-\\sum_{j=1}^{m-1} x_{i+j}$ we can define $x_{-1}, x_{-2}, \\ldots$. Thus we obtain that  $$ \\left(r_{-m}, \\ldots, r_{-1}\\right)=(0,0, \\ldots, 0,1) $$  Hence there are $m-1$ consecutive terms in the sequence $\\left(x_{i}\\right)$ that are divisible by $m$. If there were $m$ consecutive terms in the sequence $\\left(x_{i}\\right)$ divisible by $m$, then by the recurrence relation all the terms of $\\left(x_{i}\\right)$ would be divisible by $m$, which is impossible.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "21", "problem_type": "Number Theory", "exam": "IMO", "problem": "21. N2 (USA) Each positive integer $a$ undergoes the following procedure in order to obtain the number $d=d(a)$ : (1) move the last digit of $a$ to the first position to obtain the number $b$; (2) square $b$ to obtain the number $c$; (3) move the first digit of $c$ to the end to obtain the number $d$. (All the numbers in the problem are considered to be represented in base 10.) For example, for $a=2003$, we have $b=3200, c=10240000$, and $d=02400001=2400001=d(2003)$. Find all numbers $a$ for which $d(a)=a^{2}$.", "solution": "21. Let $a$ be a positive integer for which $d(a)=a^{2}$. Suppose that $a$ has $n+1$ digits, $n \\geq 0$. Denote by $s$ the last digit of $a$ and by $f$ the first digit of $c$. Then $a=\\overline{* \\ldots * s}$, where $*$ stands for a digit that is not important to us at the moment. We have $\\overline{\\ldots * s^{2}}=a^{2}=d=\\overline{* \\ldots * f}$ and $b^{2}=\\overline{s * \\ldots *}^{2}=$ $c=\\overline{f * \\ldots *}$. We cannot have $s=0$, since otherwise $c$ would have at most $2 n$ digits, while $a^{2}$ has either $2 n+1$ or $2 n+2$ digits. The following table gives all possibilities for $s$ and $f$ :  | $s$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | $f=$ last digit of $\\overline{* \\ldots * s}^{2}$ | 1 | 4 | 9 | 6 | 5 | 6 | 9 | 4 | 1 | | $f=$ first digit of $\\overline{s * \\ldots *}^{2}$ | $1,2,3$ | $4-8$ | 9,1 | 1,2 | 2,3 | 3,4 | $4,5,6$ | $6,7,8$ | 8,9 |  We obtain from the table that $s \\in\\{1,2,3\\}$ and $f=s^{2}$, and consequently $c=b^{2}$ and $d$ have exactly $2 n+1$ digits each. Put $a=10 x+s$, where $x<10^{n}$. Then $b=10^{n} s+x, c=10^{2 n} s^{2}+2 \\cdot 10^{n} s x+x^{2}$, and $d=$ $2 \\cdot 10^{n+1} s x+10 x^{2}+s^{2}$, so from $d=a^{2}$ it follows that $x=2 s \\cdot \\frac{10^{n}-1}{9}$. Thus $a=\\underbrace{6 \\ldots 6}_{n} 3, a=\\underbrace{4 \\ldots 4}_{n} 2$ or $a=\\underbrace{2 \\ldots 2}_{n} 1$. For $n \\geq 1$ we see that $a$ cannot be $a=6 \\ldots 63$ or $a=4 \\ldots 42$ (otherwise $a^{2}$ would have $2 n+2$ digits). Therefore $a$ equals $1,2,3$ or $\\underbrace{2 \\ldots 2}_{n} 1$ for $n \\geq 0$. It is easy to verify that these numbers have the required property.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "22", "problem_type": "Number Theory", "exam": "IMO", "problem": "22. N3 (BUL) ${ }^{\\mathrm{IMO} 2}$ Determine all pairs $(a, b)$ of positive integers such that  $$ \\frac{a^{2}}{2 a b^{2}-b^{3}+1} $$  is a positive integer.", "solution": "22. Let $a$ and $b$ be positive integers for which $\\frac{a^{2}}{2 a b^{2}-b^{3}+1}=k$ is a positive integer. Since $k>0$, it follows that $2 a b^{2} \\geq b^{3}$, so $2 a \\geq b$. If $2 a>b$, then from $2 a b^{2}-b^{3}+1>0$ we see that $a^{2}>b^{2}(2 a-b)+1>b^{2}$, i.e. $a>b$. Therefore, if $a \\leq b$, then $a=b / 2$. We can rewrite the given equation as a quadratic equation in $a, a^{2}-$ $2 k b^{2} a+k\\left(b^{3}-1\\right)=0$, which has two solutions, say $a_{1}$ and $a_{2}$, one of which is in $\\mathbb{N}_{0}$. From $a_{1}+a_{2}=2 k b^{2}$ and $a_{1} a_{2}=k\\left(b^{3}-1\\right)$ it follows that the other solution is also in $\\mathbb{N}_{0}$. Suppose w.l.o.g. that $a_{1} \\geq a_{2}$. Then $a_{1} \\geq k b^{2}$ and  $$ 0 \\leq a_{2}=\\frac{k\\left(b^{3}-1\\right)}{a_{1}} \\leq \\frac{k\\left(b^{3}-1\\right)}{k b^{2}}<b . $$  By the above considerations we have either $a_{2}=0$ or $a_{2}=b / 2$. If $a_{2}=0$, then $b^{3}-1=0$ and hence $a_{1}=2 k, b=1$. If $a_{2}=b / 2$, then $b=2 t$ for some $t$, and $k=b^{2} / 4, a_{1}=b^{4} / 2-b / 2$. Therefore the only solutions are  $$ (a, b) \\in\\left\\{(2 t, 1),(t, 2 t),\\left(8 t^{4}-t, 2 t\\right) \\mid t \\in \\mathbb{N}\\right\\} $$  It is easy to show that all of these pairs satisfy the given condition.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "23", "problem_type": "Number Theory", "exam": "IMO", "problem": "23. N4 (ROM) Let $b$ be an integer greater than 5 . For each positive integer $n$, consider the number  $$ x_{n}=\\underbrace{11 \\ldots 1}_{n-1} \\underbrace{22 \\ldots 2}_{n} 5, $$  written in base $b$. Prove that the following condition holds if and only if $b=10$ : There exists a positive integer $M$ such that for every integer $n$ greater than $M$, the number $x_{n}$ is a perfect square.", "solution": "23. Assume that $b \\geq 6$ has the required property. Consider the sequence $y_{n}=(b-1) x_{n}$. From the definition of $x_{n}$ we easily find that $y_{n}=b^{2 n}+$ $b^{n+1}+3 b-5$. Then $y_{n} y_{n+1}=(b-1)^{2} x_{n} x_{n+1}$ is a perfect square for all $n>M$. Also, straightforward calculation implies  $$ \\left(b^{2 n+1}+\\frac{b^{n+2}+b^{n+1}}{2}-b^{3}\\right)^{2}<y_{n} y_{n+1}<\\left(b^{2 n+1}+\\frac{b^{n+2}+b^{n+1}}{2}+b^{3}\\right)^{2} $$  Hence for every $n>M$ there is an integer $a_{n}$ such that $\\left|a_{n}\\right|<b^{3}$ and  $$ \\begin{aligned} y_{n} y_{n+1} & =\\left(b^{2 n}+b^{n+1}+3 b-5\\right)\\left(b^{2 n+2}+b^{n+2}+3 b-5\\right) \\\\ & =\\left(b^{2 n+1}+\\frac{b^{n+1}(b+1)}{2}+a_{n}\\right)^{2} . \\end{aligned} $$  Now considering this equation modulo $b^{n}$ we obtain $(3 b-5)^{2} \\equiv a_{n}^{2}$, so that assuming that $n>3$ we get $a_{n}= \\pm(3 b-5)$. If $a_{n}=3 b-5$, then substituting in (1) yields $\\frac{1}{4} b^{2 n}\\left(b^{4}-14 b^{3}+45 b^{2}-\\right.$ $52 b+20)=0$, with the unique positive integer solution $b=10$. Also, if $a_{n}=-3 b+5$, we similarly obtain $\\frac{1}{4} b^{2 n}\\left(b^{4}-14 b^{3}-3 b^{2}+28 b+20\\right)-$ $2 b^{n+1}\\left(3 b^{2}-2 b-5\\right)=0$ for each $n$, which is impossible. For $b=10$ it is easy to show that $x_{n}=\\left(\\frac{10^{n}+5}{3}\\right)^{2}$ for all $n$. This proves the statement. Second solution. In problems of this type, computing $z_{n}=\\sqrt{x_{n}}$ asymptotically usually works. From $\\lim _{n \\rightarrow \\infty} \\frac{b^{2 n}}{(b-1) x_{n}}=1$ we infer that $\\lim _{n \\rightarrow \\infty} \\frac{b^{n}}{z_{n}}=\\sqrt{b-1}$. Furthermore, from $\\left(b z_{n}+z_{n+1}\\right)\\left(b z_{n}-z_{n+1}\\right)=b^{2} x_{n}-x_{n+1}=b^{n+2}+3 b^{2}-2 b-5$ we obtain  $$ \\lim _{n \\rightarrow \\infty}\\left(b z_{n}-z_{n+1}\\right)=\\frac{b \\sqrt{b-1}}{2} $$  Since the $z_{n}$ 's are integers for all $n \\geq M$, we conclude that $b z_{n}-z_{n+1}=$ $\\frac{b \\sqrt{b-1}}{2}$ for all $n$ sufficiently large. Hence $b-1$ is a perfect square, and moreover $b$ divides $2 z_{n+1}$ for all large $n$. It follows that $b \\mid 10$; hence the only possibility is $b=10$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "24", "problem_type": "Number Theory", "exam": "IMO", "problem": "24. N5 (KOR) An integer $n$ is said to be good if $|n|$ is not the square of an integer. Determine all integers $m$ with the following property: $m$ can be represented in infinitely many ways as a sum of three distinct good integers whose product is the square of an odd integer.", "solution": "24. Suppose that $m=u+v+w$ where $u, v, w$ are good integers whose product is a perfect square of an odd integer. Since $u v w$ is an odd perfect square, we have that $u v w \\equiv 1(\\bmod 4)$. Thus either two or none of the numbers $u, v, w$ are congruent to 3 modulo 4 . In both cases $u+v+w \\equiv 3(\\bmod 4)$. Hence $m \\equiv 3(\\bmod 4)$. Now we shall prove the converse: every $m \\equiv 3(\\bmod 4)$ has infinitely many representations of the desired type. Let $m=4 k+3$. We shall represent $m$ in the form  $$ 4 k+3=x y+y z+z x, \\quad \\text { for } x, y, z \\text { odd. } $$  The product of the summands is an odd square. Set $x=1+2 l$ and $y=1-2 l$. In order to satisfy (1), $z$ must satisfy $z=2 l^{2}+2 k+1$. The summands $x y, y z, z x$ are distinct except for finitely many $l$, so it remains only to prove that for infinitely many integers $l,|x y|,|y z|$, and $|z x|$ are not perfect squares. First, observe that $|x y|=4 l^{2}-1$ is not a perfect square for any $l \\neq 0$. Let $p, q>m$ be fixed different prime numbers. The system of congruences $1+2 l \\equiv p\\left(\\bmod p^{2}\\right)$ and $1-2 l \\equiv q\\left(\\bmod q^{2}\\right)$ has infinitely many solutions $l$ by the Chinese remainder theorem. For any such $l$, the number $z=$ $2 l^{2}+2 k+1$ is divisible by neither $p$ nor $q$, and hence $|x z|$ (respectively $|y z|)$ is divisible by $p$, but not by $p^{2}$ (respectively by $q$, but not by $q^{2}$ ). Thus $x z$ and $y z$ are also good numbers.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "25", "problem_type": "Number Theory", "exam": "IMO", "problem": "25. N6 (FRA) ${ }^{\\text {IMO6 }}$ Let $p$ be a prime number. Prove that there exists a prime number $q$ such that for every integer $n$, the number $n^{p}-p$ is not divisible by $q$.", "solution": "25. Suppose that for every prime $q$, there exists an $n$ for which $n^{p} \\equiv p(\\bmod$ $q$ ). Assume that $q=k p+1$. By Fermat's theorem we deduce that $p^{k} \\equiv$ $n^{k p}=n^{q-1} \\equiv 1(\\bmod q)$, so $q \\mid p^{k}-1$. It is known that any prime $q$ such that $q \\left\\lvert\\, \\frac{p^{p}-1}{p-1}\\right.$ must satisfy $q \\equiv 1(\\bmod$ $p$ ). Indeed, from $q \\mid p^{q-1}-1$ it follows that $q \\mid p^{\\operatorname{gcd}(p, q-1)}-1$; but $q \\nmid p-1$ because $\\frac{p^{p}-1}{p-1} \\equiv 1(\\bmod p-1)$, so $\\operatorname{gcd}(p, q-1) \\neq 1$. Hence $\\operatorname{gcd}(p, q-1)=p$. Now suppose $q$ is any prime divisor of $\\frac{p^{p}-1}{p-1}$. Then $q \\mid \\operatorname{gcd}\\left(p^{k}-1, p^{p}-1\\right)=$ $p^{\\operatorname{gcd}(p, k)}-1$, which implies that $\\operatorname{gcd}(p, k)>1$, so $p \\mid k$. Consequently $q \\equiv 1$ $\\left(\\bmod p^{2}\\right)$. However, the number $\\frac{p^{p}-1}{p-1}=p^{p-1}+\\cdots+p+1$ must have at least one prime divisor that is not congruent to 1 modulo $p^{2}$. Thus we arrived at a contradiction. Remark. Taking $q \\equiv 1(\\bmod p)$ is natural, because for every other $q, n^{p}$ takes all possible residues modulo $q$ (including $p$ too). Indeed, if $p \\nmid q-1$, then there is an $r \\in \\mathbb{N}$ satisfying $p r \\equiv 1(\\bmod q-1)$; hence for any $a$ the congruence $n^{p} \\equiv a(\\bmod q)$ has the solution $n \\equiv a^{r}(\\bmod q)$. The statement of the problem itself is a special case of the Chebotarev's theorem.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "26", "problem_type": "Number Theory", "exam": "IMO", "problem": "26. N7 (BRA) The sequence $a_{0}, a_{1}, a_{2}, \\ldots$ is defined as follows:  $$ a_{0}=2, \\quad a_{k+1}=2 a_{k}^{2}-1 \\quad \\text { for } k \\geq 0 $$  Prove that if an odd prime $p$ divides $a_{n}$, then $2^{n+3}$ divides $p^{2}-1$.", "solution": "26. Define the sequence $x_{k}$ of positive reals by $a_{k}=\\cosh x_{k}$ ( $\\cosh$ is the hyperbolic cosine defined by $\\left.\\cosh t=\\frac{e^{t}+e^{-t}}{2}\\right)$. Since $\\cosh \\left(2 x_{k}\\right)=2 a_{k}^{2}-1=$ $\\cosh x_{k+1}$, it follows that $x_{k+1}=2 x_{k}$ and thus $x_{k}=\\lambda \\cdot 2^{k}$ for some $\\lambda>0$. From the condition $a_{0}=2$ we obtain $\\lambda=\\log (2+\\sqrt{3})$. Therefore  $$ a_{n}=\\frac{(2+\\sqrt{3})^{2^{n}}+(2-\\sqrt{3})^{2^{n}}}{2} $$  Let $p$ be a prime number such that $p \\mid a_{n}$. We distinguish the following two cases: (i) There exists an $m \\in \\mathbb{Z}$ such that $m^{2} \\equiv 3(\\bmod p)$. Then we have  $$ (2+m)^{2^{n}}+(2-m)^{2^{n}} \\equiv 0(\\bmod p) $$  Since $(2+m)(2-m)=4-m^{2} \\equiv 1(\\bmod p)$, multiplying both sides of $(1)$ by $(2+m)^{2^{n}}$ gives $(2+m)^{2^{n+1}} \\equiv-1(\\bmod p)$. It follows that the multiplicative order of $(2+m)$ modulo $p$ is $2^{n+2}$, or $2^{n+2} \\mid p-1$, which implies that $2^{n+3} \\mid(p-1)(p+1)=p^{2}-1$. (ii) $m^{2} \\equiv 3(\\bmod p)$ has no integer solutions. We will work in the algebraic extension $\\mathbb{Z}_{p}(\\sqrt{3})$ of the field $\\mathbb{Z}_{p}$. In this field $\\sqrt{3}$ plays the role of $m$, so as in the previous case we obtain $(2+\\sqrt{3})^{2^{n+1}}=-1$; i.e., the order of $2+\\sqrt{3}$ in the multiplicative group $\\mathbb{Z}_{p}(\\sqrt{3})^{*}$ is $2^{n+2}$. We cannot finish the proof as in the previous case: in fact, we would conclude only that $2^{n+2}$ divides the order $p^{2}-1$ of the group. However, it will be enough to find a $u \\in \\mathbb{Z}_{p}(\\sqrt{3})$ such that $u^{2}=2+\\sqrt{3}$, since then the order of $u$ is equal to $2^{n+3}$. Note that $(1+\\sqrt{3})^{2}=2(2+\\sqrt{3})$. Thus it is sufficient to prove that $\\frac{1}{2}$ is a perfect square in $\\mathbb{Z}_{p}(\\sqrt{3})$. But we know that in this field $a_{n}=$ $0=2 a_{n-1}^{2}-1$, and hence $2 a_{n-1}^{2}=1$ which implies $\\frac{1}{2}=a_{n-1}^{2}$. This completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "27", "problem_type": "Number Theory", "exam": "IMO", "problem": "27. N8 (IRN) Let $p$ be a prime number and let $A$ be a set of positive integers that satisfies the following conditions: (i) the set of prime divisors of the elements in $A$ consists of $p-1$ elements; (ii) for any nonempty subset of $A$, the product of its elements is not a perfect $p$ th power. What is the largest possible number of elements in $A$ ?", "solution": "27. Let $p_{1}, p_{2}, \\ldots, p_{r}$ be distinct primes, where $r=p-1$. Consider the sets $B_{i}=\\left\\{p_{i}, p_{i}^{p+1}, \\ldots, p_{i}^{(r-1) p+1}\\right\\}$ and $B=\\bigcup_{i=1}^{r} B_{i}$. Then $B$ has $(p-1)^{2}$ elements and satisfies (i) and (ii). Now suppose that $|A| \\geq r^{2}+1$ and that $A$ satisfies (i) and (ii), and let $\\left\\{t_{1}, \\ldots, t_{r^{2}+1}\\right\\}$ be distinct elements of $A$, where $t_{j}=p_{1}^{\\alpha_{j_{1}}} \\cdot p_{2}^{\\alpha_{j_{2}}} \\cdots p_{r}^{\\alpha_{j_{r}}}$. We shall show that the product of some elements of $A$ is a perfect $p$ th power, i.e., that there exist $\\tau_{j} \\in\\{0,1\\}\\left(1 \\leq j \\leq r^{2}+1\\right)$, not all equal to 0 , such that $T=t_{1}^{\\tau_{1}} \\cdot t_{2}^{\\tau_{2}} \\cdots t_{r^{2}+1}^{\\tau_{r^{2}+1}}$ is a $p$ th power. This is equivalent to the condition that  $$ \\sum_{j=1}^{r^{2}+1} \\alpha_{i j} \\tau_{j} \\equiv 0(\\bmod p) $$  holds for all $i=1, \\ldots, r$. By Fermat's theorem it is sufficient to find integers $x_{1}, \\ldots, x_{r^{2}+1}$, not all zero, such that the relation  $$ \\sum_{j=1}^{r^{2}+1} \\alpha_{i j} x_{j}^{r} \\equiv 0(\\bmod p) $$  is satisfied for all $i \\in\\{1, \\ldots, r\\}$. Set $F_{i}=\\sum_{j=1}^{r^{2}+1} \\alpha_{i j} x_{j}^{r}$. We want to find $x_{1}, \\ldots, x_{r}$ such that $F_{1} \\equiv F_{2} \\equiv \\cdots \\equiv F_{r} \\equiv 0(\\bmod p)$, which is by Fermat's theorem equivalent to  $$ F\\left(x_{1}, \\ldots, x_{r}\\right)=F_{1}^{r}+F_{2}^{r}+\\cdots+F_{r}^{r} \\equiv 0(\\bmod p) . $$  Of course, one solution of (1) is $(0, \\ldots, 0)$ : we are not satisfied with it because it generates the empty subset of $A$, but it tells us that (1) has at least one solution. We shall prove that the number of solutions of (1) is divisible by $p$, which will imply the existence of a nontrivial solution and thus complete the proof. To do this, consider the sum $\\sum F\\left(x_{1}, \\ldots, x_{r^{2}+1}\\right)^{r}$ taken over all vectors $\\left(x_{1}, \\ldots, x_{r^{2}+1}\\right)$ in the vector space $\\mathbb{Z}_{p}^{r^{2}+1}$. Our statement is equivalent to  $$ \\sum F\\left(x_{1}, \\ldots, x_{r^{2}+1}\\right)^{r} \\equiv 0(\\bmod p) $$  Since the degree of $F^{r}$ is $r^{2}$, in each monomial in $F^{r}$ at least one of the variables is missing. Consider any of these monomials, say $b x_{i_{1}}^{a_{1}} x_{i_{2}}^{a_{2}} \\cdots x_{i_{k}}^{a_{k}}$. Then the sum $\\sum b x_{i_{1}}^{a_{1}} x_{i_{2}}^{a_{2}} \\cdots x_{i_{k}}^{a_{k}}$, taken over the set of all vectors $\\left(x_{1}, \\ldots, x_{r^{2}+1}\\right) \\in \\mathbb{Z}_{p}^{r^{2}+1}$, is equal to  $$ p^{r^{2}+1-u} \\cdot \\sum_{\\left(x_{i_{1}}, \\ldots, x_{i_{k}}\\right) \\in \\mathbb{Z}_{p}^{k}} b x_{i_{1}}^{a_{1}} x_{i_{2}}^{a_{2}} \\cdots x_{i_{k}}^{a_{k}}, $$  which is divisible by $p$, so that (2) is proved. Thus the answer is $(p-1)^{2}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "3", "problem_type": "Algebra", "exam": "IMO", "problem": "3. A3 (GEO) Consider pairs of sequences of positive real numbers $a_{1} \\geq$ $a_{2} \\geq a_{3} \\geq \\cdots, b_{1} \\geq b_{2} \\geq b_{3} \\geq \\cdots$ and the sums $A_{n}=a_{1}+\\cdots+a_{n}$, $B_{n}=b_{1}+\\cdots+b_{n}, n=1,2, \\ldots$. For any pair define $c_{i}=\\min \\left\\{a_{i}, b_{i}\\right\\}$ and $C_{n}=c_{1}+\\cdots+c_{n}, n=1,2, \\ldots$ (a) Does there exist a pair $\\left(a_{i}\\right)_{i \\geq 1},\\left(b_{i}\\right)_{i \\geq 1}$ such that the sequences $\\left(A_{n}\\right)_{n \\geq 1}$ and $\\left(B_{n}\\right)_{n \\geq 1}$ are unbounded while the sequence $\\left(C_{n}\\right)_{n \\geq 1}$ is bounded? (b) Does the answer to question (1) change by assuming additionally that $b_{i}=1 / i, i=1,2, \\ldots ?$ Justify your answer.", "solution": "3. (a) Given any sequence $c_{n}$ (in particular, such that $C_{n}$ converges), we shall construct $a_{n}$ and $b_{n}$ such that $A_{n}$ and $B_{n}$ diverge. First, choose $n_{1}$ such that $n_{1} c_{1}>1$ and set $a_{1}=a_{2}=\\cdots=a_{n_{1}}=$ $c_{1}$ : this uniquely determines $b_{2}=c_{2}, \\ldots, b_{n_{1}}=c_{n_{1}}$. Next, choose $n_{2}$ such that $\\left(n_{2}-n_{1}\\right) c_{n_{1}+1}>1$ and set $b_{n_{1}+1}=\\cdots=b_{n_{2}}=c_{n_{1}+1}$; again $a_{n_{1}+1}, \\ldots, a_{n_{2}}$ is hereby determined. Then choose $n_{3}$ with $\\left(n_{3}-\\right.$ $\\left.n_{2}\\right) c_{n_{2}+1}>1$ and set $a_{n_{2}+1}=\\cdots=a_{n_{3}}=c_{n_{2}+1}$, and so on. It is plain that in this way we construct decreasing sequences $a_{n}, b_{n}$ such that $\\sum a_{n}$ and $\\sum b_{n}$ diverge, since they contain an infinity of subsums that exceed 1 ; on the other hand, $c_{n}=\\min \\left(a_{n}, b_{n}\\right)$ and $C_{n}$ is convergent. (b) The answer changes in this situation. Suppose to the contrary that there is such a pair of sequences $\\left(a_{n}\\right)$ and $\\left(b_{n}\\right)$. There are infinitely many indices $i$ such that $c_{i}=b_{i}$ (otherwise all but finitely many terms of the sequence $\\left(c_{n}\\right)$ would be equal to the terms of the sequence $\\left(a_{n}\\right)$, which has an unbounded sum). Thus for any $n_{0} \\in \\mathbb{N}$ there is $j \\geq 2 n_{0}$ such that $c_{j}=b_{j}$. Then we have  $$ \\sum_{k=n_{0}}^{j} c_{k} \\geq \\sum_{k=n_{0}}^{j} c_{j}=\\left(j-n_{0}\\right) \\frac{1}{j} \\geq \\frac{1}{2} $$  Hence the sequence ( $C_{n}$ ) is unbounded, a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "4", "problem_type": "Algebra", "exam": "IMO", "problem": "4. A4 (IRE) ${ }^{\\mathrm{IMO} 5}$ Let $n$ be a positive integer and let $x_{1} \\leq x_{2} \\leq \\cdots \\leq x_{n}$ be real numbers. (a) Prove that  $$ \\left(\\sum_{i, j=1}^{n}\\left|x_{i}-x_{j}\\right|\\right)^{2} \\leq \\frac{2\\left(n^{2}-1\\right)}{3} \\sum_{i, j=1}^{n}\\left(x_{i}-x_{j}\\right)^{2} $$  (b) Show that equality holds if and only if $x_{1}, \\ldots, x_{n}$ is an arithmetic progession.", "solution": "4. By the Cauchy-Schwarz inequality we have  $$ \\left(\\sum_{i, j=1}^{n}(i-j)^{2}\\right)\\left(\\sum_{i, j=1}^{n}\\left(x_{i}-x_{j}\\right)^{2}\\right) \\geq\\left(\\sum_{i, j=1}^{n}|i-j| \\cdot\\left|x_{i}-x_{j}\\right|\\right)^{2} . $$  On the other hand, it is easy to prove (for example by induction) that  $$ \\sum_{i, j=1}^{n}(i-j)^{2}=(2 n-2) \\cdot 1^{2}+(2 n-4) \\cdot 2^{2}+\\cdots+2 \\cdot(n-1)^{2}=\\frac{n^{2}\\left(n^{2}-1\\right)}{6} $$  and that  $$ \\sum_{i, j=1}^{n}|i-j| \\cdot\\left|x_{i}-x_{j}\\right|=\\frac{n}{2} \\sum_{i, j=1}^{n}\\left|x_{i}-x_{j}\\right| $$  Thus the inequality (1) becomes  $$ \\frac{n^{2}\\left(n^{2}-1\\right)}{6}\\left(\\sum_{i, j=1}^{n}\\left(x_{i}-x_{j}\\right)^{2}\\right) \\geq \\frac{n^{2}}{4}\\left(\\sum_{i, j=1}^{n}\\left|x_{i}-x_{j}\\right|\\right)^{2} $$  which is equivalent to the required one.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "5", "problem_type": "Algebra", "exam": "IMO", "problem": "5. A5 (KOR) Let $\\mathbb{R}^{+}$be the set of all positive real numbers. Find all functions $f: \\mathbb{R}^{+} \\rightarrow \\mathbb{R}^{+}$that satisfy the following conditions: (i) $f(x y z)+f(x)+f(y)+f(z)=f(\\sqrt{x y}) f(\\sqrt{y z}) f(\\sqrt{z x})$ for all $x, y, z \\in$ $\\mathbb{R}^{+}$. (ii) $f(x)<f(y)$ for all $1 \\leq x<y$.", "solution": "5. Placing $x=y=z=1$ in (i) leads to $4 f(1)=f(1)^{3}$, so by the condition $f(1)>0$ we get $f(1)=2$. Also putting $x=t s, y=\\frac{t}{s}, z=\\frac{s}{t}$ in (i) gives  $$ f(t) f(s)=f(t s)+f(t / s) $$  In particular, for $s=1$ the last equality yields $f(t)=f(1 / t)$; hence $f(t) \\geq f(1)=2$ for each $t$. It follows that there exists $g(t) \\geq 1$ such that $\\bar{f}(t)=g(t)+\\frac{1}{g(t)}$. Now it follows by induction from (1) that $g\\left(t^{n}\\right)=$ $g(t)^{n}$ for every integer $n$, and therefore $g\\left(t^{q}\\right)=g(t)^{q}$ for every rational $q$. Consequently, if $t>1$ is fixed, we have $f\\left(t^{q}\\right)=a^{q}+a^{-q}$, where $a=g(t)$. But since the set of $a^{q}(q \\in \\mathbb{Q})$ is dense in $\\mathbb{R}^{+}$and $f$ is monotone on $(0,1]$ and $[1, \\infty)$, it follows that $f\\left(t^{r}\\right)=a^{r}+a^{-r}$ for every real $r$. Therefore, if $k$ is such that $t^{k}=a$, we have  $$ f(x)=x^{k}+x^{-k} \\quad \\text { for every } x \\in \\mathbb{R} $$", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "6", "problem_type": "Algebra", "exam": "IMO", "problem": "6. A6 (USA) Let $n$ be a positive integer and let $\\left(x_{1}, \\ldots, x_{n}\\right),\\left(y_{1}, \\ldots, y_{n}\\right)$ be two sequences of positive real numbers. Suppose $\\left(z_{2}, z_{3}, \\ldots, z_{2 n}\\right)$ is a sequence of positive real numbers such that  $$ z_{i+j}^{2} \\geq x_{i} y_{j} \\quad \\text { for all } 1 \\leq i, j \\leq n $$  Let $M=\\max \\left\\{z_{2}, \\ldots, z_{2 n}\\right\\}$. Prove that  $$ \\left(\\frac{M+z_{2}+\\cdots+z_{2 n}}{2 n}\\right)^{2} \\geq\\left(\\frac{x_{1}+\\cdots+x_{n}}{n}\\right)\\left(\\frac{y_{1}+\\cdots+y_{n}}{n}\\right) . $$", "solution": "6. Set $X=\\max \\left\\{x_{1}, \\ldots, x_{n}\\right\\}$ and $Y=\\max \\left\\{y_{1}, \\ldots, y_{n}\\right\\}$. By replacing $x_{i}$ by $x_{i}^{\\prime}=\\frac{x_{i}}{X}, y_{i}$ by $y_{i}^{\\prime}=\\frac{y_{i}}{Y}$ and $z_{i}$ by $z_{i}^{\\prime}=\\frac{z_{i}}{\\sqrt{X Y}}$, we may assume that $X=Y=1$. It is sufficient to prove that  $$ M+z_{2}+\\cdots+z_{2 n} \\geq x_{1}+\\cdots+x_{n}+y_{1}+\\cdots+y_{n} $$  because this implies the result by the A-G mean inequality. To prove (1) it is enough to prove that for any $r$, the number of terms greater than $r$ on the left-hand side of (1) is at least that number on the right-hand side of (1). If $r \\geq 1$, then there are no terms on the right-hand side greater than $r$. Suppose that $r<1$ and consider the sets $A=\\left\\{i \\mid 1 \\leq i \\leq n, x_{i}>r\\right\\}$ and $B=\\left\\{i \\mid 1 \\leq i \\leq n, y_{i}>r\\right\\}$. Set $a=|A|$ and $b=|B|$. If $x_{i}>r$ and $y_{j}>r$, then $z_{i+j} \\geq \\sqrt{x_{i} y_{j}}>r$; hence  $$ C=\\left\\{k \\mid 2 \\leq k \\leq 2 n, z_{k}>r\\right\\} \\supseteq A+B=\\{\\alpha+\\beta \\mid \\alpha \\in A, \\beta \\in B\\} $$  It is easy to verify that $|A+B| \\geq|A|+|B|-1$. It follows that the number of $z_{k}$ 's greater than $r$ is $\\geq a+b-1$. But in that case $M>r$, implying that at least $a+b$ elements of the left-hand side of (1) is greater than $r$, which completes the proof.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "7", "problem_type": "Combinatorics", "exam": "IMO", "problem": "7. C1 (BRA) ${ }^{\\mathrm{IMO} 1}$ Let $A$ be a 101-element subset of the set $S=\\{1,2, \\ldots$, $1000000\\}$. Prove that there exist numbers $t_{1}, t_{2}, \\ldots, t_{100}$ in $S$ such that the sets  $$ A_{j}=\\left\\{x+t_{j} \\mid x \\in A\\right\\}, \\quad j=1,2, \\ldots, 100 $$  are pairwise disjoint.", "solution": "7. Consider the set $D=\\{x-y \\mid x, y \\in A\\}$. Obviously, the number of elements of the set $D$ is less than or equal to $101 \\cdot 100+1$. The sets $A+t_{i}$ and $A+t_{j}$ are disjoint if and only if $t_{i}-t_{j} \\notin D$. Now we shall choose inductively 100 elements $t_{1}, \\ldots, t_{100}$. Let $t_{1}$ be any element of the set $S \\backslash D$ (such an element exists, since the number of elements of $S$ is greater than the number of elements of $D$ ). Suppose now that we have chosen $k(k \\leq 99)$ elements $t_{1}, \\ldots, t_{k}$ from $D$ such that the difference of any two of the chosen elements does not belong to $D$. We can select $t_{k+1}$ to be an element of $S$ that does not belong to any of the sets $t_{1}+D, t_{2}+D, \\ldots, t_{k}+D$ (this is possible to do, since each of the previous sets has at most $101 \\cdot 100+1$ elements; hence their union has at most $99(101 \\cdot 100+1)=999999<1000000$ elements $)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "8", "problem_type": "Combinatorics", "exam": "IMO", "problem": "8. C2 (GEO) Let $D_{1}, \\ldots, D_{n}$ be closed disks in the plane. (A closed disk is a region bounded by a circle, taken jointly with this circle.) Suppose that every point in the plane is contained in at most 2003 disks $D_{i}$. Prove that there exists disk $D_{k}$ that intersects at most $7 \\cdot 2003-1$ other disks $D_{i}$.", "solution": "8. Let $S$ be the disk with the smallest radius, say $s$, and $O$ the center of that disk. Divide the plane into 7 regions: one bounded by disk $s$ and 6 regions $T_{1}, \\ldots, T_{6}$ shown in the figure. Any of the disks different from $S$, say $D_{k}$, has its center in one of the seven regions. If its center is inside $S$ then $D_{k}$ contains point $O$. Hence the number of disks different from $S$ having their centers in $S$ is at most 2002.  Consider a disk $D_{k}$ that intersects $S$ and whose center is in the region $T_{i}$. Let $P_{i}$ be the point such that $O P_{i}$ bisects the region $T_{i}$ and ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-713.jpg?height=453&width=503&top_left_y=799&top_left_x=828) $O P_{i}=s \\sqrt{3}$. We claim that $D_{k}$ contains $P_{i}$. Divide the region $T_{i}$ by a line $l_{i}$ through $P_{i}$ perpendicular to $O P_{i}$ into two regions $U_{i}$ and $V_{i}$, where $O$ and $U_{i}$ are on the same side of $l_{i}$. Let $K$ be the center of $D_{k}$. Consider two cases: (i) $K \\in U_{i}$. Since the disk with the center $P_{i}$ and radius $s$ contains $U_{i}$, we see that $K P_{i} \\leq s$. Hence $D_{k}$ contains $P_{i}$. (ii) $K \\in V_{i}$. Denote by $L$ the intersection point of the segment $K O$ with the circle $s$. We want to prove that $K L>K P_{i}$. It is enough to prove that $\\angle K P_{i} L>\\angle K L P_{i}$. However, it is obvious that $\\angle L P_{i} O \\leq 30^{\\circ}$ and $\\angle L O P_{i} \\leq 30^{\\circ}$, hence $\\angle K L P_{i} \\leq 60^{\\circ}$, while $\\angle N P_{i} L=90^{\\circ}-\\angle L P_{i} O \\geq$ $60^{\\circ}$. This implies that $\\angle K P_{i} L \\geq \\angle N P_{i} L \\geq 60^{\\circ} \\geq \\angle K L P_{i}$ ( $N$ is the point on the edge of $T_{i}$ as shown in the figure). Our claim is thus proved. Now we see that the number of disks with centers in $T_{i}$ that intersect $S$ is less than or equal to 2003, and the total number of disks that intersect $S$ is not greater than $2002+6 \\cdot 2003=7 \\cdot 2003-1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2003", "tier": "T0", "problem_label": "9", "problem_type": "Combinatorics", "exam": "IMO", "problem": "9. C3 (LIT) Let $n \\geq 5$ be a given integer. Determine the largest integer $k$ for which there exists a polygon with $n$ vertices (convex or not, with non-self-intersecting boundary) having $k$ internal right angles.", "solution": "9. Suppose that $k$ of the angles of an $n$-gon are right. Since the other $n-k$ angles are less than $360^{\\circ}$ and the sum of the angles is $(n-2) 180^{\\circ}$, we have the inequality $k \\cdot 90^{\\circ}+(n-k) 360^{\\circ}>(n-2) 180^{\\circ}$, which is equivalent to $k<\\frac{2 n+4}{3}$. Since $n$ and $k$ are integers, it follows that $k \\leq\\left[\\frac{2 n}{3}\\right]+1$. If $n=5$, then $\\left[\\frac{2 n}{3}\\right]+1=4$, but if a pentagon has four right angles, the other angle is equal to $180^{\\circ}$, which is impossible. Hence for $n=5$, $k \\leq 3$. It is easy to construct a pentagon with 3 right angles, e.g., as in the picture below. Now we shall show by induction that for $n \\geq 6$ there is an $n$-gon with $\\left[\\frac{2 n}{3}\\right]+1$ internal right angles. For $n=6,7,8$ examples are presented in the picture. Assume that there is a $(n-3)$ gon with $\\left[\\frac{2(n-3)}{3}\\right]+1=\\left[\\frac{2 n}{3}\\right]-1$ internal right angles. Then one of the internal angles, say $\\angle B A C$, is not convex. Interchange the vertex $A$ with four new vertices $A_{1}, A_{2}, A_{3}, A_{4}$ as shown in the picture such that $\\angle B A_{1} A_{2}=\\angle A_{3} A_{4} C=90^{\\circ}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-714.jpg?height=195&width=1024&top_left_y=781&top_left_x=292)", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "1", "problem_type": "Algebra", "exam": "IMO", "problem": "1. A1 (KOR) ${ }^{\\mathrm{IMO} 4}$ Let $n \\geq 3$ be an integer and $t_{1}, t_{2}, \\ldots, t_{n}$ positive real numbers such that  $$ n^{2}+1>\\left(t_{1}+t_{2}+\\cdots+t_{n}\\right)\\left(\\frac{1}{t_{1}}+\\frac{1}{t_{2}}+\\cdots+\\frac{1}{t_{n}}\\right) . $$  Show that $t_{i}, t_{j}, t_{k}$ are the side lengths of a triangle for all $i, j, k$ with $1 \\leq i<j<k \\leq n$.", "solution": "1. By symmetry, it is enough to prove that $t_{1}+t_{2}>t_{3}$. We have  $$ \\left(\\sum_{i=1}^{n} t_{i}\\right)\\left(\\sum_{i=1}^{n} \\frac{1}{t}_{i}\\right)=n^{2}+\\sum_{i<j}\\left(\\frac{t_{i}}{t_{j}}+\\frac{t_{j}}{t_{i}}-2\\right) . $$  All the summands on the RHS are positive, and therefore the RHS is not smaller than $n^{2}+T$, where $T=\\left(t_{1} / t_{3}+t_{3} / t_{1}-2\\right)+\\left(t_{2} / t_{3}+t_{3} / t_{2}-2\\right)$. We note that $T$ is increasing as a function in $t_{3}$ for $t_{3} \\geq \\max \\left\\{t_{1}, t_{2}\\right\\}$. If $t_{1}+t_{2}=t_{3}$, then $T=\\left(t_{1}+t_{2}\\right)\\left(1 / t_{1}+1 / t_{2}\\right)-1 \\geq 3$ by the Cauchy-Schwarz inequality. Hence, if $t_{1}+t_{2} \\leq t_{3}$, we have $T \\geq 1$, and consequently the RHS in (1) is greater than or equal to $n^{2}+1$, a contradiction. Remark. In can be proved, for example using Lagrange multipliers, that if $n^{2}+1$ in the problem is replaced by $(n+\\sqrt{10}-3)^{2}$, then the statement remains true. This estimate is the best possible.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "10", "problem_type": "Combinatorics", "exam": "IMO", "problem": "10. C3 (AUS) The following operation is allowed on a finite graph: Choose an arbitrary cycle of length 4 (if there is any), choose an arbitrary edge in that cycle, and delete it from the graph. For a fixed integer $n \\geq 4$, find the least number of edges of a graph that can be obtained by repeated applications of this operation from the complete graph on $n$ vertices (where each pair of vertices are joined by an edge).", "solution": "10. The least number of edges of such a graph is $n$.  We note that deleting edge $A B$ of a 4-cycle $A B C D$ from a connected and nonbipartite graph $G$ yields a connected and nonbipartite graph, say $H$. Indeed, the connectedness is obvious; also, if $H$ were bipartite with partition of the set of vertices into $P_{1}$ and $P_{2}$, then w.l.o.g. $A, C \\in P_{1}$ and $B, D \\in P_{2}$, so $G=H \\cup\\{A B\\}$ would also be bipartite with the same partition, a contradiction. Any graph that can be obtained from the complete $n$-graph in the described way is connected and has at least one cycle (otherwise it would be bipartite); hence it must have at least $n$ edges. Now consider a complete graph with vertices $V_{1}, V_{2}, \\ldots, V_{n}$. Let us remove every edge $V_{i} V_{j}$ with $3 \\leq i<j<n$ from the cycle $V_{2} V_{i} V_{j} V_{n}$. Then for $i=3, \\ldots, n-1$ we remove edges $V_{2} V_{i}$ and $V_{i} V_{n}$ from the cycles $V_{1} V_{i} V_{2} V_{n}$ and $V_{1} V_{i} V_{n} V_{2}$ respectively, thus obtaining a graph with exactly $n$ edges: $V_{1} V_{i}(i=2, \\ldots, n)$ and $V_{2} V_{n}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "11", "problem_type": "Combinatorics", "exam": "IMO", "problem": "11. C4 (POL) Consider a matrix of size $n \\times n$ whose entries are real numbers of absolute value not exceeding 1 , and the sum of all entries is 0 . Let $n$ be an even positive integer. Determine the least number $C$ such that every such matrix necessarily has a row or a column with the sum of its entries not exceeding $C$ in absolute value.", "solution": "11. Consider the matrix $A=\\left(a_{i j}\\right)_{i, j=1}^{n}$ such that $a_{i j}$ is equal to 1 if $i, j \\leq n / 2$, -1 if $i, j>n / 2$, and 0 otherwise. This matrix satisfies the conditions from the problem and all row sums and column sums are equal to $\\pm n / 2$. Hence $C \\geq n / 2$. Let us show that $C=n / 2$. Assume to the contrary that there is a matrix $B=\\left(b_{i j}\\right)_{i, j=1}^{n}$ all of whose row sums and column sums are either greater than $n / 2$ or smaller than $-n / 2$. We may assume w.l.o.g. that at least $n / 2$ row sums are positive and, permuting rows if necessary, that the first $n / 2$ rows have positive sums. The sum of entries in the $n / 2 \\times n$ submatrix $B^{\\prime}$ consisting of first $n / 2$ rows is greater than $n^{2} / 4$, and since each column of $B^{\\prime}$ has sum at most $n / 2$, it follows that more than $n / 2$ column sums of $B^{\\prime}$, and therefore also of $B$, are positive. Again, suppose w.l.o.g. that the first $n / 2$ column sums are positive. Thus the sums $R^{+}$and $C^{+}$of entries in the first $n / 2$ rows and in the first $n / 2$ columns respectively are greater than $n^{2} / 4$. Now the sum of all entries of $B$ can be written as  $$ \\sum a_{i j}=R^{+}+C^{+}+\\sum_{\\substack{i>n / 2 \\\\ j>n / 2}} a_{i j}-\\sum_{\\substack{i \\leq n / 2 \\\\ j \\leq n / 2}} a_{i j}>\\frac{n^{2}}{2}-\\frac{n^{2}}{4}-\\frac{n^{2}}{4}=0 $$  a contradiction. Hence $C=n / 2$, as claimed.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "12", "problem_type": "Combinatorics", "exam": "IMO", "problem": "12. C5 (NZL) Let $N$ be a positive integer. Two players $A$ and $B$, taking turns, write numbers from the set $\\{1, \\ldots, N\\}$ on a blackboard. $A$ begins the game by writing 1 on his first move. Then, if a player has written $n$ on a certain move, his adversary is allowed to write $n+1$ or $2 n$ (provided the number he writes does not exceed $N$ ). The player who writes $N$ wins. We say that $N$ is of type $A$ or of type $B$ according as $A$ or $B$ has a winning strategy. (a) Determine whether $N=2004$ is of type $A$ or of type $B$. (b) Find the least $N>2004$ whose type is different from that of 2004.", "solution": "12. We say that a number $n \\in\\{1,2, \\ldots, N\\}$ is winning if the player who is on turn has a winning strategy, and losing otherwise. The game is of type $A$ if and only if 1 is a losing number. Let us define $n_{0}=N, n_{i+1}=\\left[n_{i} / 2\\right]$ for $i=0,1, \\ldots$ and let $k$ be such that $n_{k}=1$. Consider the sets $A_{i}=\\left\\{n_{i+1}+1, \\ldots, n_{i}\\right\\}$. We call a set $A_{i}$ all-winning if all numbers from $A_{i}$ are winning, even-winning if even numbers are winning and odd are losing, and odd-winning if odd numbers are winning and even are losing. (i) Suppose $A_{i}$ is even-winning and consider $A_{i+1}$. Multiplying any number from $A_{i+1}$ by 2 yields an even number from $A_{i}$, which is a losing number. Thus $x \\in A_{i+1}$ is winning if and only if $x+1$ is losing, i.e., if and only if it is even. Hence $A_{i+1}$ is also even-winning. (ii) Suppose $A_{i}$ is odd-winning. Then each $k \\in A_{i+1}$ is winning, since $2 k$ is losing. Hence $A_{i+1}$ is all-winning. (iii) Suppose $A_{i}$ is all-winning. Multiplying $x \\in A_{i+1}$ by two is then a losing move, so $x$ is winning if and only if $x+1$ is losing. Since $n_{i+1}$ is losing, $A_{i+1}$ is odd-winning if $n_{i+1}$ is even and even-winning otherwise. We observe that $A_{0}$ is even-winning if $N$ is odd and odd-winning otherwise. Also, if some $A_{i}$ is even-winning, then all $A_{i+1}, A_{i+2}, \\ldots$ are evenwinning and thus 1 is losing; i.e., the game is of type $A$. The game is of type $B$ if and only if the sets $A_{0}, A_{1}, \\ldots$ are alternately odd-winning and allwinning with $A_{0}$ odd-winning, which is equivalent to $N=n_{0}, n_{2}, n_{4}, \\ldots$ all being even. Thus $N$ is of type $B$ if and only if all digits at the odd positions in the binary representation of $N$ are zeros. Since $2004=\\overline{11111010100}$ in the binary system, 2004 is of type $A$. The least $N>2004$ that is of type $B$ is $\\overline{100000000000}=2^{11}=2048$. Thus the answer to part (b) is 2048.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "13", "problem_type": "Combinatorics", "exam": "IMO", "problem": "13. C6 (IRN) For an $n \\times n$ matrix $A$, let $X_{i}$ be the set of entries in row $i$, and $Y_{j}$ the set of entries in column $j, 1 \\leq i, j \\leq n$. We say that $A$ is golden if $X_{1}, \\ldots, X_{n}, Y_{1}, \\ldots, Y_{n}$ are distinct sets. Find the least integer $n$ such that there exists a $2004 \\times 2004$ golden matrix with entries in the set $\\{1,2, \\ldots, n\\}$.", "solution": "13. Since $X_{i}, Y_{i}, i=1, \\ldots, 2004$, are 4008 distinct subsets of the set $S_{n}=$ $\\{1,2, \\ldots, n\\}$, it follows that $2^{n} \\geq 4008$, i.e. $n \\geq 12$. Suppose $n=12$. Let $\\mathcal{X}=\\left\\{X_{1}, \\ldots, X_{2004}\\right\\}, \\mathcal{Y}=\\left\\{Y_{1}, \\ldots, Y_{2004}\\right\\}, \\mathcal{A}=$ $\\mathcal{X} \\cup \\mathcal{Y}$. Exactly $2^{12}-4008=88$ subsets of $S_{n}$ do not occur in $\\mathcal{A}$. Since each row intersects each column, we have $X_{i} \\cap Y_{j} \\neq \\emptyset$ for all $i, j$. Suppose $\\left|X_{i}\\right|,\\left|Y_{j}\\right| \\leq 3$ for some indices $i, j$. Since then $\\left|X_{i} \\cup Y_{j}\\right| \\leq 5$, any of at least $2^{7}>88$ subsets of $S_{n} \\backslash\\left(X_{i} \\cap Y_{j}\\right)$ can occur in neither $\\mathcal{X}$ nor $\\mathcal{Y}$, which is impossible. Hence either in $\\mathcal{X}$ or in $\\mathcal{Y}$ all subsets are of size at least 4. Suppose w.l.o.g. that $k=\\left|X_{l}\\right|=\\min _{i}\\left|X_{i}\\right| \\geq 4$. There are  $$ n_{k}=\\binom{12-k}{0}+\\binom{12-k}{1}+\\cdots+\\binom{12-k}{k-1} $$  subsets of $S \\backslash X_{l}$ with fewer than $k$ elements, and none of them can be either in $\\mathcal{X}$ (because $\\left|X_{l}\\right|$ is minimal in $\\mathcal{X}$ ) or in $\\mathcal{Y}$. Hence we must have $n_{k} \\leq 88$. Since $n_{4}=93$ and $n_{5}=99$, it follows that $k \\geq 6$. But then none of the $\\binom{12}{0}+\\cdots+\\binom{12}{5}=1586$ subsets of $S_{n}$ is in $\\mathcal{X}$, hence at least $1586-88=1498$ of them are in $\\mathcal{Y}$. The 1498 complements of these subsets also do not occur in $\\mathcal{X}$, which adds to 3084 subsets of $S_{n}$ not occurring in $\\mathcal{X}$. This is clearly a contradiction. Now we construct a golden matrix for $n=13$. Let  $$ A_{1}=\\left[\\begin{array}{ll} 1 & 1 \\\\ 2 & 3 \\end{array}\\right] \\quad \\text { and } \\quad A_{m}=\\left[\\begin{array}{ll} A_{m-1} & A_{m-1} \\\\ A_{m-1} & B_{m-1} \\end{array}\\right] \\text { for } m=2,3, \\ldots $$  where $B_{m-1}$ is the $2^{m-1} \\times 2^{m-1}$ matrix with all entries equal to $m+2$. It can be easily proved by induction that each of the matrices $A_{m}$ is golden. Moreover, every upper-left square submatrix of $A_{m}$ of size greater than $2^{m-1}$ is also golden. Since $2^{10}<2004<2^{11}$, we thus obtain a golden matrix of size 2004 with entries in $S_{13}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "14", "problem_type": "Combinatorics", "exam": "IMO", "problem": "14. C7 (EST) ${ }^{\\mathrm{IMO} 3}$ Determine all $m \\times n$ rectangles that can be covered with hooks made up of 6 unit squares, as in the figure: ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-338.jpg?height=153&width=150&top_left_y=1988&top_left_x=722)  Rotations and reflections of hooks are allowed. The rectangle must be covered without gaps and overlaps. No part of a hook may cover area outside the rectangle.", "solution": "14. Suppose that an $m \\times n$ rectangle can be covered by \"hooks\". For any hook $H$ there is a unique hook $K$ that covers its \" inside\" square. Then also $H$ covers the inside square of $K$, so the set of hooks can be partitioned into pairs of type $\\{H, K\\}$, each of which forms one of the following two figures consisting of 12 squares: ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-730.jpg?height=252&width=979&top_left_y=972&top_left_x=306)  Thus the $m \\times n$ rectangle is covered by these tiles. It immediately follows that $12 \\mid \\mathrm{mn}$. Suppose one of $m, n$ is divisible by 4 . Let w.l.o.g. $4 \\mid m$. If $3 \\mid n$, one can easily cover the rectangle by $3 \\times 4$ rectangles and therefore by hooks. Also, if $12 \\mid m$ and $n \\notin\\{1,2,5\\}$, then there exist $k, l \\in \\mathbb{N}_{0}$ such that $n=3 k+4 l$, and thus the rectangle $m \\times n$ can be partitioned into $3 \\times 12$ and $4 \\times 12$ rectangles all of which can be covered by hooks. If $12 \\mid m$ and $n=1,2$, or 5 , then it is easy to see that covering by hooks is not possible. Now suppose that $4 \\nmid m$ and $4 \\nmid n$. Then $m, n$ are even and the number of tiles is odd. Assume that the total number of tiles of types $A_{1}$ and $B_{1}$ is odd (otherwise the total number of tiles of types $A_{2}$ and $B_{2}$ is odd, which is analogous). If we color in black all columns whose indices are divisible by 4 , we see that each tile of type $A_{1}$ or $B_{1}$ covers three black squares, which yields an odd number in total. Hence the total number of black squares covered by the tiles of types $A_{2}$ and $B_{2}$ must be odd. This is impossible, since each such tile covers two or four black squares.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "15", "problem_type": "Combinatorics", "exam": "IMO", "problem": "15. C8 (POL) For a finite graph $G$, let $f(G)$ be the number of triangles and $g(G)$ the number of tetrahedra formed by edges of $G$. Find the least constant $c$ such that  $$ g(G)^{3} \\leq c \\cdot f(G)^{4} \\text { for every graph } G $$", "solution": "15. Denote by $V_{1}, \\ldots, V_{n}$ the vertices of a graph $G$ and by $E$ the set of its edges. For each $i=1, \\ldots, n$, let $A_{i}$ be the set of vertices connected to $V_{i}$ by an edge, $G_{i}$ the subgraph of $G$ whose set of vertices is $A_{i}$, and $E_{i}$ the set of edges of $G_{i}$. Also, let $v_{i}, e_{i}$, and $t_{i}=f\\left(G_{i}\\right)$ be the numbers of vertices, edges, and triangles in $G_{i}$ respectively.  The numbers of tetrahedra and triangles one of whose vertices is $V_{i}$ are respectively equal to $t_{i}$ and $e_{i}$. Hence  $$ \\sum_{i=1}^{n} v_{i}=2|E|, \\quad \\sum_{i=1}^{n} e_{i}=3 f(G) \\quad \\text { and } \\quad \\sum_{i=1}^{n} t_{i}=4 g(G) . $$  Since $e_{i} \\leq v_{i}\\left(v_{i}-1\\right) / 2 \\leq v_{i}^{2} / 2$ and $e_{i} \\leq|E|$, we obtain $e_{i}^{2} \\leq v_{i}^{2}|E| / 2$, i.e., $e_{i} \\leq v_{i} \\sqrt{|E| / 2}$. Summing over all $i$ yields $3 f(G) \\leq 2|E| \\sqrt{|E| / 2}$, or equivalently $f(G)^{2} \\leq 2|E|^{3} / 9$. Since this relation holds for each graph $G_{i}$, it follows that  $$ t_{i}=f\\left(G_{i}\\right)=f\\left(G_{i}\\right)^{1 / 3} f\\left(G_{i}\\right)^{2 / 3} \\leq\\left(\\frac{2}{9}\\right)^{1 / 3} f(G)^{1 / 3} e_{i} $$  Summing the last inequality for $i=1, \\ldots, n$ gives us  $$ 4 g(G) \\leq 3\\left(\\frac{2}{9}\\right)^{1 / 3} f(G)^{1 / 3} \\cdot f(G), \\quad \\text { i.e. } \\quad g(G)^{3} \\leq \\frac{3}{32} f(G)^{4} $$  The constant $c=3 / 32$ is the best possible. Indeed, in a complete graph $C_{n}$ it holds that $g\\left(K_{n}\\right)^{3} / f\\left(K_{n}\\right)^{4}=\\binom{n}{4}^{3}\\binom{n}{3}^{-4} \\rightarrow \\frac{3}{32}$ as $n \\rightarrow \\infty$. Remark. Let $N_{k}$ be the number of complete $k$-subgraphs in a finite graph $G$. Continuing inductively, one can prove that $N_{k+1}^{k} \\leq \\frac{k!}{(k+1)^{k}} N_{k}^{k+1}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "16", "problem_type": "Geometry", "exam": "IMO", "problem": "16. G1 (ROM) ${ }^{\\mathrm{IMO} 1}$ Let $A B C$ be an acute-angled triangle with $A B \\neq A C$. The circle with diameter $B C$ intersects the sides $A B$ and $A C$ at $M$ and $N$, respectively. Denote by $O$ the midpoint of $B C$. The bisectors of the angles $B A C$ and $M O N$ intersect at $R$. Prove that the circumcircles of the triangles $B M R$ and $C N R$ have a common point lying on the line segment $B C$.", "solution": "16. Note that $\\triangle A N M \\sim \\triangle A B C$ and consequently $A M \\neq A N$. Since $O M=$ $O N$, it follows that $O R$ is a perpendicular bisector of $M N$. Thus, $R$ is the common point of the median of $M N$ and the bisector of $\\angle M A N$. Then it follows from a well-known fact that $R$ lies on the circumcircle of $\\triangle A M N$. Let $K$ be the intersection of $A R$ and $B C$. We then have $\\angle M R A=$ $\\angle M N A=\\angle A B K$ and $\\angle N R A=\\angle N M A=\\angle A C K$, from which we conclude that $R M B K$ and $R N C K$ are cyclic. Thus $K$ is the desired intersection of the circumcircles of $\\triangle B M R$ and $\\triangle C N R$ and it indeed lies on $B C$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "17", "problem_type": "Geometry", "exam": "IMO", "problem": "17. G2 (KAZ) The circle $\\Gamma$ and the line $\\ell$ do not intersect. Let $A B$ be the diameter of $\\Gamma$ perpendicular to $\\ell$, with $B$ closer to $\\ell$ than $A$. An arbitrary point $C \\neq A, B$ is chosen on $\\Gamma$. The line $A C$ intersects $\\ell$ at $D$. The line $D E$ is tangent to $\\Gamma$ at $E$, with $B$ and $E$ on the same side of $A C$. Let $B E$ intersect $\\ell$ at $F$, and let $A F$ intersect $\\Gamma$ at $G \\neq A$. Prove that the reflection of $G$ in $A B$ lies on the line $C F$.", "solution": "17. Let $H$ be the reflection of $G$ about $A B(G H \\| \\ell)$. Let $M$ be the intersection of $A B$ and $\\ell$. Since $\\angle F E A=\\angle F M A=90^{\\circ}$, it follows that $A E M F$ is cyclic and hence $\\angle D F E=\\angle B A E=\\angle D E F$. The last equality holds because $D E$ is tangent to $\\Gamma$. It follows that $D E=$ $D F$ and hence $D F^{2}=D E^{2}=$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-731.jpg?height=396&width=459&top_left_y=1591&top_left_x=841) $D C \\cdot D A$ (the power of $D$ with respect to $\\Gamma$ ). It then follows that $\\angle D C F=\\angle D F A=\\angle H G A=\\angle H C A$. Thus it follows that $H$ lies on $C F$ as desired.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "18", "problem_type": "Geometry", "exam": "IMO", "problem": "18. G3 (KOR) Let $O$ be the circumcenter of an acute-angled triangle $A B C$ with $\\angle B<\\angle C$. The line $A O$ meets the side $B C$ at $D$. The circumcenters of the triangles $A B D$ and $A C D$ are $E$ and $F$, respectively. Extend the sides $B A$ and $C A$ beyond $A$, and choose on the respective extension points $G$ and $H$ such that $A G=A C$ and $A H=A B$. Prove that the quadrilateral $E F G H$ is a rectangle if and only if $\\angle A C B-\\angle A B C=60^{\\circ}$.", "solution": "18. It is important to note that since $\\beta<\\gamma, \\angle A D C=90^{\\circ}-\\gamma+\\beta$ is acute. It is elementary that $\\angle C A O=90^{\\circ}-\\beta$. Let $X$ and $Y$ respectively be the intersections of $F E$ and $G H$ with $A D$. We trivially get $X \\in E F \\perp A D$ and $\\triangle A G H \\cong \\triangle A C B$. Consequently, $\\angle G A Y=\\angle O A B=90^{\\circ}-\\gamma=$ $90^{\\circ}-\\angle A G Y$. Hence, $G H \\perp A D$ and thus $G H \\| F E$. That $E F G H$ is a rectangle is now equivalent to $F X=G Y$ and $E X=H Y$. We have that $G Y=A G \\sin \\gamma=A C \\sin \\gamma$ and $F X=A F \\sin \\gamma$ (since $\\angle A F X=\\gamma$ ). Thus,  $$ F X=G Y \\Leftrightarrow C F=A F=A C \\Leftrightarrow \\angle A F C=60^{\\circ} \\Leftrightarrow \\angle A D C=30^{\\circ} . $$  Since $\\angle A D C=180^{\\circ}-\\angle D C A-\\angle D A C=180^{\\circ}-\\gamma-\\left(90^{\\circ}-\\beta\\right)$, it immediately follows that $F X=G Y \\Leftrightarrow \\gamma-\\beta=60^{\\circ}$. We similarly obtain $E X=H Y \\Leftrightarrow \\gamma-\\beta=60^{\\circ}$, proving the statement of the problem.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "19", "problem_type": null, "exam": "IMO", "problem": "19. $\\mathbf{G} 4 \\mathbf{( P O L})^{\\mathrm{IMO}}$ In a convex quadrilateral $A B C D$ the diagonal $B D$ does not bisect the angles $A B C$ and $C D A$. The point $P$ lies inside $A B C D$ and satisfies  $$ \\angle P B C=\\angle D B A \\quad \\text { and } \\quad \\angle P D C=\\angle B D A . $$  Prove that $A B C D$ is a cyclic quadrilateral if and only if $A P=C P$.", "solution": "19. Assume first that the points $A, B, C, D$ are concyclic. Let the lines $B P$ and $D P$ meet the circumcircle of $A B C D$ again at $E$ and $F$, respectively. Then it follows from the given conditions that $\\widehat{A B}=\\widehat{C F}$ and $\\widehat{A D}=\\widehat{C E}$; hence $B F \\| A C$ and $D E \\| A C$. Therefore $B F E D$ and $B F A C$ are isosceles trapezoids and thus $P=B E \\cap D F$ lies on the common bisector of segments $B F, E D, A C$. Hence $A P=C P$. Assume in turn that $A P=C P$. Let $P$ w.l.o.g. lie in the triangles $A C D$ and $B C D$. Let $B P$ and $D P$ meet $A C$ at $K$ and $L$, respectively. The points $A$ and $C$ are isogonal conjugates with respect to $\\triangle B D P$, which implies that $\\angle A P K=\\angle C P L$. Since $A P=C P$, we infer that $K$ and $L$ are symmetric with respect to the perpendicular bisector $p$ of $A C$. Let $E$ be the reflection of $D$ in $p$. Then $E$ lies on the line $B P$, and the triangles $A P D$ and $C P E$ are congruent. Thus $\\angle B D C=\\angle A D P=\\angle B E C$, which means that the points $B, C, E, D$ are concyclic. Moreover, $A, C, E, D$ are also concyclic. Hence, $A B C D$ is a cyclic quadrilateral.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "2", "problem_type": "Algebra", "exam": "IMO", "problem": "2. A2 (ROM) An infinite sequence $a_{0}, a_{1}, a_{2}, \\ldots$ of real numbers satisfies the condition  $$ a_{n}=\\left|a_{n+1}-a_{n+2}\\right| \\text { for every } n \\geq 0 $$  with $a_{0}$ and $a_{1}$ positive and distinct. Can this sequence be bounded?", "solution": "2. We claim that the sequence $\\left\\{a_{n}\\right\\}$ must be unbounded.  The condition of the sequence is equivalent to $a_{n}>0$ and $a_{n+1}=a_{n}+a_{n-1}$ or $a_{n}-a_{n-1}$. In particular, if $a_{n}<a_{n-1}$, then $a_{n+1}>\\max \\left\\{a_{n}, a_{n-1}\\right\\}$. Let us remove all $a_{n}$ such that $a_{n}<a_{n-1}$. The obtained sequence $\\left(b_{m}\\right)_{m \\in \\mathbb{N}}$ is strictly increasing. Thus the statement of the problem will follow if we prove that $b_{m+1}-b_{m} \\geq b_{m}-b_{m-1}$ for all $m \\geq 2$. Let $b_{m+1}=a_{n+2}$ for some $n$. Then $a_{n+2}>a_{n+1}$. We distinguish two cases: (i) If $a_{n+1}>a_{n}$, we have $b_{m}=a_{n+1}$ and $b_{m-1} \\geq a_{n-1}$ (since $b_{m-1}$ is either $a_{n-1}$ or $a_{n}$ ). Then $b_{m+1}-b_{m}=a_{n+2}-a_{n+1}=a_{n}=a_{n+1}-$ $a_{n-1}=b_{m}-a_{n-1} \\geq b_{m}-b_{m-1}$. (ii) If $a_{n+1}<a_{n}$, we have $b_{m}=a_{n}$ and $b_{m-1} \\geq a_{n-1}$. Consequently, $b_{m+1}-b_{m}=a_{n+2}-a_{n}=a_{n+1}=a_{n}-a_{n-1}=b_{m}-a_{n-1} \\geq b_{m}-b_{m-1}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "20", "problem_type": "Geometry", "exam": "IMO", "problem": "20. G5 (SMN) Let $A_{1} A_{2} \\ldots A_{n}$ be a regular $n$-gon. The points $B_{1}, \\ldots, B_{n-1}$ are defined as follows: (i) If $i=1$ or $i=n-1$, then $B_{i}$ is the midpoint of the side $A_{i} A_{i+1}$. (ii) If $i \\neq 1, i \\neq n-1$, and $S$ is the intersection point of $A_{1} A_{i+1}$ and $A_{n} A_{i}$, then $B_{i}$ is the intersection point of the bisector of the angle $A_{i} S A_{i+1}$ with $A_{i} A_{i+1}$. Prove the equality  $$ \\angle A_{1} B_{1} A_{n}+\\angle A_{1} B_{2} A_{n}+\\cdots+\\angle A_{1} B_{n-1} A_{n}=180^{\\circ} . $$", "solution": "20. We first establish the following lemma.  Lemma. Let $A B C D$ be an isosceles trapezoid with bases $A B$ and $C D$. The diagonals $A C$ and $B D$ intersect at $S$. Let $M$ be the midpoint of $B C$, and let the bisector of the angle $B S C$ intersect $B C$ at $N$. Then $\\angle A M D=\\angle A N D$. Proof. It suffices to show that the points $A, D, M, N$ are concyclic. The statement is trivial for $A D \\| B C$. Let us now assume that $A D$ and $B C$ meet at $X$, and let $X A=X B=a, X C=X D=b$. Since $S N$ is the bisector of $\\angle C S B$, we have  $$ \\frac{a-X N}{X N-b}=\\frac{B N}{C N}=\\frac{B S}{C S}=\\frac{A B}{C D}=\\frac{a}{b} $$  and an easy computation yields $X N=\\frac{2 a b}{a+b}$. We also have $X M=\\frac{a+b}{2}$; hence $X M \\cdot X N=X A \\cdot X D$. Therefore $A, D, M, N$ are concyclic, as needed.  Denote by $C_{i}$ the midpoint of the side $A_{i} A_{i+1}, i=1, \\ldots, n-1$. By definition $C_{1}=B_{1}$ and $C_{n-1}=B_{n-1}$. Since $A_{1} A_{i} A_{i+1} A_{n}$ is an isosceles trapezoid with $A_{1} A_{i} \\| A_{i+1} A_{n}$ for $i=2, \\ldots, n-2$, it follows from the lemma that $\\angle A_{1} B_{i} A_{n}=\\angle A_{1} C_{i} A_{n}$ for all $i$. The sum in consideration thus equals $\\angle A_{1} C_{1} A_{n}+\\angle A_{1} C_{2} A_{n}+\\cdots+$ $\\angle A_{1} C_{n-1} A_{n}$. Moreover, the triangles $A_{1} C_{i} A_{n}$ and $A_{n+2-i} C_{1} A_{n+1-i}$ are congruent (a rotation about the center of the $n$-gon carries the first one to the second), and consequently  $$ \\angle A_{1} C_{i} A_{n}=\\angle A_{n+2-i} C_{1} A_{n+1-i} $$  ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-733.jpg?height=417&width=470&top_left_y=415&top_left_x=847) for $i=2, \\ldots, n-1$. Hence $\\Sigma=\\angle A_{1} C_{1} A_{n}+\\angle A_{n} C_{1} A_{n-1}+\\cdots+\\angle A_{3} C_{1} A_{2}=\\angle A_{1} C_{1} A_{2}=180^{\\circ}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "21", "problem_type": "Geometry", "exam": "IMO", "problem": "21. G6 (GBR) Let $\\mathcal{P}$ be a convex polygon. Prove that there is a convex hexagon that is contained in $\\mathcal{P}$ and that occupies at least 75 percent of the area of $\\mathcal{P}$.", "solution": "21. Let $A B C$ be the triangle of maximum area $S$ contained in $\\mathcal{P}$ (it exists because of compactness of $\\mathcal{P}$ ). Draw parallels to $B C, C A, A B$ through $A, B, C$, respectively, and denote the triangle thus obtained by $A_{1} B_{1} C_{1}$ ( $A \\in B_{1} C_{1}$, etc.). Since each triangle with vertices in $\\mathcal{P}$ has area at most $S$, the entire polygon $\\mathcal{P}$ is contained in $A_{1} B_{1} C_{1}$. Next, draw lines of support of $\\mathcal{P}$ parallel to $B C, C A, A B$ and not intersecting the triangle $A B C$. They determine a convex hexagon $U_{a} V_{a} U_{b} V_{b} U_{c} V_{c}$ containing $\\mathcal{P}$, with $V_{b}, U_{c} \\in B_{1} C_{1}, V_{c}, U_{a} \\in C_{1} A_{1}, V_{a}, U_{b} \\in A_{1} B_{1}$. Each of the line segments $U_{a} V_{a}, U_{b} V_{b}, U_{c} V_{c}$ contains points of $\\mathcal{P}$. Choose such points $A_{0}, B_{0}, C_{0}$ on $U_{a} V_{a}, U_{b} V_{b}, U_{c} V_{c}$, respectively. The convex hexagon $A C_{0} B A_{0} C B_{0}$ is contained in $\\mathcal{P}$, because the latter is convex. We prove that $A C_{0} B A_{0} C B_{0}$ has area at least $3 / 4$ the area of $\\mathcal{P}$. Let $x, y, z$ denote the areas of triangles $U_{a} B C, U_{b} C A$, and $U_{c} A B$. Then $S_{1}=S_{A C_{0} B A_{0} C B_{0}}=S+x+y+z$. On the other hand, the triangle $A_{1} U_{a} V_{a}$ is similar to $\\triangle A_{1} B C$ with similitude $\\tau=(S-x) / S$, and hence its area is $\\tau^{2} S=(S-x)^{2} / S$. Thus the area of quadrilateral $U_{a} V_{a} C B$ is $S-(S-x)^{2} / S=2 z-z^{2} / S$. Analogous formulas hold for quadrilaterals $U_{b} V_{b} A C$ and $U_{c} V_{c} B A$. Therefore  $$ \\begin{aligned} S_{\\mathcal{P}} & \\leq S_{U_{a} V_{a} U_{b} V_{b} U_{c} V_{c}}=S+S_{U_{a} V_{a} C B}+S_{U_{b} V_{b} A C}+S_{U_{c} V_{c} B A} \\\\ & =S+2(x+y+z)-\\frac{x^{2}+y^{2}+z^{2}}{S} \\\\ & \\leq S+2(x+y+z)-\\frac{(x+y+z)^{2}}{3 S} . \\end{aligned} $$  Now $4 S_{1}-3 S_{\\mathcal{P}} \\geq=S-2(x+y+z)+(x+y+z)^{2} / S=(S-x-y-z)^{2} / S \\geq 0$; i.e., $S_{1} \\geq 3 S_{\\mathcal{P}} / 4$, as claimed.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "22", "problem_type": "Geometry", "exam": "IMO", "problem": "22. G7 (RUS) For a given triangle $A B C$, let $X$ be a variable point on the line $B C$ such that $C$ lies between $B$ and $X$ and the incircles of the triangles $A B X$ and $A C X$ intersect at two distinct points $P$ and $Q$. Prove that the line $P Q$ passes through a point independent of $X$.", "solution": "22. The proof uses the following observation:  Lemma. In a triangle $A B C$, let $K, L$ be the midpoints of the sides $A C, A B$, respectively, and let the incircle of the triangle touch $B C, C A$ at $D, E$, respectively. Then the lines $K L$ and $D E$ intersect on the bisector of the angle $A B C$. Proof. Let the bisector $\\ell_{b}$ of $\\angle A B C$ meet $D E$ at $T$. One can assume that $A B \\neq B C$, or else $T \\equiv K \\in K L$. Note that the incenter $I$ of $\\triangle A B C$ is between $B$ and $T$, and also $T \\neq E$. From the triangles $B D T$ and $D E C$ we obtain $\\angle I T D=\\alpha / 2=\\angle I A E$, which implies that $A, I, T, E$ are concyclic. Then $\\angle A T B=\\angle A E I=90^{\\circ}$. Thus $L$ is the circumcenter of $\\triangle A T B$ from which $\\angle L T B=\\angle L B T=\\angle T B C \\Rightarrow L T \\| B C \\Rightarrow T \\in$ $K L$, which is what we were supposed to prove. Let the incircles of $\\triangle A B X$ and $\\triangle A C X$ touch $B X$ at $D$ and $F$, respectively, and let them touch $A X$ at $E$ and $G$, respectively. Clearly, $D E$ and $F G$ are parallel. If the line $P Q$ intersects $B X$ and $A X$ at $M$ and $N$, respectively, then $M D^{2}=M P \\cdot M Q=M F^{2}$, i.e., $M D=M F$ and analogously $N E=N G$. It follows that $P Q$ is parallel to $D E$ and $F G$ and equidistant from them. The midpoints of $A B, A C$, and $A X$ lie on the same line $m$, parallel to $B C$. Applying the lemma to $\\triangle A B X$, we conclude that $D E$ passes through the common point $U$ of $m$ and the bisector of $\\angle A B X$. Analogously, $F G$ passes through the common point $V$ of $m$ and the bisector of $\\angle A C X$. Therefore $P Q$ passes through the midpoint $W$ of the line segment $U V$. Since $U, V$ do not depend on $X$, neither does $W$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "23", "problem_type": "Geometry", "exam": "IMO", "problem": "23. G8 (SMN) A cyclic quadrilateral $A B C D$ is given. The lines $A D$ and $B C$ intersect at $E$, with $C$ between $B$ and $E$; the diagonals $A C$ and $B D$ intersect at $F$. Let $M$ be the midpoint of the side $C D$, and let $N \\neq M$ be a point on the circumcircle of the triangle $A B M$ such that $A N / B N=$ $A M / B M$. Prove that the points $E, F$, and $N$ are collinear.", "solution": "23. To start with, note that point $N$ is uniquely determined by the imposed properties. Indeed, $f(X)=A X / B X$ is a monotone function on both arcs $A B$ of the circumcircle of $\\triangle A B M$. Denote by $P$ and $Q$ respectively the second points of intersection of the line $E F$ with the circumcircles of $\\triangle A B E$ and $\\triangle A B F$. The problem is equivalent to showing that $N \\in P Q$. In fact, we shall prove that $N$ coincides with the midpoint $\\bar{N}$ of segment $P Q$. The cyclic quadrilaterals $A P B E$, $A Q B F$, and $A B C D$ yield $\\angle A P Q=$ $180^{\\circ}-\\angle A P E=180^{\\circ}-\\angle A B E=$ $\\angle A D C$ and $\\angle A Q P=\\angle A Q F=$ $\\angle A B F=\\angle A C D$. It follows that $\\triangle A P Q \\sim \\triangle A D C$, and consequently $\\triangle A \\bar{N} P \\sim \\triangle A M D$. Analo- ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-734.jpg?height=604&width=446&top_left_y=1342&top_left_x=857) gously $\\triangle B \\bar{N} P \\sim \\triangle B M C$. Therefore $A \\bar{N} / A M=P Q / D C=B \\bar{N} / B M$, i.e., $A \\bar{N} / B \\bar{N}=A M / B M$. Moreover, $\\angle A \\bar{N} B=\\angle A \\bar{N} P+\\angle P \\bar{N} B=$ $\\angle A M D+\\angle B M C=180^{\\circ}-\\angle A M B$, which means that point $\\bar{N}$ lies on the circumcircle of $\\triangle A M B$. By the uniqueness of $N$, we conclude that $\\bar{N} \\equiv N$, which completes the solution.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "24", "problem_type": "Number Theory", "exam": "IMO", "problem": "24. N1 (BLR) Let $\\tau(n)$ denote the number of positive divisors of the positive integer $n$. Prove that there exist infinitely many positive integers $a$ such that the equation  $$ \\tau(a n)=n $$  does not have a positive integer solution $n$.", "solution": "24. Setting $m=a n$ we reduce the given equation to $m / \\tau(m)=a$.  Let us show that for $a=p^{p-1}$ the above equation has no solutions in $\\mathbb{N}$ if $p>3$ is a prime. Assume to the contrary that $m \\in \\mathbb{N}$ is such that $m=p^{p-1} \\tau(m)$. Then $p^{p-1} \\mid m$, so we may set $m=p^{\\alpha} k$, where $\\alpha, k \\in \\mathbb{N}$, $\\alpha \\geq p-1$, and $p \\nmid k$. Let $k=p_{1}^{\\alpha_{1}} \\cdots p_{r}^{\\alpha_{r}}$ be the decomposition of $k$ into primes. Then $\\tau(k)=\\left(\\alpha_{1}+1\\right) \\cdots\\left(\\alpha_{r}+1\\right)$ and $\\tau(m)=(\\alpha+1) \\tau(k)$. Our equation becomes  $$ p^{\\alpha-p+1} k=(\\alpha+1) \\tau(k) . $$  We observe that $\\alpha \\neq p-1$ : otherwise the RHS would be divisible by $p$ and the LHS would not be so. It follows that $\\alpha \\geq p$, which also easily implies that $p^{\\alpha-p+1} \\geq \\frac{p}{p+1}(\\alpha+1)$. Furthermore, since $\\alpha+1$ cannot be divisible by $p^{\\alpha-p+1}$ for any $\\alpha \\geq p$, it follows that $p \\mid \\tau(k)$. Thus if $p \\mid \\tau(k)$, then at least one $\\alpha_{i}+1$ is divisible by $p$ and consequently $\\alpha_{i} \\geq p-1$ for some $i$. Hence $k \\geq \\frac{p_{i}^{\\alpha_{i}}}{\\alpha_{i}+1} \\tau(k) \\geq \\frac{2^{p-1}}{p} \\tau(k)$. But then we have  $$ p^{\\alpha-p+1} k \\geq \\frac{p}{p+1}(\\alpha+1) \\cdot \\frac{2^{p-1}}{p} \\tau(k)>(\\alpha+1) \\tau(k), $$  contradicting (1). Therefore (1) has no solutions in $\\mathbb{N}$. Remark. There are many other values of $a$ for which the considered equation has no solutions in $\\mathbb{N}$ : for example, $a=6 p$ for a prime $p \\geq 5$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "25", "problem_type": "Number Theory", "exam": "IMO", "problem": "25. N2 (RUS) The function $\\psi$ from the set $\\mathbb{N}$ of positive integers into itself is defined by the equality  $$ \\psi(n)=\\sum_{k=1}^{n}(k, n), \\quad n \\in \\mathbb{N} $$  where $(k, n)$ denotes the greatest common divisor of $k$ and $n$. (a) Prove that $\\psi(m n)=\\psi(m) \\psi(n)$ for every two relatively prime $m, n \\in$ $\\mathbb{N}$. (b) Prove that for each $a \\in \\mathbb{N}$ the equation $\\psi(x)=a x$ has a solution. (c) Find all $a \\in \\mathbb{N}$ such that the equation $\\psi(x)=a x$ has a unique solution.", "solution": "25. Let $n$ be a natural number. For each $k=1,2, \\ldots, n$, the number $(k, n)$ is a divisor of $n$. Consider any divisor $d$ of $n$. If $(k, n)=n / d$, then $k=n l / d$ for some $l \\in \\mathbb{N}$, and $(k, n)=(l, d) n / d$, which implies that $l$ is coprime to $d$ and $l \\leq d$. It follows that $(k, n)$ is equal to $n / d$ for exactly $\\varphi(d)$ natural numbers $k \\leq n$. Therefore  $$ \\psi(n)=\\sum_{k=1}^{n}(k, n)=\\sum_{d \\mid n} \\varphi(d) \\frac{n}{d}=n \\sum_{d \\mid n} \\frac{\\varphi(d)}{d} $$  (a) Let $n, m$ be coprime. Then each divisor $f$ of $m n$ can be uniquely expressed as $f=d e$, where $d \\mid n$ and $e \\mid m$. We now have by (1)  $$ \\begin{aligned} \\psi(m n) & =m n \\sum_{f \\mid m n} \\frac{\\varphi(f)}{f}=m n \\sum_{d|n, e| m} \\frac{\\varphi(d e)}{d e} \\\\ & =m n \\sum_{d|n, e| m} \\frac{\\varphi(d)}{d} \\frac{\\varphi(e)}{e}=\\left(n \\sum_{d \\mid n} \\frac{\\varphi(d)}{d}\\right)\\left(m \\sum_{e \\mid m} \\frac{\\varphi(e)}{e}\\right) \\\\ & =\\psi(m) \\psi(n) . \\end{aligned} $$  (b) Let $n=p^{k}$, where $p$ is a prime and $k$ a positive integer. According to (1),  $$ \\frac{\\psi(n)}{n}=\\sum_{i=0}^{k} \\frac{\\varphi\\left(p^{i}\\right)}{p^{i}}=1+\\frac{k(p-1)}{p} $$  Setting $p=2$ and $k=2(a-1)$ we obtain $\\psi(n)=a n$ for $n=2^{2(a-1)}$. (c) We note that $\\psi\\left(p^{p}\\right)=p^{p+1}$ if $p$ is a prime. Hence, if $a$ has an odd prime factor $p$ and $a_{1}=a / p$, then $x=p^{p} 2^{2 a_{1}-2}$ is a solution of $\\psi(x)=a x$ different from $x=2^{2 a-2}$. Now assume that $a=2^{k}$ for some $k \\in \\mathbb{N}$. Suppose $x=2^{\\alpha} y$ is a positive integer such that $\\psi(x)=2^{k} x$. Then $2^{\\alpha+k} y=\\psi(x)=\\psi\\left(2^{\\alpha}\\right) \\psi(y)=$ $(\\alpha+2) 2^{\\alpha-1} \\psi(y)$, i.e., $2^{k+1} y=(\\alpha+2) \\psi(y)$. We notice that for each odd $y, \\psi(y)$ is (by definition) the sum of an odd number of odd summands and therefore odd. It follows that $\\psi(y) \\mid y$. On the other hand, $\\psi(y)>$ $y$ for $y>1$, so we must have $y=1$. Consequently $\\alpha=2^{k+1}-2=2 a-2$, giving us the unique solution $x=2^{2 a-2}$. Thus $\\psi(x)=a x$ has a unique solution if and only if $a$ is a power of 2 .", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "26", "problem_type": "Number Theory", "exam": "IMO", "problem": "26. N3 (IRN) A function $f$ from the set of positive integers $\\mathbb{N}$ into itself is such that for all $m, n \\in \\mathbb{N}$ the number $\\left(m^{2}+n\\right)^{2}$ is divisible by $f^{2}(m)+$ $f(n)$. Prove that $f(n)=n$ for each $n \\in \\mathbb{N}$.", "solution": "26. For $m=n=1$ we obtain that $f(1)^{2}+f(1)$ divides $\\left(1^{2}+1\\right)^{2}=4$, from which we find that $f(1)=1$. Next, we show that $f(p-1)=p-1$ for each prime $p$. By the hypothesis for $m=1$ and $n=p-1, f(p-1)+1$ divides $p^{2}$, so $f(p-1)$ equals either $p-1$ or $p^{2}-1$. If $f(p-1)=p^{2}-1$, then $f(1)+f(p-1)^{2}=p^{4}-2 p^{2}+2$ divides $\\left(1+(p-1)^{2}\\right)^{2}<p^{4}-2 p^{2}+2$, giving a contradiction. Hence $f(p-1)=p-1$. Let us now consider an arbitrary $n \\in \\mathbb{N}$. By the hypothesis for $m=p-1$, $A=f(n)+(p-1)^{2}$ divides $\\left(n+(p-1)^{2}\\right)^{2} \\equiv(n-f(n))^{2}(\\bmod A)$, and hence $A$ divides $(n-f(n))^{2}$ for any prime $p$. Taking $p$ large enough, we can obtain $A$ to be greater than $(n-f(n))^{2}$, which implies that $(n-f(n))^{2}=0$, i.e., $f(n)=n$ for every $n$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "27", "problem_type": "Number Theory", "exam": "IMO", "problem": "27. N4 (POL) Let $k$ be a fixed integer greater than 1 , and let $m=4 k^{2}-5$. Show that there exist positive integers $a$ and $b$ such that the sequence $\\left(x_{n}\\right)$ defined by  $$ x_{0}=a, \\quad x_{1}=b, \\quad x_{n+2}=x_{n+1}+x_{n} \\quad \\text { for } \\quad n=0,1,2, \\ldots $$  has all of its terms relatively prime to $m$.", "solution": "27. Set $a=1$ and assume that $b \\in \\mathbb{N}$ is such that $b^{2} \\equiv b+1(\\bmod m)$. An easy induction gives us $x_{n} \\equiv b^{n}(\\bmod m)$ for all $n \\in \\mathbb{N}_{0}$. Moreover, $b$ is obviously coprime to $m$, and hence each $x_{n}$ is coprime to $m$. It remains to show the existence of $b$. The congruence $b^{2} \\equiv b+1(\\bmod$ $m)$ is equivalent to $(2 b-1)^{2} \\equiv 5(\\bmod m)$. Taking $2 b-1 \\equiv 2 k$, i.e., $b \\equiv 2 k^{2}+k-2(\\bmod m)$, does the job. Remark. A desired $b$ exists whenever 5 is a quadratic residue modulo $m$, in particular, when $m$ is a prime of the form $10 k \\pm 1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "28", "problem_type": "Number Theory", "exam": "IMO", "problem": "28. N5 (IRN) ${ }^{\\mathrm{IMO} 6}$ We call a positive integer alternate if its decimal digits are alternately odd and even. Find all positive integers $n$ such that $n$ has an alternate multiple.", "solution": "28. If $n$ is divisible by 20 , then every multiple of $n$ has two last digits even and hence it is not alternate. We shall show that any other $n$ has an alternate multiple. (i) Let $n$ be coprime to 10 . For each $k$ there exists a number $A_{k}(n)=$ $\\overline{10 \\ldots 010 \\ldots 01 \\ldots 0 \\ldots 01}=\\frac{10^{m k}-1}{10^{k}-1}(m \\in \\mathbb{N})$ that is divisible by $n$ (by Euler's theorem, choose $\\left.m=\\varphi\\left[n\\left(10^{k}-1\\right)\\right]\\right)$. In particular, $A_{2}(n)$ is alternate. (ii) Let $n=2 \\cdot 5^{r} \\cdot n_{1}$, where $r \\geq 1$ and $\\left(n_{1}, 10\\right)=1$. We shall show by induction that, for each $k$, there exists an alternative $k$-digit odd number $M_{k}$ that is divisible by $5^{k}$. Choosing the number $10 A_{2 r}\\left(n_{1}\\right) M_{2 r}$ will then solve this case, since it is clearly alternate and divisible by $n$. We can trivially choose $M_{1}=5$. Let there be given an alternate $r$-digit multiple $M_{r}$ of $5^{r}$, and let $c \\in\\{0,1,2,3,4\\}$ be such that $M_{r} / 5^{r} \\equiv$ $-c \\cdot 2^{r}(\\bmod 5)$. Then the $(r+1)$ digit numbers $M_{r}+c \\cdot 10^{r}$ and $M_{r}+(5+c) \\cdot 10^{r}$ are respectively equal to $5^{r}\\left(M_{r} / 5^{r}+2^{r} \\cdot c\\right)$ and $5^{r}\\left(M_{r} / 5^{r}+2^{r} \\cdot c+5 \\cdot 2^{r}\\right)$, and hence they are divisible by $5^{r+1}$ and exactly one of them is alternate: we set it to be $M_{r+1}$. (iii) Let $n=2^{r} \\cdot n_{1}$, where $r \\geq 1$ and $\\left(n_{1}, 10\\right)=1$. We show that there exists an alternate $2 r$-digit number $N_{r}$ that is divisible by $2^{2 r+1}$. Choosing the number $A_{2 r}\\left(n_{1}\\right) N_{r}$ will then solve this case. We choose $N_{1}=16$, and given $N_{r}$, we can prove that one of $N_{r}+$ $m \\cdot 10^{2 r}$, for $m \\in\\{10,12,14,16\\}$, is divisible by $2^{2 r+3}$ and therefore suitable for $N_{r+1}$. Indeed, for $N_{r}=2^{2 r+1} d$ we have $N_{r}+m \\cdot 10^{2 r}=$ $2^{2 r+1}\\left(d+5^{r} m / 2\\right)$ and $d+5^{r} m / 2 \\equiv 0(\\bmod 4)$ has a solution $m / 2 \\in$ $\\{5,6,7,8\\}$ for each $d$ and $r$. Remark. The idea is essentially the same as in (SL94-24).", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "29", "problem_type": "Number Theory", "exam": "IMO", "problem": "29. N6 (IRE) Given an integer $n>1$, denote by $P_{n}$ the product of all positive integers $x$ less than $n$ and such that $n$ divides $x^{2}-1$. For each $n>1$, find the remainder of $P_{n}$ on division by $n$.", "solution": "29. Let $S_{n}=\\left\\{x \\in \\mathbb{N}|x \\leq n, n| x^{2}-1\\right\\}$. It is easy to check that $P_{n} \\equiv 1$ $(\\bmod n)$ for $n=2$ and $P_{n} \\equiv-1(\\bmod n)$ for $n \\in\\{3,4\\}$, so from now on we assume $n>4$. We note that if $x \\in S_{n}$, then also $n-x \\in S_{n}$ and $(x, n)=1$. Thus $S_{n}$ splits into pairs $\\{x, n-x\\}$, where $x \\in S_{n}$ and $x \\leq n / 2$. In each of these pairs the product of elements gives remainder -1 upon division by $n$. Therefore $P_{n} \\equiv(-1)^{m}$, where $S_{n}$ has $2 m$ elements. It remains to find the parity of $m$. Suppose first that $n>4$ is divisible by 4 . Whenever $x \\in S_{n}$, the numbers $|n / 2-x|, n-x, n-|n / 2-x|$ also belong to $S_{n}$ (indeed, $n \\mid(n / 2-x)^{2}-1=$ $n^{2} / 4-n x+x^{2}-1$ because $n \\mid n^{2} / 4$, etc.). In this way the set $S_{n}$ splits into four-element subsets $\\{x, n / 2-x, n / 2+x, n-x\\}$, where $x \\in S_{n}$ and $x<n / 4$ (elements of these subsets are different for $x \\neq n / 4$, and $n / 4$ doesn't belong to $S_{n}$ for $\\left.n>4\\right)$. Therefore $m=\\left|S_{n}\\right| / 2$ is even and $P_{n} \\equiv 1$ $(\\bmod m)$. Now let $n$ be odd. If $n \\mid x^{2}-1=(x-1)(x+1)$, then there exist natural numbers $a, b$ such that $a b=n, a|x-1, b| x+1$. Obviously $a$ and $b$ are coprime. Conversely, given any odd $a, b \\in \\mathbb{N}$ such that $(a, b)=1$ and $a b=n$, by the Chinese remainder theorem there exists $x \\in\\{1,2, \\ldots, n-1\\}$ such that $a \\mid x-1$ and $b \\mid x+1$. This gives a bijection between all ordered pairs $(a, b)$ with $a b=n$ and $(a, b)=1$ and the elements of $S_{n}$. Now if $n=p_{1}^{\\alpha_{1}} \\cdots p_{k}^{\\alpha_{k}}$ is the decomposition of $n$ into primes, the number of pairs $(a, b)$ is equal to $2^{k}$ (since for every $i$, either $p_{i}^{\\alpha_{i}} \\mid a$ or $p_{i}^{\\alpha_{i}} \\mid b$ ), and hence $m=2^{k-1}$. Thus $P_{n} \\equiv-1(\\bmod n)$ if $n$ is a power of an odd prime, and $P_{n} \\equiv 1$ otherwise. Finally, let $n$ be even but not divisible by 4 . Then $x \\in S_{n}$ if and only if $x$ or $n-x$ belongs to $S_{n / 2}$ and $x$ is odd. Since $n / 2$ is odd, for each $x \\in S_{n / 2}$ either $x$ or $x+n / 2$ belongs to $S_{n}$, and by the case of $n$ odd we have $S_{n} \\equiv \\pm 1(\\bmod n / 2)$, depending on whether or not $n / 2$ is a power of a prime. Since $S_{n}$ is odd, it follows that $P_{n} \\equiv-1(\\bmod n)$ if $n / 2$ is a power of a prime, and $P_{n} \\equiv 1$ otherwise. Second solution. Obviously $S_{n}$ is closed under multiplication modulo $n$. This implies that $S_{n}$ with multiplication modulo $n$ is a subgroup of $\\mathbb{Z}_{n}$, and therefore there exist elements $a_{1}=-1, a_{2}, \\ldots, a_{k} \\in S_{n}$ that generate $S_{n}$. In other words, since the $a_{i}$ are of order two, $S_{n}$ consists of products $\\prod_{i \\in A} a_{i}$, where $A$ runs over all subsets of $\\{1,2, \\ldots, k\\}$. Thus $S_{n}$ has $2^{k}$ elements, and the product of these elements equals $P_{n} \\equiv\\left(a_{1} a_{2} \\cdots a_{k}\\right)^{2^{k-1}}$ $(\\bmod n)$. Since $a_{i}^{2} \\equiv 1(\\bmod n)$, it follows that $P_{n} \\equiv 1$ if $k \\geq 2$, i.e., if $\\left|S_{n}\\right|>2$. Otherwise $P_{n} \\equiv-1(\\bmod n)$. We note that $\\left|S_{n}\\right|>2$ is equivalent to the existence of $a \\in S_{n}$ with $1<a<n-1$. It is easy to find that such an $a$ exists if and only if neither of $n, n / 2$ is a power of an odd prime.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "3", "problem_type": "Algebra", "exam": "IMO", "problem": "3. A3 (CAN) Does there exist a function $s: \\mathbb{Q} \\rightarrow\\{-1,1\\}$ such that if $x$ and $y$ are distinct rational numbers satisfying $x y=1$ or $x+y \\in\\{0,1\\}$, then $s(x) s(y)=-1$ ? Justify your answer.", "solution": "3. The answer is yes. Every rational number $x>0$ can be uniquely expressed as a continued fraction of the form $a_{0}+1 /\\left(a_{1}+1 /\\left(a_{2}+1 /\\left(\\cdots+1 / a_{n}\\right)\\right)\\right)$ (where $a_{0} \\in \\mathbb{N}_{0}, a_{1}, \\ldots, a_{n} \\in \\mathbb{N}$ ). Then we write $x=\\left[a_{0} ; a_{1}, a_{2}, \\ldots, a_{n}\\right]$. Since $n$ depends only on $x$, the function $s(x)=(-1)^{n}$ is well-defined. For $x<0$ we define $s(x)=-s(-x)$, and set $s(0)=1$. We claim that this $s(x)$ satisfies the requirements of the problem. The equality $s(x) s(y)=-1$ trivially holds if $x+y=0$. Suppose that $x y=1$. We may assume w.l.o.g. that $x>y>0$. Then $x>1$, so if $x=\\left[a_{0} ; a_{1}, a_{2}, \\ldots, a_{n}\\right]$, then $a_{0} \\geq 1$ and $y=0+1 / x=$ $\\left[0 ; a_{0}, a_{1}, a_{2}, \\ldots, a_{n}\\right]$. It follows that $s(x)=(-1)^{n}, s(y)=(-1)^{n+1}$, and hence $s(x) s(y)=-1$. Finally, suppose that $x+y=1$. We consider two cases: (i) Let $x, y>0$. We may assume w.l.o.g. that $x>1 / 2$. Then there exist natural numbers $a_{2}, \\ldots, a_{n}$ such that $x=\\left[0 ; 1, a_{2}, \\ldots, a_{n}\\right]=$ $1 /(1+1 / t)$, where $t=\\left[a_{2}, \\ldots, a_{n}\\right]$. Since $y=1-x=1 /(1+t)=$ $\\left[0 ; 1+a_{2}, a_{3}, \\ldots, a_{n}\\right]$, we have $s(x)=(-1)^{n}$ and $s(y)=(-1)^{n-1}$, giving us $s(x) s(y)=-1$. (ii) Let $x>0>y$. If $a_{0}, \\ldots, a_{n} \\in \\mathbb{N}$ are such that $-y=\\left[a_{0} ; a_{1}, \\ldots, a_{n}\\right]$, then $x=\\left[1+a_{0} ; a_{1}, \\ldots, a_{n}\\right]$. Thus $s(y)=-s(-y)=-(-1)^{n}$ and $s(x)=(-1)^{n}$, so again $s(x) s(y)=-1$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "30", "problem_type": "Number Theory", "exam": "IMO", "problem": "30. N7 (BUL) Let $p$ be an odd prime and $n$ a positive integer. In the coordinate plane, eight distinct points with integer coordinates lie on a circle with diameter of length $p^{n}$. Prove that there exists a triangle with vertices at three of the given points such that the squares of its side lengths are integers divisible by $p^{n+1}$.", "solution": "30. We shall denote by $k$ the given circle with diameter $p^{n}$.  Let $A, B$ be lattice points (i.e., points with integer coordinates). We shall denote by $\\mu(A B)$ the exponent of the highest power of $p$ that divides the integer $A B^{2}$. We observe that if $S$ is the area of a triangle $A B C$ where $A, B, C$ are lattice points, then $2 S$ is an integer. According to Heron's formula and the formula for the circumradius, a triangle $A B C$ whose circumcenter has diameter $p^{n}$ satisfies  $$ 2 A B^{2} B C^{2}+2 B C^{2} C A^{2}+2 C A^{2} A B^{2}-A B^{4}-B C^{4}-C A^{4}=16 S^{2} $$  and  $$ A B^{2} \\cdot B C^{2} \\cdot C A^{2}=(2 S)^{2} p^{2 n} $$  Lemma 1. Let $A, B$, and $C$ be lattice points on $k$. If none of $A B^{2}, B C^{2}$, $C A^{2}$ is divisible by $p^{n+1}$, then $\\mu(A B), \\mu(B C), \\mu(C A)$ are $0, n, n$ in some order. Proof. Let $k=\\min \\{\\mu(A B), \\mu(B C), \\mu(C A)\\}$. By (1), $(2 S)^{2}$ is divisible by $p^{2 k}$. Together with (2), this gives us $\\mu(A B)+\\mu(B C)+\\mu(C A)=$ $2 k+2 n$. On the other hand, if none of $A B^{2}, B C^{2}, C A^{2}$ is divisible by $p^{n+1}$, then $\\mu(A B)+\\mu(B C)+\\mu(C A) \\leq k+2 n$. Therefore $k=0$ and the remaining two of $\\mu(A B), \\mu(B C), \\mu(C A)$ are equal to $n$. Lemma 2. Among every four lattice points on $k$, there exist two, say $M, N$, such that $\\mu(M N) \\geq n+1$. Proof. Assume that this doesn't hold for some points $A, B, C, D$ on $k$. By Lemma $1, \\mu$ for some of the segments $A B, A C, \\ldots, C D$ is 0 , say $\\mu(A C)=0$. It easily follows by Lemma 1 that then $\\mu(B D)=0$ and $\\mu(A B)=\\mu(B C)=\\mu(C D)=\\mu(D A)=n$. Let $a, b, c, d, e, f \\in \\mathbb{N}$ be such that $A B^{2}=p^{n} a, B C^{2}=p^{n} b, C D^{2}=p^{n} c, D A^{2}=p^{n} d, A C^{2}=e$, $B D^{2}=f$. By Ptolemy's theorem we have $\\sqrt{e f}=p^{n}(\\sqrt{a c}+\\sqrt{b d})$. Taking squares, we get that $\\frac{e f}{p^{2 n}}=(\\sqrt{a c}+\\sqrt{b d})^{2}=a c+b d+2 \\sqrt{a b c d}$ is rational and hence an integer. It follows that ef is divisible by $p^{2 n}$, a contradiction. Now we consider eight lattice points $A_{1}, A_{2}, \\ldots, A_{8}$ on $k$. We color each segment $A_{i} A_{j}$ red if $\\mu\\left(A_{i} A_{j}\\right)>n$ and black otherwise, and thus obtain a graph $G$. The degree of a point $X$ will be the number of red segments with an endpoint in $X$. We distinguish three cases: (i) There is a point, say $A_{8}$, whose degree is at most 1 . We may suppose w.l.o.g. that $A_{8} A_{7}$ is red and $A_{8} A_{1}, \\ldots, A_{8} A_{6}$ black. By a well-known fact, the segments joining vertices $A_{1}, A_{2}, \\ldots, A_{6}$ determine either a red triangle, in which case there is nothing to prove, or a black triangle, say $A_{1} A_{2} A_{3}$. But in the latter case the four points $A_{1}, A_{2}, A_{3}, A_{8}$ do not determine any red segment, a contradiction to Lemma 2 . (ii) All points have degree 2. Then the set of red segments partitions into cycles. If one of these cycles has length 3 , then the proof is complete. If all the cycles have length at least 4 , then we have two possibilities: two 4 -cycles, say $A_{1} A_{2} A_{3} A_{4}$ and $A_{5} A_{6} A_{7} A_{8}$, or one 8-cycle, $A_{1} A_{2} \\ldots A_{8}$. In both cases, the four points $A_{1}, A_{3}, A_{5}, A_{7}$ do not determine any red segment, a contradiction. (iii) There is a point of degree at least 3 , say $A_{1}$. Suppose that $A_{1} A_{2}$, $A_{1} A_{3}$, and $A_{1} A_{4}$ are red. We claim that $A_{2}, A_{3}, A_{4}$ determine at least one red segment, which will complete the solution. If not, by Lemma $1, \\mu\\left(A_{2} A_{3}\\right), \\mu\\left(A_{3} A_{4}\\right), \\mu\\left(A_{4} A_{2}\\right)$ are $n, n, 0$ in some order. Assuming w.l.o.g. that $\\mu\\left(A_{2} A_{3}\\right)=0$, denote by $S$ the area of triangle $A_{1} A_{2} A_{3}$. Now by formula (1), $2 S$ is not divisible by $p$. On the other hand, since $\\mu\\left(A_{1} A_{2}\\right) \\geq n+1$ and $\\mu\\left(A_{1} A_{3}\\right) \\geq n+1$, it follows from (2) that $2 S$ is divisible by $p$, a contradiction.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "4", "problem_type": "Algebra", "exam": "IMO", "problem": "4. A4 (KOR) ${ }^{\\mathrm{IMO} 2}$ Find all polynomials $P(x)$ with real coefficients that satisfy the equality  $$ P(a-b)+P(b-c)+P(c-a)=2 P(a+b+c) $$  for all triples $a, b, c$ of real numbers such that $a b+b c+c a=0$.", "solution": "4. Let $P(x)=a_{0}+a_{1} x+\\cdots+a_{n} x^{n}$. For every $x \\in \\mathbb{R}$ the triple $(a, b, c)=$ $(6 x, 3 x,-2 x)$ satisfies the condition $a b+b c+c a=0$. Then the condition on $P$ gives us $P(3 x)+P(5 x)+P(-8 x)=2 P(7 x)$ for all $x$, implying that for all $i=0,1,2, \\ldots, n$ the following equality holds:  $$ \\left(3^{i}+5^{i}+(-8)^{i}-2 \\cdot 7^{i}\\right) a_{i}=0 $$  Suppose that $a_{i} \\neq 0$. Then $K(i)=3^{i}+5^{i}+(-8)^{i}-2 \\cdot 7^{i}=0$. But $K(i)$ is negative for $i$ odd and positive for $i=0$ or $i \\geq 6$ even. Only for $i=2$ and $i=4$ do we have $K(i)=0$. It follows that $P(x)=a_{2} x^{2}+a_{4} x^{4}$ for some real numbers $a_{2}, a_{4}$. It is easily verified that all such $P(x)$ satisfy the required condition.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "5", "problem_type": "Algebra", "exam": "IMO", "problem": "5. A5 (THA) Let $a, b, c>0$ and $a b+b c+c a=1$. Prove the inequality  $$ \\sqrt[3]{\\frac{1}{a}+6 b}+\\sqrt[3]{\\frac{1}{b}+6 c}+\\sqrt[3]{\\frac{1}{c}+6 a} \\leq \\frac{1}{a b c} $$", "solution": "5. By the general mean inequality $\\left(M_{1} \\leq M_{3}\\right)$, the LHS of the inequality to be proved does not exceed  $$ E=\\frac{3}{\\sqrt[3]{3}} \\sqrt[3]{\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+6(a+b+c)} $$  From $a b+b c+c a=1$ we obtain that $3 a b c(a+b+c)=3(a b \\cdot a c+$ $a b \\cdot b c+a c \\cdot b c) \\leq(a b+a c+b c)^{2}=1$; hence $6(a+b+c) \\leq \\frac{2}{a b c}$. Since $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=\\frac{a b+b c+c a}{a b c}=\\frac{1}{a b c}$, it follows that  $$ E \\leq \\frac{3}{\\sqrt[3]{3}} \\sqrt[3]{\\frac{3}{a b c}} \\leq \\frac{1}{a b c} $$  where the last inequality follows from the AM-GM inequality $1=a b+b c+$ $c a \\geq 3 \\sqrt[3]{(a b c)^{2}}$, i.e., $a b c \\leq 1 /(3 \\sqrt{3})$. The desired inequality now follows. Equality holds if and only if $a=b=c=1 / \\sqrt{3}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "6", "problem_type": "Algebra", "exam": "IMO", "problem": "6. A6 (RUS) Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfying the equation  $$ f\\left(x^{2}+y^{2}+2 f(x y)\\right)=(f(x+y))^{2} \\quad \\text { for all } x, y \\in \\mathbb{R} $$", "solution": "6. Let us make the substitution $z=x+y, t=x y$. Given $z, t \\in \\mathbb{R}, x, y$ are real if and only if $4 t \\leq z^{2}$. Define $g(x)=2(f(x)-x)$. Now the given functional equation transforms into  $$ f\\left(z^{2}+g(t)\\right)=(f(z))^{2} \\text { for all } t, z \\in \\mathbb{R} \\text { with } z^{2} \\geq 4 t $$  Let us set $c=g(0)=2 f(0)$. Substituting $t=0$ into (1) gives us  $$ f\\left(z^{2}+c\\right)=(f(z))^{2} \\text { for all } z \\in \\mathbb{R} $$  If $c<0$, then taking $z$ such that $z^{2}+c=0$, we obtain from (2) that $f(z)^{2}=c / 2$, which is impossible; hence $c \\geq 0$. We also observe that  $$ x>c \\quad \\text { implies } \\quad f(x) \\geq 0 $$  If $g$ is a constant function, we easily find that $c=0$ and therefore $f(x)=x$, which is indeed a solution. Suppose $g$ is nonconstant, and let $a, b \\in \\mathbb{R}$ be such that $g(a)-g(b)=d>0$. For some sufficiently large $K$ and each $u, v \\geq K$ with $v^{2}-u^{2}=d$ the equality $u^{2}+g(a)=v^{2}+g(b)$ by (1) and (3) implies $f(u)=f(v)$. This further leads to $g(u)-g(v)=2(v-u)=\\frac{d}{u+\\sqrt{u^{2}+d}}$. Therefore every value from some suitably chosen segment $[\\delta, 2 \\delta]$ can be expressed as $g(u)-g(v)$, with $u$ and $v$ bounded from above by some $M$. Consider any $x, y$ with $y>x \\geq 2 \\sqrt{M}$ and $\\delta<y^{2}-x^{2}<2 \\delta$. By the above considerations, there exist $u, v \\leq M$ such that $g(u)-g(v)=y^{2}-x^{2}$, i.e., $x^{2}+g(u)=y^{2}+g(v)$. Since $x^{2} \\geq 4 u$ and $y^{2} \\geq 4 v$, (1) leads to $f(x)^{2}=f(y)^{2}$. Moreover, if we assume w.l.o.g. that $4 M \\geq c^{2}$, we conclude from (3) that $f(x)=f(y)$. Since this holds for any $x, y \\geq 2 \\sqrt{M}$ with $y^{2}-x^{2} \\in[\\delta, 2 \\delta]$, it follows that $f(x)$ is eventually constant, say $f(x)=k$ for $x \\geq N=2 \\sqrt{M}$. Setting $x>N$ in (2) we obtain $k^{2}=k$, so $k=0$ or $k=1$ 。 By (2) we have $f(-z)= \\pm f(z)$, and thus $|f(z)| \\leq 1$ for all $z \\leq-N$. Hence $g(u)=2 f(u)-2 u \\geq-2-2 u$ for $u \\leq-N$, which implies that $g$ is unbounded. Hence for each $z$ there exists $t$ such that $z^{2}+g(t)>N$, and consequently $f(z)^{2}=f\\left(z^{2}+g(t)\\right)=k=k^{2}$. Therefore $f(z)= \\pm k$ for each $z$. If $k=0$, then $f(x) \\equiv 0$, which is clearly a solution. Assume $k=1$. Then $c=2 f(0)=2$ (because $c \\geq 0$ ), which together with (3) implies $f(x)=1$ for all $x \\geq 2$. Suppose that $f(t)=-1$ for some $t<2$. Then $t-g(t)=3 t+2>4 t$. If also $t-g(t) \\geq 0$, then for some $z \\in \\mathbb{R}$ we have $z^{2}=t-g(t)>4 t$, which by (1) leads to $f(z)^{2}=f\\left(z^{2}+g(t)\\right)=f(t)=-1$, which is impossible. Hence $t-g(t)<0$, giving us $t<-2 / 3$. On the other hand, if $X$ is any subset of $(-\\infty,-2 / 3)$, the function $f$ defined by $f(x)=-1$ for $x \\in X$ and $f(x)=1$ satisfies the requirements of the problem. To sum up, the solutions are $f(x)=x, f(x)=0$ and all functions of the form  $$ f(x)= \\begin{cases}1, & x \\notin X \\\\ -1, & x \\in X\\end{cases} $$  where $X \\subset(-\\infty,-2 / 3)$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "7", "problem_type": "Algebra", "exam": "IMO", "problem": "7. A7 (IRE) Let $a_{1}, a_{2}, \\ldots, a_{n}$ be positive real numbers, $n>1$. Denote by $g_{n}$ their geometric mean, and by $A_{1}, A_{2}, \\ldots, A_{n}$ the sequence of arithmetic means defined by $A_{k}=\\frac{a_{1}+a_{2}+\\cdots+a_{k}}{k}, k=1,2, \\ldots, n$. Let $G_{n}$ be the geometric mean of $A_{1}, A_{2}, \\ldots, A_{n}$. Prove the inequality  $$ n \\sqrt[n]{\\frac{G_{n}}{A_{n}}}+\\frac{g_{n}}{G_{n}} \\leq n+1 $$  and establish the cases of equality.", "solution": "7. Let us set $c_{k}=A_{k-1} / A_{k}$ for $k=1,2, \\ldots, n$, where we define $A_{0}=0$. We observe that $a_{k} / A_{k}=\\left(k A_{k}-(k-1) A_{k-1}\\right) / A_{k}=k-(k-1) c_{k}$. Now we can write the LHS of the inequality to be proved in terms of $c_{k}$, as follows:  $$ \\sqrt[n]{\\frac{G_{n}}{A_{n}}}=\\sqrt[n^{2}]{c_{2} c_{3}^{2} \\cdots c_{n}^{n-1}} \\text { and } \\frac{g_{n}}{G_{n}}=\\sqrt[n]{\\prod_{k=1}^{n}\\left(k-(k-1) c_{k}\\right)} $$  By the $A M-G M$ inequality we have  $$ \\begin{aligned} n \\sqrt[n^{2}]{1^{n(n+1) / 2} c_{2} c_{3}^{2} \\ldots c_{n}^{n-1}} & \\leq \\frac{1}{n}\\left(\\frac{n(n+1)}{2}+\\sum_{k=2}^{n}(k-1) c_{k}\\right) \\\\ & =\\frac{n+1}{2}+\\frac{1}{n} \\sum_{k=1}^{n}(k-1) c_{k} . \\end{aligned} $$  Also by the AM-GM inequality, we have  $$ \\sqrt[n]{\\prod_{k=1}^{n}\\left(k-(k-1) c_{k}\\right)} \\leq \\frac{n+1}{2}-\\frac{1}{n} \\sum_{k=1}^{n}(k-1) c_{k} $$  Adding (1) and (2), we obtain the desired inequality. Equality holds if and only if $a_{1}=a_{2}=\\cdots=a_{n}$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "8", "problem_type": "Combinatorics", "exam": "IMO", "problem": "8. C1 (PUR) There are 10001 students at a university. Some students join together to form several clubs (a student may belong to different clubs). Some clubs join together to form several societies (a club may belong to different societies). There are a total of $k$ societies. Suppose that the following conditions hold: (i) Each pair of students are in exactly one club. (ii) For each student and each society, the student is in exactly one club of the society. (iii) Each club has an odd number of students. In addition, a club with $2 m+1$ students ( $m$ is a positive integer) is in exactly $m$ societies. Find all possible values of $k$.", "solution": "8. Let us write $n=10001$. Denote by $\\mathcal{T}$ the set of ordered triples $(a, C, \\mathcal{S})$, where $a$ is a student, $C$ a club, and $\\mathcal{S}$ a society such that $a \\in C$ and $C \\in \\mathcal{S}$. We shall count $|\\mathcal{T}|$ in two different ways. Fix a student $a$ and a society $\\mathcal{S}$. By (ii), there is a unique club $C$ such that $(a, C, \\mathcal{S}) \\in \\mathcal{T}$. Since the ordered pair $(a, \\mathcal{S})$ can be chosen in $n k$ ways, we have that $|\\mathcal{T}|=n k$. Now fix a club $C$. By (iii), $C$ is in exactly $(|C|-1) / 2$ societies, so there are $|C|(|C|-1) / 2$ triples from $\\mathcal{T}$ with second coordinate $C$. If $\\mathcal{C}$ is the set of all clubs, we obtain $|\\mathcal{T}|=\\sum_{C \\in \\mathcal{C}} \\frac{|C|(|C|-1)}{2}$. But we also conclude from (i) that  $$ \\sum_{C \\in \\mathcal{C}} \\frac{|C|(|C|-1)}{2}=\\frac{n(n-1)}{2} $$  Therefore $n(n-1) / 2=n k$, i.e., $k=(n-1) / 2=5000$. On the other hand, for $k=(n-1) / 2$ there is a desired configuration with only one club $C$ that contains all students and $k$ identical societies with only one element (the club $C$ ). It is easy to verify that (i)-(iii) hold.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}
{"year": "2004", "tier": "T0", "problem_label": "9", "problem_type": "Combinatorics", "exam": "IMO", "problem": "9. C2 (GER) Let $n$ and $k$ be positive integers. There are given $n$ circles in the plane. Every two of them intersect at two distinct points, and all points of intersection they determine are distinct. Each intersection point must be colored with one of $n$ distinct colors so that each color is used at least once, and exactly $k$ distinct colors occur on each circle. Find all values of $n \\geq 2$ and $k$ for which such a coloring is possible.", "solution": "9. Obviously we must have $2 \\leq k \\leq n$. We shall prove that the possible values for $k$ and $n$ are $2 \\leq k \\leq n \\leq 3$ and $3 \\leq k \\leq n$. Denote all colors and circles by $1, \\ldots, n$. Let $F(i, j)$ be the set of colors of the common points of circles $i$ and $j$. Suppose that $k=2<n$. Consider the ordered pairs $(i, j)$ such that color $j$ appears on the circle $i$. Since $k=2$, clearly there are exactly $2 n$ such pairs. On the other hand, each of the $n$ colors appears on at least two circles, so there are at least $2 n$ pairs $(i, j)$, and equality holds only if each color appears on exactly 2 circles. But then at most two points receive each of the $n$ colors and there are $n(n-1)$ points, implying that $n(n-1)=2 n$, i.e., $n=3$. It is easy to find examples for $k=2$ and $n=2$ or 3 .  Next, let $k=3$. An example for $n=3$ is given by $F(i, j)=\\{i, j\\}$ for each $1 \\leq i<j \\leq 3$. Assume $n \\geq 4$. Then an example is given by $F(1,2)=$ $\\{1,2\\}, F(i, i+1)=\\{i\\}$ for $i=2, \\ldots, n-2, F(n-1, n)=\\{n-2, n-1\\}$ and $F(i, j)=n$ for all other $i, j>i$. We now prove by induction on $k$ that a desired coloring exists for each $n \\geq k \\geq 3$. Let there be given $n$ circles. By the inductive hypothesis, circles $1,2, \\ldots, n-1$ can be colored in $n-1$ colors, $k$ of which appear on each circle, such that color $i$ appears on circle $i$. Then we set $F(i, n)=\\{i, n\\}$ for $i=1, \\ldots, k$ and $F(i, n)=\\{n\\}$ for $i>n$. We thus obtain a coloring of the $n$ circles in $n$ colors, such that $k+1$ colors (including color $i$ ) appear on each circle $i$.", "metadata": {"resource_path": "IMO/segmented/en-compendium.jsonl"}}