| {"year": "2015", "tier": "T3", "problem_label": "1", "problem_type": null, "exam": "INMO", "problem": "Let $A B C$ be a right-angled triangle with $\\angle B=90^{\\circ}$. Let $B D$ be the altitude from $B$ on to $A C$. Let $P, Q$ and $I$ be the incentres of triangles $A B D, C B D$ and $A B C$ respectively. Show that the circumcentre of of the triangle $P I Q$ lies on the hypotenuse $A C$.", "solution": "We begin with the following lemma:\n\nLemma: Let $X Y Z$ be a triangle with $\\angle X Y Z=90+\\alpha$. Construct an isosceles triangle $X E Z$, externally on the side $X Z$, with base angle $\\alpha$. Then $E$ is the circumcentre of $\\triangle X Y Z$.\n\nProof of the Lemma: Draw $E D \\perp$ $X Z$. Then $D E$ is the perpendicular bisector of $X Z$. We also observe that $\\angle X E D=\\angle Z E D=90-\\alpha$. Observe that $E$ is on the perpendicular bisector of $X Z$. Construct the circumcircle of $X Y Z$. Draw perpendicular bisector of $X Y$ and let it meet $D E$ in $F$. Then $F$ is the circumcentre of $\\triangle X Y Z$. Join $X F$. Then $\\angle X F D=90-\\alpha$. But we know\n\n\nthat $\\angle X E D=90-\\alpha$. Hence $E=F$. Let $r_{1}, r_{2}$ and $r$ be the inradii of the triangles $A B D, C B D$ and $A B C$ respectively. Join $P D$ and $D Q$. Observe that $\\angle P D Q=90^{\\circ}$. Hence\n\n$$\nP Q^{2}=P D^{2}+D Q^{2}=2 r_{1}^{2}+2 r_{2}^{2}\n$$\n\nLet $s_{1}=(A B+B D+D A) / 2$. Observe that $B D=c a / b$ and $A D=$ $\\sqrt{A B^{2}-B D^{2}}=\\sqrt{c^{2}-\\left(\\frac{c a}{b}\\right)^{2}}=c^{2} / b$. This gives $s_{1}=c s / b$. But $r_{1}=$ $s_{1}-c=(c / b)(s-b)=c r / b$. Similarly, $r_{2}=a r / b$. Hence\n\n$$\nP Q^{2}=2 r^{2}\\left(\\frac{c^{2}+a^{2}}{b^{2}}\\right)=2 r^{2}\n$$\n\nConsider $\\triangle P I Q$. Observe that $\\angle P I Q=90+(B / 2)=135$. Hence $P Q$ subtends $90^{\\circ}$ on the circumference of the circumcircle of $\\triangle P I Q$. But we have seen that $\\angle P D Q=90^{\\circ}$. Now construct a circle with $P Q$ as diameter. Let it cut $A C$ again in $K$. It follows that $\\angle P K Q=90^{\\circ}$ and the points $P, D, K, Q$ are concyclic. We also notice $\\angle K P Q=\\angle K D Q=45^{\\circ}$\n\n\nand $\\angle P Q K=\\angle P D A=45^{\\circ}$.\n\nThus $P K Q$ is an isosceles right-angled triangle with $K P=K Q$. Therfore $K P^{2}+K Q^{2}=P Q^{2}=2 r^{2}$ and hence $K P=K Q=r$.\n\nNow $\\angle P I Q=90+45$ and $\\angle P K Q=2 \\times 45^{\\circ}=90^{\\circ}$ with $K P=K Q=r$.\n\nHence $K$ is the circumcentre of $\\triangle P I Q$.\n\n(Incidentally, This also shows that $K I=r$ and hence $K$ is the point of contact of the incircle of $\\triangle A B C$ with $A C$.)", "metadata": {"resource_path": "INMO/segmented/en-inmosol-15.jsonl", "problem_match": "\n1.", "solution_match": "\nSolution:"}} |
| {"year": "2015", "tier": "T3", "problem_label": "1", "problem_type": null, "exam": "INMO", "problem": "Let $A B C$ be a right-angled triangle with $\\angle B=90^{\\circ}$. Let $B D$ be the altitude from $B$ on to $A C$. Let $P, Q$ and $I$ be the incentres of triangles $A B D, C B D$ and $A B C$ respectively. Show that the circumcentre of of the triangle $P I Q$ lies on the hypotenuse $A C$.", "solution": "Here we use computation to prove that the point of contact $K$ of the incircle with $A C$ is the circumcentre of $\\triangle P I Q$. We show that $K P=K Q=r$. Let $r_{1}$ and $r_{2}$ be the inradii of triangles $A B D$ and $C B D$ respectively. Draw $P L \\perp A C$ and $Q M \\perp A C$. If $s_{1}$ is the semiperimeter of $\\triangle A B D$, then $A L=s_{1}-B D$.\n\n\n\nBut\n\n$$\ns_{1}=\\frac{A B+B D+D A}{2}, \\quad B D=\\frac{c a}{b}, \\quad A D=\\frac{c^{2}}{b}\n$$\n\nHence $s_{1}=c s / b$. This gives $r_{1}=s_{1}-c=c r / b, A L=s_{1}-B D=c(s-a) / b$. Hence $K L=A K-A L=(s-a)-\\frac{c(s-a)}{b}=\\frac{(b-c)(s-a)}{b}$. We observe that $2 r^{2}=\\frac{(c+a-b)^{2}}{2}=\\frac{c^{2}+a^{2}+b^{2}-2 b c-2 a b+2 c a}{2}=\\left(b^{2}-b a-b c+a c\\right)=(b-c)(b-a)$.\n\nThis gives\n\n$$\n\\begin{aligned}\n(s-a)(b-c)=(s-b+b-a) & (b-c)=r(b-c)+(b-a)(b-c) \\\\\n& =r(b-c)+2 r^{2}=r(b-c+c+a-b)=r a\n\\end{aligned}\n$$\n\nThus $K L=r a / b$. Finally,\n\n$$\nK P^{2}=K L^{2}+L P^{2}=\\frac{r^{2} a^{2}}{b^{2}}+\\frac{r^{2}+c^{2}}{b^{2}}=r^{2}\n$$\n\nThus $K P=r$. Similarly, $K Q=r$. This gives $K P=K I=K Q=r$ and therefore $K$ is the circumcentre of $\\triangle K I Q$.\n\n(Incidentally, this also shows that $K L=c a / b=r_{2}$ and $K M=r_{1}$.)", "metadata": {"resource_path": "INMO/segmented/en-inmosol-15.jsonl", "problem_match": "\n1.", "solution_match": "\nSolution 2:"}} |
| {"year": "2015", "tier": "T3", "problem_label": "2", "problem_type": null, "exam": "INMO", "problem": "For any natural number $n>1$, write the infinite decimal expansion of $1 / n$ (for example, we write $1 / 2=0.4 \\overline{9}$ as its infinite decimal expansion, not 0.5 ). Determine the length of the non-periodic part of the (infinite) decimal expansion of $1 / n$.", "solution": "For any prime $p$, let $\\nu_{p}(n)$ be the maximum power of $p$ dividing $n$; ie $p^{\\nu_{p}(n)}$ divides $n$ but not higher power. Let $r$ be the\nlength of the non-periodic part of the infinite decimal expansion of $1 / n$.\n\nWrite\n\n$$\n\\frac{1}{n}=0 . a_{1} a_{2} \\cdots a_{r} \\overline{b_{1} b_{2} \\cdots b_{s}}\n$$\n\nWe show that $r=\\max \\left(\\nu_{2}(n), \\nu_{5}(n)\\right)$.\n\nLet $a$ and $b$ be the numbers $a_{1} a_{2} \\cdots a_{r}$ and $b=b_{1} b_{2} \\cdots b_{s}$ respectively. (Here $a_{1}$ and $b_{1}$ can be both 0.) Then\n\n$$\n\\frac{1}{n}=\\frac{1}{10^{r}}\\left(a+\\sum_{k \\geq 1} \\frac{b}{\\left(10^{s}\\right)^{k}}\\right)=\\frac{1}{10^{r}}\\left(a+\\frac{b}{10^{s}-1}\\right)\n$$\n\nThus we get $10^{r}\\left(10^{s}-1\\right)=n\\left(\\left(10^{s}-1\\right) a+b\\right)$. It shows that $r \\geq$ $\\max \\left(\\nu_{2}(n), \\nu_{5}(n)\\right)$. Suppose $r>\\max \\left(\\nu_{2}(n), \\nu_{5}(n)\\right)$. Then 10 divides $b-a$. Hence the last digits of $a$ and $b$ are equal: $a_{r}=b_{s}$. This means\n\n$$\n\\frac{1}{n}=0 . a_{1} a_{2} \\cdots a_{r-1} \\overline{b_{s} b_{1} b_{2} \\cdots b_{s-1}}\n$$\n\nThis contradicts the definition of $r$. Therefore $r=\\max \\left(\\nu_{2}(n), \\nu_{5}(n)\\right)$.", "metadata": {"resource_path": "INMO/segmented/en-inmosol-15.jsonl", "problem_match": "\n2.", "solution_match": "\nSolution:"}} |
| {"year": "2015", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "INMO", "problem": "Find all real functions $f$ from $\\mathbb{R} \\rightarrow \\mathbb{R}$ satisfying the relation\n\n$$\nf\\left(x^{2}+y f(x)\\right)=x f(x+y)\n$$", "solution": "Put $x=0$ and we get $f(y f(0))=0$. If $f(0) \\neq 0$, then $y f(0)$ takes all real values when $y$ varies over real line. We get $f(x) \\equiv 0$. Suppose $f(0)=0$. Taking $y=-x$, we get $f\\left(x^{2}-x f(x)\\right)=0$ for all real $x$.\n\nSuppose there exists $x_{0} \\neq 0$ in $\\mathbb{R}$ such that $f\\left(x_{0}\\right)=0$. Putting $x=x_{0}$ in the given relation we get\n\n$$\nf\\left(x_{0}^{2}\\right)=x_{0} f\\left(x_{0}+y\\right)\n$$\n\nfor all $y \\in \\mathbb{R}$. Now the left side is a constant and hence it follows that $f$ is a constant function. But the only constant function which satisfies the equation is identically zero function, which is already obtained. Hence we may consider the case where $f(x) \\neq 0$ for all $x \\neq 0$.\n\nSince $f\\left(x^{2}-x f(x)\\right)=0$, we conclude that $x^{2}-x f(x)=0$ for all $x \\neq 0$. This implies that $f(x)=x$ for all $x \\neq 0$. Since $f(0)=0$, we conclude that $f(x)=x$ for all $x \\in R$.\n\nThus we have two functions: $f(x) \\equiv 0$ and $f(x)=x$ for all $x \\in \\mathbb{R}$.", "metadata": {"resource_path": "INMO/segmented/en-inmosol-15.jsonl", "problem_match": "\n3.", "solution_match": "\nSolution:"}} |
| {"year": "2015", "tier": "T3", "problem_label": "4", "problem_type": null, "exam": "INMO", "problem": "There are four basket-ball players $A, B, C, D$. Initially, the ball is with $A$. The ball is always passed from one person to a different person. In how many ways can the ball come back to $A$ after seven passes? (For example $A \\rightarrow C \\rightarrow B \\rightarrow D \\rightarrow A \\rightarrow B \\rightarrow C \\rightarrow A$ and\n$A \\rightarrow D \\rightarrow A \\rightarrow D \\rightarrow C \\rightarrow A \\rightarrow B \\rightarrow A$ are two ways in which the ball can come back to $A$ after seven passes.)", "solution": "Let $x_{n}$ be the number of ways in which $A$ can get back the ball after $n$ passes. Let $y_{n}$ be the number of ways in which the ball goes back to a fixed person other than $A$ after $n$ passes. Then\n\n$$\nx_{n}=3 y_{n-1} \\text {, }\n$$\n\nand\n\n$$\ny_{n}=x_{n-1}+2 y_{n-1}\n$$\n\nWe also have $x_{1}=0, x_{2}=3, y_{1}=1$ and $y_{2}=2$.\n\nEliminating $y_{n}$ and $y_{n-1}$, we get $x_{n+1}=3 x_{n-1}+2 x_{n}$. Thus\n\n$$\n\\begin{aligned}\n& x_{3}=3 x_{1}+2 x_{2}=2 \\times 3=6 \\\\\n& x_{4}=3 x_{2}+2 x_{3}=(3 \\times 3)+(2 \\times 6)=9+12=21 \\\\\n& x_{5}=3 x_{3}+2 x_{4}=(3 \\times 6)+(2 \\times 21)=18+42=60 \\\\\n& x_{6}=3 x_{4}+2 x_{5}=(3 \\times 21)+(2 \\times 60)=63+120=183 \\\\\n& x_{7}=3 x_{5}+2 x_{6}=(3 \\times 60)+(2 \\times 183)=180+366=546\n\\end{aligned}\n$$\n\nAlternate solution: Since the ball goes back to one of the other 3 persons, we have\n\n$$\nx_{n}+3 y_{n}=3^{n}\n$$\n\nsince there are $3^{n}$ ways of passing the ball in $n$ passes. Using $x_{n}=$ $3 y_{n-1}$, we obtain\n\n$$\nx_{n-1}+x_{n}=3^{n-1}\n$$\n\nwith $x_{1}=0$. Thus\n\n$$\n\\begin{array}{r}\nx_{7}=3^{6}-x_{6}=3^{6}-3^{5}+x_{5}=3^{6}-3^{5}+3^{4}-x_{4}=3^{6}-3^{5}+3^{4}-3^{3}+x_{3} \\\\\n=3^{6}-3^{5}+3^{4}-3^{3}+3^{2}-x_{2}=3^{6}-3^{5}+3^{4}-3^{3}+3^{2}-3 \\\\\n=\\left(2 \\times 3^{5}\\right)+\\left(2 \\times 3^{3}\\right)+(2 \\times 3)=486+54+6=546\n\\end{array}\n$$", "metadata": {"resource_path": "INMO/segmented/en-inmosol-15.jsonl", "problem_match": "\n4.", "solution_match": "\nSolution:"}} |
| {"year": "2015", "tier": "T3", "problem_label": "5", "problem_type": null, "exam": "INMO", "problem": "Let $A B C D$ be a convex quadrilateral. Let the diagonals $A C$ and $B D$ intersect in $P$. Let $P E, P F, P G$ and $P H$ be the altitudes from $P$ on to the sides $A B, B C, C D$ and $D A$ respectively. Show that $A B C D$ has an incircle if and only if\n\n$$\n\\frac{1}{P E}+\\frac{1}{P G}=\\frac{1}{P F}+\\frac{1}{P H}\n$$", "solution": "Let $A P=p, B P=q, C P=r, D P=s ; A B=a, B C=b$, $C D=c$ and $D A=d$. Let $\\angle A P B=\\angle C P D=\\theta$. Then $\\angle B P C=\\angle D P A=$ $\\pi-\\theta$. Let us also write $P E=h_{1}, P F=h_{2}, P G=h_{3}$ and $P H=h_{4}$.\n\n\n\n## Observe that\n\n$$\nh_{1} a=p q \\sin \\theta, \\quad h_{2} b=q r \\sin \\theta, \\quad h_{3} c=r s \\sin \\theta, \\quad h_{4} d=s p \\sin \\theta\n$$\n\nHence\n\n$$\n\\frac{1}{h_{1}}+\\frac{1}{h_{3}}=\\frac{1}{h_{2}}+\\frac{1}{h_{4}}\n$$\n\nis equivalent to\n\n$$\n\\frac{a}{p q}+\\frac{c}{r s}=\\frac{b}{q r}+\\frac{d}{s p}\n$$\n\nThis is the same as\n\n$$\na r s+c p q=b s p+d q r\n$$\n\nThus we have to prove that $a+c=b+d$ if and only if $a r s+c p q=b s p+d q r$. Now we can write $a+c=b+d$ as\n\n$$\na^{2}+c^{2}+2 a c=b^{2}+d^{2}+2 b d\n$$\n\nBut we know that\n\n$$\n\\begin{aligned}\n& a^{2}=p^{2}+q^{2}-2 p q \\cos \\theta, \\quad c^{2}=r^{2}+s^{2}-2 r s \\cos \\theta \\\\\n& b^{2}=q^{2}+r^{2}+2 q r \\cos \\theta, \\quad d^{2}=p^{2}+s^{2}+2 p s \\cos \\theta\n\\end{aligned}\n$$\n\nHence $a+c=b+d$ is equivalent to\n\n$$\n-p q \\cos \\theta+-r s \\cos \\theta+a c=p s \\cos \\theta+q r \\cos \\theta+b d\n$$\n\nSimilarly, by squaring ars $+c p q=b s p+d q r$ we can show that it is equivalent to\n\n$$\n-p q \\cos \\theta+-r s \\cos \\theta+a c=p s \\cos \\theta+q r \\cos \\theta+b d\n$$\n\nWe conclude that $a+c=b+d$ is equivalent to $c p q+a r s=b p s+d q r$. Hence $A B C D$ has an in circle if and only if\n\n$$\n\\frac{1}{h_{1}}+\\frac{1}{h_{3}}=\\frac{1}{h_{2}}+\\frac{1}{h_{4}}\n$$", "metadata": {"resource_path": "INMO/segmented/en-inmosol-15.jsonl", "problem_match": "\n5.", "solution_match": "\nSolution:"}} |
| {"year": "2015", "tier": "T3", "problem_label": "6", "problem_type": null, "exam": "INMO", "problem": "From a set of 11 square integers, show that one can choose 6 numbers $a^{2}, b^{2}, c^{2}, d^{2}, e^{2}, f^{2}$ such that\n\n$$\na^{2}+b^{2}+c^{2} \\equiv d^{2}+e^{2}+f^{2} \\quad(\\bmod 12)\n$$", "solution": "The first observation is that we can find 5 pairs of squares such that the two numbers in a pair have the same parity. We can see this as follows:\n\n| Odd numbers | Even numbers | Odd pairs | Even pairs | Total pairs |\n| :---: | :---: | :---: | :---: | :---: |\n| 0 | 11 | 0 | 5 | 5 |\n| 1 | 10 | 0 | 5 | 5 |\n| 2 | 9 | 1 | 4 | 5 |\n| 3 | 8 | 1 | 4 | 5 |\n| 4 | 7 | 2 | 3 | 5 |\n| 5 | 6 | 2 | 3 | 5 |\n| 6 | 5 | 3 | 2 | 5 |\n| 7 | 4 | 3 | 2 | 5 |\n| 8 | 3 | 4 | 1 | 5 |\n| 9 | 2 | 4 | 1 | 5 |\n| 10 | 1 | 5 | 0 | 5 |\n| 11 | 0 | 5 | 0 | 5 |\n\nLet us take such 5 pairs: say $\\left(x_{1}^{2}, y_{1}^{2}\\right),\\left(x_{2}^{2}, y_{2}^{2}\\right), \\ldots,\\left(x_{5}^{2}, y_{5}^{2}\\right)$. Then $x_{j}^{2}-y_{j}^{2}$ is divisible by 4 for $1 \\leq j \\leq 5$. Let $r_{j}$ be the remainder when $x_{j}^{2}-y_{j}^{2}$ is divisible by $3,1 \\leq j \\leq 3$. We have 5 remainders $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$. But these can be 0,1 or 2 . Hence either one of the remainders occur 3 times or each of the remainders occur once. If, for example $r_{1}=r_{2}=r_{3}$, then 3 divides $r_{1}+r_{2}+r_{3}$; if $r_{1}=0, r_{2}=1$ and $r_{3}=2$, then again 3 divides $r_{1}+r_{2}+r_{3}$. Thus we can always find three remainders whose sum is divisible by 3 . This means we can find 3 pairs, say, $\\left(x_{1}^{2}, y_{1}^{2}\\right),\\left(x_{2}^{2}, y_{2}^{2}\\right),\\left(x_{3}^{2}, y_{3}^{2}\\right)$ such that 3 divides $\\left(x_{1}^{2}-y_{1}^{2}\\right)+\\left(x_{2}^{2}-y_{2}^{2}\\right)+\\left(x_{3}^{2}-y_{3}^{2}\\right)$. Since each difference is divisible by 4 , we conclude that we can find 6 numbers $a^{2}, b^{2}, c^{2}, d^{2}, e^{2}, f^{2}$ such that\n\n$$\na^{2}+b^{2}+c^{2} \\equiv d^{2}+e^{2}+f^{2} \\quad(\\bmod 12)\n$$", "metadata": {"resource_path": "INMO/segmented/en-inmosol-15.jsonl", "problem_match": "\n6.", "solution_match": "\nSolution:"}} |
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