{"year": "1991", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Let $G$ be the centroid of triangle $A B C$ and $M$ be the midpoint of $B C$. Let $X$ be on $A B$ and $Y$ on $A C$ such that the points $X, Y$, and $G$ are collinear and $X Y$ and $B C$ are parallel. Suppose that $X C$ and $G B$ intersect at $Q$ and $Y B$ and $G C$ intersect at $P$. Show that triangle $M P Q$ is similar to triangle $A B C$.", "solution": "Let $R$ be the midpoint of $A C$; so $B R$ is a median and contains the centroid $G$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-1.jpg?height=558&width=647&top_left_y=663&top_left_x=673)\n\nIt is well known that $\\frac{A G}{A M}=\\frac{2}{3}$; thus the ratio of the similarity between $A X Y$ and $A B C$ is $\\frac{2}{3}$. Hence $G X=\\frac{1}{2} X Y=\\frac{1}{3} B C$.\nNow look at the similarity between triangles $Q B C$ and $Q G X$ :\n\n$$\n\\frac{Q G}{Q B}=\\frac{G X}{B C}=\\frac{1}{3} \\Longrightarrow Q B=3 Q G \\Longrightarrow Q B=\\frac{3}{4} B G=\\frac{3}{4} \\cdot \\frac{2}{3} B R=\\frac{1}{2} B R .\n$$\n\nFinally, since $\\frac{B M}{B C}=\\frac{B Q}{B R}, M Q$ is a midline in $B C R$. Therefore $M Q=\\frac{1}{2} C R=\\frac{1}{4} A C$ and $M Q \\| A C$. Similarly, $M P=\\frac{1}{4} A B$ and $M P \\| A B$. This is sufficient to establish that $M P Q$ and $A B C$ are similar (with similarity ratio $\\frac{1}{4}$ ).", "metadata": {"resource_path": "APMO/segmented/en-apmo1991_sol.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution 1"}}
{"year": "1991", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Let $G$ be the centroid of triangle $A B C$ and $M$ be the midpoint of $B C$. Let $X$ be on $A B$ and $Y$ on $A C$ such that the points $X, Y$, and $G$ are collinear and $X Y$ and $B C$ are parallel. Suppose that $X C$ and $G B$ intersect at $Q$ and $Y B$ and $G C$ intersect at $P$. Show that triangle $M P Q$ is similar to triangle $A B C$.", "solution": "Let $S$ and $R$ be the midpoints of $A B$ and $A C$, respectively. Since $G$ is the centroid, it lies in the medians $B R$ and $C S$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-1.jpg?height=558&width=647&top_left_y=1971&top_left_x=676)\n\nDue to the similarity between triangles $Q B C$ and $Q G X$ (which is true because $G X \\| B C$ ), there is an inverse homothety with center $Q$ and ratio $-\\frac{X G}{B C}=\\frac{X Y}{2 B C}$ that takes $B$ to $G$ and $C$ to $X$. This homothety takes the midpoint $M$ of $B C$ to the midpoint $K$ of $G X$.\n\nNow consider the homothety that takes $B$ to $X$ and $C$ to $G$. This new homothety, with ratio $\\frac{X Y}{2 B C}$, also takes $M$ to $K$. Hence lines $B X$ (which contains side $A B$ ), $C G$ (which contains the median $C S$ ), and $M K$ have a common point, which is $S$. Thus $Q$ lies on midline $M S$.\nThe same reasoning proves that $P$ lies on midline $M R$. Since all homothety ratios are the same, $\\frac{M Q}{M S}=\\frac{M P}{M R}$, which shows that $M P Q$ is similar to $M R S$, which in turn is similar to $A B C$, and we are done.", "metadata": {"resource_path": "APMO/segmented/en-apmo1991_sol.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution 2"}}
{"year": "1991", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "APMO", "problem": "Suppose there are 997 points given in a plane. If every two points are joined by a line segment with its midpoint coloured in red, show that there are at least 1991 red points in the plane. Can you find a special case with exactly 1991 red points?", "solution": "Embed the points in the cartesian plane such that no two points have the same $y$-coordinate. Let $P_{1}, P_{2}, \\ldots, P_{997}$ be the points and $y_{1}<y_{2}<\\ldots<y_{997}$ be their respective $y$-coordinates. Then the $y$-coordinate of the midpoint of $P_{i} P_{i+1}, i=1,2, \\ldots, 996$ is $\\frac{y_{i}+y_{i+1}}{2}$ and the $y$-coordinate of the midpoint of $P_{i} P_{i+2}, i=1,2, \\ldots, 995$ is $\\frac{y_{i}+y_{i+2}}{2}$. Since\n\n$$\n\\frac{y_{1}+y_{2}}{2}<\\frac{y_{1}+y_{3}}{2}<\\frac{y_{2}+y_{3}}{2}<\\frac{y_{2}+y_{4}}{2}<\\cdots<\\frac{y_{995}+y_{997}}{2}<\\frac{y_{996}+y_{997}}{2}\n$$\n\nthere are at least $996+995=1991$ distinct midpoints, and therefore at least 1991 red points. The equality case happens if we take $P_{i}=(0,2 i), i=1,2, \\ldots, 997$. The midpoints are $(0, i+j)$, $1 \\leq i<j \\leq 997$, which are the points $(0, k)$ with $1+2=3 \\leq k \\leq 996+997=1993$, a total of $1993-3+1=1991$ red points.", "metadata": {"resource_path": "APMO/segmented/en-apmo1991_sol.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution\n\n"}}
{"year": "1991", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "APMO", "problem": "Let $a_{1}, a_{2}, \\ldots, a_{n}, b_{1}, b_{2}, \\ldots, b_{n}$ be positive real numbers such that $a_{1}+a_{2}+\\cdots+a_{n}=b_{1}+b_{2}+$ $\\cdots+b_{n}$. Show that\n\n$$\n\\frac{a_{1}^{2}}{a_{1}+b_{1}}+\\frac{a_{2}^{2}}{a_{2}+b_{2}}+\\cdots+\\frac{a_{n}^{2}}{a_{n}+b_{n}} \\geq \\frac{a_{1}+a_{2}+\\cdots+a_{n}}{2}\n$$", "solution": "By the Cauchy-Schwartz inequality,\n\n$$\n\\left(\\frac{a_{1}^{2}}{a_{1}+b_{1}}+\\frac{a_{2}^{2}}{a_{2}+b_{2}}+\\cdots+\\frac{a_{n}^{2}}{a_{n}+b_{n}}\\right)\\left(\\left(a_{1}+b_{1}\\right)+\\left(a_{2}+b_{2}\\right)+\\cdots+\\left(a_{n}+b_{n}\\right)\\right) \\geq\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)^{2} .\n$$\n\nSince $\\left(\\left(a_{1}+b_{1}\\right)+\\left(a_{2}+b_{2}\\right)+\\cdots+\\left(a_{n}+b_{n}\\right)\\right)=2\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)$,\n\n$$\n\\frac{a_{1}^{2}}{a_{1}+b_{1}}+\\frac{a_{2}^{2}}{a_{2}+b_{2}}+\\cdots+\\frac{a_{n}^{2}}{a_{n}+b_{n}} \\geq \\frac{\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)^{2}}{2\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)}=\\frac{a_{1}+a_{2}+\\cdots+a_{n}}{2}\n$$", "metadata": {"resource_path": "APMO/segmented/en-apmo1991_sol.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution\n\n"}}
{"year": "1991", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "APMO", "problem": "During a break, $n$ children at school sit in a circle around their teacher to play a game. The teacher walks clockwise close to the children and hands out candies to some of them according to the following rule. He selects one child and gives him a candy, then he skips the next child and gives a candy to the next one, then he skips 2 and gives a candy to the next one, then he skips 3, and soon. Determine the values of $n$ for which eventually, perhaps after many rounds, all children will have at least one candy each.\n\nAnswer: All powers of 2 .", "solution": "Number the children from 0 to $n-1$. Then the teacher hands candy to children in positions $f(x)=1+2+\\cdots+x \\bmod n=\\frac{x(x+1)}{2} \\bmod n$. Our task is to find the range of $f: \\mathbb{Z}_{n} \\rightarrow \\mathbb{Z}_{n}$, and to verify whether the range is $\\mathbb{Z}_{n}$, that is, whether $f$ is a bijection.\nIf $n=2^{a} m, m>1$ odd, look at $f(x)$ modulo $m$. Since $m$ is odd, $m|f(x) \\Longleftrightarrow m| x(x+1)$. Then, for instance, $f(x) \\equiv 0(\\bmod m)$ for $x=0$ and $x=m-1$. This means that $f(x)$ is not a bijection modulo $m$, and there exists $t$ such that $f(x) \\not \\equiv t(\\bmod m)$ for all $x$. By the Chinese Remainder Theorem,\n\n$$\nf(x) \\equiv t \\quad(\\bmod n) \\Longleftrightarrow \\begin{cases}f(x) \\equiv t & \\left(\\bmod 2^{a}\\right) \\\\ f(x) \\equiv t & (\\bmod m)\\end{cases}\n$$\n\nTherefore, $f$ is not a bijection modulo $n$.\nIf $n=2^{a}$, then\n\n$$\nf(x)-f(y)=\\frac{1}{2}(x(x+1)-y(y+1))=\\frac{1}{2}\\left(x^{2}-y^{2}+x-y\\right)=\\frac{(x-y)(x+y+1)}{2} .\n$$\n\nand\n\n$$\nf(x) \\equiv f(y) \\quad\\left(\\bmod 2^{a}\\right) \\Longleftrightarrow(x-y)(x+y+1) \\equiv 0 \\quad\\left(\\bmod 2^{a+1}\\right)\n$$\n\nIf $x$ and $y$ have the same parity, $x+y+1$ is odd and $(*)$ is equivalent to $x \\equiv y\\left(\\bmod 2^{a+1}\\right)$. If $x$ and $y$ have different parity,\n\n$$\n(*) \\Longleftrightarrow x+y+1 \\equiv 0 \\quad\\left(\\bmod 2^{a+1}\\right)\n$$\n\nHowever, $1 \\leq x+y+1 \\leq 2\\left(2^{a}-1\\right)+1=2^{a+1}-1$, so $x+y+1$ is not a multiple of $2^{a+1}$. Therefore $f$ is a bijection if $n$ is a power of 2 .", "metadata": {"resource_path": "APMO/segmented/en-apmo1991_sol.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution 1"}}
{"year": "1991", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "APMO", "problem": "During a break, $n$ children at school sit in a circle around their teacher to play a game. The teacher walks clockwise close to the children and hands out candies to some of them according to the following rule. He selects one child and gives him a candy, then he skips the next child and gives a candy to the next one, then he skips 2 and gives a candy to the next one, then he skips 3, and soon. Determine the values of $n$ for which eventually, perhaps after many rounds, all children will have at least one candy each.\n\nAnswer: All powers of 2 .", "solution": "We give a full description of $a_{n}$, the size of the range of $f$.\nSince congruences modulo $n$ are defined, via Chinese Remainder Theorem, by congruences modulo $p^{\\alpha}$ for all prime divisors $p$ of $n$ and $\\alpha$ being the number of factors $p$ in the factorization of $n, a_{n}=\\prod_{p^{\\alpha} \\| n} a_{p^{\\alpha}}$.\nRefer to the first solution to check the case $p=2: a_{2^{\\alpha}}=2^{\\alpha}$.\nFor an odd prime $p$,\n\n$$\nf(x)=\\frac{x(x+1)}{2}=\\frac{(2 x+1)^{2}-1}{8}\n$$\n\nand since $p$ is odd, there is a bijection between the range of $f$ and the quadratic residues modulo $p^{\\alpha}$, namely $t \\mapsto 8 t+1$. So $a_{p^{\\alpha}}$ is the number of quadratic residues modulo $p^{\\alpha}$.\nLet $g$ be a primitive root of $p^{\\alpha}$. Then there are $\\frac{1}{2} \\phi\\left(p^{\\alpha}\\right)=\\frac{p-1}{2} \\cdot p^{\\alpha-1}$ quadratic residues that are coprime with $p: 1, g^{2}, g^{4}, \\ldots, g^{\\phi\\left(p^{n}\\right)-2}$. If $p$ divides a quadratic residue $k p$, that is, $x^{2} \\equiv k p$ $\\left(\\bmod p^{\\alpha}\\right), \\alpha \\geq 2$, then $p$ divides $x$ and, therefore, also $k$. Hence $p^{2}$ divides this quadratic residue, and these quadratic residues are $p^{2}$ times each quadratic residue of $p^{\\alpha-2}$. Thus\n\n$$\na_{p^{\\alpha}}=\\frac{p-1}{2} \\cdot p^{\\alpha-1}+a_{p^{\\alpha}-2} .\n$$\n\nSince $a_{p}=\\frac{p-1}{2}+1$ and $a_{p^{2}}=\\frac{p-1}{2} \\cdot p+1$, telescoping yields\n\n$$\na_{p^{2 t}}=\\frac{p-1}{2}\\left(p^{2 t-1}+p^{2 t-3}+\\cdots+p\\right)+1=\\frac{p\\left(p^{2 t}-1\\right)}{2(p+1)}+1\n$$\n\nand\n\n$$\na_{p^{2 t-1}}=\\frac{p-1}{2}\\left(p^{2 t-2}+p^{2 t-4}+\\cdots+1\\right)+1=\\frac{p^{2 t}-1}{2(p+1)}+1\n$$\n\nNow the problem is immediate: if $n$ is divisible by an odd prime $p, a_{p^{\\alpha}}<p^{\\alpha}$ for all $\\alpha$, and since $a_{t} \\leq t$ for all $t, a_{n}<n$.", "metadata": {"resource_path": "APMO/segmented/en-apmo1991_sol.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution 2"}}
{"year": "1991", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "APMO", "problem": "Given are two tangent circles and a point $P$ on their common tangent perpendicular to the lines joining their centres. Construct with ruler and compass all the circles that are tangent to these two circles and pass through the point $P$.", "solution": "Throughout this problem, we will assume that the given circles are externally tangent, since the problem does not have a solution otherwise.\nLet $\\Gamma_{1}$ and $\\Gamma_{2}$ be the given circles and $T$ be their tangency point. Suppose $\\omega$ is a circle that is tangent to $\\Gamma_{1}$ and $\\Gamma_{2}$ and passes through $P$.\nNow invert about point $P$, with radius $P T$. Let any line through $P$ that cuts $\\Gamma_{1}$ do so at points $X$ and $Y$. The power of $P$ with respect to $\\Gamma_{1}$ is $P T^{2}=P X \\cdot P Y$, so $X$ and $Y$ are swapped by this inversion. Therefore $\\Gamma_{1}$ is mapped to itself in this inversion. The same applies to $\\Gamma_{2}$. Since circle $\\omega$ passes through $P$, it is mapped to a line tangent to the images of $\\Gamma_{1}$ (itself) and $\\Gamma_{2}$ (also itself), that is, a common tangent line. This common tangent cannot be $P T$, as $P T$ is also mapped to itself. Since $\\Gamma_{1}$ and $\\Gamma_{2}$ have exactly other two common tangent lines, there are two solutions: the inverses of the tangent lines.\n![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-7.jpg?height=918&width=729&top_left_y=997&top_left_x=635)\n\nWe proceed with the construction with the aid of some macro constructions that will be detailed later.\n\nStep 1. Draw the common tangents to $\\Gamma_{1}$ and $\\Gamma_{2}$.\nStep 2. For each common tangent $t$, draw the projection $P_{t}$ of $P$ onto $t$.\nStep 3. Find the inverse $P_{1}$ of $P_{t}$ with respect to the circle with center $P$ and radius $P T$.\nStep 4. $\\omega_{t}$ is the circle with diameter $P P_{1}$.\nLet's work out the details for steps 1 and 3 . Steps 2 and 4 are immediate.\nStep 1. In this particular case in which $\\Gamma_{1}$ and $\\Gamma_{2}$ are externally tangent, there is a small shortcut:\n\n- Draw the circle with diameter on the two centers $O_{1}$ of $\\Gamma_{1}$ and $O_{2}$ of $\\Gamma_{2}$, and find its center $O$.\n- Let this circle meet common tangent line $O P$ at points $Q, R$. The required lines are the perpendicular to $O Q$ at $Q$ and the perpendicular to $O R$ at $R$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-8.jpg?height=572&width=729&top_left_y=282&top_left_x=635)\n\nLet's show why this construction works. Let $R_{i}$ be the radius of circle $\\Gamma_{i}$ and suppose without loss of generality that $R_{1} \\leq R_{2}$. Note that $O Q=\\frac{1}{2} O_{1} O_{2}=\\frac{R_{1}+R_{2}}{2}, O T=O O_{1}-R_{1}=\\frac{R_{2}-R_{1}}{2}$, so\n\n$$\n\\sin \\angle T Q O=\\frac{O T}{O Q}=\\frac{R_{2}-R_{1}}{R_{1}+R_{2}}\n$$\n\nwhich is also the sine of the angle between $O_{1} O_{2}$ and the common tangent lines.\nLet $t$ be the perpendicular to $O Q$ through $Q$. Then $\\angle\\left(t, O_{1} O_{2}\\right)=\\angle(O Q, Q T)=\\angle T Q O$, and $t$ is parallel to a common tangent line. Since\n\n$$\nd(O, t)=O Q=\\frac{R_{1}+R_{2}}{2}=\\frac{d\\left(O_{1}, t\\right)+d\\left(O_{2}, t\\right)}{2}\n$$\n\nand $O$ is the midpoint of $O_{1} O_{2}, O$ is also at the same distance from $t$ and the common tangent line, so these two lines coincide.\nStep 3. Finding the inverse of a point $X$ given the inversion circle $\\Omega$ with center $O$ is a well known procedure, but we describe it here for the sake of completeness.\n\n- If $X$ lies in $\\Omega$, then its inverse is $X$.\n- If $X$ lies in the interior of $\\Omega$, draw ray $O X$, then the perpendicular line $\\ell$ to $O X$ at $X$. Let $\\ell$ meet $\\Omega$ at a point $Y$. The inverse of $X$ is the intersection $X^{\\prime}$ of $O X$ and the line perpendicular to $O Y$ at $Y$. This is because $O Y X^{\\prime}$ is a right triangle with altitude $Y X$, and therefore $O X \\cdot O X^{\\prime}=O Y^{2}$.\n- If $X$ is in the exterior of $\\Omega$, draw ray $O X$ and one of the tangent lines $\\ell$ from $X$ to $\\Omega$ (just connect $X$ to one of the intersections of $\\Omega$ and the circle with diameter $O X$ ). Let $\\ell$ touch $\\Omega$ at a point $Y$. The inverse of $X$ is the projection $X^{\\prime}$ of $Y$ onto $O X$. This is because $O Y X^{\\prime}$ is a right triangle with altitude $Y X^{\\prime}$, and therefore $O X \\cdot O X^{\\prime}=O Y^{2}$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-8.jpg?height=347&width=481&top_left_y=2265&top_left_x=408)\n$X$ is inside $\\Omega$\n![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-8.jpg?height=398&width=469&top_left_y=2274&top_left_x=1096)", "metadata": {"resource_path": "APMO/segmented/en-apmo1991_sol.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution\n\n"}}