{"year": "2022", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Find all pairs $(a, b)$ of positive integers such that $a^{3}$ is a multiple of $b^{2}$ and $b-1$ is a multiple of $a-1$. Note: An integer $n$ is said to be a multiple of an integer $m$ if there is an integer $k$ such that $n=k m$.", "solution": ".1\n\nBy inspection, we see that the pairs $(a, b)$ with $a=b$ are solutions, and so too are the pairs $(a, 1)$. We will see that these are the only solutions.\n\n- Case 1. Consider the case $b<a$. Since $b-1$ is a multiple of $a-1$, it follows that $b=1$. This yields the second set of solutions described above.\n- Case 2. This leaves the case $b \\geq a$. Since the positive integer $a^{3}$ is a multiple of $b^{2}$, there is a positive integer $c$ such that $a^{3}=b^{2} c$.\nNote that $a \\equiv b \\equiv 1$ modulo $a-1$. So we have\n\n$$\n1 \\equiv a^{3}=b^{2} c \\equiv c \\quad(\\bmod a-1) .\n$$\n\nIf $c<a$, then we must have $c=1$, hence, $a^{3}=b^{2}$. So there is a positive integer $d$ such that $a=d^{2}$ and $b=d^{3}$. Now $a-1 \\mid b-1$ yields $d^{2}-1 \\mid d^{3}-1$. This implies that $d+1 \\mid d(d+1)+1$, which is impossible.\nIf $c \\geq a$, then $b^{2} c \\geq b^{2} a \\geq a^{3}=b^{2} c$. So there's equality throughout, implying $a=c=b$. This yields the first set of solutions described above.\nTherefore, the solutions described above are the only solutions.", "metadata": {"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl", "problem_match": "\nProblem 1.", "solution_match": "# Solution 1"}}
{"year": "2022", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Find all pairs $(a, b)$ of positive integers such that $a^{3}$ is a multiple of $b^{2}$ and $b-1$ is a multiple of $a-1$. Note: An integer $n$ is said to be a multiple of an integer $m$ if there is an integer $k$ such that $n=k m$.", "solution": ".2\n\nWe will start by showing that there are positive integers $x, c, d$ such that $a=x^{2} c d$ and $b=x^{3} c$. Let $g=\\operatorname{gcd}(a, b)$ so that $a=g d$ and $b=g x$ for some coprime $d$ and $x$. Then, $b^{2} \\mid a^{3}$ is equivalent to $g^{2} x^{2} \\mid g^{3} d^{3}$, which is equivalent to $x^{2} \\mid g d^{3}$. Since $x$ and $d$ are coprime, this implies $x^{2} \\mid g$. Hence, $g=x^{2} c$ for some $c$, giving $a=x^{2} c d$ and $b=x^{3} c$ as required.\nNow, it remains to find all positive integers $x, c, d$ satisfying\n\n$$\nx^{2} c d-1 \\mid x^{3} c-1\n$$\n\nThat is, $x^{3} c \\equiv 1\\left(\\bmod x^{2} c d-1\\right)$. Assuming that this congruence holds, it follows that $d \\equiv x^{3} c d \\equiv x$ $\\left(\\bmod x^{2} c d-1\\right)$. Then, either $x=d$ or $x-d \\geq x^{2} c d-1$ or $d-x \\geq x^{2} c d-1$.\n\n- If $x=d$ then $b=a$.\n- If $x-d \\geq x^{2} c d-1$, then $x-d \\geq x^{2} c d-1 \\geq x-1 \\geq x-d$. Hence, each of these inequalities must in fact be an equality. This implies that $x=c=d=1$, which implies that $a=b=1$.\n- If $d-x \\geq x^{2} c d-1$, then $d-x \\geq x^{2} c d-1 \\geq d-1 \\geq d-x$. Hence, each of these inequalities must in fact be an equality. This implies that $x=c=1$, which implies that $b=1$.\nHence the only solutions are the pairs $(a, b)$ such that $a=b$ or $b=1$. These pairs can be checked to satisfy the given conditions.", "metadata": {"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl", "problem_match": "\nProblem 1.", "solution_match": "# Solution 1"}}
{"year": "2022", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Find all pairs $(a, b)$ of positive integers such that $a^{3}$ is a multiple of $b^{2}$ and $b-1$ is a multiple of $a-1$. Note: An integer $n$ is said to be a multiple of an integer $m$ if there is an integer $k$ such that $n=k m$.", "solution": ".3\n\nAll answers are $(n, n)$ and $(n, 1)$ where $n$ is any positive integer. They all clearly work.\nTo show that these are all solutions, note that we can easily eliminate the case $a=1$ or $b=1$. Thus, assume that $a, b \\neq 1$ and $a \\neq b$. By the second divisibility, we see that $a-1 \\mid b-a$. However, $\\operatorname{gcd}(a, b) \\mid b-a$ and $a-1$ is relatively prime to $\\operatorname{gcd}(a, b)$. This implies that $(a-1) \\operatorname{gcd}(a, b) \\mid b-a$, which implies $\\operatorname{gcd}(a, b) \\left\\lvert\\, \\frac{b-1}{a-1}-1\\right.$.\nThe last relation implies that $\\operatorname{gcd}(a, b)<\\frac{b-1}{a-1}$, since the right-hand side are positive. However, due to the first divisibility,\n\n$$\n\\operatorname{gcd}(a, b)^{3}=\\operatorname{gcd}\\left(a^{3}, b^{3}\\right) \\geq \\operatorname{gcd}\\left(b^{2}, b^{3}\\right)=b^{2} .\n$$\n\nCombining these two inequalities, we get that\n\n$$\nb^{\\frac{2}{3}}<\\frac{b-1}{a-1}<2 \\frac{b}{a}\n$$\n\nThis implies $a<2 b^{\\frac{1}{3}}$. However, $b^{2} \\mid a^{3}$ gives $b \\leq a^{\\frac{3}{2}}$. This forces\n\n$$\na<2\\left(a^{\\frac{3}{2}}\\right)^{\\frac{1}{3}}=2 \\sqrt{a} \\Longrightarrow a<4 .\n$$\n\nExtracting $a=2,3$ by hand yields no additional solution.", "metadata": {"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl", "problem_match": "\nProblem 1.", "solution_match": "# Solution 1"}}
{"year": "2022", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "APMO", "problem": "Let $A B C$ be a right triangle with $\\angle B=90^{\\circ}$. Point $D$ lies on the line $C B$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $A D$ and let $F$ be the second intersection point of the circumcircle of $\\triangle A C D$ and the circumcircle of $\\triangle B D E$. Prove that as $D$ varies, the line $E F$ passes through a fixed point.\n![](https://cdn.mathpix.com/cropped/2024_11_22_59f4362abe92b4f1e49cg-2.jpg?height=909&width=1292&top_left_y=1492&top_left_x=449)", "solution": ".1\n\nLet the line $E F$ intersect the line $B C$ at $P$ and the circumcircle of $\\triangle A C D$ at $G$ distinct from $F$. We will prove that $P$ is the fixed point.\nFirst, notice that $\\triangle B E D$ is isosceles with $E B=E D$. This implies $\\angle E B C=\\angle E D P$.\nThen, $\\angle D A G=\\angle D F G=\\angle E B C=\\angle E D P$ which implies $A G \\| D C$. Hence, $A G C D$ is an isosceles trapezoid.\nAlso, $A G \\| D C$ and $A E=E D$. This implies $\\triangle A E G \\cong \\triangle D E P$ and $A G=D P$.\nSince $B$ is the foot of the perpendicular from $A$ onto the side $C D$ of the isosceles trapezoid $A G C D$, we have $P B=P D+D B=A G+D B=B C$, which does not depend on the choice of $D$. Hence, the initial statement is proven.", "metadata": {"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl", "problem_match": "\nProblem 2.", "solution_match": "# Solution 2"}}
{"year": "2022", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "APMO", "problem": "Let $A B C$ be a right triangle with $\\angle B=90^{\\circ}$. Point $D$ lies on the line $C B$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $A D$ and let $F$ be the second intersection point of the circumcircle of $\\triangle A C D$ and the circumcircle of $\\triangle B D E$. Prove that as $D$ varies, the line $E F$ passes through a fixed point.\n![](https://cdn.mathpix.com/cropped/2024_11_22_59f4362abe92b4f1e49cg-2.jpg?height=909&width=1292&top_left_y=1492&top_left_x=449)", "solution": ".2\n\nSet up a coordinate system where $B C$ is along the positive $x$-axis, $B A$ is along the positive $y$-axis, and $B$ is the origin. Take $A=(0, a), B=(0,0), C=(c, 0), D=(-d, 0)$ where $a, b, c, d>0$. Then $E=\\left(-\\frac{d}{2}, \\frac{a}{2}\\right)$. The general equation of a circle is\n\n$$\nx^{2}+y^{2}+2 f x+2 g y+h=0\n$$\n\nSubstituting the coordinates of $A, D, C$ into (1) and solving for $f, g, h$, we find that the equation of the circumcircle of $\\triangle A D C$ is\n\n$$\nx^{2}+y^{2}+(d-c) x+\\left(\\frac{c d}{a}-a\\right) y-c d=0\n$$\n\nSimilarly, the equation of the circumcircle of $\\triangle B D E$ is\n\n$$\nx^{2}+y^{2}+d x+\\left(\\frac{d^{2}}{2 a}-\\frac{a}{2}\\right) y=0\n$$\n\nThen (3)-(2) gives the equation of the line $D F$ which is\n\n$$\nc x+\\frac{a^{2}+d^{2}-2 c d}{2 a} y+c d=0\n$$\n\nSolving (3) and (4) simultaneously, we get\n\n$$\nF=\\left(\\frac{c\\left(d^{2}-a^{2}-2 c d\\right)}{a^{2}+(d-2 c)^{2}}, \\frac{2 a c(c-d)}{a^{2}+(d-2 c)^{2}}\\right)\n$$\n\nand the other solution $D=(-d, 0)$.\nFrom this we obtain the equation of the line $E F$ which is $a x+(d-2 c) y+a c=0$. It passes through $P(-c, 0)$ which is independent of $d$.", "metadata": {"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl", "problem_match": "\nProblem 2.", "solution_match": "# Solution 2"}}
{"year": "2022", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "APMO", "problem": "Find all positive integers $k<202$ for which there exists a positive integer $n$ such that\n\n$$\n\\left\\{\\frac{n}{202}\\right\\}+\\left\\{\\frac{2 n}{202}\\right\\}+\\cdots+\\left\\{\\frac{k n}{202}\\right\\}=\\frac{k}{2}\n$$\n\nwhere $\\{x\\}$ denote the fractional part of $x$.\nNote: $\\{x\\}$ denotes the real number $k$ with $0 \\leq k<1$ such that $x-k$ is an integer.", "solution": "Denote the equation in the problem statement as $\\left(^{*}\\right)$, and note that it is equivalent to the condition that the average of the remainders when dividing $n, 2 n, \\ldots, k n$ by 202 is 101 . Since $\\left\\{\\frac{i n}{202}\\right\\}$ is invariant in each residue class modulo 202 for each $1 \\leq i \\leq k$, it suffices to consider $0 \\leq n<202$.\n\nIf $n=0$, so is $\\left\\{\\frac{i n}{202}\\right\\}$, meaning that $(*)$ does not hold for any $k$. If $n=101$, then it can be checked that $\\left(^{*}\\right)$ is satisfied if and only if $k=1$. From now on, we will assume that $101 \\nmid n$.\nFor each $1 \\leq i \\leq k$, let $a_{i}=\\left\\lfloor\\frac{i n}{202}\\right\\rfloor=\\frac{i n}{202}-\\left\\{\\frac{i n}{202}\\right\\}$. Rewriting $\\left(^{*}\\right)$ and multiplying the equation by 202, we find that\n\n$$\nn(1+2+\\ldots+k)-202\\left(a_{1}+a_{2}+\\ldots+a_{k}\\right)=101 k\n$$\n\nEquivalently, letting $z=a_{1}+a_{2}+\\ldots+a_{k}$,\n\n$$\nn k(k+1)-404 z=202 k\n$$\n\nSince $n$ is not divisible by 101 , which is prime, it follows that $101 \\mid k(k+1)$. In particular, $101 \\mid k$ or $101 \\mid k+1$. This means that $k \\in\\{100,101,201\\}$. We claim that all these values of $k$ work.\n\n- If $k=201$, we may choose $n=1$. The remainders when dividing $n, 2 n, \\ldots, k n$ by 202 are 1,2 , ..., 201, which have an average of 101 .\n- If $k=100$, we may choose $n=2$. The remainders when dividing $n, 2 n, \\ldots, k n$ by 202 are 2,4 , ..., 200, which have an average of 101.\n- If $k=101$, we may choose $n=51$. To see this, note that the first four remainders are $51,102,153$, 2 , which have an average of 77 . The next four remainders $(53,104,155,4)$ are shifted upwards from the first four remainders by 2 each, and so on, until the 25 th set of the remainders ( 99 , $150,201,50)$ which have an average of 125 . Hence, the first 100 remainders have an average of $\\frac{77+125}{2}=101$. The 101th remainder is also 101 , meaning that the average of all 101 remainders is 101 .\n\nIn conclusion, all values $k \\in\\{1,100,101,201\\}$ satisfy the initial condition.", "metadata": {"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl", "problem_match": "\nProblem 3.", "solution_match": "# Solution\n\n"}}
{"year": "2022", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "APMO", "problem": "Let $n$ and $k$ be positive integers. Cathy is playing the following game. There are $n$ marbles and $k$ boxes, with the marbles labelled 1 to $n$. Initially, all marbles are placed inside one box. Each turn, Cathy chooses a box and then moves the marbles with the smallest label, say $i$, to either any empty box or the box containing marble $i+1$. Cathy wins if at any point there is a box containing only marble $n$.\nDetermine all pairs of integers $(n, k)$ such that Cathy can win this game.", "solution": "We claim Cathy can win if and only if $n \\leq 2^{k-1}$.\n\nFirst, note that each non-empty box always contains a consecutive sequence of labeled marbles. This is true since Cathy is always either removing from or placing in the lowest marble in a box. As a consequence, every move made is reversible.\n\nNext, we prove by induction that Cathy can win if $n=2^{k-1}$. The base case of $n=k=1$ is trivial. Assume a victory can be obtained for $m$ boxes and $2^{m-1}$ marbles. Consider the case of $m+1$ boxes and $2^{m}$ marbles. Cathy can first perform a sequence of moves so that only marbles $2^{m-1}, \\ldots, 2^{m}$ are left in the starting box, while keeping one box, say $B$, empty. Now move the marble $2^{m-1}$ to box $B$, then reverse all of the initial moves while treating $B$ as the starting box. At the end of that, we will have marbles $2^{m-1}+1, \\ldots, 2^{m}$ in the starting box, marbles $1,2, \\ldots, 2^{m-1}$ in box $B$, and $m-1$ empty boxes. By repeating the original sequence of moves on marbles $2^{m-1}+1, \\ldots, 2^{m}$, using the $m$ boxes that are not box $B$, we can reach a state where only marble $2^{m}$ remains in the starting box. Therefore\na victory is possible if $n=2^{k-1}$ or smaller.\n\nWe now prove by induction that Cathy loses if $n=2^{k-1}+1$. The base case of $n=2$ and $k=1$ is trivial. Assume a victory is impossible for $m$ boxes and $2^{m-1}+1$ marbles. For the sake of contradiction, suppose that victory is possible for $m+1$ boxes and $2^{m}+1$ marbles. In a winning sequence of moves, consider the last time a marble $2^{m-1}+1$ leaves the starting box, call this move $X$. After $X$, there cannot be a time when marbles $1, \\ldots, 2^{m-1}+1$ are all in the same box. Otherwise, by reversing these moves after $X$ and deleting marbles greater than $2^{m-1}+1$, it gives us a winning sequence of moves for $2^{m-1}+1$ marbles and $m$ boxes (as the original starting box is not used here), contradicting the inductive hypothesis. Hence starting from $X$, marbles 1 will never be in the same box as any marbles greater than or equal to $2^{m-1}+1$.\n\nNow delete marbles $2, \\ldots, 2^{m-1}$ and consider the winning moves starting from $X$. Marble 1 would only move from one empty box to another, while blocking other marbles from entering its box. Thus we effectively have a sequence of moves for $2^{m-1}+1$ marbles, while only able to use $m$ boxes. This again contradicts the inductive hypothesis. Therefore, a victory is not possible if $n=2^{k-1}+1$ or greater.", "metadata": {"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl", "problem_match": "\nProblem 4.", "solution_match": "# Solution\n\n"}}
{"year": "2022", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "APMO", "problem": "Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Determine the minimum value of $(a-b)(b-c)(c-d)(d-a)$ and determine all values of $(a, b, c, d)$ such that the minimum value is achieved.", "solution": "The minimum value is $-\\frac{1}{8}$. There are eight equality cases in total. The first one is\n\n$$\n\\left(\\frac{1}{4}+\\frac{\\sqrt{3}}{4},-\\frac{1}{4}-\\frac{\\sqrt{3}}{4}, \\frac{1}{4}-\\frac{\\sqrt{3}}{4},-\\frac{1}{4}+\\frac{\\sqrt{3}}{4}\\right) .\n$$\n\nCyclic shifting all the entries give three more quadruples. Moreover, flipping the sign $((a, b, c, d) \\rightarrow$ $(-a,-b,-c,-d))$ all four entries in each of the four quadruples give four more equality cases.", "metadata": {"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl", "problem_match": "\nProblem 5.", "solution_match": "# Solution\n\n"}}
{"year": "2022", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "APMO", "problem": "Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Determine the minimum value of $(a-b)(b-c)(c-d)(d-a)$ and determine all values of $(a, b, c, d)$ such that the minimum value is achieved.", "solution": ".1\n\nSince the expression is cyclic, we could WLOG $a=\\max \\{a, b, c, d\\}$. Let\n\n$$\nS(a, b, c, d)=(a-b)(b-c)(c-d)(d-a)\n$$\n\nNote that we have given $(a, b, c, d)$ such that $S(a, b, c, d)=-\\frac{1}{8}$. Therefore, to prove that $S(a, b, c, d) \\geq$ $-\\frac{1}{8}$, we just need to consider the case where $S(a, b, c, d)<0$.\n\n- Exactly 1 of $a-b, b-c, c-d, d-a$ is negative.\n\nSince $a=\\max \\{a, b, c, d\\}$, then we must have $d-a<0$. This forces $a>b>c>d$. Now, let us write\n\n$$\nS(a, b, c, d)=-(a-b)(b-c)(c-d)(a-d)\n$$\n\nWrite $a-b=y, b-c=x, c-d=w$ for some positive reals $w, x, y>0$. Plugging to the original condition, we have\n\n$$\n(d+w+x+y)^{2}+(d+w+x)^{2}+(d+w)^{2}+d^{2}-1=0(*)\n$$\n\nand we want to prove that $w x y(w+x+y) \\leq \\frac{1}{8}$. Consider the expression $(*)$ as a quadratic in $d$ :\n\n$$\n4 d^{2}+d(6 w+4 x+2 y)+\\left((w+x+y)^{2}+(w+x)^{2}+w^{2}-1\\right)=0\n$$\n\nSince $d$ is a real number, then the discriminant of the given equation has to be non-negative, i.e. we must have\n\n$$\n\\begin{aligned}\n4 & \\geq 4\\left((w+x+y)^{2}+(w+x)^{2}+w^{2}\\right)-(3 w+2 x+y)^{2} \\\\\n& =\\left(3 w^{2}+2 w y+3 y^{2}\\right)+4 x(w+x+y) \\\\\n& \\geq 8 w y+4 x(w+x+y) \\\\\n& =4(x(w+x+y)+2 w y)\n\\end{aligned}\n$$\n\nHowever, AM-GM gives us\n\n$$\nw x y(w+x+y) \\leq \\frac{1}{2}\\left(\\frac{x(w+x+y)+2 w y}{2}\\right)^{2} \\leq \\frac{1}{8}\n$$\n\nThis proves $S(a, b, c, d) \\geq-\\frac{1}{8}$ for any $a, b, c, d \\in \\mathbb{R}$ such that $a>b>c>d$. Equality holds if and only if $w=y, x(w+x+y)=2 w y$ and $w x y(w+x+y)=\\frac{1}{8}$. Solving these equations gives us $w^{4}=\\frac{1}{16}$ which forces $w=\\frac{1}{2}$ since $w>0$. Solving for $x$ gives us $x(x+1)=\\frac{1}{2}$, and we will get $x=-\\frac{1}{2}+\\frac{\\sqrt{3}}{2}$ as $x>0$. Plugging back gives us $d=-\\frac{1}{4}-\\frac{\\sqrt{3}}{4}$, and this gives us\n\n$$\n(a, b, c, d)=\\left(\\frac{1}{4}+\\frac{\\sqrt{3}}{4},-\\frac{1}{4}+\\frac{\\sqrt{3}}{4}, \\frac{1}{4}-\\frac{\\sqrt{3}}{4},-\\frac{1}{4}-\\frac{\\sqrt{3}}{4}\\right)\n$$\n\nThus, any cyclic permutation of the above solution will achieve the minimum equality.\n\n- Exactly 3 of $a-b, b-c, c-d, d-a$ are negative Since $a=\\max \\{a, b, c, d\\}$, then $a-b$ has to be positive. So we must have $b<c<d<a$. Now, note that\n\n$$\n\\begin{aligned}\nS(a, b, c, d) & =(a-b)(b-c)(c-d)(d-a) \\\\\n& =(a-d)(d-c)(c-b)(b-a) \\\\\n& =S(a, d, c, b)\n\\end{aligned}\n$$\n\nNow, note that $a>d>c>b$. By the previous case, $S(a, d, c, b) \\geq-\\frac{1}{8}$, which implies that\n\n$$\nS(a, b, c, d)=S(a, d, c, b) \\geq-\\frac{1}{8}\n$$\n\nas well. Equality holds if and only if\n\n$$\n(a, b, c, d)=\\left(\\frac{1}{4}+\\frac{\\sqrt{3}}{4},-\\frac{1}{4}-\\frac{\\sqrt{3}}{4}, \\frac{1}{4}-\\frac{\\sqrt{3}}{4},-\\frac{1}{4}+\\frac{\\sqrt{3}}{4}\\right)\n$$\n\nand its cyclic permutation.", "metadata": {"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl", "problem_match": "\nProblem 5.", "solution_match": "# Solution 5"}}
{"year": "2022", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "APMO", "problem": "Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Determine the minimum value of $(a-b)(b-c)(c-d)(d-a)$ and determine all values of $(a, b, c, d)$ such that the minimum value is achieved.", "solution": ".2\n\nThe minimum value is $-\\frac{1}{8}$. There are eight equality cases in total. The first one is\n\n$$\n\\left(\\frac{1}{4}+\\frac{\\sqrt{3}}{4},-\\frac{1}{4}-\\frac{\\sqrt{3}}{4}, \\frac{1}{4}-\\frac{\\sqrt{3}}{4},-\\frac{1}{4}+\\frac{\\sqrt{3}}{4}\\right) .\n$$\n\nCyclic shifting all the entries give three more quadruples. Moreover, flipping the sign $((a, b, c, d) \\rightarrow$ $(-a,-b,-c,-d)$ ) all four entries in each of the four quadruples give four more equality cases. We then begin the proof by the following optimization:\n\nClaim 1. In order to get the minimum value, we must have $a+b+c+d=0$.\n\nProof. Assume not, let $\\delta=\\frac{a+b+c+d}{4}$ and note that\n\n$$\n(a-\\delta)^{2}+(b-\\delta)^{2}+(c-\\delta)^{2}+(d-\\delta)^{2}<a^{2}+b^{2}+c^{2}+d^{2}\n$$\n\nso by shifting by $\\delta$ and scaling, we get an even smaller value of $(a-b)(b-c)(c-d)(d-a)$.\n\nThe key idea is to substitute the variables\n\n$$\n\\begin{aligned}\n& x=a c+b d \\\\\n& y=a b+c d \\\\\n& z=a d+b c\n\\end{aligned}\n$$\n\nso that the original expression is just $(x-y)(x-z)$. We also have the conditions $x, y, z \\geq-0.5$ because of:\n\n$$\n2 x+\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)=(a+c)^{2}+(b+d)^{2} \\geq 0\n$$\n\nMoreover, notice that\n\n$$\n0=(a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2(x+y+z) \\Longrightarrow x+y+z=\\frac{-1}{2}\n$$\n\nNow, we reduce to the following optimization problem.\nClaim 2. Let $x, y, z \\geq-0.5$ such that $x+y+z=-0.5$. Then, the minimum value of\n\n$$\n(x-y)(x-z)\n$$\n\nis $-1 / 8$. Moreover, the equality case occurs when $x=-1 / 4$ and $\\{y, z\\}=\\{1 / 4,-1 / 2\\}$.\nProof. We notice that\n\n$$\n\\begin{aligned}\n(x-y)(x-z)+\\frac{1}{8} & =\\left(2 y+z+\\frac{1}{2}\\right)\\left(2 z+y+\\frac{1}{2}\\right)+\\frac{1}{8} \\\\\n& =\\frac{1}{8}(4 y+4 z+1)^{2}+\\left(y+\\frac{1}{2}\\right)\\left(z+\\frac{1}{2}\\right) \\geq 0\n\\end{aligned}\n$$\n\nThe last inequality is true since both $y+\\frac{1}{2}$ and $z+\\frac{1}{2}$ are not less than zero.\nThe equality in the last inequality is attained when either $y+\\frac{1}{2}=0$ or $z+\\frac{1}{2}=0$, and $4 y+4 z+1=0$. This system of equations give $(y, z)=(1 / 4,-1 / 2)$ or $(y, z)=(-1 / 2,1 / 4)$ as the desired equality cases.\n\nNote: We can also prove (the weakened) Claim 2 by using Lagrange Multiplier, as follows. We first prove that, in fact, $x, y, z \\in[-0.5,0.5]$. This can be proved by considering that\n\n$$\n-2 x+\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)=(a-c)^{2}+(b-d)^{2} \\geq 0\n$$\n\nWe will prove the Claim 2, only that in this case, $x, y, z \\in[-0.5,0.5]$. This is already sufficient to prove the original question. We already have the bounded domain $[-0.5,0.5]^{3}$, so the global minimum must occur somewhere. Thus, it suffices to consider two cases:\n\n- If the global minimum lies on the boundary of $[-0.5,0.5]^{3}$. Then, one of $x, y, z$ must be -0.5 or 0.5 . By symmetry between $y$ and $z$, we split to a few more cases.\n- If $x=0.5$, then $y=z=-0.5$, so $(x-y)(x-z)=1$, not the minimum.\n- If $x=-0.5$, then both $y$ and $z$ must be greater or equal to $x$, so $(x-y)(x-z) \\geq 0$, not the minimum.\n- If $y=0.5$, then $x=z=-0.5$, so $(x-y)(x-z)=0$, not the minimum.\n- If $y=-0.5$, then $z=-x$, so\n\n$$\n(x-y)(x-z)=2 x(x+0.5)\n$$\n\nwhich obtain the minimum at $x=-1 / 4$. This gives the desired equality case.\n\n- If the global minimum lies in the interior $(-0.5,0.5)^{3}$, then we apply Lagrange multiplier:\n\n$$\n\\begin{aligned}\n& \\frac{\\partial}{\\partial x}(x-y)(x-z)=\\lambda \\frac{\\partial}{\\partial x}(x+y+z) \\\\\n& \\frac{\\partial}{\\partial y}(x-y)(x-z)=\\lambda \\frac{\\partial}{\\partial y}(x+y+z) \\\\\n& \\frac{\\partial}{\\partial z}(x-y)(x-z) \\Longrightarrow z-x=\\lambda \\frac{\\partial}{\\partial z}(x+y+z) \\\\\n& \\Longrightarrow y-x=\\lambda .\n\\end{aligned}\n$$\n\nAdding the last two equations gives $\\lambda=0$, or $x=y=z$. This gives $(x-y)(x-z)=0$, not the minimum.\n\nHaving exhausted all cases, we are done.", "metadata": {"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl", "problem_match": "\nProblem 5.", "solution_match": "# Solution 5"}}