remove empty problem

Balkan_MO/segmented/en-2020-BMO-type1.jsonl

@@ -3,6 +3,5 @@
 {"year": "2020", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "Balkan_MO", "problem": "Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies on $\\gamma$.\n![](https://cdn.mathpix.com/cropped/2024_12_10_04b2c997db02c4c6a2cdg-01.jpg?height=827&width=1193&top_left_y=649&top_left_x=458)", "solution": ". Like in the previous solutions, establish that $C$ is the midpoint of the segment $B E$.\nLet $S$ be the intersection between $B D$ and $A E$. We will first show that $S$ also lies on $C O$. Because\n\n$$\n\\angle B D C=\\angle B A D+\\angle D B A=\\angle B A D+\\angle D A E=\\angle B A E\n$$\n\n(just like in solution 1), we obtain that the triangle $\\triangle S B E$ is isosceles, so $C O$ passes through $S$, because it is the perpendicular bisector of the segment $B E$.\n\nBecause $\\angle B O C=\\angle B A S$, we obtain that the quadrilateral $A S O B$ is cyclic, so $\\angle B A L=\\angle B S O$. Denote by $L$ the intersection between $A O$ and $\\gamma$, then $\\angle L D S=\\angle B A L$. Combining these two equalities leads to $\\angle B S O=\\angle L D S$, so $L D \\| S O$.\n\nThis means that $L D$ is midline in triangle $\\triangle A O C$, so $L$, which lies on $\\gamma$, is the midpoint of the segment $A O$.\n![](https://cdn.mathpix.com/cropped/2024_12_10_04b2c997db02c4c6a2cdg-04.jpg?height=871&width=1204&top_left_y=277&top_left_x=409)", "metadata": {"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl", "problem_match": "\nProblem 1.", "solution_match": "\nSolution 3"}}
 {"year": "2020", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "Balkan_MO", "problem": "Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies on $\\gamma$.\n![](https://cdn.mathpix.com/cropped/2024_12_10_04b2c997db02c4c6a2cdg-01.jpg?height=827&width=1193&top_left_y=649&top_left_x=458)", "solution": ". Like in the previous solutions, establish that $C$ is the midpoint of the segment $B E$.\nLet $F$ be the midpoint of the segment $A B$. Then both $F$ and $C$ lie on the circle of diameter $B O$. Because the quadrilateral $B F D C$ is cyclic, it means that $D$ also lies on that circle.\n\nLet $K$ be the midpoint of $B O$. Then, $A K$ must be the perpendicular bisector of the segment $B C$, so $A K \\| O C$, which implies that $\\angle K A D=\\angle D C O$. However, $\\angle D C O=\\angle K B D$, because $D B C O$ is cyclic. From the two equalities we obtain that $\\angle K A D=\\angle K B D$, so $K$ lies on $\\gamma$. Furthermore, $A K$ is the bisector of $\\angle B A D$, so $K$ is in fact the midpoint of the $\\operatorname{arc} B D$.\n\nNow consider a reflection across $O F$. Clearly, $B$ maps to $A$. Because $O F$ is the perpendicular bisector of the segment $A B, \\gamma$ maps to itself through this reflection. Thus, $K$, the intersection between $O B$ and $\\gamma$, will map to $L$, the intersection between $O A$ and $\\gamma$. This implies that $L$ is the midpoint of the segment $A O$.", "metadata": {"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl", "problem_match": "\nProblem 1.", "solution_match": "\nSolution 4"}}
 {"year": "2020", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Balkan_MO", "problem": "Denote $\\mathbb{Z}_{>0}=\\{1,2,3, \\ldots\\}$ the set of all positive integers. Determine all functions $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ such that, for each positive integer $n$,\ni) $\\sum_{k=1}^{n} f(k)$ is a perfect square, and\nii) $f(n)$ divides $n^{3}$.", "solution": "Induct on $n$ to show that $f(n)=n^{3}$ for all positive integers $n$. It is readily checked that this $f$ satisfies the conditions in the statement. The base case, $n=1$, is clear.\n\nLet $n \\geqslant 2$ and assume that $f(m)=m^{3}$ for all positive integers $m<n$. Then $\\sum_{k=1}^{n-1} f(k)=\\frac{n^{2}(n-1)^{2}}{4}$, and reference to the first condition in the statement yields\n$f(n)=\\sum_{k=1}^{n} f(k)-\\sum_{k=1}^{n-1} f(k)=\\left(\\frac{n(n-1)}{2}+k\\right)^{2}-\\frac{n^{2}(n-1)^{2}}{4}=k\\left(n^{2}-n+k\\right)$,\nfor some positive integer $k$.\nThe divisibility condition in the statement implies $k\\left(n^{2}-n+k\\right) \\leqslant n^{3}$, which is equivalent to $(n-k)\\left(n^{2}+k\\right) \\geqslant 0$, showing that $k \\leqslant n$.\n\nOn the other hand, $n^{2}-n+k$ must also divide $n^{3}$. But, if $k<n$, then\n\n$$\nn<\\frac{n^{3}}{n^{2}-1} \\leqslant \\frac{n^{3}}{n^{2}-n+k} \\leqslant \\frac{n^{3}}{n^{2}-n+1}<\\frac{n^{3}+1}{n^{2}-n+1}=n+1\n$$\n\ntherefore $\\frac{n^{3}}{n^{2}-n+k}$ cannot be an integer.\nConsequently, $k=n$, so $f(n)=n^{3}$. This completes induction and concludes the proof.", "metadata": {"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl", "problem_match": "## 2020 BMO, Problem 2", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "Balkan_MO", "problem": "", "solution": "Let $F(n)=f(1)+f(2)+\\ldots+f(n)$. We use the following two observations:\nLemma $1 F(n) \\leq\\left(\\frac{n(n+1)}{2}\\right)^{2}$\nProof: Since $f(i) \\mid i^{3}$, for all $i$ we have $f(i) \\leq i^{3}$, and adding all up we get\n\n$$\nf(1)+f(2)+\\ldots+f(n) \\leq 1^{3}+2^{3}+\\ldots+n^{3}=\\left(\\frac{n(n+1)}{2}\\right)^{2}\n$$\n\nLemma $2 F(n) \\geq n^{2}$\nProof: Note that $F(n)$ is injective and increasing since $f(i)>0, \\forall i$. Since $F(n)$ is a perfect square for all $n$ the desired result is obtained.\n\nLemma $3 f(p)=p^{3}$ for all $p$ prime.\nProof: Since $f(p) \\mid p^{3}$, the only possible values for $f(p)$ are $1, p, p^{2}, p^{3}$. We show that $f(p)$ can not be 1 or $p$ or $p^{2}$.\nCase 1: Suppose $f(p)=1$. This implies $F(p-1)$ and $F(p)$ are two consecutive numbers, grater than 1 and perfect squares. This is impossible, contradiction.\nCase 2: Suppose $f(p)=p$. Let $F(p-1)=a^{2}$ and $F(p)=b^{2}$. Hence we have $p=(a-b)(a+b)$, so $a^{2}$ has to be $\\left(\\frac{(p-1)}{2}\\right)^{2}$ and $b^{2}$ has to be $\\left(\\frac{(p+1)}{2}\\right)^{2}$. But by Lemma 2 we know $F(p-1)=a^{2} \\geq(p-1)^{2}$, contradiction.\nCase 3: Suppose $f(p)=p^{2}$. Again we have $p^{2}=(a-b)(a+b)$ and since $a, b>0$ we have to have $a-b=1$ and $a+b=p^{2}$. This gives $F(p-1)=\\left(\\frac{p^{2}-1}{2}\\right)^{2}$. But from Lemma 2, we know $F(p-1) \\leq\\left(\\frac{p^{2}-p}{2}\\right)^{2}$, hence we get a contradiction again and $f(p)$ can not be $p^{2}$.\n\nTo finish the proof, we need to show that $f(n)=n^{3}$ for all nonprime values as well. Let $p>n$ be a prime. We know $f(p)=p^{3}$ and $f(p)=F(p)-F(p-1)=(a-b)(a+b)$. By a reasoning similar to Case 2 above, we can not have $a-b=1$ and $a+b=p^{3}$ so we have to have $a-b=p$ and $a+b=p^{2}$. This gives us $F(p-1)=f(1)+f(2)+\\ldots+f(p-1)=\\left(\\frac{p^{2}-p}{2}\\right)$, so we have equality in Lemma 1 . We know this only happens if $f(i)=i^{3}$ for all $i \\leq p-1$. This concludes the proof.", "metadata": {"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl", "problem_match": "## 2020 BMO, Problem 2", "solution_match": "# Solution 2"}}
 {"year": "2020", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "Balkan_MO", "problem": "Let $k$ be a positive integer. Determine the least integer $n \\geqslant k+1$ for which the game below can be played indefinitely:\n\nConsider $n$ boxes, labelled $b_{1}, b_{2}, \\ldots, b_{n}$. For each index $i$, box $b_{i}$ contains initially exactly $i$ coins. At each step, the following three substeps are performed in order:\n(1) Choose $k+1$ boxes;\n(2) Of these $k+1$ boxes, choose $k$ and remove at least half of the coins from each, and add to the remaining box, if labelled $b_{i}$, a number of $i$ coins.\n(3) If one of the boxes is left empty, the game ends; otherwise, go to the next step.", "solution": "The required minimum is $n=2^{k}+k-1$.\nIn this case the game can be played indefinitely by choosing the last $k+1$ boxes, $b_{2^{k}-1}, b_{2^{k}}, \\ldots, b_{2^{k}+k-1}$, at each step: At step $r$, if box $b_{2^{k}+i-1}$ has exactly $m_{i}$ coins, then $\\left\\lceil m_{i} / 2\\right\\rceil$ coins are removed from that box, unless $i \\equiv r-1(\\bmod k+1)$, in which case $2^{k}+i-1$ coins are added. Thus, after step $r$ has been performed, box $b_{2^{k}+i-1}$ contains exactly $\\left\\lfloor m_{i} / 2\\right\\rfloor$ coins, unless $i \\equiv r-1(\\bmod k+1)$, in which case it contains exactly $m_{i}+2^{k}+i-1$ coins. This game goes on indefinitely, since each time a box is supplied, at least $2^{k}-1$ coins are added, so it will then contain at least $2^{k}$ coins, good enough to survive the $k$ steps to its next supply.\n\nWe now show that no smaller value of $n$ works. So, let $n \\leqslant 2^{k}+k-2$ and suppose, if possible, that a game can be played indefinitely. Notice that a box currently containing exactly $m$ coins survives at most $w=\\left\\lfloor\\log _{2} m\\right\\rfloor$ withdrawals; this $w$ will be referred to as the weight of that box. The sum of the weigths of all boxes will referred to as the total weight. The argument hinges on the lemma below, proved at the end of the solution.\n\nLemma. Performing a step does not increase the total weight. Moreover, supplying one of the first $2^{k}-2$ boxes strictly decreases the total weight.\n\nSince the total weight cannot strictly decrease indefinitely, $n>2^{k}-2$, and from some stage on none of the first $2^{k}-2$ boxes is ever supplied. Recall that each step involves a $(k+1)$-box choice. Since $n \\leqslant 2^{k}+k-2$, from that stage on, each step involves a withdrawal from at least one of the first $2^{k}-2$ boxes. This cannot go on indefinitely, so the game must eventually come to an end, contradicting the assumption.\n\nConsequently, a game that can be played indefinitely requires $n \\geqslant 2^{k}+$ $k-1$.\n\nProof of the Lemma. Since a withdrawal from a box decreases its weight by at least 1 , it is sufficient to show that supplying a box increases its weight by at most $k$; and if the latter is amongst the first $2^{k}-2$ boxes, then its weight increases by at most $k-1$. Let the box to be supplied be $b_{i}$ and let it currently contain exactly $m_{i}$ coins, to proceed by case analysis:\n\nIf $m_{i}=1$, the weight increases by $\\left\\lfloor\\log _{2}(i+1)\\right\\rfloor \\leqslant\\left\\lfloor\\log _{2}\\left(2^{k}+k-1\\right)\\right\\rfloor \\leqslant$ $\\left\\lfloor\\log _{2}\\left(2^{k+1}-2\\right)\\right\\rfloor \\leqslant k$; and if, in addition, $i \\leqslant 2^{k}-2$, then the weight increases by $\\left\\lfloor\\log _{2}(i+1)\\right\\rfloor \\leqslant\\left\\lfloor\\log _{2}\\left(2^{k}-1\\right)\\right\\rfloor=k-1$.\n\nIf $m_{i}=2$, then the weight increases by $\\left\\lfloor\\log _{2}(i+2)\\right\\rfloor-\\left\\lfloor\\log _{2} 2\\right\\rfloor \\leqslant$ $\\left\\lfloor\\log _{2}\\left(2^{k}+k\\right)\\right\\rfloor-1 \\leqslant k-1$.\n\nIf $m_{i} \\geqslant 3$, then the weight increases by\n\n$$\n\\begin{aligned}\n\\left\\lfloor\\log _{2}\\left(i+m_{i}\\right)\\right\\rfloor-\\left\\lfloor\\log _{2} m_{i}\\right\\rfloor & \\leqslant\\left\\lfloor\\log _{2}\\left(i+m_{i}\\right)-\\log _{2} m_{i}\\right\\rfloor+1 \\\\\n& \\leqslant\\left\\lfloor\\log _{2}\\left(1+\\frac{2^{k}+k-2}{3}\\right)\\right\\rfloor+1 \\leqslant k,\n\\end{aligned}\n$$\n\nsince $1+\\frac{1}{3}\\left(2^{k}+k-2\\right)=\\frac{1}{3}\\left(2^{k}+k+1\\right)<\\frac{1}{3}\\left(2^{k}+2^{k+1}\\right)=2^{k}$.\nFinally, let $i \\leqslant 2^{k}-2$ to consider the subcases $m_{i}=3$ and $m_{i} \\geqslant 4$. In the former subcase, the weight increases by\n\n$$\n\\left\\lfloor\\log _{2}(i+3)\\right\\rfloor-\\left\\lfloor\\log _{2} 3\\right\\rfloor \\leqslant\\left\\lfloor\\log _{2}\\left(2^{k}+1\\right)\\right\\rfloor-1=k-1,\n$$\n\nand in the latter by\n\n$$\n\\begin{aligned}\n\\left\\lfloor\\log _{2}\\left(i+m_{i}\\right)\\right\\rfloor-\\left\\lfloor\\log _{2} m_{i}\\right\\rfloor & \\leqslant\\left\\lfloor\\log _{2}\\left(i+m_{i}\\right)-\\log _{2} m_{i}\\right\\rfloor+1 \\\\\n& \\leqslant\\left\\lfloor\\log _{2}\\left(1+\\frac{2^{k}-2}{4}\\right)\\right\\rfloor+1 \\leqslant k-1,\n\\end{aligned}\n$$\n\nsince $1+\\frac{1}{4}\\left(2^{k}-2\\right)=\\frac{1}{4}\\left(2^{k}+2\\right)<2^{k-2}+1$. This ends the proof and completes the solution.", "metadata": {"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl", "problem_match": "## 2020 BMO, Problem 3", "solution_match": "\nSolution."}}
 {"year": "2020", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "Balkan_MO", "problem": "Let $a_{1}=2$ and, for every positive integer $n$, let $a_{n+1}$ be the smallest integer strictly greater than $a_{n}$ that has more positive divisors than $a_{n}$. Prove that $2 a_{n+1}=3 a_{n}$ only for finitely many indices $n$.", "solution": "Begin with a mere remark on the terms of the sequence under consideration.\n\nLemma 1. Each $a_{n}$ is minimal amongst all positive integers having the same number of positive divisors as $a_{n}$.\nProof. Suppose, if possible, that for some $n$, some positive integer $b<a_{n}$ has as many positive divisors as $a_{n}$. Then $a_{m}<b \\leqslant a_{m+1}$ for some $m<n$, and the definition of the sequence forces $b=a_{m+1}$. Since $b<a_{n}$, it follows that $m+1<n$, which is a contradiction, as $a_{m+1}$ should have less positive divisors than $a_{n}$.\n\nLet $p_{1}<p_{2}<\\cdots<p_{n}<\\cdots$ be the strictly increasing sequence of prime numbers, and write canonical factorisations into primes in the form $N=\\prod_{i \\geqslant 1} p_{i}^{e_{i}}$, where $e_{i} \\geqslant 0$ for all $i$, and $e_{i}=0$ for all but finitely many indices $i$; in this notation, the number of positive divisors of $N$ is $\\tau(N)=$ $\\prod_{i \\geqslant 1}\\left(e_{i}+1\\right)$.\nLemma 2. The exponents in the canonical factorisation of each $a_{n}$ into primes form a non-strictly decreasing sequence.\nProof. Indeed, if $e_{i}<e_{j}$ for some $i<j$ in the canonical decomposition of $a_{n}$ into primes, then swapping the two exponents yields a smaller integer with the same number of positive divisors, contradicting Lemma 1.\n\nWe are now in a position to prove the required result. For convenience, a term $a_{n}$ satisfying $3 a_{n}=2 a_{n+1}$ will be referred to as a special term of the sequence.\n\nSuppose now, if possible, that the sequence has infinitely many special terms, so the latter form a strictly increasing, and hence unbounded, subsequence. To reach a contradiction, it is sufficient to show that:\n(1) The exponents of the primes in the factorisation of special terms have a common upper bound $e$; and\n(2) For all large enough primes $p$, no special term is divisible by $p$.\n\nRefer to Lemma 2 to write $a_{n}=\\prod_{i \\geqslant 1} p_{i}^{e_{i}(n)}$, where $e_{i}(n) \\geqslant e_{i+1}(n)$ for all $i$.\n\nStatement (2) is a straightforward consequence of (1) and Lemma 1. Suppose, if possible, that some special term $a_{n}$ is divisible by a prime $p_{i}>2^{e+1}$, where $e$ is the integer provided by (1). Then $e \\geqslant e_{i}(n)>0$, so $2^{e_{1}(n) e_{i}(n)+e_{i}(n)} a_{n} / p_{i}^{e_{i}(n)}$ is a positive integer with the same number of positive divisors as $a_{n}$, but smaller than $a_{n}$. This contradicts Lemma 1. Consequently, no special term is divisible by a prime exceeding $2^{e+1}$.\n\nTo prove (1), it is sufficient to show that, as $a_{n}$ runs through the special terms, the exponents $e_{1}(n)$ are bounded from above. Then, Lemma 2 shows that such an upper bound $e$ suits all primes.\n\nConsider a large enough special $a_{n}$. The condition $\\tau\\left(a_{n}\\right)<\\tau\\left(a_{n+1}\\right)$ is then equivalent to $\\left(e_{1}(n)+1\\right)\\left(e_{2}(n)+1\\right)<e_{1}(n)\\left(e_{2}(n)+2\\right)$. Alternatively, but equivalently, $e_{1}(n) \\geqslant e_{2}(n)+2$. The latter implies that $a_{n}$ is divisible by 8 , for either $e_{1}(n) \\geqslant 3$ or $a_{n}$ is a large enough power of 2 .\n\nNext, note that $9 a_{n} / 8$ is an integer strictly between $a_{n}$ and $a_{n+1}$, so $\\tau\\left(9 a_{n} / 8\\right) \\leqslant \\tau\\left(a_{n}\\right)$, which is equivalent to\n\n$$\n\\left(e_{1}(n)-2\\right)\\left(e_{2}(n)+3\\right) \\leqslant\\left(e_{1}(n)+1\\right)\\left(e_{2}(n)+1\\right),\n$$\n\nso $2 e_{1}(n) \\leqslant 3 e_{2}(n)+7$. This shows that $a_{n}$ is divisible by 3 , for otherwise, letting $a_{n}$ run through the special terms, 3 would be an upper bound for all but finitely many $e_{1}(n)$, and the special terms would therefore form a bounded sequence.\n\nThus, $4 a_{n} / 3$ is another integer strictly between $a_{n}$ and $a_{n+1}$. As before, $\\tau\\left(4 a_{n} / 3\\right) \\leqslant \\tau\\left(a_{n}\\right)$. Alternatively, but equivalently,\n\n$$\n\\left(e_{1}(n)+3\\right) e_{2}(n) \\leqslant\\left(e_{1}(n)+1\\right)\\left(e_{2}(n)+1\\right),\n$$\n\nso $2 e_{2}(n)-1 \\leqslant e_{1}(n)$. Combine this with the inequality in the previous paragraph to write $4 e_{2}(n)-2 \\leqslant 2 e_{1}(n) \\leqslant 3 e_{2}(n)+7$ and infer that $e_{2}(n) \\leqslant 9$. Consequently, $2 e_{1}(n) \\leqslant 3 e_{2}(n)+7 \\leqslant 34$, showing that $e=17$ is suitable for (1) to hold. This establishes (1) and completes the solution.", "metadata": {"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl", "problem_match": "## 2020 BMO, Problem 4", "solution_match": "\nSolution."}}

SovietUnion/segmented/en-ASU-1961-1991.jsonl

@@ -142,7 +142,6 @@
 {"year": "1970", "tier": "T1", "problem_label": "13", "problem_type": null, "exam": "AllSovietUnion", "problem": "If the numbers from 11111 to 99999 are arranged in an arbitrary order show that the resulting 444445 digit number is not a power of 2.", "solution": "Let the set of numbers be S. Define the function f on S as follows. Replace each digit i in n by 9- i for \\(0 < \\mathrm{i} < 9\\) . This gives f(n). Then f(f(n)) = n, so f is a bijection. The fixed points have only the digits 0 and 9 and so are all divisible by 9. The other points divide into pairs (n, f(n)) and the sum of each pair is divisible by 9. Hence the sum of all the numbers in S is divisible by 9.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 13", "solution_match": "# Solution"}}
 {"year": "1970", "tier": "T1", "problem_label": "14", "problem_type": null, "exam": "AllSovietUnion", "problem": "S is the set of all positive integers with n decimal digits or less and with an even digit sum. T is the set of all positive integers with n decimal digits or less and an odd digit sum. Show that the sum of the kth powers of the members of S equals the sum for T if \\(1 \\leq \\mathrm{k} < \\mathrm{n}\\) .", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 14", "solution_match": ""}}
 {"year": "1970", "tier": "T1", "problem_label": "15", "problem_type": null, "exam": "AllSovietUnion", "problem": "The vertices of a regular n- gon are colored (each vertex has only one color). Each color is applied to at least three vertices. The vertices of any given color form a regular polygon. Show that there are two colors which are applied to the same number of vertices.", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 15", "solution_match": ""}}
-{"year": "1971", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "AllSovietUnion", "problem": "", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 1", "solution_match": ""}}
 {"year": "1971", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "AllSovietUnion", "problem": "Prove that we can find a number divisible by \\(2^{n}\\) whose decimal representation uses only the digits 1 and 2.", "solution": "Induction on n. We claim that we can find N with n digits, all 1 or 2, so that N is divisible by \\(2^{n}\\) . True for \\(\\mathrm{n} = 1\\) : take \\(\\mathrm{N} = 2\\) . Suppose it is true for n. If \\(2^{n + 1}\\) divides N, then since \\(2^{n + 1}\\) divides \\(2 \\times 10^{n}\\) it also divides N' obtained from N by placing a 2 in front of it. If \\(2^{n + 1}\\) does not divide N, then \\(\\mathrm{N} = 2^{n} \\times \\mathrm{odd}\\) and \\(10^{n} = 2^{n} \\times \\mathrm{odd}\\) , so \\(\\mathrm{N} + 10^{n}\\) (in other words the \\(\\mathrm{n} + 1\\) digit number obtained by placing a 1 in front of N) is divisible by \\(2^{n + 1}\\) .", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "\nProblem 1", "solution_match": "# Solution"}}
 {"year": "1971", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "AllSovietUnion", "problem": "(1) \\(\\mathrm{A_1A_2A_3}\\) is a triangle. Points \\(\\mathrm{B_1}\\) , \\(\\mathrm{B_2}\\) , \\(\\mathrm{B_3}\\) are chosen on \\(\\mathrm{A_1A_2}\\) , \\(\\mathrm{A_2A_3}\\) , \\(\\mathrm{A_3A_1}\\) respectively and points \\(\\mathrm{D_1}\\) , \\(\\mathrm{D_2}\\) \\(\\mathrm{D_3}\\) on \\(\\mathrm{A_3A_1}\\) , \\(\\mathrm{A_1A_2}\\) , \\(\\mathrm{A_2A_3}\\) respectively, so that if parallelograms \\(\\mathrm{A_iB_iC_iD_i}\\) are formed, then the lines \\(\\mathrm{A_iC_i}\\) concur. Show that \\(\\mathrm{A_1B_1}\\) : \\(\\mathrm{A_2B_2}\\) : \\(\\mathrm{A_3B_3} = \\mathrm{A_1D_1}\\) : \\(\\mathrm{A_2D_2}\\) : \\(\\mathrm{A_3D_3}\\) .  \n\n(2) \\(\\mathrm{A_1A_2\\ldots A_n}\\) is a convex polygon. Points \\(\\mathrm{B_i}\\) are chosen on \\(\\mathrm{A_iA_{i + 1}}\\) (where we take \\(\\mathrm{A_{n + 1}}\\) to mean \\(\\mathrm{A_1}\\) ), and points \\(\\mathrm{D_i}\\) on \\(\\mathrm{A_{i - 1}A_i}\\) (where we take \\(\\mathrm{A_0}\\) to mean \\(\\mathrm{A_n}\\) ) such that if parallelograms \\(\\mathrm{A_iB_iC_iD_i}\\) are formed, then the n lines \\(\\mathrm{A_iC_i}\\) concur. Show that \\(\\prod \\mathrm{A_iB_i} = \\prod \\mathrm{A_iD_i}\\) .", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 2", "solution_match": ""}}
 {"year": "1971", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "AllSovietUnion", "problem": "(1) Player A writes down two rows of 10 positive integers, one under the other. The numbers must be chosen so that if a is under b and c is under d, then a + d = b + c. Player B is allowed to ask for the identity of the number in row i, column j. How many questions must he ask to be sure of determining all the numbers?  \n\n(2) An m x n array of positive integers is written on the blackboard. It has the property that for any four numbers a, b, c, d with a and b in the same row, c and d in the same row, a above c (in the same column) and b above d (in the same column) we have a + d = b + c. If some numbers are wiped off, how many must be left for the table to be accurately restored?", "solution": "(1) is trivial. We can write the condition as \\(\\mathrm{b - a = d - c}\\) , so the 10 numbers in the first row and 1 in the second row can all be chosen arbitrarily. Hence at least 11 questions are needed. But they are also sufficient. Having determined those numbers, the others immediately follow.  \n\n(2). The m+n-1 numbers in the first row and first column can all be chosen arbitrarily, but are sufficient to determine all the numbers. Hence at least m+n-1 numbers must survive.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
@@ -167,17 +166,11 @@
 {"year": "1972", "tier": "T1", "problem_label": "7", "problem_type": null, "exam": "AllSovietUnion", "problem": "O is the point of intersection of the diagonals of the convex quadrilateral ABCD. Prove that the line joining the centroids of ABO and CDO is perpendicular to the line joining the orthocenters of BCO and ADO.", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 7", "solution_match": ""}}
 {"year": "1972", "tier": "T1", "problem_label": "8", "problem_type": null, "exam": "AllSovietUnion", "problem": "9 lines each divide a square into two quadrilaterals with areas 2/5 and 3/5 that of the square. Show that 3 of the lines meet in a point.", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 8", "solution_match": ""}}
 {"year": "1972", "tier": "T1", "problem_label": "9", "problem_type": null, "exam": "AllSovietUnion", "problem": "A 7- gon is inscribed in a circle. The center of the circle lies inside the 7- gon. A, B, C are adjacent vertices of the 7- gon show that the sum of the angles at A, B, C is less than 450 degrees.", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 9", "solution_match": ""}}
-{"year": "1972", "tier": "T1", "problem_label": "10", "problem_type": null, "exam": "AllSovietUnion", "problem": "", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 10", "solution_match": ""}}
 {"year": "1972", "tier": "T1", "problem_label": "10", "problem_type": null, "exam": "AllSovietUnion", "problem": "Two players play the following game. At each turn the first player chooses a decimal digit, then the second player substitutes it for one of the stars in the subtraction | **** - **** |. The first player tries to end up with the largest possible result, the second player tries to end up with the smallest possible result. Show that the first player can always play so that the result is at least 4000 and that the second player can always play so that the result is at most 4000.", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "\nProblem 10", "solution_match": ""}}
-{"year": "1972", "tier": "T1", "problem_label": "11", "problem_type": null, "exam": "AllSovietUnion", "problem": "", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 11", "solution_match": ""}}
 {"year": "1972", "tier": "T1", "problem_label": "11", "problem_type": null, "exam": "AllSovietUnion", "problem": "For positive reals x, y let f(x, y) be the smallest of x, 1/y, y + 1/x. What is the maximum value of f(x, y)? What are the corresponding x, y?", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "\nProblem 11", "solution_match": ""}}
-{"year": "1972", "tier": "T1", "problem_label": "12", "problem_type": null, "exam": "AllSovietUnion", "problem": "", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 12", "solution_match": ""}}
 {"year": "1972", "tier": "T1", "problem_label": "12", "problem_type": null, "exam": "AllSovietUnion", "problem": "P is a convex polygon and X is an interior point such that for every pair of vertices A, B, the triangle XAB is isosceles. Prove that all the vertices of P lie on some circle center X.", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "\nProblem 12", "solution_match": ""}}
-{"year": "1972", "tier": "T1", "problem_label": "13", "problem_type": null, "exam": "AllSovietUnion", "problem": "", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 13", "solution_match": ""}}
 {"year": "1972", "tier": "T1", "problem_label": "13", "problem_type": null, "exam": "AllSovietUnion", "problem": "Is it possible to place the digits 0, 1, 2 into unit squares of 100 x 100 cross- lined paper such that every 3 x 4 (and every 4 x 3) rectangle contains three 0s, four 1s and five 2s?", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "\nProblem 13", "solution_match": ""}}
-{"year": "1972", "tier": "T1", "problem_label": "14", "problem_type": null, "exam": "AllSovietUnion", "problem": "", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 14", "solution_match": ""}}
 {"year": "1972", "tier": "T1", "problem_label": "14", "problem_type": null, "exam": "AllSovietUnion", "problem": "x₁, x₂, ..., xₙ are positive reals with sum 1. Let s be the largest of x₁/(1 + x₁), x₂/(1 + x₁ + x₂), ..., xₙ/(1 + x₁ + ... + xₙ). What is the smallest possible value of s? What are the corresponding xᵢ?", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "\nProblem 14", "solution_match": ""}}
-{"year": "1972", "tier": "T1", "problem_label": "15", "problem_type": null, "exam": "AllSovietUnion", "problem": "", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 15", "solution_match": ""}}
 {"year": "1972", "tier": "T1", "problem_label": "15", "problem_type": null, "exam": "AllSovietUnion", "problem": "n teams compete in a tournament. Each team plays every other team once. In each game a team gets 2 points for a win, 1 for a draw and 0 for a loss. Given any subset S of teams, one can find a team (possibly in S) whose total score in the games with teams in S was odd. Prove that n is even.", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "\nProblem 15", "solution_match": ""}}
 {"year": "1973", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "AllSovietUnion", "problem": "You are given 14 coins. It is known that genuine coins all have the same weight and that fake coins all have the same weight, but weigh less than genuine coins. You suspect that 7 particular coins are genuine and the other 7 fake. Given a balance, how can you prove this in three weighings (assuming that you turn out to be correct)?", "solution": "Let the coins you suspect to be genuine be \\(\\mathrm{G}_1\\) , \\(\\mathrm{G}_2\\) , ..., \\(\\mathrm{G}_7\\) , and the suspected fakes by \\(\\mathrm{F}_1\\) , \\(\\mathrm{F}_2\\) , ..., \\(\\mathrm{F}_7\\) . First weigh \\(\\mathrm{F}_1\\) against \\(\\mathrm{G}_1\\) . Assuming \\(\\mathrm{F}_1\\) weighs less, you have proved that \\(\\mathrm{F}_1\\) is fake and \\(\\mathrm{G}_1\\) genuine. Second, weigh \\(\\mathrm{F}_1\\) , \\(\\mathrm{G}_2\\) , \\(\\mathrm{G}_3\\) against \\(\\mathrm{G}_1\\) , \\(\\mathrm{F}_2\\) , \\(\\mathrm{F}_3\\) . Assuming the first three weigh more you have proved that they include more genuine coins than the second three. But the second three includes one genuine coin \\((\\mathrm{G}_1)\\) and the first three includes one fake \\((\\mathrm{F}_1)\\) , so you have proved that \\(\\mathrm{G}_2\\) and \\(\\mathrm{G}_3\\) are genuine and \\(\\mathrm{F}_2\\) and \\(\\mathrm{F}_3\\) fake. Finally, weigh \\(\\mathrm{F}_1\\) , \\(\\mathrm{F}_2\\) , \\(\\mathrm{F}_3\\) , \\(\\mathrm{G}_4\\) , \\(\\mathrm{G}_5\\) , \\(\\mathrm{G}_6\\) , \\(\\mathrm{G}_7\\) against \\(\\mathrm{F}_4\\) , \\(\\mathrm{F}_5\\) , \\(\\mathrm{F}_6\\) , \\(\\mathrm{F}_7\\) , \\(\\mathrm{G}_1\\) , \\(\\mathrm{G}_2\\) , \\(\\mathrm{G}_3\\) . Assuming the first group weighs more it must include more genuine coins and hence just four genuine coins. Similarly, the second group must include four fakes. So you have proved the identity of the remaining coins.", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
 {"year": "1973", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "AllSovietUnion", "problem": "Prove that a 9 digit decimal number whose digits are all different, which does not end with 5 and or contain a 0, cannot be a square.", "solution": "", "metadata": {"resource_path": "SovietUnion/segmented/en-ASU-1961-1991.jsonl", "problem_match": "# Problem 2", "solution_match": ""}}