{"year": "2013", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "RMM", "problem": "Let $P$ and $P^{\\prime}$ be two convex quadrilateral regions in the plane (regions contain their boundary). Let them intersect, with $O$ a point in the intersection. Suppose that for every line $\\ell$ through $O$ the segment $\\ell \\cap P$ is strictly longer than the segment $\\ell \\cap P^{\\prime}$. Is it possible that the ratio of the area of $P^{\\prime}$ to the area of $P$ is greater than 1.9?\n(Bulgaria) Nikolai Beluhov", "solution": "The answer is in the affirmative: Given a positive $\\epsilon<2$, the ratio in question may indeed be greater than $2-\\epsilon$.\n\nTo show this, consider a square $A B C D$ centred at $O$, and let $A^{\\prime}, B^{\\prime}$, and $C^{\\prime}$ be the reflections of $O$ in $A, B$, and $C$, respectively. Notice that, if $\\ell$ is a line through $O$, then the segments $\\ell \\cap A B C D$ and $\\ell \\cap A^{\\prime} B^{\\prime} C^{\\prime}$ have equal lengths, unless $\\ell$ is the line $A C$.\n\nNext, consider the points $M$ and $N$ on the segments $B^{\\prime} A^{\\prime}$ and $B^{\\prime} C^{\\prime}$, respectively, such that $B^{\\prime} M / B^{\\prime} A^{\\prime}=B^{\\prime} N / B^{\\prime} C^{\\prime}=(1-\\epsilon / 4)^{1 / 2}$. Finally, let $P^{\\prime}$ be the image of the convex quadrangle $B^{\\prime} M O N$ under the homothety of ratio $(1-\\epsilon / 4)^{1 / 4}$ centred at $O$. Clearly, the quadrangles $P \\equiv A B C D$ and $P^{\\prime}$ satisfy the conditions in the statement, and the ratio of the area of $P^{\\prime}$ to the area of $P$ is exactly $2-\\epsilon / 2$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_6f1684822a87e87fbe9ag-1.jpg?height=544&width=716&top_left_y=1170&top_left_x=672)\n\nRemarks. (1) With some care, one may also construct such example with a point $O$ being interior for both $P$ and $P^{\\prime}$. In our example, it is enough to replace vertex $O$ of $P^{\\prime}$ by a point on the segment $O D$ close enough to $O$. The details are left to the reader.\n(2) On the other hand, one may show that the ratio of areas of $P^{\\prime}$ and $P$ cannot exceed 2 (even if $P$ and $P^{\\prime}$ are arbitrary convex polygons rather than quadrilaterals). Further on, we denote by $[S]$ the area of $S$.\n\nIn order to see that $\\left[P^{\\prime}\\right]<2[P]$, let us fix some ray $r$ from $O$, and let $r_{\\alpha}$ be the ray from $O$ making an (oriented) angle $\\alpha$ with $r$. Denote by $X_{\\alpha}$ and $Y_{\\alpha}$ the points of $P$ and $P^{\\prime}$, respectively, lying on $r_{\\alpha}$ farthest from $O$, and denote by $f(\\alpha)$ and $g(\\alpha)$ the lengths of the segments $O X_{\\alpha}$ and $O Y_{\\alpha}$, respectively. Then\n\n$$\n[P]=\\frac{1}{2} \\int_{0}^{2 \\pi} f^{2}(\\alpha) d \\alpha=\\frac{1}{2} \\int_{0}^{\\pi}\\left(f^{2}(\\alpha)+f^{2}(\\pi+\\alpha)\\right) d \\alpha\n$$\n\nand similarly\n\n$$\n\\left[P^{\\prime}\\right]=\\frac{1}{2} \\int_{0}^{\\pi}\\left(g^{2}(\\alpha)+g^{2}(\\pi+\\alpha)\\right) d \\alpha\n$$\n\nBut $X_{\\alpha} X_{\\pi+\\alpha}>Y_{\\alpha} Y_{\\pi+\\alpha}$ yields $2 \\cdot \\frac{1}{2}\\left(f^{2}(\\alpha)+f^{2}(\\pi+\\alpha)\\right)=O X_{\\alpha}^{2}+O X_{\\pi+\\alpha}^{2} \\geq \\frac{1}{2} X_{\\alpha} X_{\\pi+\\alpha}^{2}>$ $\\frac{1}{2} Y_{\\alpha} Y_{\\pi+\\alpha}^{2} \\geq \\frac{1}{2}\\left(O Y_{\\alpha}^{2}+O Y_{\\pi+\\alpha}^{2}\\right)=\\frac{1}{2}\\left(g^{2}(\\alpha)+g^{2}(\\pi+\\alpha)\\right)$. Integration then gives us $2[P]>\\left[P^{\\prime}\\right]$, as needed.\n\nThis can also be proved via elementary methods. Actually, we will establish the following more general fact.\n\nFact. Let $P=A_{1} A_{2} A_{3} A_{4}$ and $P^{\\prime}=B_{1} B_{2} B_{3} B_{4}$ be two convex quadrangles in the plane, and let $O$ be one of their common points different from the vertices of $P^{\\prime}$. Denote by $\\ell_{i}$ the line $O B_{i}$, and assume that for every $i=1,2,3,4$ the length of segment $\\ell_{i} \\cap P$ is greater than the length of segment $\\ell_{i} \\cap P^{\\prime}$. Then $\\left[P^{\\prime}\\right]<2[P]$.\nProof. One of (possibly degenerate) quadrilaterals $O B_{1} B_{2} B_{3}$ and $O B_{1} B_{4} B_{3}$ is convex; the same holds for $O B_{2} B_{3} B_{4}$ and $O B_{2} B_{1} B_{4}$. Without loss of generality, we may (and will) assume that the quadrilaterals $O B_{1} B_{2} B_{3}$ and $O B_{2} B_{3} B_{4}$ are convex.\n\nDenote by $C_{i}$ such a point that $\\ell_{i} \\cap P^{\\prime}$ is the segment $B_{i} C_{i}$; let $a_{i}$ be the length of $\\ell_{i} \\cap P$, and let $\\alpha_{i}$ be the angle between $\\ell_{i}$ and $\\ell_{i+1}$ (hereafter, we use the cyclic notation, thus $\\ell_{5}=\\ell_{1}$ and so on). Thus $C_{2}$ and $C_{3}$ belong to the segment $B_{1} B_{4}, C_{1}$ lies on $B_{3} B_{4}$, and $C_{4}$ lies on $B_{1} B_{2}$. Assume that there exists an index $i$ such that the area of $B_{i} B_{i+1} C_{i} C_{i+1}$ is at least $\\left[P^{\\prime}\\right] / 2$; then we have\n\n$$\n\\frac{\\left[P^{\\prime}\\right]}{2} \\leq\\left[B_{i} B_{i+1} C_{i} C_{i+1}\\right]=\\frac{B_{i} C_{i} \\cdot B_{i+1} C_{i+1} \\cdot \\sin \\alpha_{i}}{2}<\\frac{a_{i} a_{i+1} \\sin \\alpha_{i}}{2} \\leq[P]\n$$\n\nas desired. Assume, to the contrary, that such index does not exist. Two cases are possible.\n![](https://cdn.mathpix.com/cropped/2024_11_22_6f1684822a87e87fbe9ag-2.jpg?height=468&width=1271&top_left_y=1189&top_left_x=397)\n\nCase 1. Assume that the rays $B_{1} B_{2}$ and $B_{4} B_{3}$ do not intersect (see the left figure above). This means, in particular, that $d\\left(B_{1}, B_{3} B_{4}\\right) \\leq d\\left(B_{2}, B_{3} B_{4}\\right)$.\n\nSince the ray $B_{3} O$ lies in the angle $B_{1} B_{3} B_{4}$, we obtain $d\\left(B_{1}, B_{3} C_{3}\\right) \\leq d\\left(C_{4}, B_{3} C_{3}\\right)$; hence $\\left[B_{3} B_{4} B_{1}\\right] \\leq\\left[B_{3} B_{4} C_{3} C_{4}\\right]<\\left[P^{\\prime}\\right] / 2$. Similarly, $\\left[B_{1} B_{2} B_{4}\\right] \\leq\\left[B_{1} B_{2} C_{1} C_{2}\\right]<\\left[P^{\\prime}\\right] / 2$. Thus,\n\n$$\n\\begin{aligned}\n{\\left[B_{2} B_{3} C_{2} C_{3}\\right] } & =\\left[P^{\\prime}\\right]-\\left[B_{1} B_{2} C_{3}\\right]-\\left[B_{3} B_{4} C_{2}\\right]=\\left[P^{\\prime}\\right]-\\frac{B_{1} C_{3}}{B_{1} B_{4}} \\cdot\\left[B_{1} B_{2} B_{4}\\right]-\\frac{B_{4} C_{2}}{B_{1} B_{4}} \\cdot\\left[B_{3} B_{4} B_{1}\\right] \\\\\n& >\\left[P^{\\prime}\\right]\\left(1-\\frac{B_{1} C_{3}+B_{4} C_{2}}{2 B_{1} B_{4}}\\right) \\geq \\frac{\\left[P^{\\prime}\\right]}{2} .\n\\end{aligned}\n$$\n\nA contradiction.\nCase 2. Assume now that the rays $B_{1} B_{2}$ and $B_{4} B_{3}$ intersect at some point (see the right figure above). Denote by $L$ the common point of $B_{2} C_{1}$ and $B_{3} C_{4}$. We have $\\left[B_{2} C_{4} C_{1}\\right] \\geq\\left[B_{2} C_{4} B_{3}\\right]$, hence $\\left[C_{1} C_{4} L\\right] \\geq\\left[B_{2} B_{3} L\\right]$. Thus we have\n\n$$\n\\begin{aligned}\n{\\left[P^{\\prime}\\right]>\\left[B_{1} B_{2} C_{1} C_{2}\\right]+\\left[B_{3} B_{4} C_{3} C_{4}\\right] } & =\\left[P^{\\prime}\\right]+\\left[L C_{1} C_{2} C_{3} C_{4}\\right]-\\left[B_{2} B_{3} L\\right] \\\\\n& \\geq\\left[P^{\\prime}\\right]+\\left[C_{1} C_{4} L\\right]-\\left[B_{2} B_{3} L\\right] \\geq\\left[P^{\\prime}\\right] .\n\\end{aligned}\n$$\n\nA final contradiction.", "metadata": {"resource_path": "RMM/segmented/en-2013-Solutions2013-2.jsonl", "problem_match": "\nProblem 4.", "solution_match": "\nSolution."}}
{"year": "2013", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "RMM", "problem": "Given an integer $k \\geq 2$, set $a_{1}=1$ and, for every integer $n \\geq 2$, let $a_{n}$ be the smallest $x>a_{n-1}$ such that:\n\n$$\nx=1+\\sum_{i=1}^{n-1}\\left\\lfloor\\sqrt[k]{\\frac{x}{a_{i}}}\\right\\rfloor .\n$$\n\nProve that every prime occurs in the sequence $a_{1}, a_{2}, \\ldots$.\n(Bulgaria) Alexander Ivanov", "solution": ". We prove that the $a_{n}$ are precisely the $k$ th-power-free positive integers, that is, those divisible by the $k$ th power of no prime. The conclusion then follows.\n\nLet $B$ denote the set of all $k$ th-power-free positive integers. We first show that, given a positive integer $c$,\n\n$$\n\\sum_{b \\in B, b \\leq c}\\left\\lfloor\\sqrt[k]{\\frac{c}{b}}\\right\\rfloor=c\n$$\n\nTo this end, notice that every positive integer has a unique representation as a product of an element in $B$ and a $k$ th power. Consequently, the set of all positive integers less than or equal to $c$ splits into\n\n$$\nC_{b}=\\left\\{x: x \\in \\mathbb{Z}_{>0}, x \\leq c, \\text { and } x / b \\text { is a } k \\text { th power }\\right\\}, \\quad b \\in B, b \\leq c .\n$$\n\nClearly, $\\left|C_{b}\\right|=\\lfloor\\sqrt[k]{c / b}\\rfloor$, whence the desired equality.\nFinally, enumerate $B$ according to the natural order: $1=b_{1}<b_{2}<\\cdots<b_{n}<\\cdots$. We prove by induction on $n$ that $a_{n}=b_{n}$. Clearly, $a_{1}=b_{1}=1$, so let $n \\geq 2$ and assume $a_{m}=b_{m}$ for all indices $m<n$. Since $b_{n}>b_{n-1}=a_{n-1}$ and\n\n$$\nb_{n}=\\sum_{i=1}^{n}\\left\\lfloor\\sqrt[k]{\\frac{b_{n}}{b_{i}}}\\right\\rfloor=\\sum_{i=1}^{n-1}\\left\\lfloor\\sqrt[k]{\\frac{b_{n}}{b_{i}}}\\right\\rfloor+1=\\sum_{i=1}^{n-1}\\left\\lfloor\\sqrt[k]{\\frac{b_{n}}{a_{i}}}\\right\\rfloor+1\n$$\n\nthe definition of $a_{n}$ forces $a_{n} \\leq b_{n}$. Were $a_{n}<b_{n}$, a contradiction would follow:\n\n$$\na_{n}=\\sum_{i=1}^{n-1}\\left\\lfloor\\sqrt[k]{\\frac{a_{n}}{b_{i}}}\\right\\rfloor=\\sum_{i=1}^{n-1}\\left\\lfloor\\sqrt[k]{\\frac{a_{n}}{a_{i}}}\\right\\rfloor=a_{n}-1\n$$\n\nConsequently, $a_{n}=b_{n}$. This completes the proof.", "metadata": {"resource_path": "RMM/segmented/en-2013-Solutions2013-2.jsonl", "problem_match": "\nProblem 5.", "solution_match": "\nSolution 1"}}
{"year": "2013", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "RMM", "problem": "Given an integer $k \\geq 2$, set $a_{1}=1$ and, for every integer $n \\geq 2$, let $a_{n}$ be the smallest $x>a_{n-1}$ such that:\n\n$$\nx=1+\\sum_{i=1}^{n-1}\\left\\lfloor\\sqrt[k]{\\frac{x}{a_{i}}}\\right\\rfloor .\n$$\n\nProve that every prime occurs in the sequence $a_{1}, a_{2}, \\ldots$.\n(Bulgaria) Alexander Ivanov", "solution": ". (Ilya Bogdanov) For every $n=1,2,3, \\ldots$, introduce the function\n\n$$\nf_{n}(x)=x-1-\\sum_{i=1}^{n-1}\\left\\lfloor\\sqrt[k]{\\frac{x}{a_{i}}}\\right\\rfloor\n$$\n\nDenote also by $g_{n}(x)$ the number of the indices $i \\leq n$ such that $x / a_{i}$ is the $k$ th power of an integer. Then $f_{n}(x+1)-f_{n}(x)=1-g_{n}(x)$ for every integer $x \\geq a_{n}$; hence $f_{n}(x)+1 \\geq f_{n}(x+1)$. Moreover, $f_{n}\\left(a_{n-1}\\right)=-1$ (since $f_{n-1}\\left(a_{n-1}\\right)=0$ ). Now a straightforward induction shows that $f_{n}(x)<0$ for all integers $x \\in\\left[a_{n-1}, a_{n}\\right)$.\n\nNext, if $g_{n}(x)>0$ then $f_{n}(x) \\leq f_{n}(x-1)$; this means that such an $x$ cannot equal $a_{n}$. Thus $a_{j} / a_{i}$ is never the $k$ th power of an integer if $j>i$.\n\nNow we are prepared to prove by induction on $n$ that $a_{1}, a_{2}, \\ldots, a_{n}$ are exactly all $k$ th-power-free integers in $\\left[1, a_{n}\\right]$. The base case $n=1$ is trivial.\n\nAssume that all the $k$ th-power-free integers on $\\left[1, a_{n}\\right]$ are exactly $a_{1}, \\ldots, a_{n}$. Let $b$ be the least integer larger than $a_{n}$ such that $g_{n}(b)=0$. We claim that: (1) $b=a_{n+1}$; and (2) $b$ is the least $k$ th-power-free number greater than $a_{n}$.\n\nTo prove (1), notice first that all the numbers of the form $a_{j} / a_{i}$ with $1 \\leq i<j \\leq n$ are not $k$ th powers of rational numbers since $a_{i}$ and $a_{j}$ are $k$ th-power-free. This means that for every integer $x \\in\\left(a_{n}, b\\right)$ there exists exactly one index $i \\leq n$ such that $x / a_{i}$ is the $k$ th power of an integer (certainly, $x$ is not $k$ th-power-free). Hence $f_{n+1}(x)=f_{n+1}(x-1)$ for each such $x$, so $f_{n+1}(b-1)=f_{n+1}\\left(a_{n}\\right)=-1$. Next, since $b / a_{i}$ is not the $k$ th power of an integer, we have $f_{n+1}(b)=f_{n+1}(b-1)+1=0$, thus $b=a_{n+1}$. This establishes (1).\n\nFinally, since all integers in $\\left(a_{n}, b\\right)$ are not $k$ th-power-free, we are left to prove that $b$ is $k$ th-power-free to establish (2). Otherwise, let $y>1$ be the greatest integer such that $y^{k} \\mid b$; then $b / y^{k}$ is $k$ th-power-free and hence $b / y^{k}=a_{i}$ for some $i \\leq n$. So $b / a_{i}$ is the $k$ th power of an integer, which contradicts the definition of $b$.\n\nThus $a_{1}, a_{2}, \\ldots$ are exactly all $k$ th-power-free positive integers; consequently all primes are contained in this sequence.", "metadata": {"resource_path": "RMM/segmented/en-2013-Solutions2013-2.jsonl", "problem_match": "\nProblem 5.", "solution_match": "\nSolution 2"}}
{"year": "2013", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "RMM", "problem": "$2 n$ distinct tokens are placed at the vertices of a regular $2 n$-gon, with one token placed at each vertex. A move consists of choosing an edge of the $2 n$-gon and interchanging the two tokens at the endpoints of that edge. Suppose that after a finite number of moves, every pair of tokens have been interchanged exactly once. Prove that some edge has never been chosen.\n(Russia) Alexander Gribalko", "solution": "Step 1. Enumerate all the tokens in the initial arrangement in clockwise circular order; also enumerate the vertices of the $2 n$-gon accordingly. Consider any three tokens $i<j<k$. At each moment, their cyclic order may be either $i, j, k$ or $i, k, j$, counted clockwise. This order changes exactly when two of these three tokens have been switched. Hence the order has been reversed thrice, and in the final arrangement token $k$ stands on the arc passing clockwise from token $i$ to token $j$. Thus, at the end, token $i+1$ is a counter-clockwise neighbor of token $i$ for all $i=1,2, \\ldots, 2 n-1$, so the tokens in the final arrangement are numbered successively in counter-clockwise circular order.\n\nThis means that the final arrangement of tokens can be obtained from the initial one by reflection in some line $\\ell$.\n\nStep 2. Notice that each token was involved into $2 n-1$ switchings, so its initial and final vertices have different parity. Hence $\\ell$ passes through the midpoints of two opposite sides of a $2 n$-gon; we may assume that these are the sides $a$ and $b$ connecting $2 n$ with 1 and $n$ with $n+1$, respectively.\n\nDuring the process, each token $x$ has crossed $\\ell$ at least once; thus one of its switchings has been made at edge $a$ or at edge $b$. Assume that some two its switchings were performed at $a$ and at $b$; we may (and will) assume that the one at $a$ was earlier, and $x \\leq n$. Then the total movement of token $x$ consisted at least of: (i) moving from vertex $x$ to $a$ and crossing $\\ell$ along $a$; (ii) moving from $a$ to $b$ and crossing $\\ell$ along $b$; (iii) coming to vertex $2 n+1-x$. This tales at least $x+n+(n-x)=2 n$ switchings, which is impossible.\n\nThus, each token had a switching at exactly one of the edges $a$ and $b$.\nStep 3. Finally, let us show that either each token has been switched at $a$, or each token has been switched at $b$ (then the other edge has never been used, as desired). To the contrary, assume that there were switchings at both $a$ and at $b$. Consider the first such switchings, and let $x$ and $y$ be the tokens which were moved clockwise during these switchings and crossed $\\ell$ at $a$ and $b$, respectively. By Step $2, x \\neq y$. Then tokens $x$ and $y$ initially were on opposite sides of $\\ell$.\n\nNow consider the switching of tokens $x$ and $y$; there was exactly one such switching, and we assume that it has been made on the same side of $\\ell$ as vertex $y$. Then this switching has been made after token $x$ had traced $a$. From this point on, token $x$ is on the clockwise arc from token $y$ to $b$, and it has no way to leave out from this arc. But this is impossible, since token $y$ should trace $b$ after that moment. A contradiction.\n\nRemark. The same statement for $(2 n-1)$-gon is also valid. The problem is stated for a polygon with an even number of sides only to avoid case consideration.\n\nLet us outline the solution in the case of a $(2 n-1)$-gon. We prove the existence of line $\\ell$ as in Step 1. This line passes through some vertex $x$, and through the midpoint of the opposite edge $a$. Then each token either passes through $x$, or crosses $\\ell$ along $a$ (but not both; this can be shown as in Step 2). Finally, since a token is involved into an even number of moves, it passes through $x$ but not through $a$, and $a$ is never used.", "metadata": {"resource_path": "RMM/segmented/en-2013-Solutions2013-2.jsonl", "problem_match": "\nProblem 6.", "solution_match": "\nSolution."}}