| {"year": "2016", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "RMM", "problem": "Let $x$ and $y$ be positive real numbers such that $x+y^{2016} \\geq 1$. Prove that $x^{2016}+y>$ 1 - 1/100.", "solution": "If $x \\geq 1-1 /(100 \\cdot 2016)$, then\n\n$$\nx^{2016} \\geq\\left(1-\\frac{1}{100 \\cdot 2016}\\right)^{2016}>1-2016 \\cdot \\frac{1}{100 \\cdot 2016}=1-\\frac{1}{100}\n$$\n\nby Bernoulli's inequality, whence the conclusion.\nIf $x<1-1 /(100 \\cdot 2016)$, then $y \\geq(1-x)^{1 / 2016}>(100 \\cdot 2016)^{-1 / 2016}$, and it is sufficient to show that the latter is greater than $1-1 / 100=99 / 100$; alternatively, but equivalently, that\n\n$$\n\\left(1+\\frac{1}{99}\\right)^{2016}>100 \\cdot 2016\n$$\n\nTo establish the latter, refer again to Bernoulli's inequality to write\n\n$$\n\\left(1+\\frac{1}{99}\\right)^{2016}>\\left(1+\\frac{1}{99}\\right)^{99 \\cdot 20}>\\left(1+99 \\cdot \\frac{1}{99}\\right)^{20}=2^{20}>100 \\cdot 2016\n$$\n\nRemarks. (1) Although the constant $1 / 100$ is not sharp, it cannot be replaced by the smaller constant $1 / 400$, as the values $x=1-1 / 210$ and $y=1-1 / 380$ show.\n(2) It is natural to ask whether $x^{n}+y \\geq 1-1 / k$, whenever $x$ and $y$ are positive real numbers such that $x+y^{n} \\geq 1$, and $k$ and $n$ are large. Using the inequality $\\left(1+\\frac{1}{k-1}\\right)^{k}>\\mathrm{e}$, it can be shown along the lines in the solution that this is indeed the case if $k \\leq \\frac{n}{2 \\log n}(1+o(1))$. It seems that this estimate differs from the best one by a constant factor.", "metadata": {"resource_path": "RMM/segmented/en-2016-Solutions_RMM2016-2.jsonl", "problem_match": "\nProblem 4.", "solution_match": "\nSolution."}} |
| {"year": "2016", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "RMM", "problem": "A convex hexagon $A_{1} B_{1} A_{2} B_{2} A_{3} B_{3}$ is inscribed in a circle $\\Omega$ of radius $R$. The diagonals $A_{1} B_{2}, A_{2} B_{3}$, and $A_{3} B_{1}$ concur at $X$. For $i=1,2,3$, let $\\omega_{i}$ be the circle tangent to the segments $X A_{i}$ and $X B_{i}$, and to the arc $A_{i} B_{i}$ of $\\Omega$ not containing other vertices of the hexagon; let $r_{i}$ be the radius of $\\omega_{i}$.\n(a) Prove that $R \\geq r_{1}+r_{2}+r_{3}$.\n(b) If $R=r_{1}+r_{2}+r_{3}$, prove that the six points where the circles $\\omega_{i}$ touch the diagonals $A_{1} B_{2}, A_{2} B_{3}, A_{3} B_{1}$ are concyclic.", "solution": "(a) Let $\\ell_{1}$ be the tangent to $\\Omega$ parallel to $A_{2} B_{3}$, lying on the same side of $A_{2} B_{3}$ as $\\omega_{1}$. The tangents $\\ell_{2}$ and $\\ell_{3}$ are defined similarly. The lines $\\ell_{1}$ and $\\ell_{2}, \\ell_{2}$ and $\\ell_{3}, \\ell_{3}$ and $\\ell_{1}$ meet at $C_{3}, C_{1}, C_{2}$, respectively (see Fig. 1). Finally, the line $C_{2} C_{3}$ meets the rays $X A_{1}$ and $X B_{1}$ emanating from $X$ at $S_{1}$ and $T_{1}$, respectively; the points $S_{2}, T_{2}$, and $S_{3}, T_{3}$ are defined similarly.\n\nEach of the triangles $\\Delta_{1}=\\triangle X S_{1} T_{1}, \\Delta_{2}=\\triangle T_{2} X S_{2}$, and $\\Delta_{3}=\\triangle S_{3} T_{3} X$ is similar to $\\Delta=\\triangle C_{1} C_{2} C_{3}$, since their corresponding sides are parallel. Let $k_{i}$ be the ratio of similitude of $\\Delta_{i}$ and $\\Delta$ (e.g., $k_{1}=X S_{1} / C_{1} C_{2}$ and the like). Since $S_{1} X=C_{2} T_{3}$ and $X T_{2}=S_{3} C_{1}$, it follows that $k_{1}+k_{2}+k_{3}=1$, so, if $\\rho_{i}$ is the inradius of $\\Delta_{i}$, then $\\rho_{1}+\\rho_{2}+\\rho_{3}=R$.\n\nFinally, notice that $\\omega_{i}$ is interior to $\\Delta_{i}$, so $r_{i} \\leq \\rho_{i}$, and the conclusion follows by the preceding.\n\n\nFig. 1\n\n\nFig. 2\n(b) By part (a), the equality $R=r_{1}+r_{2}+r_{3}$ holds if and only if $r_{i}=\\rho_{i}$ for all $i$, which implies in turn that $\\omega_{i}$ is the incircle of $\\Delta_{i}$. Let $K_{i}, L_{i}, M_{i}$ be the points where $\\omega_{i}$ touches the sides $X S_{i}, X T_{i}, S_{i} T_{i}$, respectively. We claim that the six points $K_{i}$ and $L_{i}(i=1,2,3)$ are equidistant from $X$.\n\nClearly, $X K_{i}=X L_{i}$, and we are to prove that $X K_{2}=X L_{1}$ and $X K_{3}=X L_{2}$. By similarity, $\\angle T_{1} M_{1} L_{1}=\\angle C_{3} M_{1} M_{2}$ and $\\angle S_{2} M_{2} K_{2}=\\angle C_{3} M_{2} M_{1}$, so the points $M_{1}, M_{2}, L_{1}, K_{2}$ are collinear. Consequently, $\\angle X K_{2} L_{1}=\\angle C_{3} M_{1} M_{2}=\\angle C_{3} M_{2} M_{1}=\\angle X L_{1} K_{2}$, so $X K_{2}=X L_{1}$. Similarly, $X K_{3}=X L_{2}$.\n\nRemark. Under the assumption in part (b), the point $M_{i}$ is the centre of a homothety mapping $\\omega_{i}$ to $\\Omega$. Since this homothety maps $X$ to $C_{i}$, the points $M_{i}, C_{i}, X$ are collinear, so $X$ is the Gergonne point of the triangle $C_{1} C_{2} C_{3}$. This condition is in fact equivalent to $R=r_{1}+r_{2}+r_{3}$.", "metadata": {"resource_path": "RMM/segmented/en-2016-Solutions_RMM2016-2.jsonl", "problem_match": "\nProblem 5.", "solution_match": "\nSolution."}} |
| {"year": "2016", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "RMM", "problem": "A set of $n$ points in Euclidean 3-dimensional space, no four of which are coplanar, is partitioned into two subsets $\\mathcal{A}$ and $\\mathcal{B}$. An $\\mathcal{A B}$-tree is a configuration of $n-1$ segments, each of which has an endpoint in $\\mathcal{A}$ and the other in $\\mathcal{B}$, and such that no segments form a closed polyline. An $\\mathcal{A B}$-tree is transformed into another as follows: choose three distinct segments $A_{1} B_{1}, B_{1} A_{2}$ and $A_{2} B_{2}$ in the $\\mathcal{A B}$-tree such that $A_{1}$ is in $\\mathcal{A}$ and $A_{1} B_{1}+A_{2} B_{2}>A_{1} B_{2}+A_{2} B_{1}$, and remove the segment $A_{1} B_{1}$ to replace it by the segment $A_{1} B_{2}$. Given any $\\mathcal{A B}$-tree, prove that every sequence of successive transformations comes to an end (no further transformation is possible) after finitely many steps.", "solution": "The configurations of segments under consideration are all bipartite geometric trees on the points $n$ whose vertex-parts are $\\mathcal{A}$ and $\\mathcal{B}$, and transforming one into another preserves the degree of any vertex in $\\mathcal{A}$, but not necessarily that of a vertex in $\\mathcal{B}$.\n\nThe idea is to devise a strict semi-invariant of the process, i.e., assign each $\\mathcal{A B}$-tree a real number strictly decreasing under a transformation. Since the number of trees on the $n$ points is finite, the conclusion follows.\n\nTo describe the assignment, consider an $\\mathcal{A B}$-tree $\\mathcal{T}=(\\mathcal{A} \\sqcup \\mathcal{B}, \\mathcal{E})$. Removal of an edge $e$ of $\\mathcal{T}$ splits the graph into exactly two components. Let $p_{\\mathcal{T}}(e)$ be the number of vertices in $\\mathcal{A}$ lying in the component of $\\mathcal{T}-e$ containing the $\\mathcal{A}$-endpoint of $e$; since $\\mathcal{T}$ is a tree, $p_{\\mathcal{T}}(e)$ counts the number of paths in $\\mathcal{T}-e$ from the $\\mathcal{A}$-endpoint of $e$ to vertices in $\\mathcal{A}$ (including the one-vertex path). Define $f(\\mathcal{T})=\\sum_{e \\in \\mathcal{E}} p_{\\mathcal{T}}(e)|e|$, where $|e|$ is the Euclidean length of $e$.\n\nWe claim that $f$ strictly decreases under a transformation. To prove this, let $\\mathcal{T}^{\\prime}$ be obtained from $\\mathcal{T}$ by a transformation involving the polyline $A_{1} B_{1} A_{2} B_{2}$; that is, $A_{1}$ and $A_{2}$ are in $\\mathcal{A}, B_{1}$\nand $B_{2}$ are in $\\mathcal{B}, A_{1} B_{1}+A_{2} B_{2}>A_{1} B_{2}+A_{2} B_{1}$, and $\\mathcal{T}^{\\prime}=\\mathcal{T}-A_{1} B_{1}+A_{1} B_{2}$. It is readily checked that $p_{\\mathcal{T}^{\\prime}}(e)=p_{\\mathcal{T}}(e)$ for every edge $e$ of $\\mathcal{T}$ different from $A_{1} B_{1}, A_{2} B_{1}$ and $A_{2} B_{2}, p_{\\mathcal{T}^{\\prime}}\\left(A_{1} B_{2}\\right)=$ $p_{\\mathcal{T}}\\left(A_{1} B_{1}\\right), p_{\\mathcal{T}^{\\prime}}\\left(A_{2} B_{1}\\right)=p_{\\mathcal{T}}\\left(A_{2} B_{1}\\right)+p_{\\mathcal{T}}\\left(A_{1} B_{1}\\right)$, and $p_{\\mathcal{T}^{\\prime}}\\left(A_{2} B b_{2}\\right)=p_{\\mathcal{T}}\\left(A_{2} B_{2}\\right)-p_{\\mathcal{T}}\\left(A_{1} B_{1}\\right)$. Consequently,\n\n$$\n\\begin{aligned}\nf\\left(\\mathcal{T}^{\\prime}\\right)-f(\\mathcal{T})= & p_{\\mathcal{T}^{\\prime}}\\left(A_{1} B_{2}\\right) \\cdot A_{1} B_{2}+\\left(p_{\\mathcal{T}^{\\prime}}\\left(A_{2} B_{1}\\right)-p_{\\mathcal{T}}\\left(A_{2} B_{1}\\right)\\right) \\cdot A_{2} B_{1}+ \\\\\n& \\left(p_{\\mathcal{T}^{\\prime}}\\left(A_{2} B_{2}\\right)-p_{\\mathcal{T}}\\left(A_{2} B_{2}\\right)\\right) \\cdot A_{2} B_{2}-p_{\\mathcal{T}}\\left(A_{1} B_{1}\\right) \\cdot A_{1} B_{1} \\\\\n= & p_{\\mathcal{T}}\\left(A_{1} B_{1}\\right)\\left(A_{1} B_{2}+A_{2} B_{1}-A_{2} B_{2}-A_{1} B_{1}\\right)<0\n\\end{aligned}\n$$\n\nRemarks. (1) The solution above does not involve the geometric structure of the configurations, so the conclusion still holds if the Euclidean length (distance) is replaced by any real-valued function on $\\mathcal{A} \\times \\mathcal{B}$.\n(2) There are infinitely many strict semi-invariants that can be used to establish the conclusion, as we are presently going to show. The idea is to devise a non-strict real-valued semiinvariant $f_{A}$ for each $A$ in $\\mathcal{A}$ (i.e., $f_{A}$ does not increase under a transformation) such that $\\sum_{A \\in \\mathcal{A}} f_{A}=f$. It then follows that any linear combination of the $f_{A}$ with positive coefficients is a strict semi-invariant.\n\nTo describe $f_{A}$, where $A$ is a fixed vertex in $\\mathcal{A}$, let $\\mathcal{T}$ be an $\\mathcal{A B}$-tree. Since $\\mathcal{T}$ is a tree, by orienting all paths in $\\mathcal{T}$ with an endpoint at $A$ away from $A$, every edge of $\\mathcal{T}$ comes out with a unique orientation so that the in-degree of every vertex of $\\mathcal{T}$ other than $A$ is 1 . Define $f_{A}(\\mathcal{T})$ to be the sum of the Euclidean lengths of all out-going edges from $\\mathcal{A}$. It can be shown that $f_{A}$ does not increase under a transformation, and it strictly decreases if the paths from $A$ to each of $A_{1}$, $A_{2}, B_{1}, B_{2}$ all pass through $A_{1}$ - i.e., of these four vertices, $A_{1}$ is combinatorially nearest to $A$. In particular, this is the case if $A_{1}=A$, i.e., the edge-switch in the transformation occurs at $A$. It is not hard to prove that $\\sum_{A \\in \\mathcal{A}} f_{A}(\\mathcal{T})=f(\\mathcal{T})$.\n\nThe conclusion of the problem can also be established by resorting to a single carefully chosen $f_{A}$. Suppose, if possible, that the process is infinite, so some tree $\\mathcal{T}$ occurs (at least) twice. Let $A$ be the vertex in $\\mathcal{A}$ at which the edge-switch occurs in the transformation of the first occurrence of $\\mathcal{T}$. By the preceding paragraph, consideration of $f_{A}$ shows that $\\mathcal{T}$ can never occur again.\n(3) Recall that the degree of any vertex in $\\mathcal{A}$ is invariant under a transformation, so the linear combination $\\sum_{A \\in \\mathcal{A}}(\\operatorname{deg} A-1) f_{A}$ is a strict semi-invariant for $\\mathcal{A B}$-trees $\\mathcal{T}$ whose vertices in $\\mathcal{A}$ all have degrees exceeding 1. Up to a factor, this semi-invariant can alternatively, but equivalently be described as follows. Fix a vertex $*$ and assign each vertex $X$ a number $g(X)$ so that $g(*)=0$, and $g(A)-g(B)=A B$ for every $A$ in $\\mathcal{A}$ and every $B$ in $\\mathcal{B}$ joined by an edge. Next, let $\\beta(\\mathcal{T})=\\frac{1}{|\\mathcal{B}|} \\sum_{B \\in \\mathcal{B}} g(B)$, let $\\alpha(\\mathcal{T})=\\frac{1}{|\\mathcal{E}|-|\\mathcal{A}|} \\sum_{A \\in \\mathcal{A}}(\\operatorname{deg} A-1) g(A)$, where $\\mathcal{E}$ is the edge-set of $\\mathcal{T}$, and set $\\mu(\\mathcal{T})=\\beta(\\mathcal{T})-\\alpha(\\mathcal{T})$. It can be shown that $\\mu$ strictly decreases under a transformation; in fact, $\\mu$ and $\\sum_{A \\in \\mathcal{A}}(\\operatorname{deg} A-1) f_{A}$ are proportional to one another.", "metadata": {"resource_path": "RMM/segmented/en-2016-Solutions_RMM2016-2.jsonl", "problem_match": "\nProblem 6.", "solution_match": "\nSolution."}} |
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