| {"year": "2011", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Let $a, b, c$ be positive real numbers such that $a^{2}+b^{2}+c^{2}+(a+b+c)^{2} \\leq 4$. Prove that $$ \\frac{a b+1}{(a+b)^{2}}+\\frac{b c+1}{(b+c)^{2}}+\\frac{c a+1}{(c+a)^{2}} \\geq 3 $$", "solution": " The condition becomes $2 \\geq a^{2}+b^{2}+c^{2}+a b+b c+c a$. Therefore, $$ \\begin{aligned} \\sum_{\\mathrm{cyc}} \\frac{2 a b+2}{(a+b)^{2}} & \\geq \\sum_{\\mathrm{cyc}} \\frac{2 a b+\\left(a^{2}+b^{2}+c^{2}+a b+b c+c a\\right)}{(a+b)^{2}} \\\\ & =\\sum_{\\mathrm{cyc}} \\frac{(a+b)^{2}+(c+a)(c+b)}{(a+b)^{2}} \\\\ & =3+\\sum_{\\mathrm{cyc}} \\frac{(c+a)(c+b)}{(a+b)^{2}} \\\\ & \\geq 3+3 \\sqrt[3]{\\prod_{\\text {cyc }} \\frac{(c+a)(c+b)}{(a+b)^{2}}}=3+3=6 \\end{aligned} $$ with the last line by AM-GM. This completes the proof.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2011-notes.jsonl"}} |
| {"year": "2011", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "An integer is assigned to each vertex of a regular pentagon so that the sum of the five integers is 2011. A turn of a solitaire game consists of subtracting an integer $m$ (not necessarily positive) from each of the integers at two neighboring vertices and adding $2 m$ to the opposite vertex, which is not adjacent to either of the first two vertices. (The amount $m$ and the vertices chosen can vary from turn to turn.) The game is won at a certain vertex if, after some number of turns, that vertex has the number 2011 and the other four vertices have the number 0. Prove that for any choice of the initial integers, there is exactly one vertex at which the game can be won.", "solution": " Call the vertices $0,1,2,3,4$ in order. First, notice that the quantity $$ S:=N_{1}+2 N_{2}+3 N_{3}+4 N_{4} \\quad(\\bmod 5) $$ is invariant, where $N_{i}$ is the amount at vertex $i$. This immediately implies that at most one vertex can win, since in a winning situation all $N_{i}$ are 0 except for one, which is 2011. (For example, if $S \\equiv 3(\\bmod 5)$, any victory must occur at the vertex 3 , via $N_{3}=2011$, $N_{0}=N_{1}=N_{2}=N_{4}=0$.) Now we prove we can win on this unique vertex. Let $a_{i}, x_{i}$ denote the number initially at $i$ and $x_{i}$ denote $\\sum m$ over all $m$ where vertex $i$ gains $2 m$. WLOG the possible vertex is 0 , meaning $a_{1}+2 a_{2}+3 a_{3}+4 a_{4} \\equiv 0(\\bmod 5)$. Moreover we want $$ \\begin{aligned} 2011 & =a_{0}+2 x_{0}-x_{2}-x_{3} \\\\ 0 & =a_{1}+2 x_{1}-x_{3}-x_{4} \\\\ 0 & =a_{2}+2 x_{2}-x_{4}-x_{0} \\\\ 0 & =a_{3}+2 x_{3}-x_{0}-x_{1} \\\\ 0 & =a_{4}+2 x_{4}-x_{1}-x_{2} \\end{aligned} $$ We can ignore the first equation since its the sum of the other four, and we can WLOG shift $x_{0} \\rightarrow 0$ by shifting each $x_{i}$ by a fixed amount. We will now solve the resulting system of equations. First, we have $$ x_{4}=2 x_{2}+a_{2} \\text { and } x_{1}=2 x_{3}+a_{3} $$ Using these to remove all instances of $x_{1}$ and $x_{4}$ gives $$ 2 x_{2}-3 x_{3}=2 a_{3}+a_{1}-a_{2} \\text { and } 2 x_{3}-3 x_{2}=2 a_{2}+a_{4}-a_{3} $$ whence we have a two-variable system of equations! To verify its solution is integral, note that $$ x_{2}-x_{3}=\\frac{a_{1}-3 a_{2}+3 a_{3}-a_{4}}{5} $$ is an integer, since $$ a_{1}-3 a_{2}+3 a_{3}-a_{4} \\equiv a_{1}+2 a_{2}+3 a_{3}+4 a_{4} \\equiv 0 \\quad(\\bmod 5) $$ Abbreviating $\\frac{a_{1}-3 a_{2}+3 a_{3}-a_{4}}{5}$ as $k$, we obtain the desired $x_{i}$ : $$ \\begin{aligned} x_{2} & =2 a_{3}+a_{1}-a_{2}+2 k \\\\ x_{3} & =x_{2}+k \\\\ x_{1} & =2 x_{3}+a_{3} \\\\ x_{4} & =2 x_{2}+a_{2} \\\\ x_{0} & =0 . \\end{aligned} $$ Remark. In principle, you could unwind all the definitions above to explicitly write every $x_{i}$ as a function of $a_{1}, a_{2}, a_{3}, a_{4}$. If you did this, you could get the long equations $$ \\begin{aligned} & x_{0}=0 \\\\ & x_{1}=-\\frac{1}{5}\\left(6 a_{1}+2 a_{2}+3 a_{3}+4 a_{4}\\right) \\\\ & x_{2}=-\\frac{1}{5}\\left(2 a_{1}+4 a_{2}+a_{3}+3 a_{4}\\right) \\\\ & x_{3}=-\\frac{1}{5}\\left(3 a_{1}+a_{2}+4 a_{3}+2 a_{4}\\right) \\\\ & x_{4}=-\\frac{1}{5}\\left(4 a_{1}+3 a_{2}+2 a_{3}+6 a_{4}\\right) \\end{aligned} $$ which indeed are all integers whenever $a_{1}+2 a_{2}+3 a_{3}+4 a_{4} \\equiv 0(\\bmod 5)$. However, this is quite tedious and also unnecessary to solve the problem. That's because we only care that the $x_{i}$ are integers, and do not need to actually know the values. This lets us work more indirectly to avoid long calculation, as we did above.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2011-notes.jsonl"}} |
| {"year": "2011", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "In hexagon $A B C D E F$, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy $\\angle A=3 \\angle D, \\angle C=3 \\angle F$, and $\\angle E=3 \\angle B$. Furthermore $A B=D E, B C=E F$, and $C D=F A$. Prove that diagonals $\\overline{A D}, \\overline{B E}$, and $\\overline{C F}$ are concurrent.", "solution": " Main idea: Claim - In a satisfying hexagon, $B, D, F$ are reflections of $A, C, E$ across the sides of $\\triangle A C E$. (This claim looks plausible because every excellent hexagon is satisfying, and both configuration spaces are three-dimensional.) Call a hexagon of this shape \"excellent\"; in a excellent hexagon the diagonals clearly concur (at the orthocenter). Set $\\beta=\\angle B, \\delta=\\angle D, \\varphi=\\angle F$. Now given a satisfying hexagon $A B C D E F$, construct a \"phantom hexagon\" $A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime} E^{\\prime} F^{\\prime}$ with the same angles which is excellent (see figure). This is possible since $\\beta+\\delta+\\varphi=180^{\\circ}$.  Then it would suffice to prove that: ## Lemma A satisfying hexagon is uniquely determined by its angles up to similarity. That is, at most one hexagon (up to similarity) has angles $\\beta, \\delta, \\gamma$ as above. To do this, consider the unit complex numbers in the directions $\\overrightarrow{B A}$ and $\\overrightarrow{D E}$ respectively and let $\\vec{x}$ denote their sum. Define $\\vec{y}, \\vec{z}$ similarly. Note that the condition $\\overline{A B} \\nVdash \\overline{D E}$ implies $\\vec{x} \\neq 0$, and similarly. Then we have the identities $$ A B \\cdot \\vec{x}+C D \\cdot \\vec{y}+E F \\cdot \\vec{z}=A^{\\prime} B^{\\prime} \\cdot \\vec{x}+C^{\\prime} D^{\\prime} \\cdot \\vec{y}+E^{\\prime} F^{\\prime} \\cdot \\vec{z}=0 . $$ So we would obtain $A B: C D: E F=A^{\\prime} B^{\\prime}: C^{\\prime} D^{\\prime}: E^{\\prime} F^{\\prime}$ if only we could show that $\\vec{x}, \\vec{y}, \\vec{z}$ are not multiples of each other (linear dependency reasons). This is a tiresome computation with arguments, but here it is. First note that none of $\\beta, \\delta, \\varphi$ can be $90^{\\circ}$, since otherwise we get a pair of parallel sides. Now work in the complex plane, fix a reference such that $\\vec{A}-\\vec{B}$ has argument 0 , and assume $A B C D E F$ are labelled counterclockwise. Then - $\\vec{B}-\\vec{C}$ has argument $\\pi-\\beta$ - $\\vec{C}-\\vec{D}$ has argument $-(\\beta+3 \\varphi)$ - $\\vec{D}-\\vec{E}$ has argument $\\pi-(\\beta+3 \\varphi+\\delta)$ - $\\vec{E}-\\vec{F}$ has argument $-(4 \\beta+3 \\varphi+\\delta)$ So the argument of $\\vec{x}$ has argument $\\frac{\\pi-(\\beta+3 \\varphi+\\delta)}{2}(\\bmod \\pi)$. The argument of $\\vec{y}$ has argument $\\frac{\\pi-(5 \\beta+3 \\varphi+\\delta)}{2}(\\bmod \\pi)$. Their difference is $2 \\beta(\\bmod \\pi)$, and since $\\beta \\neq 90^{\\circ}$, it follows that $\\vec{x}$ and $\\vec{y}$ are not multiples of each other; the other cases are similar. Then the lemma implies $A B C D E F \\sim A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime} E^{\\prime} F$ and we're done. Remark. This problem turned out to be known already. It appears in this reference: Nikolai Beluhov, Matematika, 2008, issue 6, problem 3. It was reprinted as Kvant, 2009, issue 2, problem M2130; the reprint is available at http: //kvant.ras.ru/pdf/2009/2009-02.pdf. Remark. The vector perspective also shows the condition about parallel sides cannot be dropped. Here is a counterexample from Ryan Kim in the event that it is.  By adjusting the figure above so that the triangles are right isosceles (instead of just right), one also finds an example of a hexagon which is satisfying and whose diagonals are concurrent, but which is not excellent.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2011-notes.jsonl"}} |
| {"year": "2011", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "Consider the assertion that for each positive integer $n \\geq 2$, the remainder upon dividing $2^{2^{n}}$ by $2^{n}-1$ is a power of 4 . Either prove the assertion or find (with proof) a counterexample.", "solution": " We claim $n=25$ is a counterexample. Since $2^{25} \\equiv 2^{0}\\left(\\bmod 2^{25}-1\\right)$, we have $$ 2^{2^{25}} \\equiv 2^{2^{25} \\bmod 25} \\equiv 2^{7} \\bmod 2^{25}-1 $$ and the right-hand side is actually the remainder, since $0<2^{7}<2^{25}$. But $2^{7}$ is not a power of 4 . Remark. Really, the problem is just equivalent for asking $2^{n}$ to have odd remainder when divided by $n$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2011-notes.jsonl"}} |
| {"year": "2011", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "Let $P$ be a point inside convex quadrilateral $A B C D$. Points $Q_{1}$ and $Q_{2}$ are located within $A B C D$ such that $$ \\begin{array}{ll} \\angle Q_{1} B C=\\angle A B P, & \\angle Q_{1} C B=\\angle D C P, \\\\ \\angle Q_{2} A D=\\angle B A P, & \\angle Q_{2} D A=\\angle C D P . \\end{array} $$ Prove that $\\overline{Q_{1} Q_{2}} \\| \\overline{A B}$ if and only if $\\overline{Q_{1} Q_{2}} \\| \\overline{C D}$.", "solution": null, "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2011-notes.jsonl"}} |
| {"year": "2011", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "USAMO", "problem": "Let $A$ be a set with $|A|=225$, meaning that $A$ has 225 elements. Suppose further that there are eleven subsets $A_{1}, \\ldots, A_{11}$ of $A$ such that $\\left|A_{i}\\right|=45$ for $1 \\leq i \\leq 11$ and $\\left|A_{i} \\cap A_{j}\\right|=9$ for $1 \\leq i<j \\leq 11$. Prove that $\\left|A_{1} \\cup A_{2} \\cup \\cdots \\cup A_{11}\\right| \\geq 165$, and give an example for which equality holds.", "solution": " Ignore the 225 - it is irrelevant. Denote the elements of $A_{1} \\cup \\cdots \\cup A_{11}$ by $a_{1}, \\ldots, a_{n}$, and suppose that $a_{i}$ appears $x_{i}$ times among $A_{i}$ for each $1 \\leq i \\leq n$ (so $1 \\leq x_{i} \\leq 11$ ). Then we have $$ \\sum_{i=1}^{11} x_{i}=\\sum_{i=1}^{11}\\left|A_{i}\\right|=45 \\cdot 11 $$ and $$ \\sum_{i=1}^{11}\\binom{x_{i}}{2}=\\sum_{1 \\leq i<j \\leq 11}\\left|A_{i} \\cap A_{j}\\right|=\\binom{11}{2} \\cdot 9 $$ Therefore, we deduce that $\\sum x_{i}=495$ and $\\sum_{i} x_{i}^{2}=1485$. Now, by Cauchy Schwarz $$ n\\left(\\sum_{i} x_{i}^{2}\\right) \\geq\\left(\\sum x_{i}\\right)^{2} $$ which implies $n \\geq \\frac{495^{2}}{1485}=165$. Equality occurs if we let $A$ consist of the 165 three-element subsets of $\\{1, \\ldots, 11\\}$ (plus 60 of your favorite reptiles if you really insist $|A|=225$ ). Then we let $A_{i}$ denote those subsets containing $i$, of which there are $\\binom{10}{2}=45$, and so that $\\left|A_{i} \\cap A_{j}\\right|=\\binom{9}{1}=9$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2011-notes.jsonl"}} |
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