{"year": "2021", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Rectangles $B C C_{1} B_{2}, C A A_{1} C_{2}$, and $A B B_{1} A_{2}$ are erected outside an acute triangle $A B C$. Suppose that  $$ \\angle B C_{1} C+\\angle C A_{1} A+\\angle A B_{1} B=180^{\\circ} . $$  Prove that lines $B_{1} C_{2}, C_{1} A_{2}$, and $A_{1} B_{2}$ are concurrent.", "solution": "  The angle condition implies the circumcircles of the three rectangles concur at a single point $P$. ![](https://cdn.mathpix.com/cropped/2024_11_19_128c4918e9847b51d7d1g-04.jpg?height=803&width=786&top_left_y=1072&top_left_x=635)  Then $\\measuredangle C P B_{2}=\\measuredangle C P A_{1}=90^{\\circ}$, hence $P$ lies on $A_{1} B_{2}$ etc., so we're done. Remark. As one might guess from the two-sentence solution, the entire difficulty of the problem is getting the characterization of the concurrence point.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2021-notes.jsonl"}}
{"year": "2021", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "The Planar National Park is a undirected 3-regular planar graph (i.e. all vertices have degree 3). A visitor walks through the park as follows: she begins at a vertex and starts walking along an edge. When she reaches the other endpoint, she turns left. On the next vertex she turns right, and so on, alternating left and right turns at each vertex. She does this until she gets back to the vertex where she started. What is the largest possible number of times she could have entered any vertex during her walk, over all possible layouts of the park?", "solution": "  The answer is 3. We consider the trajectory of the visitor as an ordered sequence of turns. A turn is defined by specifying a vertex, the incoming edge, and the outgoing edge. Hence there are six possible turns for each vertex.  Claim - Given one turn in the sequence, one can reconstruct the entire sequence of turns.   This implies already that the trajectory of the visitor, when extended to an infinite sequence, is totally periodic (not just eventually periodic), because there are finitely many possible turns, so some turn must be repeated. So, any turn appears at most once in the period of the sequence, giving a naïve bound of 6 for the original problem.  However, the following claim improves the bound to 3 . Claim - It is impossible for both of the turns $a \\rightarrow b \\rightarrow c$ and $c \\rightarrow b \\rightarrow a$ to occur in the same trajectory.  ![](https://cdn.mathpix.com/cropped/2024_11_19_128c4918e9847b51d7d1g-06.jpg?height=806&width=432&top_left_y=245&top_left_x=818)  However, we assumed for contradiction the red and blue paths were part of the same trajectory, yet they clearly never meet.  It remains to give a construction showing 3 can be achieved. There are many, many valid constructions. One construction due to Danielle Wang is given here, who provided the following motivation: \"I was lying in bed and drew the first thing I could think of\". The path is CAHIFGDBAHEFGJBAC which visits $A$ three times. ![](https://cdn.mathpix.com/cropped/2024_11_19_128c4918e9847b51d7d1g-06.jpg?height=952&width=1009&top_left_y=1409&top_left_x=523)  Remark. As the above example shows it is possible to transverse an edge more than once even in the same direction, as in edge $A H$ above.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2021-notes.jsonl"}}
{"year": "2021", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "Let $n \\geq 2$ be an integer. An $n \\times n$ board is initially empty. Each minute, you may perform one of three moves:  - If there is an L-shaped tromino region of three cells without stones on the board (see figure; rotations not allowed), you may place a stone in each of those cells. ![](https://cdn.mathpix.com/cropped/2024_11_19_128c4918e9847b51d7d1g-02.jpg?height=161&width=164&top_left_y=1250&top_left_x=1043) - If all cells in a column have a stone, you may remove all stones from that column. - If all cells in a row have a stone, you may remove all stones from that row.  For which $n$ is it possible that, after some non-zero number of moves, the board has no stones?", "solution": "  The answer is $3 \\mid n$. Construction: For $n=3$, the construction is fairly straightforward, shown below. ![](https://cdn.mathpix.com/cropped/2024_11_19_128c4918e9847b51d7d1g-07.jpg?height=222&width=1012&top_left_y=1493&top_left_x=523)  This can be extended to any $3 \\mid n$.  $$ \\begin{aligned} \\operatorname{deg}_{x} P & \\leq n-2 \\\\ \\operatorname{deg}_{y} P & \\leq n-2 \\\\ (1+x+y) P(x, y) & \\in\\left\\langle 1+x+\\cdots+x^{n-1}, 1+y+\\cdots+y^{n-1}\\right\\rangle \\end{aligned} $$  (Here $\\langle\\ldots\\rangle$ is an ideal of $\\mathbb{R}[x, y]$.) In particular, if $S$ is the set of $n$th roots of unity other than 1 , we should have  $$ \\left(1+z_{1}+z_{2}\\right) P\\left(z_{1}, z_{2}\\right)=0 $$  for any $z_{1}, z_{2} \\in S$. Since $3 \\nmid n$, it follows that $1+z_{1}+z_{2} \\neq 0$ always. So $P$ vanishes on $S \\times S$, a contradiction to the bounds on $\\operatorname{deg} P$ (by, say, combinatorial nullstellensatz on any nonzero term).  cell, the number of times it is operated on entirely as a function of $r_{j}$. For example, a hypothetical illustration with $n=6$ is partially drawn below, with the number in each cell denoting how many times it was the corner of an $L$.  $$ \\left[\\begin{array}{cccccc} 0 & 0 & 0 & 0 & 0 & 0 \\\\ c_{1} & c_{2} & c_{3}=r_{3} & c_{4}=r_{5}-r_{4} & c_{5}=r_{5} & 0 \\\\ \\vdots & \\vdots & r_{2}+r_{3}-r_{5} & r_{5}-r_{3} & r_{4} & 0 \\\\ \\vdots & \\vdots & r_{1}+r_{2}+r_{3}-r_{4}-r_{5} & r_{5}-r_{2} & r_{3} & 0 \\\\ \\vdots & \\vdots & r_{1}+r_{2}+r_{4}-r_{5} & r_{5}-r_{1} & r_{2} & 0 \\\\ \\vdots & \\vdots & r_{1}+r_{4}-r_{5} & r_{5} & r_{1} & 0 \\end{array}\\right] $$  Let $a_{i, j}$ be the expression in $(i, j)$. It will also be helpful to define $c_{i}$ in the obvious way as well.  Claim - We have $c_{n}=r_{n}=0, a_{n-1, j}=r_{j}$ and $a_{i, n-1}=c_{i}$.   Claim - For $1 \\leq i, j \\leq n-1$, the following recursion holds:  $$ a_{i, j}+a_{i+1, j}+a_{i+1, j-1}=r_{j}+c_{i+1} $$   We can coerce the table above into matrix form now as follows. Define  $$ K=\\left[\\begin{array}{cccccccc} -1 & -1 & 0 & 0 & \\ldots & 0 & 0 & 0 \\\\ 0 & -1 & -1 & 0 & \\ldots & 0 & 0 & 0 \\\\ 0 & 0 & -1 & -1 & \\ldots & 0 & 0 & 0 \\\\ \\vdots & \\vdots & \\vdots & \\vdots & \\ddots & \\vdots & \\vdots & \\vdots \\\\ 0 & 0 & 0 & 0 & \\ldots & -1 & -1 & 0 \\\\ 0 & 0 & 0 & 0 & \\ldots & 0 & -1 & -1 \\\\ 1 & 1 & 1 & 1 & \\ldots & 1 & 1 & 0 \\end{array}\\right] $$  Then define a sequence of matrices $M_{i}$ recursively by $M_{n-1}=\\mathrm{id}$, and  $$ M_{i}=\\mathrm{id}+K M_{i+1}=\\mathrm{id}+K+\\cdots+K^{n-1-i} $$  The matrices are chosen so that, by construction,  $$ \\left\\langle r_{1}, \\ldots, r_{n-1}\\right\\rangle M_{i}=\\left\\langle a_{i, 1}, \\ldots, a_{i, n-1}\\right\\rangle $$  for $i=1,2, \\ldots, n-1$. On the other hand, we can extend the recursion one level deeper and obtain  $$ \\left\\langle r_{1}, \\ldots, r_{n-1}\\right\\rangle M_{0}=\\langle 0, \\ldots, 0\\rangle . $$   Claim - The eigenvalues of $K$ are exactly $-\\left(1+e^{\\frac{2 \\pi i k}{n}}\\right)$ for $k=1,2, \\ldots, n-1$.  However, we are told that apparently  $$ 0=\\operatorname{det} M_{0}=\\operatorname{det}\\left(\\mathrm{id}+K+K^{2}+\\cdots+K^{n-1}\\right) $$  which means $\\operatorname{det}\\left(K^{n}-\\mathrm{id}\\right)=0$. This can only happen if $K^{n}$ has eigenvalue 1 , meaning that  $$ [-(1+\\omega)]^{n}=1 $$  for $\\omega$ some $n^{\\text {th }}$ root of unity, not necessarily primitive. This can only happen if $|1+\\omega|=1$, ergo $3 \\mid n$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2021-notes.jsonl"}}
{"year": "2021", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "A finite set $S$ of positive integers has the property that, for each $s \\in S$, and each positive integer divisor $d$ of $s$, there exists a unique element $t \\in S$ satisfying $\\operatorname{gcd}(s, t)=d$. (The elements $s$ and $t$ could be equal.)  Given this information, find all possible values for the number of elements of $S$.", "solution": "  The answer is that $|S|$ must be a power of 2 (including 1 ), or $|S|=0$ (a trivial case we do not discuss further).  【 Construction. For any nonnegative integer $k$, a construction for $|S|=2^{k}$ is given by  $$ S=\\left\\{\\left(p_{1} \\text { or } q_{1}\\right) \\times\\left(p_{2} \\text { or } q_{2}\\right) \\times \\cdots \\times\\left(p_{k} \\text { or } q_{k}\\right)\\right\\} $$  for $2 k$ distinct primes $p_{1}, \\ldots, p_{k}, q_{1}, \\ldots, q_{k}$. ■ Converse. The main claim is as follows. Claim - In any valid set $S$, for any prime $p$ and $x \\in S, \\nu_{p}(x) \\leq 1$.   - On the one hand, by taking $x$ in the statement, we see $\\frac{e}{e+1}$ of the elements of $S$ are divisible by $p$. - On the other hand, consider a $y \\in S$ such that $\\nu_{p}(y)=1$ which must exist (say if $\\operatorname{gcd}(x, y)=p$ ). Taking $y$ in the statement, we see $\\frac{1}{2}$ of the elements of $S$ are divisible by $p$.  So $e=1$, contradiction. Now since $|S|$ equals the number of divisors of any element of $S$, we are done.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2021-notes.jsonl"}}
{"year": "2021", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "Let $n \\geq 4$ be an integer. Find all positive real solutions to the following system of $2 n$ equations:  $$ \\begin{aligned} a_{1} & =\\frac{1}{a_{2 n}}+\\frac{1}{a_{2}}, & a_{2} & =a_{1}+a_{3}, \\\\ a_{3} & =\\frac{1}{a_{2}}+\\frac{1}{a_{4}}, & a_{4} & =a_{3}+a_{5}, \\\\ a_{5} & =\\frac{1}{a_{4}}+\\frac{1}{a_{6}}, & a_{6} & =a_{5}+a_{7}, \\\\ & \\vdots & & \\vdots \\\\ a_{2 n-1} & =\\frac{1}{a_{2 n-2}}+\\frac{1}{a_{2 n}}, & a_{2 n} & =a_{2 n-1}+a_{1} . \\end{aligned} $$", "solution": "  We will prove $a_{2 k}$ is a constant sequence, at which point the result is obvious. 【 First approach (Andrew Gu). Apparently, with indices modulo $2 n$, we should have  $$ a_{2 k}=\\frac{1}{a_{2 k-2}}+\\frac{2}{a_{2 k}}+\\frac{1}{a_{2 k+2}} $$  for every index $k$ (this eliminates all $a_{\\text {odd }}$ 's). Define  $$ m=\\min _{k} a_{2 k} \\quad \\text { and } \\quad M=\\max _{k} a_{2 k} . $$  Look at the indices $i$ and $j$ achieving $m$ and $M$ to respectively get  $$ \\begin{aligned} & m=\\frac{2}{m}+\\frac{1}{a_{2 i-2}}+\\frac{1}{a_{2 i+2}} \\geq \\frac{2}{m}+\\frac{1}{M}+\\frac{1}{M}=\\frac{2}{m}+\\frac{2}{M} \\\\ & M=\\frac{2}{M}+\\frac{1}{a_{2 j-2}}+\\frac{1}{a_{2 j+2}} \\leq \\frac{2}{M}+\\frac{1}{m}+\\frac{1}{m}=\\frac{2}{m}+\\frac{2}{M} \\end{aligned} $$  Together this gives $m \\geq M$, so $m=M$. That means $a_{2 i}$ is constant as $i$ varies, solving the problem.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2021-notes.jsonl"}}
{"year": "2021", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "Let $n \\geq 4$ be an integer. Find all positive real solutions to the following system of $2 n$ equations:  $$ \\begin{aligned} a_{1} & =\\frac{1}{a_{2 n}}+\\frac{1}{a_{2}}, & a_{2} & =a_{1}+a_{3}, \\\\ a_{3} & =\\frac{1}{a_{2}}+\\frac{1}{a_{4}}, & a_{4} & =a_{3}+a_{5}, \\\\ a_{5} & =\\frac{1}{a_{4}}+\\frac{1}{a_{6}}, & a_{6} & =a_{5}+a_{7}, \\\\ & \\vdots & & \\vdots \\\\ a_{2 n-1} & =\\frac{1}{a_{2 n-2}}+\\frac{1}{a_{2 n}}, & a_{2 n} & =a_{2 n-1}+a_{1} . \\end{aligned} $$", "solution": "  We will prove $a_{2 k}$ is a constant sequence, at which point the result is obvious. ब Second approach (author's solution). As before, we have  $$ a_{2 k}=\\frac{1}{a_{2 k-2}}+\\frac{2}{a_{2 k}}+\\frac{1}{a_{2 k+2}} $$   - Define  $$ S=\\sum_{k} a_{2 k}, \\quad \\text { and } \\quad T=\\sum_{k} \\frac{1}{a_{2 k}} . $$  Summing gives $S=4 T$. On the other hand, Cauchy-Schwarz says $S \\cdot T \\geq n^{2}$, so $T \\geq \\frac{1}{2} n$.  - On the other hand,  $$ 1=\\frac{1}{a_{2 k-2} a_{2 k}}+\\frac{2}{a_{2 k}^{2}}+\\frac{1}{a_{2 k} a_{2 k+2}} $$  Sum this modified statement to obtain  $$ n=\\sum_{k}\\left(\\frac{1}{a_{2 k}}+\\frac{1}{a_{2 k+2}}\\right)^{2} \\stackrel{\\text { QM-AM }}{\\geq} \\frac{1}{n}\\left(\\sum_{k} \\frac{1}{a_{2 k}}+\\frac{1}{a_{2 k+2}}\\right)^{2}=\\frac{1}{n}(2 T)^{2} $$  So $T \\leq \\frac{1}{2} n$.  - Since $T \\leq \\frac{1}{2} n$ and $T \\geq \\frac{1}{2} n$, we must have equality everywhere above. This means $a_{2 k}$ is a constant sequence.  Remark. The problem is likely intractable over $\\mathbb{C}$, in the sense that one gets a high-degree polynomial which almost certainly has many complex roots. So it seems likely that most solutions must involve some sort of inequality, using the fact we are over $\\mathbb{R}_{>0}$ instead.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2021-notes.jsonl"}}
{"year": "2021", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "USAMO", "problem": "Let $A B C D E F$ be a convex hexagon satisfying $\\overline{A B}\\|\\overline{D E}, \\overline{B C}\\| \\overline{E F}, \\overline{C D} \\| \\overline{F A}$, and  $$ A B \\cdot D E=B C \\cdot E F=C D \\cdot F A $$  Let $X, Y$, and $Z$ be the midpoints of $\\overline{A D}, \\overline{B E}$, and $\\overline{C F}$. Prove that the circumcenter of $\\triangle A C E$, the circumcenter of $\\triangle B D F$, and the orthocenter of $\\triangle X Y Z$ are collinear.", "solution": "   ![](https://cdn.mathpix.com/cropped/2024_11_19_128c4918e9847b51d7d1g-13.jpg?height=777&width=809&top_left_y=1279&top_left_x=629)  Claim - If $A B \\cdot D E=B C \\cdot E F=C D \\cdot F A=k$, then the circumcenters of $A C E$ and $A^{\\prime} C^{\\prime} E^{\\prime}$ coincide.  ![](https://cdn.mathpix.com/cropped/2024_11_19_128c4918e9847b51d7d1g-14.jpg?height=992&width=495&top_left_y=246&top_left_x=780)  Claim - Triangle $X Y Z$ is the vector average of the (congruent) medial triangles of triangles $A^{\\prime} C^{\\prime} E^{\\prime}$ and $B^{\\prime} D^{\\prime} F^{\\prime}$.   $$ \\begin{aligned} \\frac{\\vec{M}+\\vec{N}}{2} & =\\frac{\\frac{\\vec{C}^{\\prime}+\\vec{E}^{\\prime}}{2}+\\frac{\\vec{B}^{\\prime}+\\vec{F}^{\\prime}}{2}}{2} \\\\ & =\\frac{\\overrightarrow{C^{\\prime}}+\\vec{E}^{\\prime}+\\vec{B}^{\\prime}+\\vec{F}^{\\prime}}{4} \\\\ & =\\frac{(\\vec{A}+\\vec{E}-\\vec{F})+(\\vec{C}+\\vec{A}-\\vec{B})+(\\vec{D}+\\vec{F}-\\vec{E})+(\\vec{B}+\\vec{D}-\\vec{C})}{4} \\\\ & =\\frac{\\vec{A}+\\vec{D}}{2}=\\vec{X} . \\end{aligned} $$  Hence the orthocenter of $X Y Z$ is the midpoint of the orthocenters of the medial triangles of $A^{\\prime} C^{\\prime} E^{\\prime}$ and $B^{\\prime} D^{\\prime} F^{\\prime}$ - that is, their circumcenters.  ![](https://cdn.mathpix.com/cropped/2024_11_19_128c4918e9847b51d7d1g-15.jpg?height=966&width=1100&top_left_y=248&top_left_x=478)  Claim - In trapezoid $A B D E$, the perpendicular bisector of $\\overline{X Y}$ is the same as the perpendicular bisector of the midline $\\overline{W N}$.   Claim - The points $V, W, M, N$ are cyclic.   $$ W Y \\cdot Y N=\\frac{1}{2} D E \\cdot \\frac{1}{2} A B=\\frac{1}{2} E F \\cdot \\frac{1}{2} B C=V Y \\cdot Y M $$  Applying all the cyclic variations of the above two claims, it follows that all six points $U, V, W, M, N, P$ are concyclic, and the center of that circle coincides with the circumcenter of $\\triangle X Y Z$.  Remark. It is also possible to implement ideas from the first solution here, by showing all six midpoints have equal power to $(X Y Z)$.  Claim - The orthocenter of $X Y Z$ is the midpoint of the circumcenters of $\\triangle A C E$ and $\\triangle B D F$.   $$ \\begin{aligned} \\operatorname{orthocenter}(X Y Z) & =x+y+z=\\frac{a+b+c+d+e+f}{2} \\\\ \\operatorname{circumcenter}(A C E) & =\\operatorname{orthocenter}(M N P) \\end{aligned} $$  $$ =m+n+p=\\frac{c+e}{2}+\\frac{e+a}{2}+\\frac{a+c}{2}=a+c+e $$  circumcenter $(B D F)=$ orthocenter $(U V W)$  $$ =u+v+w=\\frac{d+f}{2}+\\frac{f+b}{2}+\\frac{b+d}{2}=b+d+f $$", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2021-notes.jsonl"}}