{"year": "2024", "tier": "T3", "problem_label": "1", "problem_type": null, "exam": "USAJMO", "problem": "Let $A B C D$ be a cyclic quadrilateral with $A B=7$ and $C D=8$. Points $P$ and $Q$ are selected on line segment $A B$ so that $A P=B Q=3$. Points $R$ and $S$ are selected on line segment $C D$ so that $C R=D S=2$. Prove that $P Q R S$ is a cyclic quadrilateral.", "solution": " \\l The one-liner. The four points $P, Q, R, S$ have equal power -12 with respect to $(A B C D)$. So in fact they're on a circle concentric with $(A B C D)$.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2024-notes.jsonl"}}
{"year": "2024", "tier": "T3", "problem_label": "1", "problem_type": null, "exam": "USAJMO", "problem": "Let $A B C D$ be a cyclic quadrilateral with $A B=7$ and $C D=8$. Points $P$ and $Q$ are selected on line segment $A B$ so that $A P=B Q=3$. Points $R$ and $S$ are selected on line segment $C D$ so that $C R=D S=2$. Prove that $P Q R S$ is a cyclic quadrilateral.", "solution": " \\ The external power solution. We distinguish between two cases.  Case where $A B$ and $C D$ are not parallel. We let lines $A B$ and $C D$ meet at $T$. Without loss of generality, $A$ lies between $B$ and $T$ and $D$ lies between $C$ and $T$. Let $x=T A$ and $y=T D$, as shown below. ![](https://cdn.mathpix.com/cropped/2024_11_19_805564b9ec1a761a333dg-03.jpg?height=541&width=812&top_left_y=1346&top_left_x=633)  By power of a point,  $$ \\begin{aligned} A B C D \\text { cyclic } \\Longleftrightarrow x(x+7) & =y(y+8) \\\\ P Q R S \\text { cyclic } \\Longleftrightarrow(x+3)(x+4) & =(y+2)(y+6) . \\end{aligned} $$  However, the latter equation is just the former with 12 added to both sides. (That is, $(x+3)(x+4)=x(x+7)+12$ while $(y+2)(y+6)=y(y+8)+12$.$) So the conclusion is$ immediate.  Case where $A B$ and $C D$ are parallel. In that case $A B C D$ is an isosceles trapezoid. Then the entire picture is symmetric around the common perpendicular bisector of the lines $A B$ and $C D$. Now $P Q R S$ is also an isosceles trapezoid, so it's cyclic too. ![](https://cdn.mathpix.com/cropped/2024_11_19_805564b9ec1a761a333dg-04.jpg?height=420&width=420&top_left_y=244&top_left_x=818)", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2024-notes.jsonl"}}
{"year": "2024", "tier": "T3", "problem_label": "2", "problem_type": null, "exam": "USAJMO", "problem": "Let $m$ and $n$ be positive integers. Let $S$ be the set of lattice points $(x, y)$ with $1 \\leq x \\leq 2 m$ and $1 \\leq y \\leq 2 n$. A configuration of $m n$ rectangles is called happy if each point of $S$ is the vertex of exactly one rectangle. Prove that the number of happy configurations is odd.", "solution": " I Original proposer's solution. To this end, let's denote by $f(2 m, 2 n)$ the number of happy configurations for a $2 m \\times 2 n$ grid of lattice points (not necessarily equally spaced - this doesn't change the count). We already have the following easy case.  Claim - We have $f(2,2 n)=(2 n-1)!!=(2 n-1) \\cdot(2 n-3) \\cdots \\cdots 3 \\cdot 1$.  We will prove that: Claim - Assume $m, n \\geq 1$. When $f(2 m, 2 n) \\equiv f(2 m-2,2 n)(\\bmod 2)$.  Now configurations fixed by $\\tau$ can be described readily: this occurs if and only if the last two columns are self-contained, meaning every rectangle with a vertex in these columns is completely contained in these two columns. ![](https://cdn.mathpix.com/cropped/2024_11_19_805564b9ec1a761a333dg-06.jpg?height=809&width=847&top_left_y=241&top_left_x=596)  Hence it follows that  $$ f(2 m, 2 n)=2(\\text { number of pairs })+f(2 m-2,2 n) \\cdot f(2,2 n) $$  Taking modulo 2 gives the result. By the same token $f(2 m, 2 n) \\equiv f(2 m, 2 n-2)(\\bmod 2)$. So all $f$-values have the same parity, and from $f(2,2)=1$ we're done.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2024-notes.jsonl"}}
{"year": "2024", "tier": "T3", "problem_label": "2", "problem_type": null, "exam": "USAJMO", "problem": "Let $m$ and $n$ be positive integers. Let $S$ be the set of lattice points $(x, y)$ with $1 \\leq x \\leq 2 m$ and $1 \\leq y \\leq 2 n$. A configuration of $m n$ rectangles is called happy if each point of $S$ is the vertex of exactly one rectangle. Prove that the number of happy configurations is odd.", "solution": " \\l Evan's permutation-based solution. Retain the notation $f(2 m, 2 n)$ from before. Given a happy configuration, consider all the rectangles whose left edge is in the first column. Highlight every column containing the right edge of such a rectangle. For example, in the figure below, there are two highlighted columns. (The rectangles are drawn crooked so one can tell them apart.) ![](https://cdn.mathpix.com/cropped/2024_11_19_805564b9ec1a761a333dg-06.jpg?height=515&width=832&top_left_y=1861&top_left_x=612)  We organize happy configurations based on the set of highlighted columns. Specifically, define the relation $\\sim$ on configurations by saying that $\\mathcal{C} \\sim \\mathcal{C}^{\\prime}$ if they differ by any permutation of the highlighted columns. This is an equivalence relation. And in general, if there are $k$ highlighted columns, its equivalence class under $\\sim$ has $k$ ! elements.  Then  Claim - $f(2 m, 2 n)$ has the same parity as the number of happy configurations with exactly one highlighted column.  There are $2 m-1$ ways to pick a single highlighted column, and then $f(2,2 n)=(2 n-1)!!$ ways to use the left column and highlighted column. So the count in the claim is exactly given by  $$ (2 m-1) \\cdot(2 n-1)!!f(2 m-2,2 n) $$  This implies $f(2 m, 2 n) \\equiv f(2 m-2,2 n)(\\bmod 2)$ and proceeding by induction as before solves the problem.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2024-notes.jsonl"}}
{"year": "2024", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "USAJMO", "problem": "A sequence $a_{1}, a_{2}, \\ldots$ of positive integers is defined recursively by $a_{1}=2$ and  $$ a_{n+1}=a_{n}^{n+1}-1 \\quad \\text { for } n \\geq 1 $$  Prove that for every odd prime $p$ and integer $k$, some term of the sequence is divisible by $p^{k}$.", "solution": "  We start with the following. Claim - Assume $n$ is a positive integer divisible by $p-1$. Then either $a_{n-1} \\equiv 0$ $(\\bmod p)$ or $a_{n} \\equiv 0(\\bmod p)$.   $$ a_{n}=a_{n-1}^{n}-1 \\equiv x^{p-1}-1 \\equiv 0 \\quad(\\bmod p) $$  where $x:=a_{n-1}^{\\frac{n}{p-1}}$ is an integer not divisible by $p$. Claim - If $n \\geq 2$ is even, then  $$ a_{n}^{n+1} \\mid a_{n+2} $$   By considering multiples $n$ of $p-1$ which are larger than $k$, we see that if $a_{n} \\equiv 0$ $(\\bmod p)$ ever happens, we are done by combining the two previous claims. So the difficult case of the problem is the bad situation where $a_{n-1} \\equiv 0(\\bmod p)$ occurs for almost all $n \\equiv 0(\\bmod p-1)$.  To resolve the difficult case and finish the problem, we zoom in on specific $n$ that will let us use lifting the exponent on $a_{n-1}$.  Claim - Suppose $n$ is an (even) integer satisfying  $$ \\begin{aligned} & n \\equiv 0 \\quad(\\bmod p-1) \\\\ & n \\equiv 1 \\quad\\left(\\bmod p^{k-1}\\right) \\end{aligned} $$  If $a_{n-1} \\equiv 0(\\bmod p)$, then in fact $p^{k} \\mid a_{n-1}$.  $$ 0 \\equiv \\frac{1}{a_{n-2}}-1 \\quad(\\bmod p) \\Longrightarrow a_{n-2} \\equiv 1 \\quad(\\bmod p) $$  Hence lifting the exponent applies and we get  $$ \\nu_{p}\\left(a_{n-1}\\right) \\equiv \\nu_{p}\\left(a_{n-2}^{n-1}-1\\right)=\\nu_{p}\\left(a_{n-2}-1\\right)+\\nu_{p}(n-1) \\geq 1+(k-1)=k $$  as desired.  Remark. The first few terms are $a_{1}=2, a_{2}=3, a_{3}=26, a_{4}=456975$, and $a_{5}=$ 19927930852449199486318359374, ... No element of the sequence is divisible by 4: the residues modulo 4 alternate between 2 and 3 .  Remark. The second claim is important for the solution to work once $k \\geq 2$. One could imagine a variation of the first claim that states if $n$ is divisible by $\\varphi\\left(p^{k}\\right)=p^{k-1}(p-1)$, then either $a_{n-1} \\equiv 0(\\bmod p)$ or $a_{n} \\equiv 0\\left(\\bmod p^{k}\\right)$. However this gives an obstruction (for $k \\geq 2)$ where we are guaranteed to have $n-1 \\not \\equiv 0(\\bmod p)$ now, so lifting the exponent will never give additional factors of $p$ we want.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2024-notes.jsonl"}}
{"year": "2024", "tier": "T3", "problem_label": "4", "problem_type": null, "exam": "USAJMO", "problem": "Let $n \\geq 3$ be an integer. Rowan and Colin play a game on an $n \\times n$ grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if:  - no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and - no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. In terms of $n$, how many orderly colorings are there?", "solution": "  The answer is $2 n!+2$. In fact, we can describe all the orderly colorings as follows:  - The all-blue coloring. - The all-red coloring. - Each of the $n$ ! colorings where every row/column has exactly one red cell. - Each of the $n$ ! colorings where every row/column has exactly one blue cell.  These obviously work; we turn our attention to proving these are the only ones. For the other direction, fix a orderly coloring $\\mathcal{A}$. Consider any particular column $C$ in $\\mathcal{A}$ and let $m$ denote the number of red cells that $C$ has. Any row permutation (say $\\sigma$ ) that Rowan chooses will transform $C$ into some column $\\sigma(C)$, and our assumption requires $\\sigma(C)$ has to appear somewhere in the original assignment $\\mathcal{A}$. An example for $n=7, m=2$, and a random $\\sigma$ is shown below. ![](https://cdn.mathpix.com/cropped/2024_11_19_805564b9ec1a761a333dg-10.jpg?height=692&width=815&top_left_y=1910&top_left_x=618)  On the other hand, the number of possible patterns of $\\sigma(C)$ is easily seen to be exactly ( $\\binom{n}{m}$ - and they must all appear. In particular, if $m \\notin\\{0,1, n-1, n\\}$, then we immediately get a contradiction because $\\mathcal{A}$ would need too many columns (there are only $n$ columns in $\\mathcal{A}$, which is fewer than $\\binom{n}{m}$ ). Moreover, if either $m=1$ or $m=n-1$, these columns are all the columns of $\\mathcal{A}$; these lead to the $2 n$ ! main cases we found before.  The only remaining case is when $m \\in\\{0, n\\}$ for every column, i.e. every column is monochromatic. It's easy to see in that case the columns must all be the same color.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2024-notes.jsonl"}}
{"year": "2024", "tier": "T3", "problem_label": "5", "problem_type": null, "exam": "USAJMO", "problem": "Solve over $\\mathbb{R}$ the functional equation $f\\left(x^{2}-y\\right)+2 y f(x)=f(f(x))+f(y)$.", "solution": "  The answer is $f(x) \\equiv x^{2}, f(x) \\equiv 0, f(x) \\equiv-x^{2}$, which obviously work. Let $P(x, y)$ be the usual assertion. Claim - We have $f(0)=0$ and $f$ even.   Claim $-f(x) \\in\\left\\{-x^{2}, 0, x^{2}\\right\\}$ for every $x \\in \\mathbb{R}$.  $$ x^{2} f(x)=f\\left(x^{2}\\right)=f(f(x)) $$  Repeating this key identity several times gives  $$ \\begin{aligned} f(f(f(x))) & =f\\left(f\\left(x^{2}\\right)\\right)=f\\left(x^{4}\\right)=x^{4} f\\left(x^{2}\\right) \\\\ & =f(x)^{2} \\cdot f(f(x))=f(x)^{2} f\\left(x^{2}\\right)=f(x)^{3} x^{2} \\end{aligned} $$  Suppose $t \\neq 0$ is such that $f\\left(t^{2}\\right) \\neq 0$. Then the above equalities imply  $$ t^{4} f\\left(t^{2}\\right)=f(t)^{2} f\\left(t^{2}\\right) \\Longrightarrow f(t)= \\pm t^{2} $$  and then  $$ f(t)^{2} f\\left(t^{2}\\right)=f(t)^{3} t^{2} \\Longrightarrow f\\left(t^{2}\\right)= \\pm t^{2} $$  Together with $f$ even, we get the desired result.   Now, note that $P(1, y)$ gives  $$ f(1-y)+2 y \\cdot f(1)=f(1)+f(y) $$  We consider cases on $f(1)$ and show that $f$ matches the desired form.  - If $f(1)=1$, then $f(1-y)+(2 y-1)=f(y)$. Consider the nine possibilities that arise:  $$ \\begin{array}{lll} (1-y)^{2}+(2 y-1)=y^{2} & 0+(2 y-1)=y^{2} & -(1-y)^{2}+(2 y-1)=y^{2} \\\\ (1-y)^{2}+(2 y-1)=0 & 0+(2 y-1)=0 & -(1-y)^{2}+(2 y-1)=0 \\\\ (1-y)^{2}+(2 y-1)=-y^{2} & 0+(2 y-1)=-y^{2} & -(1-y)^{2}+(2 y-1)=-y^{2} \\end{array} $$  Each of the last eight equations is a nontrivial polynomial equation. Hence, there is some constant $C>100$ such that the latter eight equations are all false for any real number $y>C$. Consequently, $f(y)=y^{2}$ for $y>C$. Finally, for any real number $z>0$, take $x, y>C$ such that $x^{2}-y=z$; then $P(x, y)$ proves $f(z)=z^{2}$ too.  - Note that (as $f$ is even), $f$ works iff $-f$ works, so the case $f(1)=-1$ is analogous. - If $f(1)=0$, then $f(1-y)=f(y)$. Hence for any $y$ such that $|1-y| \\neq|y|$, we conclude $f(y)=0$. Then take $P(2,7 / 2) \\Longrightarrow f(1 / 2)=0$.   $$ P\\left(x, y^{2}\\right) \\Longrightarrow f\\left(x^{2}-y^{2}\\right)+2 y^{2} f(x)=f(f(x))+f(f(y)) $$  Since $f$ is even gives $f\\left(x^{2}-y^{2}\\right)=f\\left(y^{2}-x^{2}\\right)$, one can swap the roles of $x$ and $y$ to get $2 y^{2} f(x)=2 x^{2} f(y)$. Set $y=1$ to finish.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2024-notes.jsonl"}}
{"year": "2024", "tier": "T3", "problem_label": "6", "problem_type": null, "exam": "USAJMO", "problem": "Point $D$ is selected inside acute triangle $A B C$ so that $\\angle D A C=\\angle A C B$ and $\\angle B D C=90^{\\circ}+\\angle B A C$. Point $E$ is chosen on ray $B D$ so that $A E=E C$. Let $M$ be the midpoint of $B C$. Show that line $A B$ is tangent to the circumcircle of triangle $B E M$.", "solution": " ब The author's original solution. Complete isosceles trapezoid $A B Q C$ (so $D \\in \\overline{A Q}$ ). Reflect $B$ across $E$ to point $F$. ![](https://cdn.mathpix.com/cropped/2024_11_19_805564b9ec1a761a333dg-14.jpg?height=663&width=718&top_left_y=959&top_left_x=675)  Claim - We have $D Q C F$ is cyclic.  $$ \\begin{aligned} \\measuredangle F D C & =-\\measuredangle C D B=180^{\\circ}-\\left(90^{\\circ}+\\measuredangle C A B\\right)=90^{\\circ}-\\measuredangle C A B \\\\ & =90^{\\circ}-\\measuredangle Q C A=\\measuredangle F Q C . \\end{aligned} $$  To conclude, note that  $$ \\measuredangle B E M=\\measuredangle B F C=\\measuredangle D F C=\\measuredangle D Q C=\\measuredangle A Q C=\\measuredangle A B C=\\measuredangle A B M $$  Remark (Motivation). Here is one possible way to come up with the construction of point $F$ (at least this is what led Evan to find it). If one directs all the angles in the obvious way, there are really two points $D$ and $D^{\\prime}$ that are possible, although one is outside the triangle; they give corresponding points $E$ and $E^{\\prime}$. The circles $B E M$ and $B E^{\\prime} M$ must then actually coincide since they are both alleged to be tangent to line $A B$. See the figure below. ![](https://cdn.mathpix.com/cropped/2024_11_19_805564b9ec1a761a333dg-15.jpg?height=1112&width=1047&top_left_y=238&top_left_x=513)  One can already prove using angle chasing that $\\overline{A B}$ is tangent to $\\left(B E E^{\\prime}\\right)$. So the point of the problem is to show that $M$ lies on this circle too. However, from looking at the diagram, one may realize that in fact it seems  $$ \\triangle M E E^{\\prime} \\stackrel{ }{\\sim} \\triangle C D D^{\\prime} $$  is going to be true from just those marked in the figure (and this would certainly imply the desired concyclic conclusion). Since $M$ is a midpoint, it makes sense to dilate $\\triangle E M E^{\\prime}$ from $B$ by a factor of 2 to get $\\triangle F C F^{\\prime}$ so that the desired similarity is actually a spiral similarity at $C$. Then the spiral similarity lemma says that the desired similarity is equivalent to requiring $\\overline{D D^{\\prime}} \\cap \\overline{F F^{\\prime}}=Q$ to lie on both $(C D F)$ and $\\left(C D^{\\prime} F^{\\prime}\\right)$. Hence the key construction and claim from the solution are both discovered naturally, and we find the solution above. (The points $D^{\\prime}, E^{\\prime}, F^{\\prime}$ can then be deleted to hide the motivation.)", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2024-notes.jsonl"}}
{"year": "2024", "tier": "T3", "problem_label": "6", "problem_type": null, "exam": "USAJMO", "problem": "Point $D$ is selected inside acute triangle $A B C$ so that $\\angle D A C=\\angle A C B$ and $\\angle B D C=90^{\\circ}+\\angle B A C$. Point $E$ is chosen on ray $B D$ so that $A E=E C$. Let $M$ be the midpoint of $B C$. Show that line $A B$ is tangent to the circumcircle of triangle $B E M$.", "solution": " Another short solution. Let $Z$ be on line $B D E$ such that $\\angle B A Z=90^{\\circ}$. This lets us interpret the angle condition as follows:  $$ \\text { Claim - Points } A, D, Z, C \\text { are cyclic. } $$  ![](https://cdn.mathpix.com/cropped/2024_11_19_805564b9ec1a761a333dg-16.jpg?height=804&width=701&top_left_y=249&top_left_x=683)  Define $W$ as the midpoint of $\\overline{B Z}$, so $\\overline{M W} \\| \\overline{C Z}$. And let $O$ denote the center of $(A B C)$.  Claim - Points $M, E, O, W$ are cyclic.   $$ \\begin{aligned} \\measuredangle M O E & =\\measuredangle(\\overline{O M}, \\overline{B C})+\\measuredangle(\\overline{B C}, \\overline{A C})+\\measuredangle(\\overline{A C}, \\overline{O E}) \\\\ & =90^{\\circ}+\\measuredangle B C A+90^{\\circ} \\\\ & =\\measuredangle B C A=\\measuredangle C A D=\\measuredangle C Z D=\\measuredangle M W D=\\measuredangle M W E . \\end{aligned} $$  To finish, note  $$ \\begin{aligned} \\measuredangle M E B & =\\measuredangle M E W=\\measuredangle M O W \\\\ & =\\measuredangle(\\overline{M O}, \\overline{B C})+\\measuredangle(\\overline{B C}, \\overline{A B})+\\measuredangle(\\overline{A B}, \\overline{O W}) \\\\ & =90^{\\circ}+\\measuredangle C B A+90^{\\circ}=\\measuredangle C B A=\\measuredangle M B A . \\end{aligned} $$  This implies the desired tangency.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2024-notes.jsonl"}}
{"year": "2024", "tier": "T3", "problem_label": "6", "problem_type": null, "exam": "USAJMO", "problem": "Point $D$ is selected inside acute triangle $A B C$ so that $\\angle D A C=\\angle A C B$ and $\\angle B D C=90^{\\circ}+\\angle B A C$. Point $E$ is chosen on ray $B D$ so that $A E=E C$. Let $M$ be the midpoint of $B C$. Show that line $A B$ is tangent to the circumcircle of triangle $B E M$.", "solution": " I A Menelaus-based approach (Kevin Ren). Let $P$ be on $\\overline{B C}$ with $A P=P C$. Let $Y$ be the point on line $A B$ such that $\\angle A C Y=90^{\\circ}$; as $\\angle A Y C=90^{\\circ}-A$ it follows $B D Y C$ is cyclic. Let $K=\\overline{A P} \\cap \\overline{C Y}$, so $\\triangle A C K$ is a right triangle with $P$ the midpoint of its hypotenuse. ![](https://cdn.mathpix.com/cropped/2024_11_19_805564b9ec1a761a333dg-17.jpg?height=1192&width=1075&top_left_y=249&top_left_x=496)  Claim - Triangles BPE and DYK are similar.  Claim - Triangles $B E M$ and $Y D C$ are similar.   $$ \\frac{B P}{B C} \\frac{Y C}{Y K} \\frac{A K}{A P}=1 $$  Since $A K / A P=2$ (note that $P$ is the midpoint of the hypotenuse of right triangle $A C K)$ and $B C=2 B M$, this simplifies to  $$ \\frac{B P}{B M}=\\frac{Y K}{Y C} $$  To finish, note that  $$ \\measuredangle D B A=\\measuredangle D B Y=\\measuredangle D C Y=\\measuredangle B M E $$  implying the desired tangency.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2024-notes.jsonl"}}
{"year": "2024", "tier": "T3", "problem_label": "6", "problem_type": null, "exam": "USAJMO", "problem": "Point $D$ is selected inside acute triangle $A B C$ so that $\\angle D A C=\\angle A C B$ and $\\angle B D C=90^{\\circ}+\\angle B A C$. Point $E$ is chosen on ray $B D$ so that $A E=E C$. Let $M$ be the midpoint of $B C$. Show that line $A B$ is tangent to the circumcircle of triangle $B E M$.", "solution": " 『 A spiral similarity approach (Hans $\\mathbf{Y u}$ ). As in the previous solution, let $Y$ be the point on line $A B$ such that $\\angle A C Y=90^{\\circ}$; so $B D Y C$ is cyclic. Let $\\Gamma$ be the circle through $B$ and $M$ tangent to $\\overline{A B}$, and let $\\Omega:=(B C Y D)$. We need to show $E \\in \\Gamma$. ![](https://cdn.mathpix.com/cropped/2024_11_19_805564b9ec1a761a333dg-18.jpg?height=1101&width=1029&top_left_y=429&top_left_x=519)  Denote by $S$ the second intersection of $\\Gamma$ and $\\Omega$. The main idea behind is to consider the spiral similarity  $$ \\Psi: \\Omega \\rightarrow \\Gamma \\quad C \\mapsto M \\text { and } Y \\mapsto B $$  centered at $S$ (due to the spiral similarity lemma), and show that $\\Psi(D)=E$. The spiral similarity lemma already promises $\\Psi(D)$ lies on line $B D$.  Claim - We have $\\Psi(A)=O$, the circumcenter of $A B C$.   Claim - $\\Psi$ maps line $A D$ to line $O P$.   $$ \\measuredangle(\\overline{A D}, \\overline{O P})=\\measuredangle A P O=\\measuredangle O P C=\\measuredangle Y C P=\\measuredangle(\\overline{Y C}, \\overline{B M}) $$  As $\\Psi$ maps line $Y C$ to line $B M$ and $\\Psi(A)=O$, we're done. Hence $\\Psi(D)$ should not only lie on $B D$ but also line $O P$. This proves $\\Psi(D)=E$, so $E \\in \\Gamma$ as needed.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2024-notes.jsonl"}}