\documentclass[11pt]{scrartcl}
\usepackage{evan}

\begin{document}
\title{Mexico Q3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 138}
\maketitle

\section*{Problem}
Let $\Gamma_1$ and $\Gamma_2$ be circles intersecting at points $A$ and $B$.
A line through $A$ intersects $\Gamma_1$ and $\Gamma_2$ at $C$ and $D$ respectively.
Let $P$ be the intersection of the lines tangent to $\Gamma_1$ at $A$ and $C$,
and let $Q$ be the intersection of the lines tangent to $\Gamma_2$ at $A$ and $D$.
Let $X$ be the second intersection point of the circumcircles of $BCP$ and $BDQ$,
and let $Y$ be the intersection of lines $AB$ and $PQ$.
Prove that $C$, $D$, $X$ and $Y$ are concyclic.

\section*{Video}
\href{https://www.youtube.com/watch?v=c5NGYpfv8OQ&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/c5NGYpfv8OQ}}

\section*{External Link}
\url{https://aops.com/community/p14889633}

\newpage

\section*{Solution}
Ignore the point $X$ for now and focus on getting control over $Y$.
We start with the following claim:
\begin{claim*}
  Let $Z$ be the reflection of $A$ over $Y$. Then $Z$ lies on $(BCD)$.
\end{claim*}
\begin{center}
\begin{asy}
/*
Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
pair A = (1.63598,3.31936);
pair B = (1.63598,-3.31936);
pair C = (-2.96050,2.22037);
pair D = (8.39061,4.93434);
pair P = (-1.11818,4.67679);
pair Q = (4.35247,6.89075);
pair Y = (1.63598,5.79140);
pair Z = (1.63598,8.26344);
pair Pp = (-3.87236,6.03422);
pair Qp = (7.06895,10.46215);
pair K = (-0.08328,7.56765);

import graph;
size(9cm);
pen xfqqff = rgb(0.49803,0.,1.);
pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw(circle((0.,0.), 3.70062), linewidth(1.0) + blue);
draw(circle((6.,0.), 5.48295), linewidth(1.0) + xfqqff);
draw(C--D, linewidth(0.6));
draw(P--A, linewidth(0.6) + blue);
draw(P--C, linewidth(0.6) + blue);
draw(Q--A, linewidth(0.6) + xfqqff);
draw(Q--D, linewidth(0.6) + xfqqff);
draw(P--Q, linewidth(0.6));
draw(Y--B, linewidth(0.6));
draw(B--C, linewidth(0.6) + blue);
draw(B--D, linewidth(0.6) + xfqqff);
draw(circle((2.97932,2.47204), 5.94515), deepgreen + dashed);
draw(Y--Z, linewidth(0.6));
draw(circle(P, 3.07052), linewidth(0.6) + linetype("2 2") + blue);
draw(circle(Q, 4.48710), linewidth(0.6) + linetype("2 2") + xfqqff);
draw(Pp--Qp, linewidth(0.6));
draw(P--Pp, linewidth(0.6) + blue);
draw(Q--Qp, linewidth(0.6) + xfqqff);

dot("$A$", A, dir((5.914, 14.784)));
dot("$B$", B, dir((-20.697, -63.496)));
dot("$C$", C, dir((-53.352, -24.635)));
dot("$D$", D, dir((6.076, 14.430)));
dot("$P$", P, dir((6.350, 12.096)));
dot("$Q$", Q, dir((-26.235, 24.286)));
dot("$Y$", Y, dir((5.914, 11.514)));
dot("$Z$", Z, dir((-14.783, 20.072)));
dot("$P'$", Pp, dir((-49.392, 12.365)));
dot("$Q'$", Qp, dir((5.187, 11.611)));
dot("$K$", K, dir((-38.004, 33.472)));
\end{asy}
\end{center}
\begin{proof}
  Reflect $A$ over $P$ and $Q$ to get $P'$ and $Q'$, collinear with $Z$.
  Then let $K$ be the foot from $A$ to $P'Q'$,
  which lies on the circles with diameter $\ol{AP'}$ and $\ol{AQ'}$
  (that pass through $C$ and $D$ respectively).
  Now,
  \[ \dang BCK = \dang BCD + \dang DCK = \dang ZAP + \dang AP'K
  = \dang AZK = \dang BZK \]
  so $C$ lies on $(BZK)$. Then so does $D$, and we're done.
\end{proof}

We then trade in $Z$ for the midpoints $M$ and $N$ of $\ol{CA}$ and $\ol{AD}$.
\begin{claim*}
  Point $Y$ lies on both $(CBN)$ and $(DBM)$.
\end{claim*}
\begin{proof}
  Follows by power of a point from $A$ on the previous claim.
\end{proof}

Bring back point $X$ now.

\begin{center}
\begin{asy}
/*
Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
pair A = (1.63598,3.31936);
pair B = (1.63598,-3.31936);
pair C = (-2.96050,2.22037);
pair D = (8.39061,4.93434);
pair P = (-1.11818,4.67679);
pair Q = (4.35247,6.89075);
pair Y = (1.63598,5.79140);
pair X = (0.33953,5.26672);
pair M = (-0.66225,2.76986);
pair N = (5.01329,4.12685);

import graph;
size(12cm);
pen xfqqff = rgb(0.49803,0.,1.);
pen qqwuqq = rgb(0.,0.39215,0.);
pen ffqqff = rgb(1.,0.,1.);
pen qqffff = rgb(0.,1.,1.);
pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw(circle((0.,0.), 3.70062), linewidth(1.2) + blue);
draw(circle((6.,0.), 5.48295), linewidth(1.2) + xfqqff);
draw(C--D, linewidth(0.6));
draw(P--A, linewidth(0.6) + blue);
draw(P--C, linewidth(0.6) + blue);
draw(Q--A, linewidth(0.6) + xfqqff);
draw(Q--D, linewidth(0.6) + xfqqff);
draw(P--Q, linewidth(0.6));
draw(Y--B, linewidth(0.6));
draw(circle((1.21481,1.00796), 4.34777), linewidth(0.6) + qqwuqq);
draw(circle((4.21481,1.46095), 5.43154), linewidth(0.6) + qqwuqq);
draw(B--C, linewidth(0.6) + blue);
draw(B--D, linewidth(0.6) + xfqqff);
draw(circle((4.48966,1.23602), 5.37540), linewidth(0.6) + linetype("2 2") + ffqqff);
draw(circle((1.48966,1.23602), 4.55773), linewidth(0.6) + linetype("2 2") + qqffff);

dot("$A$", A, dir((5.914, 14.784)));
dot("$B$", B, dir((-20.697, -63.496)));
dot("$C$", C, dir((-53.352, -24.635)));
dot("$D$", D, dir((6.076, 14.430)));
dot("$P$", P, dir((6.350, 12.096)));
dot("$Q$", Q, dir((-26.235, 24.286)));
dot("$Y$", Y, dir((5.914, 11.514)));
dot("$X$", X, dir((-21.150, 19.631)));
dot("$M$", M, dir((6.587, 12.076)));
dot("$N$", N, dir((5.256, 12.389)));
\end{asy}
\end{center}

Now, for the main calculation, we have from symmedian theory that
\begin{align*}
  \dang CXB &= \dang CPB = \dang BMA = \dang BMD = \dang BYD \\
  \dang BXD &= \dang BQD = \dang ANB = \dang CNB = \dang CYB.
\end{align*}
So summing gives $\dang CXD = \dang CYD$, as needed.

\begin{remark*}
  Note that $X$ also lies on line $PQ$ because
  $\dang PXB = \dang PCB = \dang CAB = \dang DAB = \dang QDB = \dang QXB$.
\end{remark*}

\end{document}
