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

\begin{document}
\title{Iberoamerican 2021/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 87}
\maketitle

\section*{Problem}
Consider an acute-angled triangle $ABC$, with $AC>AB$,
and let $\Gamma$ be its circumcircle.
Let $E$ and $F$ be the midpoints of the sides $AC$ and $AB$,
respectively.
The circumcircle of the triangle $CEF$ and $\Gamma$
meet at $X$ and $C$, with $X\neq C$.
The line $BX$ and the tangent to $\Gamma$ through $A$ meet at $Y$.
Let $P$ be the point on segment $AB$ so that $YP = YA$,
with $P\neq A$, and let $Q$ be the point where $AB$
and the parallel to $BC$ through $Y$ meet each other.
Show that $F$ is the midpoint of $PQ$.

\section*{Video}
\href{https://www.youtube.com/watch?v=vdvTg5tLx4I&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/vdvTg5tLx4I}}

\section*{External Link}
\url{https://aops.com/community/p23437637}

\newpage

\section*{Solution}
Let $T = \ol{AA} \cap \ol{BC}$.
\begin{center}
\begin{asy}
import graph; size(12cm);
pen zzttqq = rgb(0.6,0.2,0.); pen yqqqqq = rgb(0.50196,0.,0.); pen qqwuqq = rgb(0.,0.39215,0.); pen yqqqyq = rgb(0.50196,0.,0.50196); pen xdxdff = rgb(0.49019,0.49019,1.); pen cqcqcq = rgb(0.75294,0.75294,0.75294);
pair A = (-4.55630,4.74594), B = (-6.,-1.), C = (2.,-1.), F = (-5.27815,1.87297), O = (-2.,1.04932), X = (-6.48864,1.27690), Y = (-6.88720,3.13406), Q = (-4.96129,3.13406), P = (-5.59500,0.61187), T = (-12.86540,-1.), Z = (-1.49456,-3.41657);
draw(A--B--C--cycle, linewidth(0.6) + zzttqq);

draw(A--B, linewidth(0.6) + zzttqq);
draw(B--C, linewidth(0.6) + zzttqq);
draw(C--A, linewidth(0.6) + zzttqq);
draw(circle((-2.,1.04932), 4.49441), linewidth(0.6) + yqqqqq);
draw(Y--Q, linewidth(0.6) + qqwuqq);
draw(B--Y, linewidth(0.6) + qqwuqq);
draw(C--X, linewidth(0.6) + yqqqyq);
draw(Y--P, linewidth(0.6) + qqwuqq);
draw(T--B, linewidth(0.6) + yqqqqq);
draw(X--(0.55630,4.74594), linewidth(0.6) + xdxdff);
draw(A--T, linewidth(0.6) + red);
draw((0.55630,4.74594)--Z, linewidth(0.6) + xdxdff);
draw((-1.27815,1.87297)--F, linewidth(0.6) + yqqqyq);
dot("$A$", A, dir((5.364, 14.027)));
dot("$B$", B, dir((-24.637, -60.663)));
dot("$C$", C, dir((16.031, -39.399)));
dot("$E$", (-1.27815,1.87297), dir((5.027, 10.705)));
dot("$F$", F, dir((26.514, -36.078)));
dot("$O$", O, dir((6.330, 10.846)));
dot("$X$", X, dir((-40.985, -38.847)));
dot("$Y$", Y, dir((-49.329, 9.349)));
dot("$Q$", Q, dir((17.510, -16.168)));
dot("$P$", P, dir((19.922, -17.709)));
dot("$T$", T, dir((-44.089, -57.828)));
dot("$Z$", Z, dir((-8.773, -64.261)));
dot("$A'$", (0.55630,4.74594), dir((5.876, 14.027)));
\end{asy}
\end{center}

\begin{claim*}
  [Main claim]
  \[ \frac{AX}{BX} = \frac{AT}{AB}. \]
\end{claim*}
\begin{proof}
  Let $A'$ denote the point such that $ABCA'$ is an isosceles trapezoid
  with $AA' \parallel BC$.
  Note that $A'$ lies on line $FX$,
  because $\dang CXF = \dang CEF = \dang AEF = \dang ACB$.

  Now let $Z$ be the harmonic conjugate of $X$ with respect to $AB$.
  Projecting $AXBZ$ through $A'$ onto line $AB$ we find
  $\ol{A'Z} \parallel \ol{AB}$.

  Finally from $\triangle BAT \sim \triangle A'AC$, we get
  \[ \frac{AX}{XB} = \frac{AZ}{ZB} = \frac{BA'}{AA'}
    = \frac{AC}{AA'} = \frac{AT}{AB}. \qedhere \]
\end{proof}

Since
\[ \dang APY = \dang YAP = \dang YAB
  = \dang ACB = \dang AXB = \dang AXY \]
we get $YXPA$ is cyclic and hence
\[ BP \cdot BA = BX \cdot BY
  = BX \cdot \frac{AB \cdot AY}{AX} \]
with the last equality due to $\triangle BYA \sim \triangle AYX$.
Meanwhile,
\[ AQ = \frac{AY}{AT} \cdot AF = \frac{AY}{AT} \cdot AB. \]
Combining with the main claim, we have $AQ = BP$ as needed.

\end{document}
