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

\begin{document}
\title{SMO 2020/3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 29}
\maketitle

\section*{Problem}
Let $\triangle ABC$ be an acute scalene triangle
with incenter $I$ and incircle $\omega$.
Two points $X$ and $Y$ are chosen on minor arcs $AB$ and $AC$,
respectively, of the circumcircle of triangle $\triangle ABC$
such that $XY$ is tangent to $\omega$ at $P$ and $\ol{XY} \perp \ol{AI}$.
Let $\omega$ be tangent to sides $AC$ and $AB$ at $E$ and $F$, respectively.
Denote the intersection of lines $XF$ and $YE$ as $T$.

Prove that if the
circumcircles of triangles $\triangle TEF$ and $\triangle ABC$
are tangent at some point $Q$,
then lines $PQ$, $XE$, and $YF$ are concurrent.

\section*{Video}
\href{https://www.youtube.com/watch?v=rScZrRpRkR0&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/rScZrRpRkR0}}



\newpage

\section*{Solution}
We ignore the condition that $(TEF)$ and $(ABC)$
are tangent until the very end.

Instead, we define $Q$ as the intersection
of the circles $(TEF)$, $(XPF)$, $(YPE)$ ---
this point exists by Miquel's theorem on $\triangle TXY$
with $E \in \ol{TY}$, $F \in \ol{TX}$, $P \in \ol{XY}$.

\begin{center}
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\begin{claim*}
  Point $Q$ lies on $\ol{PS}$.
\end{claim*}
\begin{proof}
  It suffices to show $S$ lies on the radical axis
  of $(XPF)$ and $(YPE)$.
  Let $\ol{ESX}$ meet $(XPF)$ again at $Z_1$
  and let $\ol{ESY}$ meet $(YPE)$ again at $Z_2$.
  Then
  \[ \dang SZ_1F = \dang XZ_1F = \dang XPF = \dang YPE
    = \dang YZ_2E = \dang SZ_2E \]
  which implies $Z_1$, $Z_2$, $E$, $F$ are cyclic.

  On the other hand, since $\ol{EF} \parallel \ol{XY}$
  (both are perpendicular to $\ol{AI}$),
  it follows that $SX : SY = SE : SF$.
  So we are able to get $SZ_1 \cdot  SX = SZ_2 \cdot SY$ as needed.
\end{proof}

Let $W$ denote the circumcenter of $TFEQ$ which lies on $\ol{API}$.
\begin{claim*}
  Line $XY$ passes through the pole $K$ of line $QT$ with respect to $(TFEQ)$.
\end{claim*}
\begin{proof}
  We have
  $\dang QPW = \dang QPX + 90\dg = \dang QFX + 90\dg = \dang QFT + 90\dg$
  which can be checked to equal $\dang QTW$ is computable.
  Hence $QPWT$ cyclic.
  This circle also passes through the pole of $\ol{QT}$,
  and since $\ol{WP} \perp \ol{XY}$, the conclusion follows.
\end{proof}

So if the two circles are tangent,
then the tangency point must coincide with the $Q$ we found.
And $Q$ already lies on $PS$.

\end{document}
