\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{SIME 2020/15}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 22}
\maketitle
\section*{Problem}
Triangle $ABC$ has side lengths $AB = 13$, $BC = 14$, and $AC = 15$.
Suppose $M$ and $N$ are
the midpoints of $\ol{AB}$ and $\ol{AC}$, respectively.
Let $P$ be a point on $\ol{MN}$,
such that if the circumcircles of triangles
$\triangle BMP$ and $\triangle CNP$ intersect at
a second point $Q$ distinct from $P$,
then $PQ$ is parallel to $AB$.
Calculate $AP^2$.
\section*{Video}
\href{https://www.youtube.com/watch?v=EIz39iVoGcg&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/EIz39iVoGcg}}
\newpage
\section*{Solution}
We ignore the condition that $\ol{PQ} \parallel \ol{AB}$ momentarily,
and prove the following two claims.
\begin{claim*}
The point $Q$ lies on $(ABC)$.
\end{claim*}
\begin{proof}
Miquel point of $P$, $C$, $B$ with respect to $\triangle AMN$.
\end{proof}
\begin{claim*}
The line $PQ$ passes through the reflection of $A$
over the perpendicular bisector of line $BC$.
\end{claim*}
\begin{proof}
Because $\dang BQP = \dang BMP = \dang NMA = \dang CBA = \dang
BCA' = \dang BQA'$.
\end{proof}
We now bring in the condition of the problem.
We have $AMPA'$ is a parallelogram in the given problem,
so $MP = AA'$, essentially locating $P$.
To compute $AA'$, we use Ptolemy's theorem by
\[ AA' \cdot 14 + 13^2 = 15^2 \implies MP = AA' = 4 \]
and a calculation on $\triangle AMN$
now gives $AP^2 = 153/4$.
\end{document}