\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}

\begin{center}
\begin{asy}
import graph; size(10cm);
pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215,0.); pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw((-5.,11.)--(-10.,-5.)--(11.,-5.)--cycle, linewidth(0.6) + zzttqq);

draw((-5.,11.)--(-10.,-5.), linewidth(0.6) + zzttqq);
draw((-10.,-5.)--(11.,-5.), linewidth(0.6) + zzttqq);
draw((11.,-5.)--(-5.,11.), linewidth(0.6) + zzttqq);
draw(circle((0.5,0.5), 11.85326), linewidth(0.6));
draw(circle((-3.6,-2.60937), 6.83191), linewidth(0.6) + blue);
draw(circle((1.65,-6.35), 9.44695), linewidth(0.6) + blue);
draw((-7.62825,-8.12737)--(6.,11.), linewidth(0.6) + qqwuqq);
dot("$A$", (-5.,11.), dir((15.837, 39.800)));
dot("$B$", (-10.,-5.), dir((-99.285, -26.323)));
dot("$C$", (11.,-5.), dir((37.150, -83.354)));
dot("$M$", (-7.5,3.), dir((-108.939, 29.143)));
dot("$N$", (3.,3.), dir((14.272, 33.217)));
dot("$P$", (0.3,3.), dir((-17.178, 69.880)));
dot("$A'$", (6.,11.), dir((15.722, 39.800)));
dot("$Q$", (-7.62825,-8.12737), dir((-75.746, -108.730)));
\end{asy}
\end{center}

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}
