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

\begin{document}
\title{SIME 2020/13}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 25}
\maketitle

\section*{Problem}
In acute triangle $\triangle ABC$, $AB=20$ and $AC=21$.
Let the feet of the perpendiculars from $A$
to the angle bisectors of $\angle ACB$ and $\angle ABC$ be $X$ and $Y$, respectively.
Let $M$ be the midpoint of $\ol{XY}$.
Suppose $P$ is the point on side $BC$ such that $MP$
is parallel to the angle bisector of $\angle BAC$.
Given that $BP = 11$, find the length of $BC$.

\section*{Video}
\href{https://www.youtube.com/watch?v=hQ4lFgub7tU&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/hQ4lFgub7tU}}



\newpage

\section*{Solution}
The answer is $BC = 205/9$.
Introduce the intouch triangle $DEF$.

\begin{center}
\begin{asy}
  import graph; size(10cm);
pen zzttqq = rgb(0.6,0.2,0.); pen ffxfqq = rgb(1.,0.49803,0.); pen qqwuqq = rgb(0.,0.39215,0.); pen qqqqcc = rgb(0.,0.,0.8); pen ffqqtt = rgb(1.,0.,0.2); pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw((0.86,7.76)--(-1.,-6.)--(17.2,-6.02)--cycle, linewidth(0.6) + zzttqq);
draw((0.86,7.76)--(-1.65741,0.88174)--(4.35515,-6.00588)--(6.87256,0.87237)--cycle, linewidth(0.6) + qqqqcc);

draw((0.86,7.76)--(-1.,-6.), linewidth(0.6) + zzttqq);
draw((-1.,-6.)--(17.2,-6.02), linewidth(0.6) + zzttqq);
draw((17.2,-6.02)--(0.86,7.76), linewidth(0.6) + zzttqq);
draw((-1.65741,0.88174)--(6.87256,0.87237), linewidth(0.6));
draw((2.60757,0.87705)--(5.26109,-6.00688), linewidth(0.6) + ffxfqq);
draw((-1.,-6.)--(6.87256,0.87237), linewidth(0.6));
draw((17.2,-6.02)--(-1.65741,0.88174), linewidth(0.6));
draw(circle((2.61014,3.21964), 4.86599), linewidth(0.6) + qqwuqq);
draw((0.86,7.76)--(-1.65741,0.88174), linewidth(0.6) + qqqqcc);
draw((-1.65741,0.88174)--(4.35515,-6.00588), linewidth(0.6) + qqqqcc);
draw((4.35515,-6.00588)--(6.87256,0.87237), linewidth(0.6) + qqqqcc);
draw((6.87256,0.87237)--(0.86,7.76), linewidth(0.6) + qqqqcc);
draw((0.86,7.76)--(4.35515,-6.00588), linewidth(0.6));
draw(circle((4.36029,-1.32071), 4.68517), linewidth(0.6) + ffqqtt);
draw((0.86,7.76)--(6.16703,-6.00787), linewidth(0.6) + ffxfqq);
dot("$A$", (0.86,7.76), dir((8.000, 20.000)));
dot("$B$", (-1.,-6.), dir((-54.000, -54.000)));
dot("$C$", (17.2,-6.02), dir((8.000, -64.000)));
dot("$I$", (4.36029,-1.32071), dir((1.971, 40.071)));
dot("$X$", (-1.65741,0.88174), dir((-66.259, -18.174)));
dot("$Y$", (6.87256,0.87237), dir((34.744, -15.237)));
dot("$M$", (2.60757,0.87705), dir((7.243, 16.295)));
dot("$P$", (5.26109,-6.00688), dir((-14.109, -71.312)));
dot("$R$", (3.51351,0.87606), dir((8.649, 16.394)));
dot("$D$", (4.35515,-6.00588), dir((-17.515, -69.412)));
dot("$F$", (-0.28264,-0.69311), dir((-67.736, -44.689)));
dot("$E$", (7.38074,2.26086), dir((7.926, 15.914)));
dot("$K$", (6.16703,-6.00787), dir((-14.703, -65.213)));
\end{asy}
\end{center}

\begin{claim*}
  Point $M$ is the midpoint of $\ol{AD}$.
\end{claim*}
\begin{proof}
  By EGMO Lemma 1.45, we have $X$ on line $DF$.
  Since $\ol{DFX} \perp \ol{BI}$ and $\ol{BI} \perp \ol{AY}$,
  we have $\ol{DX} \parallel \ol{AY}$.
  Similarly, $\ol{DY} \parallel \ol{AX}$,
  and it follows $DXAY$ is a parallelogram.
\end{proof}
Let $K$ be the foot of angle bisector.
Then it follows that $BP = \half(BD+BK)$.
Thus we may write in the usual notation
\begin{align*}
  11 &= BP = \frac{(s-b) + \frac{c}{c+b} \cdot a}{2}
  = \frac{\frac{a-1}{2} + \frac{20}{41} a}{2} \\
  \implies \frac{45}{2} &= \frac{81}{82} \cdot a \implies a = \frac{205}{9}.
\end{align*}

\end{document}
