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

\begin{document}
\title{Twitch 124.3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 124}
\maketitle

\section*{Problem}
Let $\triangle ABC$ be a right triangle with right angle $B$
with incenter $I$ and intouch triangle $\triangle DEF$.
Let the circle with center $B$ containing $D$ and $F$
intersect the circumcircle at $M$ and $N$.
Let $X$ and $Y$ be the intersection point of $\ol{MN}$
with $\ol{ID}$ and $\ol{IF}$ respectively.
Show that $X$, $Y$, the midpoint of $AB$,
and the midpoint of $BC$, are concyclic.

\section*{Video}
\href{https://www.youtube.com/watch?v=f4F7tDglpzA&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/f4F7tDglpzA}}



\newpage

\section*{Solution}
Let $M_a$, $O$, $M_c$ be the midpoints of $ABC$.
Then let $T$ be the midpoint of minor arc $\arc{BC}$.

\begin{center}
\begin{asy}
/*
    Converted from GeoGebra by User:Azjps using Evan's magic cleaner
    https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
pair A = (3.,5.);
pair B = (3.,-3.);
pair C = (14.,-3.);
pair I = (5.69926,-0.30073);
pair D = (5.69926,-3.);
pair F = (3.,-0.30073);
pair E = (7.28689,1.88225);
pair X = (1.69926,-0.30073);
pair Y = (5.69926,-5.80073);
pair M_a = (8.5,-3.);
pair M_c = (3.,1.);
pair O = (8.5,1.);
pair T = (8.5,-5.80073);

import graph;
size(9cm);
pen zzttqq = rgb(0.6,0.2,0.);
pen yqqqqq = rgb(0.50196,0.,0.);
pen qqwuqq = rgb(0.,0.39215,0.);
pen cqcqcq = rgb(0.75294,0.75294,0.75294);
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(I, 2.69926), linewidth(0.6) + yqqqqq);
draw(circle(O, 6.80073), linewidth(0.6) + yqqqqq);
draw(circle(B, 2.69926), linewidth(0.6) + linetype("2 2") + blue);
draw(I--X, linewidth(0.6) + qqwuqq);
draw(I--Y, linewidth(0.6) + qqwuqq);
draw(X--Y, linewidth(0.6) + qqwuqq);
draw(Y--T, linewidth(0.6) + red);
draw(O--T, linewidth(0.6) + yqqqqq);

dot("$A$", A, dir((-44.099, 27.745)));
dot("$B$", B, dir((-47.639, -55.391)));
dot("$C$", C, dir((25.748, -62.470)));
dot("$I$", I, dir((6.311, 14.486)));
dot("$D$", D, dir((2.772, -66.010)));
dot("$F$", F, dir((-42.329, 12.716)));
dot("$E$", E, dir((6.831, 13.874)));
dot("$X$", X, dir((-57.380, 18.026)));
dot("$Y$", Y, dir((-25.545, -63.798)));
dot("$M_a$", M_a, dir((7.638, -62.470)));
dot("$M_c$", M_c, dir((-79.495, 17.148)));
dot("$O$", O, dir((7.638, 13.608)));
dot("$T$", T, dir((-18.910, -56.719)));
\end{asy}
\end{center}

\begin{claim*}
$DM_a = M_aT$.
\end{claim*}
\begin{proof}
In the usual notation,
$DM_a = BM_a - BD = \frac a2 - \frac{b+c-a}{2} = \frac{b-c}{2}$,
while $M_aT = OT - OM_a = \frac b2 - \frac c2$.
\end{proof}

\begin{claim*}
$DYTM_a$ is a square.
\end{claim*}
\begin{proof}
Let $Y'$ be the point such that $DM_aTY'$ is a square.
Then $Y'T^2 = Y'D^2$,
while $Y'T$ is automatically tangent to $(ABC)$
and $Y'D$ is automatically tangent to the circle centered at $B$
with radius $BD = BF$.
So $Y' = Y$.
\end{proof}

Hence $DY = DM_a$, and $Y$ is the reflection of $M_a$ around the line $\ol{DF}$.
The same is true for $X$,
so $XM_cM_aY$ is an isosceles trapezoid symmetric around line $\ol{DF}$.

\end{document}
