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

\begin{document}
\title{XOOK 2023/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 144}
\maketitle

\section*{Problem}
Let $ABC$ be a triangle with incenter $I$, which has foot $D$ to $BC$,
and let $T$ be the $A$-mixtillinear touch point.
Let $M$ be the midpoint o f $BC$, which has foot $X$ to line $AI$.
Let $AT$ intersect $(DXT)$ again at $Q$.
Show that $(AXQ)$ is tangent to the incircle.

\section*{Video}
\href{https://www.youtube.com/watch?v=cuUQkgZ4xVc&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/cuUQkgZ4xVc}}

\section*{External Link}
\url{https://aops.com/community/p29571593}

\newpage

\section*{Solution}
Let $E$ be the contact point of the $A$-excircle,
so $\ol{ATQ}$ and $\ol{AE}$ are isogonal.
The circle centered at $M$ with radius $MD=ME$ intersects
the incircle again at a point $W$ such that $\angle DWE = 90\dg$;
this implies $A$, $W$, $E$ are collinear, $MW = MD = ME$,
and $\ol{MW}$ is tangent to the incircle.
(See USA TST 2015/1 for a figure with the same circle centered at $M$.)

\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 = (-0.04783,4.09061);
pair B = (-0.5,0.5);
pair C = (7.,0.5);
pair I = (1.10459,1.91520);
pair D = (1.10459,0.5);
pair M = (3.25,0.5);
pair T = (-0.10636,-0.29880);
pair X = (2.16016,-0.07734);
pair Q = (-0.11899,-1.24610);
pair W = (2.40557,2.47222);
pair E = (5.39540,0.5);

import graph;
size(12cm);
pen zzttqq = rgb(0.6,0.2,0.);
pen yqqqqq = rgb(0.50196,0.,0.);
pen qqffff = rgb(0.,1.,1.);
pen qqwuqq = rgb(0.,0.39215,0.);
pen ffxfqq = rgb(1.,0.49803,0.);
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((3.25,1.85154), 3.98612), linewidth(0.6) + yqqqqq);
draw(circle(I, 1.41520), linewidth(0.6) + yqqqqq);
draw(circle((1.08556,-0.78843), 1.28857), linewidth(0.6) + qqffff);
draw(circle((-0.04783,1.42178), 2.66883), linewidth(0.6) + qqwuqq);
draw(A--(3.25,-2.13457), linewidth(0.6) + yqqqqq);
draw(A--E, linewidth(0.6) + gray);
draw(M--W, linewidth(0.6));
draw(circle(M, 2.14540), linewidth(0.6) + blue);
draw(circle((2.17729,1.20760), 1.28506), linewidth(0.6) + ffxfqq);
draw(T--(6.54783,4.09061), linewidth(0.6) + gray);
draw(X--W, linewidth(0.6));
draw(A--Q, linewidth(0.6) + gray);

dot("$A$", A, dir((-23.070, 15.654)));
dot("$B$", B, dir((-32.230, -26.525)));
dot("$C$", C, dir((7.066, -31.468)));
dot("$I$", I, dir((-22.967, -5.736)));
dot("$D$", D, dir((-20.495, -38.883)));
dot("$M$", M, dir((-1.048, -42.179)));
dot("$T$", T, dir((-29.575, -10.908)));
dot("$X$", X, dir((20.603, -19.048)));
dot("$Q$", Q, dir((-34.079, -26.581)));
dot("$W$", W, dir(70));
dot("$E$", E, dir((11.809, -29.820)));
\end{asy}
\end{center}

\begin{claim*}
  $WIDXM$ is cyclic with diameter $\ol{IM}$.
\end{claim*}
\begin{proof}
  Because $\dang IDM = \dang IXM = \dang IWM = 90\dg$.
\end{proof}

\begin{claim*}
  $\ol{MW}$ is also tangent to $(AWX)$.
\end{claim*}
\begin{proof}
  $\dang WXA = \dang WXI = \dang WDI = \dang WED = \dang MWE = \dang MWA$.
\end{proof}

\begin{claim*}
  $Q$ lies on $(AXW)$.
\end{claim*}
\begin{proof}
  Note that
  \begin{align*}
    \dang XQT = \dang XDT &= \dang XDM + \dang(\ol{MD}, \ol{DT})
    = \dang XWM + \dang(\ol{AE}, \ol{EM}) \\
    &= \dang XWM + \dang WEM = \dang XWM + \dang MWE \\
    &= \dang XWE = \dang XWA. \qedhere
  \end{align*}
\end{proof}

\end{document}
