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

\begin{document}
\title{Shortlist 2007 G2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 2}
\maketitle

\section*{Problem}
Denote by $M$ midpoint of side $BC$ in an
isosceles triangle $\triangle ABC$ with $AC = AB$.
Take a point $X$ on a smaller arc $\widehat{MA}$
of circumcircle of triangle $\triangle ABM$.
Denote by $T$ point inside of angle $BMA$
such that $\angle TMX = 90\dg$ and $TX = BX$.

Prove that $\angle MTB - \angle CTM$ does not depend on choice of $X$.

\section*{Video}
\href{https://www.youtube.com/watch?v=vA1iwW41Jmw&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/vA1iwW41Jmw}}

\section*{External Link}
\url{https://aops.com/community/p1143932}

\newpage

\section*{Solution}
Construct parallelogram $CTBS$ whose diagonals meet at $M$.
Also, let $N$ be the midpoint of $\ol{BT}$.

\begin{center}
\begin{asy}
pair A = dir(180);
pair B = dir(0);
pair M = dir(35);
pair C = 2*M-B;
pair X = dir(100);

filldraw(A--B--C--cycle, opacity(0.1)+lightred, red);
draw(A--M, red);
filldraw(unitcircle, opacity(0.1)+cyan, blue);

pair Y = -X;
draw(CP(X, B), mediumgreen);
pair T = IP(M--Y, CP(X, B));
pair S = 2*M-T;
draw(S--T, mediumgreen);
draw(B--X--M, mediumgreen);

pair N = midpoint(B--T);
draw(B--T, deepgreen);
draw(T--X--S, mediumgreen);
draw(X--N, deepgreen);

filldraw(circumcircle(X, M, T), opacity(0.1)+yellow, dotted+deepgreen);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$M$", M, dir(M));
dot("$C$", C, dir(C));
dot("$X$", X, dir(X));
dot("$T$", T, dir(T));
dot("$S$", S, dir(S));
dot("$N$", N, dir(N));

/* TSQ Source:

A = dir 180
B = dir 0
M = dir 35
C = 2*M-B
X = dir 100

A--B--C--cycle 0.1 lightred / red
A--M red
unitcircle 0.1 cyan / blue

Y := -X
CP X B mediumgreen
T = IP M--Y CP X B
S = 2*M-T
S--T mediumgreen
B--X--M mediumgreen

N = midpoint B--T
B--T deepgreen
T--X--S mediumgreen
X--N deepgreen

circumcircle X M T 0.1 yellow / dotted deepgreen

*/
\end{asy}
\end{center}

We first eliminate $C$ from the diagram by noting that
\[ \angle CTM = \angle MSB = \angle TSB = \half \angle TXB = \angle NXB.  \]
Also, noting that $XMTN$ are cyclic
(as $\angle XMT = \angle XNT = 90\dg$), we have
\[ \angle MTB = \angle MTN = \angle NXM. \]
Thus $\angle MTB - \angle CTM = \angle NXM - \angle NXB = \angle MXB$
which is fixed.

\end{document}
