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

\begin{document}
\title{IMO 1997/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 101}
\maketitle

\section*{Problem}
It is known that $ \angle BAC$
is the smallest angle in the triangle $ ABC$.
The points $ B$ and $ C$ divide
the circumcircle of the triangle into two arcs.
Let $ U$ be an interior point of the arc
between $ B$ and $ C$ which does not contain $ A$.
The perpendicular bisectors of $ AB$ and $ AC$
meet the line $ AU$ at $ V$ and $ W$, respectively.
The lines $ BV$ and $ CW$ meet at $ T$.

Show that $ AU = TB + TC$.

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

\section*{External Link}
\url{https://aops.com/community/p356701}

\newpage

\section*{Solution}
Let $\ol{BTV}$ meet the circle again at $U_1$,
so that $AU_1 UB$ is an isosceles trapezoid.
Define $U_2$ similarly.

\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(230);
pair C = dir(310);
pair U = dir(250);
pair U_1 = A*B/U;
pair U_2 = A*C/U;
pair T = extension(B, U_1, C, U_2);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
filldraw(A--B--C--cycle, opacity(0.1)+lightred, red);
draw(A--U, deepgreen);
draw(C--U_2, deepgreen);
draw(B--U_1, deepgreen);
draw(B--U_2, red);
draw(U_1--U--U_2, red);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$U$", U, dir(U));
dot("$U_1$", U_1, dir(U_1));
dot("$U_2$", U_2, dir(U_2));
dot("$T$", T, dir(T));

/* TSQ Source:

A = dir 110
B = dir 230
C = dir 310
U = dir 250
U_1 = A*B/U
U_2 = A*C/U
T = extension B U_1 C U_2
unitcircle 0.1 lightcyan / blue
A--B--C--cycle 0.1 lightred / red
A--U deepgreen
C--U_2 deepgreen
B--U_1 deepgreen

*/
\end{asy}
\end{center}
Now from the isosceles trapezoids we get
\[ AU = BU_1 = BT + TU_1 = BT + TC \]
as desired.

\end{document}
