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

\begin{document}
\title{BAMO 2019/3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 121}
\maketitle

\section*{Problem}
In $\triangle ABC$, mark points $A'$, $B'$, and $C'$
on segments $\ol{BC}$, $\ol{CA}$, and $\ol{AB}$ respectively,
so that $\ol{AA'}$ is an altitude, $\ol{BB'}$ is a median,
and $\ol{CC'}$ is an angle bisector.
Prove that if $\triangle A'B'C'$ is equilateral
if and only if $\triangle ABC$ is equilateral.

\section*{Video}
\href{https://www.youtube.com/watch?v=4VQxYmE2tEQ&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/4VQxYmE2tEQ}}

\section*{External Link}
\url{https://aops.com/community/p13025889}

\newpage

\section*{Solution}
If $\triangle ABC$ is equilateral,
it's obvious because $A'$, $B'$, $C'$ are the midpoints.

Conversely, consider any triangle.
Note that $A'$ unconditionally lies on the circle with diameter $\ol{AC}$
centered at $B'$, meaning $A'B' = B'A = B'C$.

\begin{center}
\begin{asy}
pair A = dir(120);
pair B = dir(200);
pair C = dir(340);

pair A_prime = foot(A, B, C);
pair B_prime = midpoint(A--C);
pair C_prime = midpoint(A--B);

filldraw(A--B--C--cycle, opacity(0.1)+cyan, blue);

draw(C_prime--B_prime--A_prime, red);
draw(circumcircle(A, A_prime, C), red);
draw(C--C_prime, deepcyan);
draw(A--A_prime, blue);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$A'$", A_prime, dir(A_prime));
dot("$B'$", B_prime, dir(B_prime));
dot("$C'$", C_prime, dir(C_prime));

/* -----------------------------------------------------------------+
|                 TSQX: by CJ Quines and Evan Chen                  |
| https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py |
+-------------------------------------------------------------------+
A = dir 120
B = dir 200
C = dir 340
A' = foot A B C
B' = midpoint A--C
C' = midpoint A--B
A--B--C--cycle / 0.1 cyan / blue
C'--B'--A' / red
circumcircle A A' C / red
C--C' / deepcyan
A--A' / blue
*/
\end{asy}
\end{center}

So if $\triangle A'B'C'$ was equilateral, then $C'B' = B'A'$,
then $C'$ would lie on this circle as well.
This means $\angle AC'C = 90\dg$.
As $C'$ was the foot of the bisector, it follows $AC = BC$.

To finish, note that
\[ 60\dg = \angle C'B'A' = 2 \angle C'CA' = \angle ACB. \]

\end{document}
