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

\begin{document}
\title{USAMO 2021/1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 65}
\maketitle

\section*{Problem}
Rectangles $BCC_1B_2$, $CAA_1C_2$, and $ABB_1A_2$ are erected
outside an acute triangle $ABC$. Suppose that
\[ \angle BC_1C + \angle CA_1A + \angle AB_1B = 180^\circ. \]
Prove that lines $B_1C_2$, $C_1A_2$, and $A_1B_2$ are concurrent.

\section*{Video}
\href{https://www.youtube.com/watch?v=9WNgDETHOlI&t=890}{\texttt{https://youtu.be/9WNgDETHOlI}}

\section*{External Link}
\url{https://aops.com/community/p21498558}

\newpage

\section*{Solution}
The angle condition implies the circumcircles of the three
rectangles concur at a single point $P$.
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair P = 0.2*dir(190);

filldraw(A--B--C--cycle, opacity(0.2)+lightcyan, blue);

pair X = circumcenter(P, B, C);
pair Y = circumcenter(P, C, A);
pair Z = circumcenter(P, A, B);

pair C_1 = 2*X-B;
pair B_2 = 2*X-C;

pair A_1 = 2*Y-C;
pair C_2 = 2*Y-A;

pair B_1 = 2*Z-A;
pair A_2 = 2*Z-B;

filldraw(circumcircle(A, B, P), opacity(0.05)+yellow, red);
filldraw(circumcircle(B, C, P), opacity(0.05)+yellow, red);
filldraw(circumcircle(C, A, P), opacity(0.05)+yellow, red);

draw(B_1--C_2, deepgreen+dashed);
draw(C_1--A_2, deepgreen+dashed);
draw(A_1--B_2, deepgreen+dashed);

draw(C--C_1--B_2--B, red);
draw(A--A_1--C_2--C, red);
draw(B--B_1--A_2--A, red);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(280));
dot("$C_1$", C_1, dir(C_1));
dot("$B_2$", B_2, dir(B_2));
dot("$A_1$", A_1, dir(A_1));
dot("$C_2$", C_2, dir(C_2));
dot("$B_1$", B_1, dir(B_1));
dot("$A_2$", A_2, dir(A_2));

/* -----------------------------------------------------------------+
|                 TSQX: by CJ Quines and Evan Chen                  |
| https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py |
+-------------------------------------------------------------------+
A = dir 110
B = dir 210
C = dir 330
P 280 = 0.2*dir(190)
A--B--C--cycle / 0.2 lightcyan / blue
X := circumcenter P B C
Y := circumcenter P C A
Z := circumcenter P A B
C_1 = 2*X-B
B_2 = 2*X-C
A_1 = 2*Y-C
C_2 = 2*Y-A
B_1 = 2*Z-A
A_2 = 2*Z-B
circumcircle A B P / 0.1 yellow / red
circumcircle B C P / 0.1 yellow / red
circumcircle C A P / 0.1 yellow / red
B_1--C_2 / deepgreen
C_1--A_2 / deepgreen
A_1--B_2 / deepgreen
C--C_1--B_2--B / red
A--A_1--C_2--C / red
B--B_1--A_2--A / red
*/
\end{asy}
\end{center}
Then $\dang C P B_2 = \dang C P A_1 = 90\dg$,
hence $P$ lies on $A_1 B_2$ etc., so we're done.

\begin{remark*}
  As one might guess from the two-sentence solution,
  the entire difficulty of the problem
  is getting the characterization of the concurrence point.
\end{remark*}

\end{document}
