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

\begin{document}
\title{Taiwan MO 2024/4}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 145}
\maketitle

\section*{Problem}
Suppose $O$ is the circumcenter of $\triangle ABC$,
and $E$ and $F$ are points on segments $CA$ and $AB$ respectively with $E, F \neq A$.
Let $P$ be a point such that $PB = PF$ and $PC = PE$.
Let $OP$ intersect $CA$ and $AB$ at points $Q$ and $R$ respectively.
Let the line passing through $P$ and perpendicular to $EF$ intersect $CA$ and $AB$
at points $S$ and $T$ respectively.
Prove that points $Q$, $R$, $S$, and $T$ are concyclic.

\section*{Video}
\href{https://www.youtube.com/watch?v=eiHul2kJINE&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/eiHul2kJINE}}

\section*{External Link}
\url{https://aops.com/community/p29778110}

\newpage

\section*{Solution}
Note that
\begin{align*}
  \dang QST &= \dang(\ol{PST}, \ol{AC}) = 90\dg + \dang(\ol{EF}, \ol{AC}) \\
  \dang QRT &= \dang(\ol{AB}, \ol{OP}).
\end{align*}
So we can erase all the points $Q$, $R$, $S$, $T$ from the picture
and focus just on proving the claim that:
\begin{claim*}
  $\dang(\ol{AB}, \ol{OP}) = \dang(\ol{EF}, \ol{AC}) + 90\dg$.
\end{claim*}

\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 = (-2.,2.);
pair B = (-4.,-2.5);
pair C = (2.5,-2.5);
pair F = (-3.45808,-1.28069);
pair E = (0.33045,-0.33045);
pair P = (-0.49655,-3.32700);
pair O = (-0.75,-1.25);
pair Q = (-1.02796,1.02796);
pair R = (-1.33041,3.50656);
pair S = (-1.77664,1.77664);
pair T = (-1.89303,2.24067);
pair M = (1.41522,-1.41522);
pair N = (-3.72904,-1.89034);

import graph;
size(9.14447cm);
pen zzttqq = rgb(0.6,0.2,0.);
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(O, 3.48209), linewidth(0.6));
draw(P--R, linewidth(0.6));
draw(A--R, linewidth(0.6));
draw(P--T, linewidth(0.6));
draw(E--F, linewidth(0.6));
draw(P--M, linewidth(0.6));
draw(P--N, linewidth(0.6));

dot("$A$", A, dir((-25.918, 4.569)));
dot("$B$", B, dir((-27.115, -23.952)));
dot("$C$", C, dir((4.415, -24.858)));
dot("$F$", F, dir((-22.398, 17.250)));
dot("$E$", E, dir((3.860, 9.229)));
dot("$P$", P, dir((-9.506, -31.881)));
dot("$O$", O, dir((4.057, 6.930)));
dot("$Q$", Q, dir((3.758, 7.519)));
dot("$R$", R, dir((3.189, 7.076)));
dot("$S$", S, dir((-16.534, -26.566)));
dot("$T$", T, dir((-7.614, 24.004)));
dot("$M$", M, dir((3.232, 7.139)));
dot("$N$", N, dir((-15.240, 19.306)));
\end{asy}
\end{center}

\begin{proof}
  Use complex numbers with $ABC$ the unit circle.
  Let $P$ be any point, with coordinate $p$.
  Then the foot from $P$ to $\ol{AC}$ has coordinates
  \[ M \coloneqq \half \left( p + a + c - ac \ol p \right) \]
  and so (since $M$ is the midpoint of $\ol{EC}$) we get
  \[ E = 2 \cdot M - c = p + a - ac \ol p. \]
  Analogously
  \[ F = p + a - ab \ol p. \]
  So then we have the ratio
  \[ z \coloneqq \frac{E-F}{a-c} \div \frac{a-b}{p-0}
    = \frac{a\ol p(b-c) \cdot p}{(a-b)(a-c)}
    = p \bar p \cdot \frac{a(b-c)}{(a-b)(a-c)}. \]
  The complex conjugate is
  \[ \bar p p \cdot \frac{\frac 1a\left( \frac 1b- \frac 1c \right)}
  {\left( \frac 1a - \frac 1b \right)\left( \frac 1a - \frac 1c \right)} = -z. \]
  so $z$ is pure imaginary and we're done.
\end{proof}

\end{document}
