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

\begin{document}
\title{JMO 2026/3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 178}
\maketitle

\section*{Problem}
Let $ABC$ be an acute scalene triangle with no angle equal to $60^{\circ}$.
Let $\omega$ be the circumcircle of $ABC$.
Let $\Delta_B$ be the equilateral triangle with three vertices on $\omega$,
one of which is $B$.
Let $\ell_B$ be the line through the two vertices of $\Delta_B$ other than $B$.
Let $\Delta_C$ and $\ell_C$ be defined analogously.
Let $Y$ be the intersection of $AC$ and $\ell_B$,
and let $Z$ be the intersection of $AB$ and $\ell_C$.

Suppose that the circumcircle of $AYZ$ intersects $\omega$ at $P\neq A$,
$BC$ intersects $YZ$ at $D$, and $PA$ intersects $YZ$ at $E$.
Prove that $PE=PD$.

\section*{Video}
\href{https://youtu.be/fGN283EHQaU}{\texttt{https://youtu.be/fGN283EHQaU}}

\section*{External Link}
\url{https://aops.com/community/p37578105}

\newpage

\section*{Solution}
Because $P$ is the Miquel point of $BZYC$, it follows $DBZP$ is cyclic.
Hence
\begin{align*}
  \dang PDE &= \dang PDZ = \dang PBZ = \dang PBA \\
  \dang DEP &= \dang (\ol{YZ}, \ol{AP}).
\end{align*}
So the problem is solved if we can show
\[ \dang PBA = \dang (\ol{YZ}, \ol{AP}). \qquad(\bigstar) \]

\begin{center}
\begin{asy}
size(11cm);
pair A = dir(95);
pair B = dir(225);
pair C = dir(315);

pair w = dir(120);
pair Y = extension(A, C, B*w, B/w);
pair Z = extension(A, B, C*w, C/w);
pair p_formula = (B*Y-C*Z)/(B+Y-C-Z);
pair P = p_formula;
pair D = extension(B, C, Y, Z);
pair E = extension(P, A, Y, Z);

filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(A--B--C--cycle, blue);
draw(D--E--P, deepgreen);
draw(B--D, blue);
filldraw(circumcircle(A, Y, Z), opacity(0.1)+yellow, blue);
draw(circumcircle(B, Z, D), gray);

dot("$A$", A, dir(A));
dot("$B$", B, dir(270));
dot("$C$", C, dir(C));
dot("$Y$", Y, dir(270));
dot("$Z$", Z, dir(270));
dot("$P$", P, dir(P));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));

/* --------------------------------+
| TSQX: by Evan Chen and CJ Quines |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(11cm);
A = dir 95
B 270 = dir 225
C = dir 315

w := dir 120
Y 270 = extension A C B*w B/w
Z 270 = extension A B C*w C/w
!pair p_formula = (B*Y-C*Z)/(B+Y-C-Z);
P = p_formula
D = extension B C Y Z
E = extension P A Y Z

unitcircle / 0.1 lightcyan / blue
A--B--C--cycle / blue
D--E--P / deepgreen
B--D / blue
circumcircle A Y Z / 0.1 yellow / blue
circumcircle B Z D / gray
*/
\end{asy}
\end{center}


We use complex numbers to prove $(\bigstar)$.
Let $\omega = e^{2\pi i/3}$.
Since $Y$ is the intersection of $a$, $c$, $\omega b$ and $\omega^2 b$, we have
\begin{align*}
  y &= \frac{b^2(a+c) - ac(\omega b + \omega^2 b)}{b^2-ac}
  = \frac{b \cdot (ab+bc+ca)}{b^2-ac}. \\
  z &= \frac{c \cdot (ab+bc+ca)}{c^2-ab}. \\
\end{align*}
We compute the point $P$ now.
\begin{claim*}
  We have \[ p = \frac{ab+bc+ca}{a+b+c}. \]
\end{claim*}
\begin{proof}
  Note that
  \[ \frac{p-z}{p-y} = \frac{p-b}{p-c} \iff p = \frac{by-cz}{b+y-c-z}. \]
  The numerator and denominator are respectively
  \begin{align*}
    by-cz &= (ab+bc+ca) \cdot \left( \frac{b^2}{b^2-ac} - \frac{c^2}{c^2-ab} \right) \\
    &= (ab+bc+ca) \cdot \frac{a(c^3-b^3)}{(b^2-ac)(c^2-ab)} \\
    &= -(ab+bc+ca) \cdot \frac{a(b-c)(b^2+bc+c^2)}{(b^2-ac)(c^2-ab)} \\
    b+y-c-z &= \frac{(b-c)[(b^2-ac)(c^2-ab)-(ab+bc+ca)^2]}{(b^2-ac)(c^2-ab)} \\
    &= \frac{(b-c)[
      -ab^3-ac^3+a^2bc
      -a^2b^2-c^2a^2-2abc(a+b+c)
    ]}{(b^2-ac)(c^2-ab)} \\
    &= \frac{-a(b-c)[a(b^2+bc+c^2) + b^3+c^3+2bc(b+c)]}{(b^2-ac)(c^2-ab)} \\
    &= \frac{-a(b-c)(a+b+c)(b^2+bc+c^2)}{(b^2-ac)(c^2-ab)}.
  \end{align*}
  Dividing gives the conclusion.
\end{proof}
The desired conclusion $(\bigstar)$ is encoded as
\[
  \RR\ni  \frac{p}{p+a} \div \frac{y-z}{p-a}
  = \frac{p-a}{p+a} \cdot \frac{p}{y-z}
  \iff \frac{p}{y-z} \in i \RR
\]
since $\frac{p-a}{p+a}$ is obviously pure imaginary
(for $|a|=|p|=1$).
We compute
\begin{align*}
  y-z
  &= (ab+bc+ca) \cdot \frac{b(c^2-ab)-c(b^2-ac)}{(b^2-ac)(c^2-ab)} \\
  &= (ab+bc+ca) \cdot \frac{-(b-c)(bc+ab+ac)}{(b^2-ac)(c^2-ab)} \\
  &= -(ab+bc+ca)^2 \cdot \frac{b-c}{(b^2-ac)(c^2-ab)} \\
\end{align*}
Hence
\[ \frac{p}{y-z} = \frac{(b^2-ac)(c^2-ab)}{(b-c)(a+b+c)(ab+bc+ca)}. \]
The conjugate is
\[ \ol{\left(\frac{p}{y-z}\right)}
  = \frac{\left( \frac{1}{b^2}-\frac{1}{ac} \right) \left( \frac{1}{c^2}-\frac{1}{ab} \right)}
    {\left( \frac 1b - \frac 1c \right) \cdot \frac{ab+bc+ca}{abc} \cdot \frac{a+b+c}{abc}}
  = - \frac{p}{y-z} \]
as desired.

\end{document}