\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{IMO 1979/3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 21}
\maketitle
\section*{Problem}
Two circles in a plane intersect and $A$ is one of the points of intersection.
Starting simultaneously from $A$ two points move with constant speed,
each travelling along its own circle in the same direction.
The two points return to $A$ simultaneously after one revolution.
Prove that there is a fixed point $P$ in the plane such that the two points
are always equidistant from $P$.
\section*{Video}
\href{https://www.youtube.com/watch?v=cavQ_A0QS8o&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/cavQ\_A0QS8o}}
\newpage
\section*{Solution}
Let $B$ and $C$ be the antipodes of $A$ on the two circles
and let $D$ be the foot of the altitude from $A$,
which is the other intersection point of the two circles.
Also, let $M$ be the midpoint of $\ol{BC}$,
and construct rectangle $ADMZ$.
Our claim is that $Z$ is the fixed point.
We let $X$ and $Y$ be the two points;
by the condition the angles are the same.
So we have a spiral similarity \[ \triangle AXY \sim \triangle ABC. \]
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair D = foot(A, B, C);
pair M = midpoint(B--C);
pair Z = A+M-D;
pair X = midpoint(A--B)+dir(200)*abs(A-B)/2;
pair Y = -D+2*foot(midpoint(A--C), X, D);
draw(circumcircle(A, B, D), blue);
draw(circumcircle(A, C, D), blue);
draw(circumcircle(A, D, M), dashed+deepgreen);
pair N = midpoint(X--Y);
filldraw(A--X--Y--cycle, opacity(0.1)+red, purple);
filldraw(A--B--C--cycle, opacity(0.1)+yellow, blue);
draw(A--D, blue);
draw(A--Z--M, deepgreen);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$M$", M, dir(M));
dot("$Z$", Z, dir(Z));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$N$", N, dir(N));
/* TSQ Source:
A = dir 110
B = dir 210
C = dir 330
D = foot A B C
M = midpoint B--C
Z = A+M-D
X = midpoint(A--B)+dir(200)*abs(A-B)/2
Y = -D+2*foot midpoint A--C X D
circumcircle A B D blue
circumcircle A C D blue
circumcircle A D M dashed deepgreen
N = midpoint X--Y
A--X--Y--cycle 0.1 red / purple
A--B--C--cycle 0.1 yellow / blue
A--D blue
A--Z--M deepgreen
*/
\end{asy}
\end{center}
Now let $N$ be the midpoint of $\ol{XY}$.
By spiral similarity, since $N$ maps to $M$,
it follows $M$, $N$, $A$, and $D$ are cyclic too.
So actually $N$ lies on the circumcircle of rectangle $ADMZ$,
meaning $\ol{ZN} \perp \ol{XY}$,
hence $ZX = ZY$ as needed.
\begin{remark*}
The special point $Z$ can be identified
by selecting the special case $X \to A$, $Y \to A$
and $X=B$, $Y=C$.
\end{remark*}
\end{document}