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

\begin{document}
\title{JMO 2026/6}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 176}
\maketitle

\section*{Problem}
Emily has a red sheet of paper.
She draws $2026$ circles (not necessarily of equal size) on the piece of paper.
She chooses a circle to color black, then cuts the paper around the circumference of all $2026$ circles.
She then separates the pieces of paper, into at least $2$ black pieces and some number of red pieces.
Is it possible that all black pieces are congruent?

\section*{Video}
\href{https://youtu.be/LZ4J5GQwrgA}{\texttt{https://youtu.be/LZ4J5GQwrgA}}

\section*{External Link}
\url{https://aops.com/community/p37586342}

\newpage

\section*{Solution}
No, this is impossible.

Let $\Gamma$ be the black circle,
and for each point inside it consider the number of the $2026$ circles it's contained in.
Let $X$ be a point for which that number achieves its maximal value,
and consider the piece $\mathcal{P}$ containing $X$.

\begin{center}
\begin{asy}
  usepackage("amssymb");
  filldraw(unitcircle, lightgray, black+1.4);
  path g1 = circle((-0.7,-0.8), 1.4);
  path g2 = circle(( 0.5, 0.3), 0.9);
  path g3 = circle(( 0.1,-0.6), 1.1);
  draw(g1, lightblue);
  draw(g2, lightblue);
  draw(g3, lightblue);
  path piece = buildcycle(g3,g2,g1);
  filldraw(piece, paleyellow, blue+1);
  label("$\mathcal{P}$", (0.15,-0.05), blue);
\end{asy}
\end{center}


\begin{claim*}
  All the curves that bound $\mathcal{P}$ point outwards.
\end{claim*}
\begin{proof}
  If not, then one could replace $X$ with a point $Y$
  just across an inwards-pacing curve bounding $\mathcal{P}$.
\end{proof}

However, it's clear that if there are more than two black pieces,
then some piece has a curve that points inwards.
So it's impossible that the pieces are all congruent.

\end{document}