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

\begin{document}
\title{USAMTS 4/3/34}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 111}
\maketitle

\section*{Problem}
Let $\omega$ be a circle with center $O$ and radius $10$,
and let $H$ be a point such that $OH = 6$.
A point $P$ is called \emph{snug} if, for all triangles $ABC$
with circumcircle $\omega$ and orthocenter $H$,
we have that $P$ lies on $\triangle ABC$ or in the interior of $\triangle ABC$.
Find the area of the region consisting of all snug points.

\section*{Video}
\href{https://www.youtube.com/watch?v=V58ZweJKKGA&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/V58ZweJKKGA}}

\section*{External Link}
\url{https://aops.com/community/p26832931}

\newpage

\section*{Solution}
We claim that the locus is the ellipse $\mathcal E$
whose foci are at $O$ and $H$
and which contains every point $S$ satisfying $OS+SH \le 10$
(i.e.\ has major axis of length $10$).

\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 O = (0.,0.);
pair H = (6.,0.);
pair N = (3.,0.);
pair D = (4.15930,-4.86374);
pair A = (8.17800,5.75501);
pair X = (2.31860,-9.72749);
pair B = (-9.98797,0.49033);
pair C = (7.80996,-6.24535);
pair P = (0.86184,-3.61581);

size(12cm);
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(circle(O, 10.), linewidth(0.6));
draw(A--B, linewidth(0.6) + zzttqq);
draw(B--C, linewidth(0.6) + zzttqq);
draw(C--A, linewidth(0.6) + zzttqq);
draw(A--X, linewidth(0.6)); draw(shift(N)*rotate(0.)*xscale(5.)*yscale(4.)*unitcircle, linewidth(0.6) + green);
draw(O--X, linewidth(0.6));

dot("$O$", O, dir((11.100, 25.903)));
dot("$H$", H, dir((36.537, -63.800)));
dot("$N$", N, dir((10.114, 20.920)));
dot("$D$", D, dir(290));
dot("$A$", A, dir(A));
dot("$X$", X, dir(X));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir((-40.199, -80.970)));
\end{asy}
\end{center}

\begin{claim*}
  $\mathcal E$ is always an in-ellipse of $\triangle ABC$,
  and hence every point on or inside $\mathcal E$ is snug.
\end{claim*}
\begin{proof}
  Since $H$ is inside $\omega$, it follows $ABC$ is acute.
  Now, let $X$ be the reflection of $H$ over $BC$,
  and let $P = \ol{BC} \cap \ol{OX}$.
  Then
  \[ OP + PH = OP + PX = OX = 10 \]
  so $P$ lies on $\mathcal E$.
  Moreover, because of the ``river reflection'' problem
  (or just noting that $\dang OPB = \dang CPH$),
  it follows that line $BC$ is actually \emph{tangent} to $\mathcal E$ at $P$.

  Similarly, $\mathcal E$ is tangent to lines $CA$ and $AB$.
  So $\mathcal E$ is an in-ellipse of $\triangle ABC$.
\end{proof}

Conversely, suppose $S$ is a point not inside the ellipse.
Then line segment $NS$ intersects the ellipse at some point $P$.
For continuity reasons,
one can choose $ABC$ such that $P$ is the tangency point of $BC$.
Then line $BC$ separates $N$ and $P$, and $N$ is obviously inside $ABC$,
so $P$ is outside $ABC$ and hence not snug.

\end{document}
