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

\begin{document}
\title{OMMC 2023/9}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 131}
\maketitle

\section*{Problem}
Let $ABC$ be a triangle with incircle $\omega$.
Let $\omega_1$, $\omega_2$, and $\omega_3$ be three circles
centered at $A$, $B$, and $C$ respectively tangent to $\omega$
at points $D$, $E$, and $F$ respectively.
Show there exists a circle $\Gamma \neq \omega$ tangent to circles
$\omega_1$, $\omega_2$, and $\omega_3$ centered on the Euler line of $\triangle DEF$.

\section*{Video}
\href{https://www.youtube.com/watch?v=tGv-hnYaK0s&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/tGv-hnYaK0s}}

\section*{External Link}
\url{https://aops.com/community/p27839254}

\newpage

\section*{Solution}
Let $\triangle XYZ$ be the intouch triangle of $ABC$.
Let $\triangle P_1 P_2 P_3$ be so that $\triangle DEF$ is its intouch triangle.
Let $K$ be the radical center of $\omega_1$, $\omega_2$, $\omega_3$.

\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 = (-4.29366,5.57723);
pair B = (-5.,-1.);
pair C = (2.5,-1.);
pair I = (-2.67041,1.09279);
pair D = (-3.38272,3.06064);
pair E = (-4.22725,-0.30579);
pair F = (-0.73050,0.30759);
pair X = (-2.67041,-1.);
pair Y = (-1.21473,2.59638);
pair Z = (-4.75125,1.31626);
pair H = (-2.99964,0.87684);
pair P_1 = (-1.98779,-2.79863);
pair P_2 = (1.03041,4.65808);
pair P_3 = (-6.30217,2.00388);
pair K = (-1.98779,1.54055);

import graph;
size(15cm);
pen zzttqq = rgb(0.6,0.2,0.);
pen ffqqff = rgb(1.,0.,1.);
pen qqwuqq = rgb(0.,0.39215,0.);
pen wqwqwq = rgb(0.37647,0.37647,0.37647);
pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw(A--B--C--cycle, linewidth(0.6) + zzttqq);
draw(D--E--F--cycle, linewidth(0.6) + qqwuqq);
draw(P_1--P_2--P_3--cycle, linewidth(0.6) + wqwqwq);

draw(A--B, linewidth(0.6) + zzttqq);
draw(B--C, linewidth(0.6) + zzttqq);
draw(C--A, linewidth(0.6) + zzttqq);
draw(circle(I, 2.09279), linewidth(0.6) + blue);
draw(A--I, linewidth(0.6) + gray);
draw(B--I, linewidth(0.6) + gray);
draw(C--I, linewidth(0.6) + gray);
draw(circle(A, 2.67638), linewidth(0.6) + ffqqff);
draw(circle(B, 1.03877), linewidth(0.6) + ffqqff);
draw(circle(C, 3.48510), linewidth(0.6) + ffqqff);
draw(D--E, linewidth(0.6) + qqwuqq);
draw(E--F, linewidth(0.6) + qqwuqq);
draw(F--D, linewidth(0.6) + qqwuqq);
draw(D--X, linewidth(0.6) + qqwuqq);
draw(E--Y, linewidth(0.6) + qqwuqq);
draw(F--Z, linewidth(0.6) + qqwuqq);
draw(K--P_1, linewidth(0.6));
draw(K--P_2, linewidth(0.6));
draw(K--P_3, linewidth(0.6));
draw(circle(K, 4.33919), linewidth(0.6));
draw(P_1--P_2, linewidth(0.6) + wqwqwq);
draw(P_2--P_3, linewidth(0.6) + wqwqwq);
draw(P_3--P_1, linewidth(0.6) + wqwqwq);

dot("$A$", A, dir(A-I));
dot("$B$", B, dir(B-I));
dot("$C$", C, dir(C-I));
dot("$I$", I, dir((14.754, 3.864)));
dot("$D$", D, dir((4.039, 7.705)));
dot("$E$", E, dir((-31.130, -4.156)));
dot("$F$", F, dir((-1.217, -34.411)));
dot("$X$", X, dir((-5.026, -30.809)));
dot("$Y$", Y, dir((-6.483, 20.223)));
dot("$Z$", Z, dir((-24.884, 4.123)));
dot("$H$", H, dir((-36.153, 16.040)));
dot("$P_1$", P_1, dir((-13.006, -22.373)));
dot("$P_2$", P_2, dir((3.537, 7.143)));
dot("$P_3$", P_3, dir((-35.567, 9.772)));
dot("$K$", K, dir((-27.135, -27.725)));
\end{asy}
\end{center}

\begin{claim*}
  $\ol{KP_1}$ is perpendicular to $\ol{BXC}$, etc.
\end{claim*}
\begin{proof}
  $P_1$ lies on the radical axis of $\omega_2$ and $\omega_3$,
  so $\ol{P_1K}$ is their radical axis and perpendicular to the line
  through the centers $B$ and $C$.
\end{proof}

\begin{claim*}
  $K$ is actually the circumcenter of $\triangle P_1 P_2 P_3$.
\end{claim*}
\begin{proof}
  Note that $\ol{BC}$ and $\ol{P_2P_3}$ are antiparallel
  with respect to $\angle P_1 P_2 P_3$, from $\ol{DX} \perp \ol{EF}$.
\end{proof}

Consequently, $K$ lies on the Euler line of $\triangle DEF$,
because the inverse of $(P_1P_2P_3)$ with respect to $(DEF)$
is exactly the nine-point circle of $\triangle DEF$.

Finally, it's known that if $\Gamma \neq \omega$ is
tangent to all three $\omega_i$, then the line through
the centers of $\Gamma$ and $\omega$ passes through the radical center
(in fact the radical center is either the insimilicenter or exsimilicenter).
So that solves the problem.

\end{document}
