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

\begin{document}
\title{Florida 2023B}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 124}
\maketitle

\section*{Problem}
Given a fixed acute triangle, a variable point $P$ lies on arc $BC$
not containing $A$ of the circumcircle of $\triangle ABC$.
Let $D$ and $E$ be the incenters of $ABP$ and $ACP$, respectively.
As $P$ varies on arc $BC$, show that the midpoint of $\ol{DE}$
lies on a fixed circle.

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



\newpage

\section*{Solution}
Let $I$ and $J$ denote the incenters of triangles $ABC$ and $PBC$.
\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.,5.);
pair B = (1.,-3.);
pair C = (11.,-3.);
pair I = (4.95692,-0.25784);
pair Z = (0.68492,1.68065);
pair Y = (10.27200,3.42550);
pair P = (3.60948,-5.46110);
pair D = (2.46271,-2.66069);
pair E = (6.39294,-1.74846);
pair M = (4.42782,-2.20457);
pair J = (3.89872,-4.15131);

import graph;
size(8cm);
pen zzttqq = rgb(0.6,0.2,0.);
pen yqqqqq = rgb(0.50196,0.,0.);
pen ffdxqq = rgb(1.,0.84313,0.);
pen qqwuqq = rgb(0.,0.39215,0.);
pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw(A--B--C--cycle, linewidth(0.6) + zzttqq);
draw(I--D--J--E--cycle, linewidth(0.6) + qqwuqq);

draw(A--B, linewidth(0.6) + zzttqq);
draw(B--C, linewidth(0.6) + zzttqq);
draw(C--A, linewidth(0.6) + zzttqq);
draw(circle((6.,-0.3125), 5.67650), linewidth(0.6));
draw(B--P, linewidth(0.6) + yqqqqq);
draw(C--P, linewidth(0.6) + yqqqqq);
draw(D--E, linewidth(0.6) + blue);
draw(circle((5.47846,2.55307), 4.87228), linewidth(0.6) + ffdxqq);
draw(I--D, linewidth(0.6) + qqwuqq);
draw(D--J, linewidth(0.6) + qqwuqq);
draw(J--E, linewidth(0.6) + qqwuqq);
draw(E--I, linewidth(0.6) + qqwuqq);

dot("$A$", A, dir((8.000, 20.000)));
dot("$B$", B, dir((-60.000, -30.000)));
dot("$C$", C, dir((30.000, -34.000)));
dot("$I$", I, dir((8.308, 15.784)));
dot("$Z$", Z, dir((-64.492, -12.065)));
dot("$Y$", Y, dir((8.800, 15.450)));
dot("$P$", P, dir((-12.948, -65.890)));
dot("$D$", D, dir((-46.271, 0.069)));
dot("$E$", E, dir((8.706, 16.846)));
dot("$M$", M, dir((7.218, 16.457)));
dot("$J$", J, dir((22.128, -40.869)));
\end{asy}
\end{center}

We appeal to the following result,
available
\href{https://www.cut-the-knot.org/Curriculum/Geometry/CyclicQuadrilateral.shtml#explanation}{here}
or
\href{https://blog.evanchen.cc/2018/02/28/revisiting-arc-midpoints-in-complex-numbers/}{here}:
\begin{theorem*}
  [Japanese theorem for cyclic quadrilaterals]
  $DIEJ$ is a rectangle.
\end{theorem*}
Now $I$ is fixed, and $J$ moves on a fixed circle
(because $\angle BJC = 90\dg + \half \angle BPC$ is fixed).
So the midpoint of $\ol{IJ}$ moves along on a circle, as needed.

\end{document}
