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

\begin{document}
\title{ELMO Revenge 2022/1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 112}
\maketitle

\section*{Problem}
Let $ABC$ and $DBC$ be triangles with incircles touching at a point $P$ on
$BC$. Points $A$, $D$ lie on the same side of $BC$ and $DB < AB < DC < AC$.
The bisector of $\angle BDC$ meets line $AP$ at $X$,
and the altitude from $A$ meets $DP$ at $Y$.
Point $Z$ lies on line $XY$ so $ZP \perp BC$.
Show the reflection of $A$ over $BC$ is on line $ZD$.

\section*{Video}
\href{https://www.youtube.com/watch?v=QVD4kFNyCHQ&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/QVD4kFNyCHQ}}

\section*{External Link}
\url{https://aops.com/community/c6h2881590p25619349}

\newpage

\section*{Solution}
\paragraph{Intended solution.}
Let $A'$ be the reflection of $A$ over $BC$.
Then the points $A$, $D$, $P$, $A'$
lie on a hyperbola with foci at $B$ and $C$ by construction.
Then apply Pascal to $DDPPAA'$.

\paragraph{Synthetic solution.}
Let $E$ be the reflection of $A$ across $\ol{BC}$.
Let $I$ and $J$ denote the incenters of $\triangle EBC$ and $\triangle DBC$.
Redefine $Z = \ol{DE} \cap \ol{IPJ}$;
then the problem is to show $X$, $Y$, $Z$ are collinear.

\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 = (10.00216,0.58182);
pair B = (9.,-3.);
pair C = (45.,-3.);
pair P = (11.26936,-3.);
pair J = (11.26936,-2.03424);
pair D = (10.88129,-1.04465);
pair Y = (10.00216,3.38498);
pair Z = (11.26936,1.39956);
pair K = (10.00216,-3.);
pair E = (10.00216,-6.58182);
pair I = (11.26936,-4.72156);
pair W = (10.78721,-1.63719);

size(16cm);
pen zzttqq = rgb(0.6,0.2,0.);
pen yqqqqq = rgb(0.50196,0.,0.);
pen qqffff = rgb(0.,1.,1.);
pen xfqqff = rgb(0.49803,0.,1.);
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(B--D--C--cycle, linewidth(0.6) + qqffff);
draw(B--E--C--cycle, linewidth(0.6) + zzttqq);

draw(A--B, linewidth(0.6) + zzttqq);
draw(B--C, linewidth(0.6) + zzttqq);
draw(C--A, linewidth(0.6) + zzttqq);
draw(circle((11.26936,-1.27843), 1.72156), linewidth(0.6) + yqqqqq);
draw(circle(J, 0.96575), linewidth(0.6) + blue);
draw(B--D, linewidth(0.6) + qqffff);
draw(D--C, linewidth(0.6) + qqffff);
draw(C--B, linewidth(0.6) + qqffff);
draw(P--Y, linewidth(0.6) + xfqqff);
draw(circle(I, 1.72156), linewidth(0.6) + yqqqqq);
draw(B--E, linewidth(0.6) + zzttqq);
draw(E--C, linewidth(0.6) + zzttqq);
draw(C--B, linewidth(0.6) + zzttqq);
draw(E--Z, linewidth(0.6) + qqwuqq);
draw(I--Z, linewidth(0.6) + yqqqqq);
draw(A--E, linewidth(0.6));
draw(A--P, linewidth(0.6));
draw(A--Y, linewidth(0.6));

dot("$A$", A, dir(160));
dot("$B$", B, dir(180));
dot("$C$", C, dir((-3794.550, 716.315)));
dot("$P$", P, dir((4.522, 8.332)));
dot("$J$", J, dir(0));
dot("$D$", D, dir(150));
dot("$Y$", Y, dir((4.454, 7.830)));
dot("$Z$", Z, dir((4.522, 8.583)));
dot("$E$", E, dir(270));
dot("$I$", I, dir(-45));
dot("$W$", W, dir(200));

clip(box( (8,-10), (15,10) ) );
\end{asy}
\end{center}

To interpret the condition about $P$:
\begin{claim*}
  $(IJ;PZ) = -1$.
\end{claim*}
\begin{proof}
  Because the condition on $P$ forces $BD+CE=BE+CD$,
  there is an incircle for $BDCE$. Hence by Monge theorem,
  $Z$ coincides with the exsimilicenter of $(I)$ and $(J)$.
  As $I$ is the insimilicenter, the conclusion follows.
\end{proof}

Next, let $W = \ol{ZDE} \cap \ol{AP}$.
\begin{claim*}
  Triangles $DJP$ and $AWE$ are perspective,
  i.e.\ the lines $\ol{DA}$, $\ol{JW}$, $\ol{PE}$ are concurrent.
\end{claim*}
\begin{proof}
  We have $-1 = (ZP;JI) \overset{W}{=} (E,A;\ol{JW}\cap\ol{AE},Y)$.
  Then it follows by looking at complete quadrilateral $ADPEWY$.
\end{proof}

Now by Desargue's theorem, it follows $\ol{DJ} \cap \ol{AW} = X$,
$\ol{DP} \cap \ol{AE} = Y$, $\ol{JP} \cap \ol{WE} = Z$ are collinear.

\end{document}
