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

\begin{document}
\title{Twitch 033.1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 33}
\maketitle

\section*{Problem}
Let $H$ be the orthocenter of acute triangle $\triangle ABC$.
Two points $X$ and $Y$ are on the circumcircle of triangle $\triangle ABC$
such that $H$ lies on chord $XY$.
Let $P$ and $Q$ be the feet of the altitudes from $H$ onto $AX$ and $AY$, respectively,
and let line $PQ$ intersect line $XY$ at $T$.
Let $E$ and $F$ are the feet of the altitudes from $B$ and $C$ to $AC$ and $AB$.

Prove that as the chord $XY$ containing $H$ varies,
point $T$ traces out part of a circle whose center lies on line $EF$.

\section*{Video}
\href{https://www.youtube.com/watch?v=zMN8WeefUuI&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/zMN8WeefUuI}}



\newpage

\section*{Solution}
We introduce $G = (AEF) \cap (ABC) \neq A$.
\begin{claim*}
  The point $G$ is the Miquel point of complete quadrilateral $XPQY$.
\end{claim*}
\begin{proof}
  Clear.
\end{proof}

\begin{center}
\begin{asy}
import graph; size(14cm);
pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215,0.); pen ffdxqq = rgb(1.,0.84313,0.); pen qqffff = rgb(0.,1.,1.); pen yqqqqq = rgb(0.50196,0.,0.); pen cqcqcq = rgb(0.75294,0.75294,0.75294);
pair A = (10.,5.), B = (6.,-10.), C = (24.,-10.), H = (10.,-6.26666), X = (4.76880,-7.20517), Y = (25.57984,-3.47153), P = (5.92047,-4.51816), Q = (14.72826,2.42900), T = (1.84988,-7.72885), F = (7.19502,-5.51867), G = (4.59879,-2.23376);
draw(A--B--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((15.,-4.36666), 10.61764), linewidth(0.6) + red);
draw(X--Y, linewidth(0.6) + blue);
draw(circle((10.,-0.63333), 5.63333), linewidth(0.6) + blue);
draw(A--X, linewidth(0.6) + blue);
draw(A--Y, linewidth(0.6) + blue);
draw(B--(15.61995,-1.02137), linewidth(0.6) + zzttqq);
draw(C--F, linewidth(0.6) + zzttqq);
draw(T--Q, linewidth(0.6) + qqwuqq);
draw(X--T, linewidth(0.6) + qqwuqq);
draw(circle((5.79377,-6.26666), 4.20622), linewidth(0.6) + ffdxqq);
draw(circle((2.82780,-4.78294), 3.10398), linewidth(0.6));
draw(circle((14.62721,-10.68557), 13.11497), linewidth(0.6));
draw(A--H, linewidth(0.6) + yqqqqq);
dot("$A$", A, dir((8.508, 20.120)));
dot("$B$", B, dir((8.982, 19.861)));
dot("$C$", C, dir((8.886, 19.861)));
dot("$H$", H, dir((8.508, 15.477)));
dot("$X$", X, dir((7.760, 19.638)));
dot("$Y$", Y, dir((7.859, 21.339)));
dot("$P$", P, dir((-44.217, -14.647)));
dot("$Q$", Q, dir((8.590, 16.305)));
dot("$T$", T, dir((8.161, 16.969)));
dot("$E$", (15.61995,-1.02137), dir((9.110, 16.854)));
dot("$F$", F, dir((7.707, 16.098)));
dot("$G$", G, dir((8.454, 15.789)));
// dot("$S$", (5.79377,-6.26666), dir((7.183, 15.477)));
\end{asy}
\end{center}

By this point, we already have
\[ \dang GTH = \dang GQY = \dang GQY = \dang GHA \]
which is fixed as the chord varies.
This already shows $T$ lies on a fixed circle
and we need only to show its center lies on line $EF$.

In fact, the conditions imply that $(GHT)$ is orthogonal to $(AEF)$.
Now we prove that:
\begin{claim*}
  $(GH;FE) = -1$.
\end{claim*}
\begin{proof}
  Since $GF / GE = BH / HC = HF / HE$ by spiral similarity.
  (Alternatively, note that $GH$, $FF$, $EE$ bisect $BC$.)
\end{proof}
The harmonic quadrilateral, together with the orthogonality,
imply that $(GHT)$ is exactly the Apollonian circle through $G$ and $H$
with respect to $E$ and $F$, so its center lies on line $EF$.

\end{document}
