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

\begin{document}
\title{Taiwan TST 2019/2J/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 13}
\maketitle

\section*{Problem}
Let $ABC$ be a triangle with incircle $\omega$ and circumcircle $\Omega$.
Assume that $\omega$ is tangent to sides $AB$, $AC$ at $F$, $E$, respectively.
Line $EF$ meets $\Omega$ at points $P$ and $Q$.
Let $M$ be the midpoint of $\ol{BC}$.

Take a point $R$ on the circumcircle of $\triangle MPQ$, say $\Gamma$,
such that $\ol{MR} \perp \ol{EF}$.
Prove that line $AR$, $\omega$ and $\Gamma$ intersect at one point.

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

\section*{External Link}
\url{https://aops.com/community/p14533530}

\newpage

\section*{Solution}
Let $D$ be the foot from $I$ to $\ol{BC}$.
\begin{claim*}
  Quadrilateral $MDPQ$ is cyclic.
\end{claim*}
\begin{proof}
  Let $X = PQ \cap BC$.
  Then $(XD;BC) = -1$ by Ceva/Menelaus with the Gergonne point.
  Hence, $XD \cdot XM = XB \cdot XC = XP \cdot XQ$ as needed.
\end{proof}

We let $G$ be the intersection of the $D$-altitude
to $\ol{EF}$ with the incircle.
Let line $AG$ meet the incircle again at $H$.

\begin{center}
\begin{asy}
size(12cm);
pair A = dir(130);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A, B, C);
pair D = foot(I, B, C);
pair E = foot(I, C, A);
pair F = foot(I, A, B);
pair M = midpoint(B--C);
pair Z = orthocenter(D, E, F);
pair K = foot(D, E, F);
pair N = midpoint(E--F);
pair G = -Z+2*K;
pair H = -G+2*foot(I, A, G);

filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, blue);
filldraw(incircle(A, B, C), opacity(0.1)+lightcyan, blue);

pair X = extension(E, F, B, C);
pair S = foot(A, I, -A);

pair Y = extension(H, D, E, F);
filldraw(unitcircle, opacity(0.1)+palecyan, blue);
pair P = IP(circumcircle(H, D, M), unitcircle);
pair Q = OP(circumcircle(H, D, M), unitcircle);
filldraw(circumcircle(H, D, M), opacity(0.1)+yellow, orange);
filldraw(circumcircle(A, E, F), opacity(0.1)+lightgreen, deepgreen);
draw(E--X--P, blue);
draw(X--B, blue);
pair Rp = extension(M, foot(M, P, Q), A, H);
draw(H--Rp--M, red);
draw(S--I--A, deepgreen);
draw(A--Y--D, lightred);
draw(D--G, lightblue);

dot("$A$", A, dir(A));
dot("$B$", B, dir(240));
dot("$C$", C, dir(300));
dot("$I$", I, dir(I));
dot("$D$", D, dir(D));
dot("$E$", E, dir(350));
dot("$F$", F, dir(F));
dot("$M$", M, dir(M));
dot("$K$", K, dir(K));
dot("$N$", N, dir(N));
dot("$G$", G, dir(G));
dot("$H$", H, dir(H));
dot("$X$", X, dir(X));
dot("$S$", S, dir(S));
dot("$Y$", Y, dir(140));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(160));
dot("$R'$", Rp, dir(Rp));

/* TSQ Source:

!size(12cm);
A = dir 130
B = dir 210 R240
C = dir 330 R300
I = incenter A B C
D = foot I B C
E = foot I C A R350
F = foot I A B
M = midpoint B--C
Z := orthocenter D E F
K = foot D E F
N = midpoint E--F
G = -Z+2*K
H = -G+2*foot I A G

A--B--C--cycle 0.1 lightcyan / blue
incircle A B C 0.1 lightcyan / blue

X = extension E F B C
S = foot A I -A

Y = extension H D E F R140
unitcircle 0.1 palecyan / blue
P = IP circumcircle H D M unitcircle
Q = OP circumcircle H D M unitcircle R160
circumcircle H D M 0.1 yellow / orange
circumcircle A E F 0.1 lightgreen / deepgreen
E--X--P blue
X--B blue
R' = extension M foot M P Q A H
H--Rp--M red
S--I--A deepgreen
A--Y--D lightred
D--G lightblue

*/
\end{asy}
\end{center}

\begin{claim*}
  Quadrilateral $HDPQ$ is cyclic.
\end{claim*}
\begin{proof}[Proof $HDPQ$ cyclic]
  For this proof, we need to introduce several points:
  \begin{itemize}
    \ii Let $K$ be the foot from $D$ to $\ol{EF}$.
    \ii Let $N$ be the midpoint of $\ol{EF}$.
    \ii Let $S$ be the inverse of $K$ with respect
    to the incircle,
    which is known to satisfy $\angle AIS = 90\dg$.
    \ii Let $Y = \ol{HD} \cap \ol{EF}$.
  \end{itemize}
  We have $-1 = (EF;GH) \overset{D}{=} (EF;KY)$.
  Since $\ol{SK}$ bisects $\angle FSE$,
  this implies $Y$ also lies on line $AS$.
  We can then calculate
  \[ YH \cdot YD = YF \cdot YE = YK \cdot YN
    = YS \cdot YA = YP \cdot YQ \]
  which implies the concyclic condition.
\end{proof}

We let $R'$ be a point on line $\ol{AGH}$ such that
$\ol{R'M} \perp \ol{PQ}$. Then
\[ \dang HR'M = \dang HGD = \dang HDB = \dang HDM \]
so $R' = R$ and the problem is solved
(the concurrence point in the problem is $H$).

\end{document}
