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

\begin{document}
\title{Japan 2019/4}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 12}
\maketitle

\section*{Problem}
Let $ABC$ be a triangle with its incenter $I$,
incircle $\omega$, and let $M$ be a midpoint of the $\ol{BC}$.
The line through $A$ perpendicular to $\ol{BC}$
and the line through $M$ perpendicular to $\ol{AI}$ meet at $K$.
Show that the circle with diameter $\ol{AK}$ is tangent to $\omega$.

\section*{Video}
\href{https://www.youtube.com/watch?v=cm8svlDoPfs&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/cm8svlDoPfs}}

\section*{External Link}
\url{https://aops.com/community/p11754885}

\newpage

\section*{Solution}
We make the following preliminary setup.
\begin{itemize}
\ii Let $E$ be the tangency point of the $A$-excircle,
so $\ol{AE}$ passes through the antipode of $D$ on $\omega$.
\ii Thus $\ol{AE}$ intersects the incircle again at $T$,
the foot from $D$ to $\ol{AE}$.
Since $DM = ME$ and $\dang DTE = 90\dg$,
it follows that $MD = MT = ME$
so in fact $\ol{MT}$ is also tangent to $\omega$.
\ii Let $\ol{AF}$ be the $A$-altitude.
\end{itemize}


\begin{center}
\begin{asy}
pair A = dir(135);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A, B, C);
pair M = (B+C)/2;
filldraw(incircle(A, B, C), opacity(0.1)+lightcyan, deepcyan);
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, deepcyan);
pair D = foot(I, B, C);
pair E = 2*M-D;
pair H = 2*I-D;
pair T = foot(D, A, E);
pair F = foot(A, B, C);
pair Kp = extension(A, F, T, D);
pair L = foot(A, M, Kp);

filldraw(circumcircle(T, M, D), opacity(0.1)+yellow, orange);
draw(A--T--Kp, lightred);
draw(T--E, lightred);
draw(A--Kp, lightblue);
draw(M--Kp, deepgreen);
draw(I--L, deepgreen);
draw(A--I, deepgreen+dotted);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(I));
dot("$M$", M, dir(M));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot(H);
dot("$T$", T, dir(10));
dot("$F$", F, dir(235));
dot("$K'$", Kp, dir(Kp));
dot("$L$", L, dir(L));

/* TSQ Source:

A = dir 135
B = dir 210
C = dir 330
I = incenter A B C
M = (B+C)/2
incircle A B C 0.1 lightcyan / deepcyan
A--B--C--cycle 0.1 lightcyan / deepcyan
D = foot I B C
E = 2*M-D
H .= 2*I-D
T = foot D A E R10
F = foot A B C R235
K' = extension A F T D
L = foot A M Kp

circumcircle T M D 0.1 yellow / orange
A--T--Kp lightred
T--E lightred
A--Kp lightblue
M--Kp deepgreen
I--L deepgreen
A--I deepgreen dotted

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

Let $K'$ be the intersection of $\ol{TD}$ with $A$-altitude.
By homothety, the circle with diameter $\ol{AK'}$
is certainly tangent to $\omega$.
We are going to prove $K' = K$.

Let $L$ denote the second intersection of $\ol{KM}$
with $(IDMT)$, so $\dang MLI = 90\dg$.

\begin{claim*}
  Quadrilateral $ATLK'$ is cyclic.
\end{claim*}
\begin{proof}
  Since $\dang K'LT = \dang MLT = \dang MDT = \dang FDT = \dang FAT =
  \dang K'AT$.
\end{proof}

Thus $\dang MLA = \dang ALK' = \dang ATK' = 90\dg$ as well.
Thus we conclude $\ol{AI}$ and $\ol{MK'}$ are perpendicular at $L$
as desired.

\end{document}
