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

\begin{document}
\title{China TST 2019/4/1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 29}
\maketitle

\section*{Problem}
Let $ABCD$ be a cyclic quadrilateral
inscribed in a circle with center $O$.
Points $M$ and $N$ are the midpoints of $\ol{BC}$ and $\ol{CD}$,
and points $E$ and $F$ lie on $AB$ and $AD$ respectively
such that $EF$ passes through $O$ and $EO=OF$.
Lines $EN$ meet $FM$ at $P$.
Let $S$ denote the circumcenter of $\triangle PEF$.
Line $PO$ intersects $AD$ and $BA$ at $Q$ and $R$ respectively.
Suppose $OSPC$ is a parallelogram.
Prove that $AQ=AR$.

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

\section*{External Link}
\url{https://aops.com/community/p12139797}

\newpage

\section*{Solution}
We let $H$ denote the orthocenter of $\triangle PEF$.
\begin{center}
\begin{asy}
pair E = dir(142.15);
pair F = dir(37.85);
pair H = dir(275);
pair P = E+H+F;
pair M = extension(F, P, E, H);
pair N = extension(E, P, F, H);
pair C = midpoint(H--P);
pair D = 2*N-C;
pair B = 2*M-C;
pair A = extension(B, E, D, F);

filldraw(E--F--H--cycle, opacity(0.1)+lightgreen, deepgreen);
pair O = midpoint(E--F);
filldraw(circumcircle(C, M, N), opacity(0.1)+lightred, red);
draw(CP(O, C), blue);
draw(circumcircle(P, E, F), blue);
filldraw(A--B--C--D--cycle, opacity(0.1)+lightcyan, deepcyan);
pair S = circumcenter(P, E, F);

draw(M--F, lightcyan);
draw(N--E, lightcyan);

dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$H$", H, dir(H));
dot("$P$", P, dir(P));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$B$", B, dir(B));
dot("$A$", A, dir(A));
dot("$O$", O, dir(O));
dot("$S$", S, dir(210));

/* TSQ Source:

E = dir 142.15
F = dir 37.85
H = dir 275
P = E+H+F
M = extension F P E H
N = extension E P F H
C = midpoint H--P
D = 2*N-C
B = 2*M-C
A = extension B E D F

E--F--H--cycle 0.1 lightgreen / deepgreen
O = midpoint E--F
circumcircle C M N 0.1 lightred / red
CP O C blue
circumcircle P E F blue
A--B--C--D--cycle 0.1 lightcyan / deepcyan
S = circumcenter P E F R210

M--F lightcyan
N--E lightcyan

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


Let $\omega$ denote the circle with diameter $\ol{OC}$,
passing through $M$ and $N$.
\begin{claim*}
  The circle $\omega$ is the nine-point circle
  of $\triangle PEF$ (or $\triangle HEF$ if you prefer).
\end{claim*}
\begin{proof}
  We observe a few facts:
  \begin{itemize}
    \ii Clearly $\omega$ has radius half that of $(O)$.
    Since $SP = OC$, the circles $(S)$ and $(S)$ are congruent,
    hence the radius of $\omega$ is half that of $(PEF)$ too.

    \ii Point $O$ is the midpoint of $\ol{EF}$,

    \ii The antipode of $O$ --- namely $C$ ---
    is known to lie on the $P$-altitude
    (because $\ol{SO} \perp \ol{CP}$ and $\ol{SO} \parallel \ol{EF}$).
    \qedhere
  \end{itemize}
\end{proof}

\begin{claim*}
  $\ol{AC}$ bisects $\angle BAD$.
\end{claim*}
\begin{proof}
  We have $OM = ON$, so $BC = CD$.
\end{proof}

\begin{claim*}
  We have $\ol{CA} \parallel \ol{OP}$.
\end{claim*}
\begin{proof}
  From $ABCD$ is cyclic,
  we can compute
  \[ \dang EAF = \dang BAD = \dang BCD = \dang MCN =
  \dang MON = 2 \dang FHE \]
  so $A$ lies on the circle through $E$, $F$,
  and the circumcenter of $\triangle HEF$.
  Denote this circumcenter by $W$.
  As $WE = WF$, so this implies $\ol{AW}$ bisects $\angle EAF$,
  and hence $AWC$ are collinear.
  Since $OWCP$ is a parallelogram, this completes the proof.
\end{proof}

This completes the solution.

\end{document}
