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

\begin{document}
\title{Mexico 2019/6}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 21}
\maketitle

\section*{Problem}
Let $ABC$ be a triangle such that $\angle BAC = 45^{\circ}$.
Let $H$ and $O$ be the orthocenter and circumcenter of $ABC$, respectively.
Let $\omega$ be the circumcircle of $ABC$ and $P$ the point on $\omega$
such that the circumcircle of $PBH$ is tangent to $BC$.
Let $X$ and $Y$ be the circumcenters of $PHB$ and $PHC$ respectively.
Let $O_1,O_2$ be the circumcenters of $PXO$ and $PYO$ respectively.
Prove that $O_1$ and $O_2$ lie on $AB$ and $AC$, respectively.

\section*{Video}
\href{https://www.youtube.com/watch?v=lxuqQaN59tU&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/lxuqQaN59tU}}

\section*{External Link}
\url{https://aops.com/community/p13469495}

\newpage

\section*{Solution}
We let $\ol{BK}$ and $\ol{CL}$ be the altitudes of $\triangle ABC$.
The circle with diameter $\ol{BC}$ will be denoted by $\gamma$;
and we'll denote the center by $M$.

\begin{claim*}
  The circle $(PHC)$ is also tangent to line $BC$.
\end{claim*}
\begin{proof}
  We are given $\dang PBC = \dang PHB$.
  Since $\dang BHC = -\dang BAC = -\dang BPC$,
  it follows $\dang PCB = \dang PHC$ too.
\end{proof}

\begin{claim*}
  Points $P$, $H$, $M$ are collinear.
  Actually, $P$ is the inverse of $H$ with respect to $\gamma$.
\end{claim*}
\begin{proof}
  Line $PH$ bisects $\ol{BC}$ by radical axis
  on $(BPH)$ and $(CPH)$.
  Also, $MH \cdot MP = MB^2 = MC^2$.
\end{proof}

\begin{center}
\begin{asy}
pair B = dir(180);
pair C = dir(0);
pair K = dir(65);
pair L = dir(90)*K;
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
filldraw(B--C--K--L--cycle, opacity(0.1)+lightcyan, blue);
pair A = extension(B, L, C, K);
pair H = extension(B, K, C, L);
draw(B--K, blue);
draw(C--L, blue);
pair O = circumcenter(A, B, C);

pair P = 1/conj(H);
pair M = origin;
filldraw(circumcircle(P, B, H), opacity(0.1)+yellow, orange);
pair X = circumcenter(P, H, B);
pair O_1 = circumcenter(P, X, O);
draw(L--A--K, dotted+deepcyan);

draw(circumcircle(X, O, L), dashed+deepgreen);
draw(P--M, blue);
draw(CP(O_1, P), dotted+grey);

dot("$B$", B, dir(225));
dot("$C$", C, dir(315));
dot("$K$", K, dir(K));
dot("$L$", L, dir(L));
dot("$A$", A, dir(A));
dot("$H$", H, dir(110));
dot("$O$", O, dir(60));
dot("$P$", P, dir(P));
dot("$M$", M, dir(-90));
dot("$X$", X, dir(X));
dot("$O_1$", O_1, dir(O_1));

/* TSQ Source:

B = dir 180 R225
C = dir 0 R315
K = dir 65
L = dir(90)*K
unitcircle 0.1 lightcyan / blue
B--C--K--L--cycle 0.1 lightcyan / blue
A = extension B L C K
H = extension B K C L R110
B--K blue
C--L blue
O = circumcenter A B C R60

P = 1/conj(H)
M = origin R-90
circumcircle P B H 0.1 yellow / orange
X = circumcenter P H B
O_1 = circumcenter P X O
L--A--K dotted deepcyan

circumcircle X O L dashed deepgreen
P--M blue
CP O_1 P dotted grey

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

We now actually use the condition
that $\dang BAC = 45\dg$,
which is equivalent to $\dang BHC = 135\dg$
and $\dang BOC = 90\dg$.
This means that $O$ is the arc midpoint of $(BC)$.

\begin{claim*}
  $LOKH$ is a parallelogram.
\end{claim*}
\begin{proof}
  Since arcs $KL$ and $OB$ of $\gamma$ measure $90\dg$,
  the arcs $BL$ and $OK$ are equal,
  hence $\ol{LO} \parallel \ol{BK}$.
  Similarly $\ol{KO} \parallel \ol{CL}$.
\end{proof}

\begin{claim*}
  We have $\dang XPO = 45\dg$.
\end{claim*}
\begin{proof}
  Since $\triangle HLK \sim \triangle HBC$,
  and $\ol{HO}$ bisects $\ol{LK}$ by previous claim,
  it follows \[ \dang BHM = \dang OHL. \]
  Now
  \begin{align*}
    \dang XPO &= \dang XPH + \dang MPO
    = (90\dg - \dang HBP) + \dang HOM \\
    &= 90\dg - \dang MBP + \dang MBH + \dang HOM \\
    &= 90\dg - \dang BHM + \dang MBH + \dang HOM \\
    &= 90\dg - \dang OHL + \dang MBH + \dang HOM \\
    &= 90\dg + \dang(LH, HO) + \dang(MB, BH) + \dang(HO,MO) \\
    &= 90\dg + \dang(LH, BH) + \dang(MB, MO) = 45\dg. \qedhere
  \end{align*}
\end{proof}

\begin{claim*}
  We have $X$, $O_1$, $O$, $L$, $H$ are concyclic,
  in the circle with diameter $\ol{XO}$.
\end{claim*}
\begin{proof}
  First, since $\triangle LBH$ is a $45\dg$-$45\dg$-$90\dg$ triangle and $XB=XH$,
  it follows that $\ol{XL}$ is the perpendicular bisector of $\ol{BH}$.
  Hence, $\dang XLO = 90\dg$.

  On the other hand, since $\dang XPO = 45\dg$,
  we have $\dang XO_1O = 90\dg$.
  This implies concyclic.
\end{proof}

Finally, $\dang BLO = 135\dg$ and
\[ \dang OLO_1 = \dang OXO_1 = 90\dg - \dang XPO = 45\dg \]
this implies $O_1$ lies on line $BL$,
ergo lies on line $AB$.

\end{document}
