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

\begin{document}
\title{USEMO 2019/1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 10}
\maketitle

\section*{Problem}
Let $ABCD$ be a cyclic quadrilateral.
A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$
again at points $E$ and $F$ (distinct from $A$, $B$, $C$).
Let $H$ denote the orthocenter of triangle $DEF$.
Prove that if lines $AC$, $DO$, $EF$ are concurrent,
then triangles $ABC$ and $EHF$ are similar.

\section*{Video}
\href{https://www.youtube.com/watch?v=V2TNgUwbs6A&list=PLi6h8GM1FA6yfhMMNqJi1mht4KmZnXpj6}{\texttt{https://youtu.be/V2TNgUwbs6A}}

\section*{External Link}
\url{https://aops.com/community/p15412066}

\newpage

\section*{Solution}
Define $G$ as the intersection of $\ol{AC}$ and $\ol{EF}$.

\begin{claim*}
  Quadrilateral $DCGF$ is cyclic.
\end{claim*}
\begin{proof}[First proof]
  Because $\dang DCG = \dang DCA = \dang DBA
  = \dang DBE = \dang DFE = \dang DFG$
\end{proof}
\begin{proof}[Second proof]
  Follows since $D$ is Miquel point of $GABF$.
\end{proof}
\begin{center}
\begin{asy}
pair A = dir(110);
pair D = dir(190);
pair B = dir(-10);
pair C = dir(-B*D/A);
pair G = 0.33*C+0.67*A;
pair O = extension(D, G, dir(90), dir(270));
pair F = -B+2*foot(O, B, C);
pair E = -B+2*foot(O, B, A);
filldraw(circumcircle(B, E, D), opacity(0.1)+lightred, red);
filldraw(unitcircle, opacity(0.1)+cyan, cyan);
draw(F--E--B, deepgreen);
draw(A--C--B, deepgreen);
filldraw(A--G--F--B--cycle, opacity(0.1)+yellow, green);
pair H = orthocenter(D, E, F);
draw(A--D--C, blue+dashed);
draw(B--D, blue+dashed);
draw(D--O, red);
draw(E--H--F, lightgreen);

dot("$A$", A, dir(A));
dot("$D$", D, dir(D));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$G$", G, dir(170));
dot("$O$", O, dir(170));
dot("$F$", F, dir(F));
dot("$E$", E, dir(E));
dot("$H$", H, dir(H));

/* TSQ Source:

A = dir 110
D = dir 190
B = dir -10
C = dir -B*D/A
G = 0.33*C+0.67*A R170
O = extension D G dir 90 dir 270 R170
F = -B+2*foot O B C
E = -B+2*foot O B A
circumcircle B E D 0.1 lightred / red
unitcircle 0.1 cyan / cyan
F--E--B deepgreen
A--C--B deepgreen
A--G--F--B--cycle 0.1 yellow / green
H = orthocenter D E F
A--D--C blue dashed
B--D blue dashed
D--O red
E--H--F lightgreen

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

\begin{claim*}
  If $G$ lies on line $\ol{DO}$, we have $\ol{AC} \perp \ol{BD}$.
\end{claim*}
\begin{proof}
  We have
  \begin{align*}
    \dang BDG &= \dang BDO = 90\dg - \dang DEB = 90\dg - \dang DFB \\
    &= 90\dg - \dang DFC = 90\dg - \dang DGC. \qedhere
  \end{align*}
\end{proof}

To finish,
\[ \dang HEF = 90\dg - \dang EFD = 90\dg - \dang EBD = 90\dg - \dang ABD = \dang CAB. \]
Similarly $\dang HFE =\dang ACB$ and the proof is done.

\begin{remark*}
  The original version of this problem was in the converse direction:
  showing that $\ol{AC} \perp \ol{BD}$ implied the concurrence.
  Unfortunately, this turns out to be susceptible to Cartesian coordinates
  by setting the $x$ and $y$ axes along these lines,
  as well as complex methods.

  Interestingly, it does not appear to be easy to show directly that
  the converse of the problem implies the original statement
  (other than actually solving the problem, and adapting the proof).
  Note in particular that the case where $E=A$ and $F=C$
  is a counterexample to the converse direction as stated.
\end{remark*}

\end{document}
