\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{USEMO 2021/4}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 90}
\maketitle
\section*{Problem}
Let $ABC$ be a triangle with circumcircle $\omega$,
and let $X$ be the reflection of $A$ in $B$.
Line $CX$ meets $\omega$ again at $D$.
Lines $BD$ and $AC$ meet at $E$, and lines $AD$ and $BC$ meet at $F$.
Let $M$ and $N$ denote the midpoints of $AB$ and $AC$.
Can line $EF$ share a point with the circumcircle of triangle $AMN$?
\section*{Video}
\href{https://youtu.be/PhNIee2CzdY}{\texttt{https://youtu.be/V-9UBJr7aDI}}
\newpage
\section*{Solution}
The answer is no, they never intersect.
\paragraph{Classical solution, by author}
Let $P$ denote the midpoint of $\ol{AD}$, which
\begin{itemize}
\ii lies on $\ol{BN}$, since $\ol{BN} \parallel \ol{CX}$; and
\ii lies on $(AMN)$,
since it's homothetic to $(ABC)$ through $A$ with factor $\half$.
\end{itemize}
\begin{center}
\begin{asy}
size(10cm);
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair X = 2*B-A;
pair D = -C+2*foot(origin, C, X);
pair P = midpoint(A--D);
pair E = extension(B, D, C, A);
pair F = extension(A, D, B, C);
filldraw(unitcircle, opacity(0.1)+lightred, red);
draw(B--X, red);
draw(C--E, red);
draw(A--B--C--cycle, red);
draw(B--E, orange);
draw(A--D, orange);
draw(C--X, blue);
pair M = midpoint(A--B);
pair N = midpoint(A--C);
filldraw(circumcircle(A, M, N), opacity(0.1)+lightcyan, mediumblue);
draw(B--N, blue);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(10));
dot("$X$", X, dir(X));
dot("$D$", D, dir(D));
dot("$P$", P, dir(310));
dot("$E$", E, dir(E));
dot("$F$", F, dir(230));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
/* TSQ Source:
!size(10cm);
A = dir 110
B = dir 210
C = dir 330 R10
X = 2*B-A
D = -C+2*foot origin C X
P = midpoint A--D R310
E = extension B D C A
F = extension A D B C R230
unitcircle 0.1 lightred / red
B--X red
C--E red
A--B--C--cycle red
B--E orange
A--D orange
C--X blue
M = midpoint A--B
N = midpoint A--C
circumcircle A M N 0.1 lightcyan / mediumblue
B--N blue
*/
\end{asy}
\end{center}
Now, note that
\begin{align*}
\dang FBP = \dang CBN = \dang BCD = \dang BAD = \dang BAF
&\implies FB^2 = FP \cdot FA \\
\dang EBN = \dang EDC = \dang BDC = \dang BAC = \dang BAE
&\implies EB^2 = EN \cdot EA.
\end{align*}
This means that line $EF$ is the radical axis
of the circle centered at $B$ with radius zero,
and the circumcircle of triangle $AMN$.
Since $B$ obviously lies outside $(AMN)$,
the disjointness conclusion follows.
\paragraph{Projective solution, by Ankit Bisain}
In this approach we are still going to prove that $\ol{EF}$
is the radical axis of $(AMN)$ and the circle of radius zero at $B$,
but we are not going to use the point $P$, or even points $E$ and $F$.
Instead, let $Y = \ol{EF} \cap \ol{AB}$,
which by Brokard's theorem on $ABDC$ satisfies $(AB;XY) = -1$.
Since $XB=XA$, it follows that $AY:YB = 2$.
From here it is straightforward to verify that
\[ YB^2 = \frac{1}{9}AB^2 = YM \cdot YA. \]
Thus $Y$ lies on the radical axis.
Finally, by Brokard's theorem again,
if $O$ is the center of $\omega$ then $\ol{OX} \perp \ol{EF}$.
Taking a homothety with scale factor $2$ at $A$,
it follows that the line through $B$ and the center of $(AMN)$
is perpendicular to $\ol{EF}$.
Since $\ol{EF}$ contains $Y$,
it now follows that $\ol{EF}$ is the radical axis, as claimed.
\paragraph{Solution with inversion, projective, and Cartesian coordinates, by Ankan Bhattacharya}
In what follows,
let $O$ be the center of $\omega$.
Note that Brokard's theorem gives that $\ol{EF}$ is the polar of $X$.
Note that since none of $E$, $F$, $X$ are points at infinity,
$O$ is different from all three.
We consider inversion in $\omega$ to eliminate the polar:
\begin{itemize}
\ii The circumcircle of $\triangle AMN$,
i.e.\ the circle with diameter $\ol{AO}$,
is sent to the line $\ell$ tangent to $\omega$ at $A$.
\ii The line $EF$, as the polar of $X$,
is sent to the circle with diameter $\ol{OX}$.
(It is indeed a circle, because $O$ does not lie on line $EF$.)
\end{itemize}
Thus, if the posed question is true,
then we see that $\ell$ intersects $(OX)$.
We claim this is impossible.
Establish Cartesian coordinates with $A = (0, 0)$
and $O = (2, 0)$, so $\ell$ is the $y$-axis.
Let $T$ be the center of $(OX)$: the midpoint of $\ol{OX}$.
Observe:
\begin{itemize}
\ii $B$ lies on the circle with center $(2, 0)$ and radius $2$.
\ii $X$ lies on the circle with center $(4, 0)$ and radius $4$.
\ii $T$ lies on the circle with center $(3, 0)$ and radius $2$.
\end{itemize}
Thus, let the coordinates of $T$ be $(x, y)$,
with $(x-3)^2 + y^2 = 4$.
The intersection of $\ell$ and $(OX)$ being nonempty is equivalent to
\begin{align*}
d(T, \ell)^2 & \le OT^2\\
\iff x^2 & \le (x - 2)^2 + y^2\\
\iff x^2 & \le (x - 2)^2 + [4 - (x - 3)^2]\\
\iff (x-1)^2 & \le 0,
\end{align*}
or $x = 1$ (which forces $y = 0$); i.e. $T = (1, 0)$.
However, this forces
\[ B = (0, 0) = A, \]
which is not permitted.
Thus, line $\ell$ cannot share a point with $(OX)$,
and so line $EF$ cannot share a point with $(AMN)$.
\end{document}