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

\begin{document}
\title{Twitch 106.1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 106}
\maketitle

\section*{Problem}
Given an equilateral triangle $ABC$ and a point $X$,
let the reflection of $X$ about $AB$ be $X_C$,
and the other reflections defined similarly.
Prove or disprove whether or not
$AX_A$, $BX_B$, and $CX_C$ are concurrent or parallel.

\section*{Video}
\href{https://www.youtube.com/watch?v=9iE3J43Sozk&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/9iE3J43Sozk}}



\newpage

\section*{Solution}
Use complex numbers with $(ABC)$ the unit circle.
Ignore for now the condition that $ABC$ is equilateral to retain symmetry.
We know that
\[ x_A = b + c - bc \ol x. \]
The equation $Z \in AX_A$ is then given by
\[
  0 = \det \begin{bmatrix}
    a & \ol a & 1 \\
    b+c-bc \ol x & \ol b + \ol c - \ol b \ol c x & 1 \\
    z & \ol z & 1
  \end{bmatrix}
\]
or
\[
  (\ol a b + \ol a c - \ol a b c \ol x) - (a \ol b + a \ol c - a \ol b \ol c x)
  = (\ol a - \ol b - \ol c + \ol b \ol c x) z
  - (a  - b - c + bc \ol x)\ol z
\]
Consider the analogous equations for $BX_B$ and $CX_C$
to get a $3 \times 3$ determinant.
\begin{claim*}
  When $a = 1$, $b = e^{\frac23 \pi i}$, $c = e^{\frac43 \pi i}$,
  this determinant vanishes.
  In fact, the sum of the rows is the zero vector.
\end{claim*}
\begin{proof}
  Note that $abc=1$, $a\inv+b\inv+c\inv=a+b+c=a^2+b^2+c^2=0$.
  \begin{align*}
    \sum_{\text{cyc}} \ol a b + \ol a c
    &= \sum_{\text{cyc}} \ol a (-a) = -3 \\
    \sum_{\text{cyc}} \ol a b c \ol x
    &= \sum_{\text{cyc}} a^{-2} \ol x = 0 \\
    \sum_{\text{cyc}} a \ol b \ol c x
    &= \sum_{\text{cyc}} a^2 x = 0 \\
    \sum_{\text{cyc}} (\ol a - \ol b - \ol c)
    &= \sum_{\text{cyc}} 2 \ol a = 0 \\
    \sum_{\text{cyc}} (a - b - c)
    &= \sum_{\text{cyc}} 2 a = 0 \\
    \sum_{\text{cyc}} (\ol b \ol c) x
    &= \sum_{\text{cyc}} ax = 0 \\
    \sum_{\text{cyc}} bc \ol x
    &= \sum_{\text{cyc}} a^{-1} x = 0. \qedhere
  \end{align*}
\end{proof}

\end{document}
