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

\begin{document}
\title{China TST 2012/1/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 27}
\maketitle

\section*{Problem}
Let $ABC$ be a scalene triangle
whose incircle touches $\ol{BC}$, $\ol{CA}$, $\ol{AB}$ at $D$, $E$, $F$.
Let $L$, $M$, $N$ denote the reflections of $D$ across $\ol{EF}$,
$E$ across $\ol{FD}$, $F$ across $\ol{DE}$.
Let $P = \ol{AL} \cap \ol{BC}$, $Q = \ol{BM} \cap \ol{CA}$,
$R = \ol{CN} \cap \ol{AB}$. Prove that $P$, $Q$, $R$ are collinear.

\section*{Video}
\href{https://www.youtube.com/watch?v=X3Nw-GWac9k&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/X3Nw-GWac9k}}

\section*{External Link}
\url{https://aops.com/community/p2628489}

\newpage

\section*{Solution}
We use complex numbers with $\triangle DEF$ as the unit circle.
To compute $P$, we know that \[ p + d^2 \ol p = 2d \]
and
\begin{align*}
  0 &= \det \begin{bmatrix}
    p & \ol p & 1 \\
    \frac{2ef}{e+f} & \frac{2}{e+f} & 1 \\
    e + f - \frac{ef}{d} & \frac 1e + \frac 1f - \frac{d}{ef} & 1 \\
  \end{bmatrix} \\
  \implies 0 &= \det \begin{bmatrix}
    p & \ol p & 1 \\
    2ef & 2 & e+f \\
    ef[d(e+f)-ef] & d(e+f-d) & def \\
  \end{bmatrix} \\
  &= d\left[ d(e+f)-e^2-f^2 \right]p \\
  &\qquad - ef\left[ ef(e+f)-d(e^2+f^2) \right] \ol p + 2ef\left[ ef-d^2 \right].
\end{align*}
Eliminating $\ol p$ then lets one explicitly compute $p$:
\begin{align*}
  p &=
  \frac{\left[ (d^2-ef) \cdot 2ef \right] \cdot d^2 +
    ef\left[ ef(e+f)-d(e^2+f^2) \right] \cdot 2d}
    {d\left[ d(e+f)-e^2-f^2 \right] \cdot d^2
      + ef\left[ ef(e+f)-d(e^2+f^2) \right] \cdot 1} \\
  &= 2def \cdot \frac{d(d^2-ef) +
    ef(e+f)-d(e^2+f^2)}
    {d^3\left[ d(e+f)-e^2-f^2 \right]
      + ef\left[ ef(e+f)-d(e^2+f^2) \right]} \\
  &= 2def \cdot \frac{d^3-(e^2+ef+f^2)d+ef(e+f)}
    {(e+f)d^4 - (e^2+f^2)d^3 - ef(e^2+f^2)d + e^2f^2(e+f)} \\
\end{align*}
Now
\[
  p - \frac{2df}{d+f} = 2df \left[
    \frac{e(d^3-(e^2+ef+f^2)d+ef(e+f))}
    {(e+f)d^4 - (e^2+f^2)d^3 - ef(e^2+f^2)d + e^2f^2(e+f)}
    - \frac{1}{d+f} \right] \]
The bulk of the work is expanding the numerator
of the resulting fraction which equals
\begin{align*}
  & e(d+f)(d^3-(e^2+ef+f^2)d+ef(e+f)) \\
    &\qquad - \left[(e+f)d^4 - (e^2+f^2)d^3 - ef(e^2+f^2)d +
    e^2f^2(e+f)\right] \\
  &= -f \cdot d^4 + (e^2+ef+f^2)d^3 -
    e(e^2+ef+f^2)d^2 + e^3f d \\
  &= d\left[ f(e^3-d^3) + d(e^2+ef+f^2)(d-e) \right] \\
  &= d\left[ (d-e)(d(e^2+ef+f^2)-f(d^2+de+e^2)) \right] \\
  &= d(d-e) \left( de^2+df^2-fd^2-fe^2 \right) \\
  &= d(d-e)(d-f)(e^2-df).
\end{align*}
Thus, we have that
\[ p - b = \frac{2df}{d+f} \cdot \frac{d(d-e)(d-f)(e^2-df)}
  {(e+f)d^4 - (e^2+f^2)d^3 - ef(e^2+f^2)d + e^2f^2(e+f)} \]
Now, it follows that
\[ \frac{p-b}{p-c} = \frac{f(d+e)(e^2-df)} {e(d+f)(f^2-de)}. \]
Multiplying cyclically gives
$\frac{p-b}{p-c} \frac{q-c}{q-a} \frac{r-a}{r-b} = 1$.
Therefore, by Menelaus theorem, the problem is solved.

\begin{remark*}
  Actually, it turns out that $P$, $Q$, $R$
  lie on the Euler line of triangle $DEF$,
  and there is a factor of $d+e+f$ hidden in the numerator of $P$.
  So a simpler approach would be to factor this out from $P$,
  and then notice the resulting quotient is real.
\end{remark*}

\end{document}
