\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{All-Lincoln 2022/6}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 109}
\maketitle
\section*{Problem}
Consider acute triangle $ABC$. Let $D$, $E$, $F$ be the $A$, $B$, $C$ intouch
points of $ABC$, and $X$, $Y$, $Z$ as the arc midpoints of $BC$, $CA$, $AB$ in
the circumcircle of $ABC$. Prove that the triangle bounded by the lines $XE$,
$YF$, $ZD$ has area at most half of the area of $ABC$.
\section*{Video}
\href{https://www.youtube.com/watch?v=GEa2nOS1PBM&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/GEa2nOS1PBM}}
\newpage
\section*{Solution}
The following stronger claim is true:
\begin{claim*}
Let $DEF$ be any triangle.
Let $XYZ$ be a triangle obtained from a homothety of ratio $\rho \ge 1$
whose center lies inside $\triangle DEF$.
Then the triangle bounded by the lines $XE$, $YF$, $ZD$
has area at most $\rho$ of the area of $DEF$.
\end{claim*}
\begin{proof}
Brute-force bary on $\triangle DEF$.
Let $\lambda = \rho - 1 \ge 0$, and $\mu = \lambda\inv$.
Also, let the homothety center be $(u,v,w)$ for $u,v,w > 0$ and $u+v+w=1$.
Then
\begin{align*}
X &= \left( \lambda(v+w)+1, -\lambda v, -\lambda w \right). \\
&= \left( v+w+\mu: -v: -w \right) \\
Y &= (-u : w+u+\mu: -w) \\
Z &= (-u : -v : u+v+\mu) \\
DZ \cap EX &= ( (u+v+\mu)(v+w+\mu) : wv : -w(u+v+\mu)) \\
EX \cap FY &= \left( -u(v+w+\mu) : (v+w+\mu)(w+u+\mu) : uw \right) \\
FY \cap DZ &= \left( uv : -v(w+u+\mu) : (w+u+\mu)(u+v+\mu) \right).
\end{align*}
Direct computation gives that
\[
\frac{\opname{Area}(DZ\cap EX, EX \cap FY, FY \cap DZ)}{[DEF]}
=
\frac{(uvw + \prod_{\text{cyc}} (u+v+\mu))^2}{\prod_{\text{cyc}}
\left( \mu^2+(u+2v)\mu + v(u+v+w) \right)}.
\]
Therefore, since $\rho = \mu\inv + 1$, we need to show
\begin{align*}
\left( uvw + \prod_{\text{cyc}} (u+v+\mu) \right)^2
&\le
\left( 1 + \frac{u+v+w}{\mu} \right) \\
& \cdot \prod_{\text{cyc}}
\left( \mu^2+(u+2v)\mu + v(u+v+w) \right).
\end{align*}
However, using Sage reveals that
\begin{align*}
\text{RHS}-\text{LHS}
&= \mu^4 (uv+vw+wu) \\
&+ \mu^3 \left( 4(uv^2+vw^2+wu^2) + 2(u^2v+v^2w+w^2u) + 9uvw \right) \\
&+ \mu^2 \sum_{\text{cyc}} (u^3v+6u^2v^2+6uv^3+20u^2vw) \\
&+ \mu \sum_{\text{cyc}} (2u^3v^2+4uv^4+6u^2v^3+19u^3vw+32u^2v^2w) \\
&+ \sum_{\text{cyc}}
(u^3v^3+u^5w+2u^2v^4+8u^4vw +19u^2vw^3+20u^2v^3w + 12u^2v^2w^2) \\
&+ \frac{1}{\mu} \sum_{\text{cyc}}
\left( u^5vw+4u^2v^4w+4u^2vw^4+6u^3v^3w+12u^2v^2w^3 \right) \\
&\ge 0.
\qedhere
\end{align*}
\end{proof}
\begin{remark*}
Note that equality occurs if say $D=X$, which corresponds to $v=w=0$.
\end{remark*}
\begin{proof}
[Less terrible proof of the claim, sent by Darij Grinberg]
Let $P$ be the center of the homothety, and let $U = DZ \cap EX$ and $V =
FY \cap DZ$ and $W = EX \cap FY$. We must show that $[UVW] \leq \rho [DEF]$.
The line $FP$ intersects both (closed) segments $EP$ and $ED$, so it also
intersects the closed segment $EX$ (since $X$ is on the segment $DP$). In
other words, $W$ lies on the segment $EX$. On the other hand, the point $U$
lies on the extension of this segment beyond $X$, since it lies between $X$
and $EU \cap FD$ (because the point $Z$ lies between $P$ and $F$). Hence,
the point $X$ lies on the segment $WU$. Similarly, $Y$ lies on the segment
$VW$, and $Z$ lies on the segment $UV$.
As $P$ lies inside $\triangle XYZ$,
\[ [UVW] = [PXWY] + [PYVZ] + [PZUX]. \]
But since $W$ and $Y$ lie on the segments $EU$ and $EP$, we have
\[ [PXWY] \leq [PXE] = \rho [PDE] \]
and similarly (cyclically)
\[ [PYVZ] \leq \rho [PEF], \qquad [PZUX] \leq \rho [PFD]. \]
Summing these three inequalities yields
\[ [PXWY] + [PYVZ] + [PZUX]
\leq \rho [PDE] + \rho [PEF] + \rho [PFD] = \rho [DEF] \]
as desired.
\end{proof}
We now use the following theorem.
\begin{theorem*}
[Apparently not well-known]
We have $[DEF]/[ABC] = 2R/r$, where $r$ and $R$ are the inradius and
circumradius.
\end{theorem*}
(This is possibly a bit overkill, as all that's needed is $R/r \ge 2$ here.)
Note that in the original problem, $\triangle DEF$ and $\triangle XYZ$
are homothetic with ratio $\frac{YZ}{EF} = \frac{R}{r}$.
Their homothety center is the concurrence point $X_{56}$
of lines $DX$, $EY$ and $FZ$,
so we'd be done upon showing:
\begin{claim*}
[Annoying interior analysis]
When $\triangle ABC$ is acute, $X_{56}$ lies inside $\triangle DEF$.
\end{claim*}
\begin{proof}
Let $I$ denote the incenter,
so $I$ is the orthocenter of acute triangle $XYZ$
and in particular lies inside acute triangle $DEF$.
Then $\ol{YZ}$ is the perpendicular bisector of $\ol{AI}$,
while $\ol{EF}$ is perpendicular to $\ol{AI}$ at a point
closer to $I$ than $A$ (because $\angle A < 90\dg \implies \angle EIF > 90\dg$).
Hence $F = \ol{EF} \cap \ol{DF}$ lies inside $\triangle XYZ$,
and so $\ol{ZF}$ is an internal cevian of $\triangle XYZ$.
The same is true for $\ol{DX}$ and $\ol{EY}$, and we're done.
\end{proof}
\end{document}