2009 Zhautykov International Olympiad Problems/Problem 3

Problem

For a convex hexagon $ABCDEF$ with an area $S$, prove that:

$AC\cdot(BD + BF - DF) + CE\cdot(BD + DF - BF) + AE\cdot(BF + DF - BD)\geq 2\sqrt {3}$.

Lemma

First, let's set aside the original problem and introduce the "Gergonne point" and its distance to the vertices.

Let the incircle of $\triangle ABC$ touch $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. By Ceva's theorem, it is easy to see that $AD$, $BE$, $CF$ are concurrent at a point $Q$, which we call the Gergonne point of $\triangle ABC$.

1509090255f169edf94b63df3e.gif

Let $AE=AF=x$, $BF=BD=y$, $CD=CE=z$. As shown in the figure, by Stewart's theorem, we have \[AD^2=\frac{(z+x)^2y+(x+y)^2z}{y+z}-yz=\frac{x(4yz+zx+xy)}{y+z},\] In $\triangle ACD$, by Menelaus' theorem, we have \[\frac{AQ}{AD-AQ}\cdot \frac y{y+z}\cdot \frac zx=1,\] Solving this, we get \[AQ=\frac{x(y+z)}{yz+zx+xy}\cdot AD=\frac{x\sqrt{x(y+z)(4yz+zx+xy)}}{yz+zx+xy}.\]

Noting that by AM-GM inequality, we have x(y+z)(4yz+zx+xy)=133x(y+z)(4yz+zx+xy)3x(y+z)+4yz+zx+xy23=2(yz+zx+xy)3, we get \[AQ\leqslant \frac2{\sqrt3}x,\] Thus, we obtain the following lemma.

Lemma: Let $Q$ be the Gergonne point of $\triangle ABC$, then \[AB+AC-BC\geqslant \sqrt3AQ.\]

Returning to the original problem

1509090255851ede02a626c22a.gif

Let $K$ be the Gergonne point of $\triangle BDF$. As shown in the figure, by the lemma, we have \[AC\cdot(BD + BF - DF)\geqslant AC\cdot\sqrt3BK\geqslant 2\sqrt3S_{ABCK},\] Similarly, we have $CE\cdot(BD + DF - BF)\geqslant 2\sqrt3S_{CDEK}$, $AE\cdot(BF + DF - BD)\geqslant 2\sqrt3S_{EFAK}$, adding these together, we get the desired result.