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Difference between revisions of "2009 Zhautykov International Olympiad Problems/Problem 3"

(Created page with "First, let's set aside the original problem and introduce the "Gergonne point" and its distance to the vertices. Let the incircle of <math>\triangle ABC</math> touch <math>BC...")
 
 
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==Problem==
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For a convex hexagon <math>ABCDEF</math> with an area <math>S</math>, prove that:
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<center><math>AC\cdot(BD + BF - DF) + CE\cdot(BD + DF - BF) + AE\cdot(BF + DF - BD)\geq 2\sqrt {3}</math>.</center>
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==Lemma==
 
First, let's set aside the original problem and introduce the "Gergonne point" and its distance to the vertices.
 
First, let's set aside the original problem and introduce the "Gergonne point" and its distance to the vertices.
  
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<cmath>AB+AC-BC\geqslant \sqrt3AQ.</cmath>
 
<cmath>AB+AC-BC\geqslant \sqrt3AQ.</cmath>
  
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==Returning to the original problem==
 
[[File:1509090255851ede02a626c22a.gif]]
 
[[File:1509090255851ede02a626c22a.gif]]
  
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Returning to the original problem, let <math>K</math> be the Gergonne point of <math>\triangle BDF</math>. As shown in the figure, by the lemma, we have
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Let <math>K</math> be the Gergonne point of <math>\triangle BDF</math>. As shown in the figure, by the lemma, we have
 
<cmath>AC\cdot(BD + BF - DF)\geqslant AC\cdot\sqrt3BK\geqslant 2\sqrt3S_{ABCK},</cmath>
 
<cmath>AC\cdot(BD + BF - DF)\geqslant AC\cdot\sqrt3BK\geqslant 2\sqrt3S_{ABCK},</cmath>
 
Similarly, we have <math>CE\cdot(BD + DF - BF)\geqslant 2\sqrt3S_{CDEK}</math>, <math>AE\cdot(BF + DF - BD)\geqslant 2\sqrt3S_{EFAK}</math>, adding these together, we get the desired result.
 
Similarly, we have <math>CE\cdot(BD + DF - BF)\geqslant 2\sqrt3S_{CDEK}</math>, <math>AE\cdot(BF + DF - BD)\geqslant 2\sqrt3S_{EFAK}</math>, adding these together, we get the desired result.

Latest revision as of 20:03, 9 November 2024

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 \begin{align*} \sqrt{x(y+z)(4yz+zx+xy)}&=\frac1{\sqrt3}\sqrt{3x(y+z)(4yz+zx+xy)} \\ & \leqslant \frac{3x(y+z)+4yz+zx+xy}{2\sqrt3} \\ & =\frac{2(yz+zx+xy)}{\sqrt3}, \end{align*} 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.

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