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== | ||
+ | For a convex hexagon <math>ABCDEF</math> with an area <math>S</math>, prove that: | ||
+ | |||
+ | <center><math>AC\cdot(BD + BF - DF) + CE\cdot(BD + DF - BF) + AE\cdot(BF + DF - BD)\geq 2\sqrt {3}</math>.</center> | ||
+ | |||
+ | ==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> | ||
+ | ==Returning to the original problem== | ||
[[File:1509090255851ede02a626c22a.gif]] | [[File:1509090255851ede02a626c22a.gif]] | ||
− | + | 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 with an area
, prove that:

Lemma
First, let's set aside the original problem and introduce the "Gergonne point" and its distance to the vertices.
Let the incircle of touch
,
,
at
,
,
respectively. By Ceva's theorem, it is easy to see that
,
,
are concurrent at a point
, which we call the Gergonne point of
.
Let ,
,
. As shown in the figure, by Stewart's theorem, we have
In
, by Menelaus' theorem, we have
Solving this, we get
Noting that by AM-GM inequality, we have
Thus, we obtain the following lemma.
Lemma: Let be the Gergonne point of
, then
Returning to the original problem
Let be the Gergonne point of
. As shown in the figure, by the lemma, we have
Similarly, we have
,
, adding these together, we get the desired result.