# Difference between revisions of "1995 IMO Problems/Problem 5"

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Let <math>ABCDEF</math> be a convex hexagon with <math>AB=BC=CD</math> and <math>DE=EF=FA</math>, such that <math>\angle BCD=\angle EFA=\frac{\pi}{3}</math>. Suppose <math>G</math> and <math>H</math> are points in the interior of the hexagon such that <math>\angle AGB=\angle DHE=\frac{2\pi}{3}</math>. Prove that <math>AG+GB+GH+DH+HE\ge CF</math>. | Let <math>ABCDEF</math> be a convex hexagon with <math>AB=BC=CD</math> and <math>DE=EF=FA</math>, such that <math>\angle BCD=\angle EFA=\frac{\pi}{3}</math>. Suppose <math>G</math> and <math>H</math> are points in the interior of the hexagon such that <math>\angle AGB=\angle DHE=\frac{2\pi}{3}</math>. Prove that <math>AG+GB+GH+DH+HE\ge CF</math>. | ||

==Solution== | ==Solution== | ||

+ | Draw <math>AE</math> and <math>BD</math> to make equilateral <math>\triangle EFA</math> and <math>\triangle BCD</math>, and draw points <math>I</math> and <math>J</math> such that <math>IA=IB</math>, <math>JD=JE</math>, directed angle <math>\measuredangle IAB=-\measuredangle CDB</math>, and directed angle <math>\measuredangle JDE=-\measuredangle FAE</math> to make equilateral <math>\triangle AIB</math> and <math>\triangle DJE</math>. Notice that <math>G</math> is on the circumcircle of <math>\triangle AIB</math> and <math>H</math> is on the circumcircle of <math>\triangle DJE</math>. By Ptolemy, <math>GA+GB=GI</math> and <math>HD+HE=HJ</math>. So, <cmath>AG+GB+GH+DH+HE=IG+GH+HJ</cmath>. Notice that octagon <math>AIBCDJEF</math> is symmetric about <math>\overline{BE}</math>. So, <math>IG+GH+HJ\ge IJ=CF</math>. |

## Revision as of 20:18, 5 July 2020

## Problem

Let be a convex hexagon with and , such that . Suppose and are points in the interior of the hexagon such that . Prove that .

## Solution

Draw and to make equilateral and , and draw points and such that , , directed angle , and directed angle to make equilateral and . Notice that is on the circumcircle of and is on the circumcircle of . By Ptolemy, and . So, . Notice that octagon is symmetric about . So, .