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

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==Problem==
 
==Problem==
 
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>.
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==Solution==
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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>. --tigerzhang
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==See Also==
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{{IMO box|year=1995|num-b=4|num-a=6}}

Latest revision as of 20:24, 4 July 2024

Problem

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

Solution

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

See Also

1995 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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