1995 IMO Problems/Problem 5

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

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

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