1995 IMO Problems/Problem 5

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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

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