Problem 5

Let

ABCDEF

be a convex hexagon with

AB=BC=CD

and

DE=EF=FA

, such that

\angle BCD=\angle EFA=60^\circ

. Suppose that

G

and

H

are points in the interior of the hexagon such that

\angle AGB=\angle DHE=120^\circ

. Prove that

AG+GB+GH+DH+HE\ge CF

Step 3 of 4: Apply Ptolemy's inequality

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In plain words

The 120-degree conditions make the broken paths dominate the reflected diagonals.

C'G\le AG+GB,\quad HF'\le DH+HE

Detailed analysis

Apply Ptolemy's inequality to the equilateral triangle $C'AB$ together with $G$ and to $F'DE$ together with $H$ . Using $\angle AGB=\angle DHE=120^\circ$ gives $C'G\le AG+GB$ and $HF'\le DH+HE$ .