IMO Shortlist 1995 problem G5
Dodao/la:
arhiva2. travnja 2012. Let
![ABCDEF](/media/m/9/f/e/9fe205b534135e3a700ffb54d8b96cb0.png)
be a convex hexagon with
![AB = BC = CD](/media/m/5/0/e/50ed9bba77ff13f40fc7bf7b77a4395c.png)
and
![DE = EF = FA](/media/m/3/f/8/3f868f004f461485be77a276b6e6e4cd.png)
, such that
![\angle BCD = \angle EFA = \frac {\pi}{3}](/media/m/c/4/2/c42ba7726a40573ae11ef647a633fd1f.png)
. Suppose
![G](/media/m/f/e/b/feb7f8fc95cee3c3a479382202e06a86.png)
and
![H](/media/m/4/c/0/4c0872a89da410a25f00b86366efece7.png)
are points in the interior of the hexagon such that
![\angle AGB = \angle DHE = \frac {2\pi}{3}](/media/m/2/1/1/21133100a7a4030022f269168c9ddcbf.png)
. Prove that
![AG + GB + GH + DH + HE \geq CF](/media/m/e/e/0/ee0e206292be911949388d47d6defae5.png)
.
%V0
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 $AG + GB + GH + DH + HE \geq CF$.
Izvor: Međunarodna matematička olimpijada, shortlist 1995