Convex hexagon $A_1A_2A_3A_4A_5A_6$ lies in the interior of convex hexagon $B_1B_2B_3B_4B_5B_6$ such that $A_1A_2 \parallel B_1B_2$, $A_2A_3 \parallel B_2B_3$,..., $A_6A_1 \parallel B_6B_1$. Prove that the areas of simple hexagons $A_1B_2A_3B_4A_5B_6$ and $B_1A_2B_3A_4B_5A_6$ are equal. (A simple hexagon is a hexagon which does not intersect itself.) Proposed by Hirad Aalipanah - Mahdi Etesamifard
Problem
Source:
Tags: IGO, 2018 igo, Iran, geometry
InternetPerson10
20.09.2018 12:51
The two hexagons have common area $A_1A_2A_3A_4A_5A_6$. Subtract this area from both of the given hexagons.
We are now left with six triangles from each hexagon. Show that the sum of the areas of the six triangles have equal area.
I suggest you try this out first. If you're really stuck, here's the hint.Draw in lines $A_1B_1$, $A_2B_2$, $\cdots$, $A_6B_6$. You now have six trapezoids.
Please try out the problem first. If you're really stuck, use the big hint. SolutionNotice that triangles $A_1A_2B_1$ and $A_1A_2B_2$ have equal area - as they have the same values for base and height (they're part of trapezoid $A_1A_2B_2B_1$. Repeating this process for the other five trapezoids, we find that the areas $A_1B_2A_3B_4A_5B_6-A_1A_2A_3A_4A_5A_6$ and $B_1A_2B_3A_4B_5A_6-A_1A_2A_3A_4A_5A_6$, each composed of six triangles, are of equal area. Thus the two simple hexagons have the same area.