TITLE: Geometry question with a lot of triangles
QUESTION [6 upvotes]: Let $ABCDEF$ be a convex hexagon, and denote by $P, Q, R, S, T, U$ the midpoints of the sides AB, BC, CD, DE, EF, FA respectively. Suppose that the areas of the triangles $ABR, BCS, CDT, DEU, EFP$ and $FAQ$ are 12, 34, 56, 12, 34 and 56 respectively. Find the area of the hexagon.
I tried to draw the complete diagram at first but it turned out to have a lot of triangles so it is a bit confusing.
Then I tried to draw pairs of triangles with the same areas only but that doesn't really help.
Thank you

REPLY [2 votes]: Since $R$ is the midpoint of the segment $CD$ we have 
$$ \textrm{Area} ABR = 1/2 ( \textrm{Area} (ABC) + \textrm{Area} (ABD))$$
Similarly for all the other triangles considered above. Now sum up all the areas given in the problem. Using the formulas just stated, this is $1/2$ multiplied by the sum of all the areas on the right hand side. This sum will be an integer multiple of the area of the hexagon. To see quickly what this multiple is, assume that the hexagon is regular. Hence, we get the sum of all the areas on the right hand side $= 6 \times 1/2 \cdot \textrm{Area hexagon}$. Therefore
$$12+ 34+ 56+12+34 + 56 = 1/2 \times 6 \times 1/2 \textrm{Area hexagon}=3/2 \cdot\textrm{Area hexagon} $$.