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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

Plato
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1 Answers1

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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 $$12+ 34+ 56+12+34 + 56 = 1/2 \times 6 \times 1/2 \textrm{Area hexagon}=3/2 \cdot\textrm{Area hexagon} $$.

orangeskid
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  • Hello, why since R is the midpoint of the segment CD we have that equation? – Plato Sep 08 '14 at 16:51
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    Draw the heights $h_C$, $h_R$ and $h_D$ from $C$, $R$ and $D$ onto the line $AB$. We obtain a trapezoid with bases $h_C$, $h_D$ and midline $h_R$. – orangeskid Sep 08 '14 at 16:55
  • It's not clear why the sum of all the areas on the right hand side will be an integer multiple of the area of the hexagon. – Christian Blatter Sep 08 '14 at 17:41
  • A change of notation will make it clearer. Let $A_i$ the vertices of the hexagon and $M_i$ the middle of the size $A_i A_{i+1}$. $$(A_i A_{i+1} M_{i+2}) = 1/2( (A_i A_{i+1} A_{i+2}) + (A_i A_{i+1} A_{i+3})$$ sum up, on the right hand side get $$\frac{1}{2} ( \sum_i (A_i A_{i+1} A_{i+2}) + \sum ( (A_i A_{i+1} A_{i+3})) = $$ $$=\frac{1}{2} ( \sum_i (A_{i+1} A_{i+2} A_{i+3}) + \sum_i (A_{i} A_{i+1} A_{i+3}))= 1/2 \sum_i (A_i A_{i+1} A_{i+2}A_{i+3})$$ – orangeskid Sep 08 '14 at 22:44