# Area of convex hexagon consisting of midpoints of diagonals of unit area convex hexagon

We have A1A2A3A4A5A6 unit area convex hexagon. The midpoints of A1A3, A2A4, A3A5, A4A6, A5A1, A6A2 diagonals also make a convex hexagon. What's the area of the latter hexagon?

Sorry for my english, it is not my first language, but it is hopefully understandabe. Thanks for any help in advance!

EDIT: The hexagon unfortunately is not regular.

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It might not be relevant, but is the hexagon regular? – Aryabhata Apr 6 '12 at 7:39
No its not. That is exactly my problem that its not. – Gabor Magyar Apr 6 '12 at 8:14
What happens if the smaller hexagon is self-intersecting? – Henry Apr 6 '12 at 9:26
The task says that in our case its not. – Gabor Magyar Apr 6 '12 at 9:38

If the co-ordinates of the hexagon are $(x_i,y_i)$ then the area of the larger hexagon is $$\frac{| x_1 y_2 - x_2 y_1 + x_2 y_3 - x_3 y_2 + x_3 y_4 - x_4 y_3 + x_4 y_5 - x_5 y_4 + x_5 y_6 - x_6 y_5 + x_6 y_1 - x_1 y_6 |}{2}.$$

The area of the smaller hexagon can be calculated similarly but replacing $x_i$ by $\frac{x_{i-1}+x_{i+1}}{2}$ and $y_i$ by $\frac{y_{i-1}+y_{i+1}}{2}$ making the obvious adjustments for indices above 6 or below 1.

Dividing the area of the smaller hexagon by that of the larger hexagon and tidying up will give the answer you want of $\frac14$.

This presupposes that the smaller hexagon is simple enough for this to be meaningful. There is probably a simpler counterexample, but starting with the points $$(20,20),(20,72),(164,76),(126,24)(88,8),(42,6)$$ the larger hexagon is convex, while the smaller hexagon is self-intersecting.

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A "simpler" example having a self-intersecting second/derived ("subdiagonal midpoint") hexagon could be $\{(1,0),(2,0),(4,1),(1,4),(0,2),(0,1)\}$. – bgins Apr 6 '12 at 10:00
@bgins: that works well, though multiplying everything by 2 would make the midpoints have integer values too. – Henry Apr 6 '12 at 10:44

Choose an origin $O$ in the interior of the smaller hexagon and define vectors $a_i:=OA_i$ numbered cyclically and counterclockwise. Then the area $A$ of the large hexagon is given by $$A={1\over2}\sum_i (a_{i-1}\wedge a_i)\ ,$$ and the area $A'$ of the small hexagon is given by $$A'={1\over2}\sum_i\Bigl({a_{i-1}+a_{i+1}\over2}\wedge{a_i+a_{i+2}\over2}\Bigr)\ .$$ In the second sum there is some cancellation, and it will finally turn out that $A'={1\over4} A$.

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That's great and thank you. but could you please explain it without wedging? I'm not supposed to know what that is yet. (I just checked wikipedia, so I get it, but still, I can't hand it in.) Or if there is an other solution without vectors that would be great too. – Gabor Magyar Apr 6 '12 at 9:42
@Gabor: My reply is essentially the same but does not use vectors – Henry Apr 6 '12 at 10:45

I assume that hexagon is regular .

Let's denote side of small hexagon as $a$ (see picture below) and side of large hexagon as $b$ .Since triangles $\Delta A_1A_2A_3$ and $\Delta A_4A_5A_6$ are isoscales triangles with two angles of $30^{\circ}$ it follows :

$2a=2b-2\cdot b \cdot \sin 30^{\circ}=b \Rightarrow a=\frac{b}{2}$

Hence :

$A=\frac{3}{2} \cdot a^2 \sqrt 3=\frac{3 \sqrt 3}{8}\cdot b^2=\frac{1}{4}$

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The question does not say that the hexagon is regular... – Aryabhata Apr 6 '12 at 7:39
Thanks, this is great, but the hexagon is not regular. Only convex. – Gabor Magyar Apr 6 '12 at 8:14