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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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    The task says that i$n$ our case its not.2012-04-06

3 Answers 3

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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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    @bgins: that works well, though multiplying everything by 2 would make the midpoints have integer values too.2012-04-06
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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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    @Gabor: My reply is essentially the same but does not use vectors2012-04-06
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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} $

enter image description here

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    Thanks, this is great, but the hexagon is not regular. Only convex.2012-04-06