The corners of a five-pointed star lie on the sides of a square ABCD with the side length 1, that two corners of the star coincide with the corner points A and D. Further corner points of the star lie in each case in the interior of the edges AB, BC and CD.

The area of the middle pentagon is $\frac{1}{12}$. Calculate the sum of the areas of the gray-colored triangles.

Image of the figure

I've got absolutly no idea how to solve this.


Shaded triangles around a pentagon

Label the vertices as shown in diagram above. Let $A_g$ be the area in grey and $A_p = \frac{1}{12}$ be the area of central pentagon. It is easy to see $$A_g + 2A_p = \verb/Area/(AEDH) + \verb/Area/(HFG)$$

Since $DG \parallel AF$, $$\verb/Area/(AFD) = \verb/Area/(AFG) \quad\implies\quad \verb/Area/(AHD) = \verb/Area/(HFG)$$ i.e. the two triangles outlined in red has same area. As a result, $$\begin{align} & A_g + 2A_p = \verb/Area/(AEDH) + \verb/Area/(AHD) = \verb/Area/(AED) = \frac12\\ \implies & A_g = \frac12 - 2A_p = \frac12 - \frac{2}{12} = \frac13 \end{align} $$


The idea is to use the several way to compute the grey areas.

I don't want to multiply the notations to name all the lines intersections.

I will call however $A_g$ the sum of all grey areas, $A_{ext}$ the sum of all white exterior areas and $A_m$ the area of the middle pentagon.

If you calculate the area of each of the grey triangles based on the triangles and quadrilateres using the sides of the square, you notice that each exterior area will be used twice.

In fact you will have every time grey triangle = large triangle - exterior parts in the said triangle (in some cases we are dealing with quadrilateres, but it does not change anything).

Then you have $A_g=\frac{3}{2}-2A_{ext}$

Since you have the total area of the square being $1$, you can write it like this


But $1-A_{ext}=A_g+A_m$!

Thus $A_g=\frac 12-2A_m=\frac13$


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