# Area of the intersection of regular pentagons

I was playing around in Geogebra and came up with an interesting problem.

Take an arbitrary point $A$ and draw a circle with center $A$. Then draw any line through $A$. Call the points where the circle intersects with the line $B$ and $C$. Construct two regular pentagons with base $AB$ and $AC$, such that the pentagons intersect in a kite-shaped region. What is the area of the region divided by the area of either of the pentagons?

I created a diagram with the circle removed: Now, using Mathematica, trigonometry, and analytical geometry, I was able to come up with a solution:

$$2-\frac{4 \sqrt{5}}{5}$$

I won't reproduce the entire solution here, I'll just give a quick sketch. Referring to the above figure, I considered $A$ to be the center of the coordinate system, and let each pentagon have side length one. I found the $y$-coordinate of $H$ by adding the apothem and the radius of the pentagon. The slope of line $HG$ must be $\tan 144^\circ$, so I was able to find the equation of the line, and therefore, the $y$-coordinate of $J$ which is equal to $JA$. I found $GF = DF - DG$. After that, I had the two diagonals of the kite so I found the area. Finally, I divided by the area of the pentagon.

As you might imagine, this solution isn't really feasible by hand.

Is there a solution which uses only geometry (or clean trigonometry, where "clean" is subjectively defined)?

• You do not need the circle, but could start with $B$ and $C$, having $A$ as their midpoint. Mar 14, 2014 at 21:25
• It is not difficult to show that if the pentagon has side $1$ then its area is $\frac{5}{4\tan(36^\circ)}$ while the area of the kite is $\frac{\sin(18^\circ)\sin(108^\circ)}{\sin(126^\circ)}=\frac{\tan(36^\circ)}{2}$. So the ratio is $\frac25 \tan^2(36^\circ) = \frac25\frac{(3-\phi)}{(1+\phi)}$. This will lead to your result. Mar 14, 2014 at 22:59
• @Henry, thanks, that works. Can you post it as an answer, or should I? Mar 15, 2014 at 16:56

It is not difficult to show that if the pentagon has side 1 then its area is $\frac{5}{4\tan(36^\circ)}$ while the area of the kite is $\frac{\sin(18^\circ)\sin(108^\circ)}{\sin(126^\circ)}=\frac{\tan(36^\circ)}{2}$. So the ratio is $$\frac25 \tan^2(36^\circ) = \frac25\frac{(3-\phi)}{(1+\phi)}.$$ This will lead to your result.