2021 WSMO Team Round/Problem 6
Since this is a dodecagon, there are 
sides.
Thus, the angle measure for an angle on the regular polygon is:
Next, we can draw a two lines down from 
and from 
that are perpendicular to 
.
We can denote these intersection point as 
and 
, respectively.
Since the angle measure of 
is 
and 
, we can say that the shaded area is equal to ![$4(2 \cdot [LA A_1] + [ABB_1 A_1])$](http://latex.artofproblemsolving.com/9/6/c/96cba5daa70bfbc05ba1f2f0093c52108adfa66e.png)
The area of 
\frac{5}{2} \cdot \frac{5\sqrt{3}}{2} = \frac{25\sqrt{3}}{4}$because it is a 30-60-90 triangle.
Next,$ (Error compiling LaTeX. Unknown error_msg)[AB B_1 A_1] = 5 \cdot \frac{5}{2} = \frac{25}{2}
(\frac{25\sqrt{3}}{4}) \cdot 4 + (\frac{25}{2}) \cdot 4 = 25\sqrt{3} + 50 = 50 + 25\sqrt{3}
a + b + c = 50 + 25 + 3 = 103$
