2021 WSMO Team Round/Problem 6

Since this is a dodecagon, there are $12$ sides.

Thus, the angle measure for an angle on the regular polygon is:

$\frac{10 \cdot 180}{12} = 150$

Next, we can draw a two lines down from $A$ and from $B$ that are perpendicular to $LC$ .

We can denote these intersection point as $A_1$ and $B_1$ , respectively.

Since the angle measure of $LA A_1$ is $\angle{60}$ and $m \angle{BA A_1} = 90$ , we can say that the shaded area is equal to $4(2 \cdot [LA A_1] + [ABB_1 A_1])$

$[LA A_1]$ is equal to $\frac{5}{2} \cdot \frac{5\sqrt{3}}{2} = \frac{25\sqrt{3}}{4}$ because it is a 30-60-90 triangle.

Next, $[AB B_1 A_1] = 5 \cdot \frac{5}{2} = \frac{25}{2}$

So the whole shaded area is equal to $(\frac{25\sqrt{3}}{4}) \cdot 4 + (\frac{25}{2}) \cdot 4 = 25\sqrt{3} + 50 = 50 + 25\sqrt{3}$

Thus, $a + b + c = 50 + 25 + 3 = 78$

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2021 WSMO Team Round/Problem 6