Summer Problem Solving Marathon Solution #18

Call the triangle ABC, and the center of the circumscribed circle O. Draw the lines OA and OB. The result is the triangle OAB. Angle O is a 120 degree angle (because it cuts off a third of the circle), and side AB is length p/3.

Now drop a perpendicular from O to AB. Call the point of intersection D. Consider triangle ODA. This is a 30-60-90 right triangle. (It is a right triangle because OD is a perpendicular to AB, and since OAB is isoceles, the perpendicular bisects angle O, so DOA is a 60 degree angle).

In this 30-60-90 right triangle, side AD is of length p/6. This side corresponds to the 60 degree angle. So side OA is (p/6)(2/√3) = p/(3√3).

But OA is a radius of the circumscribed circle. So the area of that circle is π(p/3√3)² = (πp²)/27.