1989 AMC 12 Problem 19

Question

A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths

3,

4,

and

5.

What is the area of the triangle?

6

\frac{18}{\pi^2}

\frac9{\pi^2}(\sqrt3-1)

\frac9{\pi^2}(\sqrt3+1)

\frac9{\pi^2}(\sqrt3+3)

The circumference is

12,

so first find the radius and the three central angles

Split the triangle into three triangles having the circle’s center as a common vertex

The radius is

R=\frac{12}{2\pi}=\frac{6}{\pi}.

The arc lengths

3,4,5

give central angles

\frac{\pi}{2},\frac{2\pi}{3},\frac{5\pi}{6}.

Their sines sum to

1+\frac{\sqrt3}{2}+\frac{1}{2}=\frac{3+\sqrt3}{2}.

Splitting the triangle at the center, its area is

\begin{aligned} K&=\frac12\cdot\frac{36}{\pi^2} \cdot\frac{3+\sqrt3}{2}\\ &=\frac9{\pi^2}(\sqrt3+3). \end{aligned}

Thus the correct answer is E.

Problem 19 in other years