1989 AHSME Problems/Problem 19

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?

$\mathrm{(A) \ 6 } \qquad \mathrm{(B) \frac{18}{\pi^2} } \qquad \mathrm{(C) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(D) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(E) \frac{9}{\pi^2}(\sqrt{3}+3) }$

Solution

The three arcs make up the entire circle, so the circumference of the circle is $3+4+5=12$ and the radius is $\frac{12}{2\pi}=\frac{6}{\pi}$ . Also, the lengths of the arcs are proportional to their corresponding central angles. Thus, we can write the values of the arcs as $3\theta$ , $4\theta$ , and $5\theta$ for some $\theta$ . By Circle Angle Sum, we obtain $3\theta+4\theta+5\theta=360$ . Solving yields $\theta=30$ . Thus, the angles of the triangle are $90$ , $120$ , and $150$ . Using $[ABC]=\frac{1}{2}ab\sin{C}$ , we obtain $\frac{r^2}{2}(\sin{90}+\sin{120}+\sin{150})$ . Substituting $\frac{6}{\pi}$ for $r$ and evaluating yields $\frac{9}{\pi}(\sqrt{3}+3)\implies{\boxed{E}}$ .

-Solution by thecmd999

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1989 AHSME Problems/Problem 19

Solution