Mock AIME 4 2006-2007 Problems/Problem 4

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Problem

Points $A$, $B$, and $C$ are on the circumference of a unit circle so that the measure of $\widehat{AB}$ is $72^{\circ}$, the measure of $\widehat{BC}$ is $36^{\circ}$, and the measure of $\widehat{AC}$ is $108^\circ$. The area of the triangular shape bounded by $\widehat{BC}$ and line segments $\overline{AB}$ and $\overline{AC}$ can be written in the form $\frac{m}{n} \cdot \pi$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution


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Let the center of the circle be $O$. The area of the desired region is easily seen to be that of sector $BOC$ plus the area of triangle $AOB$ minus the area of triangle $AOC$. Using the area formula $K_{\triangle XYZ} = \frac{1}{2} XY \cdot YZ \cdot \sin Y$ to compute the areas of the two triangles, this is $\pi \cdot \frac{36}{360} + \frac{1}{2}\sin 72^\circ - \frac{1}{2}\sin108^{\circ} = \frac{1}{10}\cdot \pi$, so the answer is $1 + 10 = 011$.