1983 AIME Problems/Problem 15

Problem

The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the minor arc $AB$ is a rational number. If this fraction is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$ ?

Solution

Let $A$ be any fixed point on circle $O$ and let $AD$ be a chord of circle $O$ . The locus of midpoints $N$ of the chord $AD$ is a circle $P$ , with diameter $AD$ . Generally, the circle $P$ can intersect the chord $BC$ at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle $P$ is tangent to BC at point N.

Let M be the midpoint of the chord $BC$ such that $BC=3$ . From right angle triangle $OMB$ , $OM = \sqrt{OB^2 - BM^2} =4$ . Thus, $\tan \angle BOM = \frac{BM}{OM} = \frac 3 4$ .

Notice that the distance $OM$ equals $PN + PO \cos AOM = r(1 + \cos AOM)$ (Where $r$ is the radius of circle P). Evaluating this, $\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5$ . From $\cos \angle AOM$ , we see that $\tan \angle AOM = \displaystyle \frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3$

Next, notice that $\angle AOB = \angle AOM - \angle BOM$ . We can therefore apply the tangent subtraction formula to obtain , $\tan AOB = \displaystyle \frac{\tan AOM - \tan BOM}{1 + \tan AOM \cdot \tan AOM} = \displaystyle \frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}$ . It follows that $\sin AOB =\displaystyle \frac{7^2}{\sqrt{7^2+24^2}} = \frac{7}{25}$ , resulting in an answer of $7 \cdot 25=175$ .

1983 AIME (Problems • Answer Key • Resources)
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1983 AIME Problems/Problem 15

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