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1983 AIME Problems/Problem 15

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Problem

The adjoining figure shows two intersecting chords in a circle, with on minor arc . Suppose that the radius of the circle is , that , and that is bisected by . Suppose further that is the only chord starting at which is bisected by . It follows that the sine of the minor arc is a rational number. If this fraction is expressed as a fraction in lowest terms, what is the product ?

Image:1983_AIME-15.png

Solution

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

Let M be the midpoint of the chord such that . From right triangle , . Thus, \tan \angle BOM = \frac{BM}{OM} = \frac 3 4.

Notice that the distance equals PN + PO \cos AOM = r(1 + \cos AOM) (Where 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 , we see that \tan \angle AOM =\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 =\frac{\tan AOM - \tan BOM}{1 + \tan AOM \cdot \tan AOM} =\frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}. It follows that \sin AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}, resulting in an answer of .

See also

1983 AIME (ProblemsResources)
Preceded by
Problem 14
Followed by
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