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 
?
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, 
.
Notice that the distance 
equals 
(Where 
is the radius of circle P). Evaluating this, 
. From 
, we see that 
Next, notice that 
. We can therefore apply the tangent subtraction formula to obtain , 
. It follows that 
, resulting in an answer of 
.
See also
| 1983 AIME (Problems • Resources) | ||
| Preceded by Problem 14 | Followed by Last question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
