1983 AIME Problems/Problem 15
Contents
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 . 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 .
Solution 2
The above solution works, but is quite messy and somewhat difficult to follow. This solution provides a diagram to scale, and the motivation behind the solution.
First of all, where did the statement " is the only chord starting at and bisected by " come from? What is its significance in this problem? What is the criterion for this statement to be true?
We consider the locus of midpoints of the chords from . It is well known that this is the circle with diameter , where is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with ratio with center . Thus, the locus is the result of the dilation with ratio of circle with center . Let the center of this circle be .
Aha! Now we see. is bisected by if they cross at some point on the circle. Moreover, since is the only chord, must be tangent to the circle .
The rest of this problem is straight forward.
Our goal is to find where is the midpoint of . Then we have and . Let be the projection of onto , and similarly be the projection of onto . Then it remains to find so we can use the sine addition formula.
As is a radius of circle , , and similarly, . Since , . Thus, .
From here, we see that is a dilation of about center with ratio , so .
Lastly, we apply the formula:
Thus, our answer is .
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |