1995 AIME Problems/Problem 14
Problem
In a circle of radius , two chords of length intersect at a point whose distance from the center is . The two chords divide the interior of the circle into four regions. Two of these regions are bordered by segments of unequal lengths, and the area of either of them can be expressed uniquely in the form where and are positive integers and is not divisible by the square of any prime number. Find
Solution
Let the center of the circle be , and the two chords be and intersecting at , such that . Let be the midpoint of . Then .
By the Pythagorean Theorem, , and . Then is a right triangle, so . Thus , and by the Law of Cosines,
It follows that is an equilateral triangle, so . The desired area can be broken up into two regions, and the region bounded by and minor arc . The former can be found by Heron's formula to be . The latter is the difference between the area of sector and the equilateral , or .
Thus, the desired area is , and .
Note: the area of can be more easily found by using the sine method .
-NL008
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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