2001 AIME II Problems/Problem 6
Problem
Square is inscribed in a circle. Square has vertices and on and vertices and on the circle. If the area of square is , then the area of square can be expressed as where and are relatively prime positive integers and . Find .
Solution 1(Pythagorean Theorem)
Let be the center of the circle, and be the side length of , be the side length of . By the Pythagorean Theorem, the radius of .
Now consider right triangle , where is the midpoint of . Then, by the Pythagorean Theorem,
Thus (since lengths are positive, we discard the other root). The ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, so , and the answer is .
Another way to proceed from is to note that is the quantity we need; thus, we divide by to get
This is a quadratic in , and solving it gives . The negative solution is extraneous, and so the ratio of the areas is and the answer is .
Solution 2 (Coordinates)
Let point be the top-left corner of square and the rest of the vertices be arranged, in alphabetical order, in a clockwise arrangement from there. Let have coordinates and the side length of square be . Let = and diameter go through the midpoint of . Since a diameter always bisects a chord perpendicular to it, = and since and must be symmetric around the diameter, and it follows that Hence the side of square has length . has coordinates and has coordinates We know that point must be on the circle - hence it must satisfy the circle equation. Since the center of the circle is at the center of the square and has radius , half the diagonal of the square, follows as the circle equation. Then substituting coordinates of into the equation, . Simplifying and factoring, we get Since would imply , and in the problem, we must use the other factor. We get , meaning the ratio of areas = = = Then .
See also
2001 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.