2001 AIME II Problems/Problem 6
Problem
Square is inscribed in a circle. Square has vertices and on and vertices and on the circle. The ratio of the area of square to the area of square can be expressed as where and are relatively prime positive integers and . Find .
Solution
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 .
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 |
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