1999 AIME Problems/Problem 4
Problem
The two squares shown share the same center and have sides of length 1. The length of is and the area of octagon is where and are relatively prime positive integers. Find
Solution
Simple Solution
Triangles , , , etc. are congruent by symmetry (you can prove it rigorously by using the power of a point to argue that exactly two chords of length in the circumcircle of the squares pass through , etc.), and each area is . Since the area of a triangle is , the area of all of them is and the answer is .
Other Solution
Define the two possible distances from one of the labeled points and the corners of the square upon which the point lies as and . The area of the octagon (by subtraction of areas) is .
By the Pythagorean theorem,
Also,
Substituting,
Thus, the area of the octagon is , so .
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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