Difference between revisions of "1999 AIME Problems/Problem 4"
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By the [[Pythagorean theorem]], | By the [[Pythagorean theorem]], | ||
− | :<math> | + | :<math>x^2 + y^2 = \left(\frac{43}{99}\right)^2</math> |
Also, | Also, | ||
Line 18: | Line 18: | ||
Thus, the area of the octagon is <math>1 - \frac{13}{99} = \frac{86}{99}</math>, so <math>m + n = 185</math>. | Thus, the area of the octagon is <math>1 - \frac{13}{99} = \frac{86}{99}</math>, so <math>m + n = 185</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | Each of the triangle <math>AOB</math>, <math>BOC</math>, <math>COD</math>, etc. are congruent, and their areas are <math>(43/99\cdot2)/2</math>, since the area of a triangle is bh/2, so the area of all 8 of them is 86/99 and the answer is 185. | ||
== See also == | == See also == |
Revision as of 08:51, 24 February 2008
Contents
[hide]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
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 .
Solution 2
Each of the triangle , , , etc. are congruent, and their areas are , since the area of a triangle is bh/2, so the area of all 8 of them is 86/99 and the answer is 185.
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 | ||
All AIME Problems and Solutions |