Difference between revisions of "1999 AIME Problems/Problem 4"
m (→Solution: \left( \right)) |
m (→Solution: \left(\right) again) |
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Line 14: | Line 14: | ||
Substituting, | Substituting, | ||
− | :<math>(\frac{43}{99})^2 + 2xy = \left(\frac{56}{99}\right)^2</math> | + | :<math>\left(\frac{43}{99}\right)^2 + 2xy = \left(\frac{56}{99}\right)^2</math> |
:<math>2xy = \frac{(56 + 43)(56 - 43)}{99^2} = \frac{13}{99}</math> | :<math>2xy = \frac{(56 + 43)(56 - 43)}{99^2} = \frac{13}{99}</math> | ||
Revision as of 16:53, 9 September 2007
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
.
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 |