Difference between revisions of "1999 AIME Problems/Problem 4"
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[[Image:AIME_1999_Problem_4.png]] | [[Image:AIME_1999_Problem_4.png]] | ||
== Solution == | == Solution == | ||
| − | Define the two possible [[distance]]s from one of the labeled points and the [[vertex|corners]] of the square upon which the point lies as <math>x</math> and <math>y</math>. The area of the [[octagon]] (by [[subtraction]] of areas) is <math>1 - 4(\frac{1}{2}xy) = 1 - 2xy</math>. | + | Define the two possible [[distance]]s from one of the labeled points and the [[vertex|corners]] of the square upon which the point lies as <math>x</math> and <math>y</math>. The area of the [[octagon]] (by [[subtraction]] of areas) is <math>1 - 4\left(\frac{1}{2}xy\right) = 1 - 2xy</math>. |
By the [[Pythagorean theorem]], | By the [[Pythagorean theorem]], | ||
| − | :<math>x^2 + y^2 = (\frac{43}{99})^2</math> | + | :<math>\displaystyle x^2 + y^2 = \left(\frac{43}{99}\right)^2 \displaystyle </math> |
Also, | Also, | ||
Revision as of 14:09, 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 | ||
