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 distances from one of the labeled points and the 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>. | ||
| + | |||
| + | By the [[Pythagorean theorem]], | ||
| + | :<math>x^2 + y^2 = (\frac{43}{99})^2</math> | ||
| + | |||
| + | Also, | ||
| + | :<math>x + y + \frac{43}{99} = 1</math> | ||
| + | :<math>x^2 + 2xy + y^2 = (\frac{56}{99})^2</math> | ||
| + | |||
| + | Substituting, | ||
| + | :<math>(\frac{43}{99})^2 + 2xy = (\frac{56}{99})^2</math> | ||
| + | :<math>2xy = \frac{(56 + 43)(56 - 43)}{99^2} = \frac{13}{99}</math> | ||
| + | |||
| + | Thus, the area of the octagon is <math>1 - \frac{13}{99} = \frac{86}{99}</math>, so <math>m + n = 185</math>. | ||
== See also == | == See also == | ||
| − | |||
| − | |||
* [[1999 AIME Problems]] | * [[1999 AIME Problems]] | ||
| + | |||
| + | {{AIME box|year=1999|num-b=3|num-a=5}} | ||
Revision as of 22:38, 9 February 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 | ||
