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Difference between revisions of "1999 AIME Problems/Problem 4"

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== Problem ==
 
== Problem ==
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The two squares shown share the same center <math>\displaystyle O_{}</math> and have sides of length 1. The length of <math>\displaystyle \overline{AB}</math> is <math>\displaystyle 43/99</math> and the area of octagon <math>\displaystyle ABCDEFGH</math> is <math>\displaystyle m/n,</math> where <math>\displaystyle m_{}</math> and <math>\displaystyle n_{}</math> are relatively prime positive integers.  Find <math>\displaystyle m+n.</math>
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The two [[square]]s shown share the same [[center]] <math>\displaystyle O_{}</math> and have sides of length 1. The length of <math>\displaystyle \overline{AB}</math> is <math>\displaystyle 43/99</math> and the [[area]] of octagon <math>\displaystyle ABCDEFGH</math> is <math>\displaystyle m/n,</math> where <math>\displaystyle m_{}</math> and <math>\displaystyle n_{}</math> are [[relatively prime]] [[positive]] [[integer]]s.  Find <math>\displaystyle m+n.</math>
  
 
[[Image:AIME_1999_Problem_4.png]]
 
[[Image:AIME_1999_Problem_4.png]]
 
== Solution ==
 
== Solution ==
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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>.  
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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>.  
  
 
By the [[Pythagorean theorem]],
 
By the [[Pythagorean theorem]],
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Also,
 
Also,
 
:<math>x + y + \frac{43}{99} = 1</math>
 
:<math>x + y + \frac{43}{99} = 1</math>
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:<math>x^2 + 2xy + y^2 = (\frac{56}{99})^2</math>
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:<math>x^2 + 2xy + y^2 = \left(\frac{56}{99}\right)^2</math>
  
 
Substituting,
 
Substituting,
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:<math>(\frac{43}{99})^2 + 2xy = (\frac{56}{99})^2</math>
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:<math>(\frac{43}{99})^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>
  
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== See also ==
 
== See also ==
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* [[1999 AIME Problems]]
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{{AIME box|year=1999|num-b=3|num-a=5}}
  
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{{AIME box|year=1999|num-b=3|num-a=5}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 18:55, 8 March 2007

Problem

The two squares shown share the same center $\displaystyle O_{}$ and have sides of length 1. The length of $\displaystyle \overline{AB}$ is $\displaystyle 43/99$ and the area of octagon $\displaystyle ABCDEFGH$ is $\displaystyle m/n,$ where $\displaystyle m_{}$ and $\displaystyle n_{}$ are relatively prime positive integers. Find $\displaystyle m+n.$

AIME 1999 Problem 4.png

Solution

Define the two possible distances from one of the labeled points and the corners of the square upon which the point lies as $x$ and $y$. The area of the octagon (by subtraction of areas) is $1 - 4(\frac{1}{2}xy) = 1 - 2xy$.

By the Pythagorean theorem,

$x^2 + y^2 = (\frac{43}{99})^2$

Also,

$x + y + \frac{43}{99} = 1$
$x^2 + 2xy + y^2 = \left(\frac{56}{99}\right)^2$

Substituting,

$(\frac{43}{99})^2 + 2xy = \left(\frac{56}{99}\right)^2$
$2xy = \frac{(56 + 43)(56 - 43)}{99^2} = \frac{13}{99}$

Thus, the area of the octagon is $1 - \frac{13}{99} = \frac{86}{99}$, so $m + n = 185$.

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
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