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

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By the [[Pythagorean theorem]],
 
By the [[Pythagorean theorem]],
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:<math>\displaystyle x^2 + y^2 = \left(\frac{43}{99}\right)^2 \displaystyle </math>
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:<math>x^2 + y^2 = \left(\frac{43}{99}\right)^2</math>
  
 
Also,
 
Also,
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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>.
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 +
== Solution 2 ==
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 +
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

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 $1 - 4\left(\frac{1}{2}xy\right) = 1 - 2xy$.

By the Pythagorean theorem,

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

Also,

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

Substituting,

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

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

Solution 2

Each of the triangle $AOB$, $BOC$, $COD$, etc. are congruent, and their areas are $(43/99\cdot2)/2$, 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
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