1999 AIME Problems/Problem 4

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 $\displaystyle m+n.$

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 $((43/99)\cdot(1/2))/2$ , 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.

1999 AIME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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1999 AIME Problems/Problem 4

Contents

Problem

Solution

Solution 2

See also