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

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=== Solution 2 ===
 
=== 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>\frac{43}{99}\cdot(1/2))/2</math>, since the area of a triangle is <math>bh/2</math>, so the area of all <math>8</math> of them is <math>\frac{86}{99}</math> and the answer is <math>\boxed{185}</math>.
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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>\frac{\frac{43}{99}\cdot\frac{1}{2}}{2}</math>, since the area of a triangle is <math>bh/2</math>, so the area of all <math>8</math> of them is <math>\frac{86}{99}</math> and the answer is <math>\boxed{185}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 22:29, 24 March 2011

Problem

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

AIME 1999 Problem 4.png

Solution

Solution 1

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, \begin{align*}x + y + \frac{43}{99} &= 1\\ x^2 + 2xy + y^2 &= \left(\frac{56}{99}\right)^2\end{align*}

Substituting, \begin{align*}\left(\frac{43}{99}\right)^2 + 2xy &= \left(\frac{56}{99}\right)^2 \\ 2xy = \frac{(56 + 43)(56 - 43)}{99^2} &= \frac{13}{99} \end{align*}

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

Solution 2

Each of the triangle $AOB$, $BOC$, $COD$, etc. are congruent, and their areas are $\frac{\frac{43}{99}\cdot\frac{1}{2}}{2}$, since the area of a triangle is $bh/2$, so the area of all $8$ of them is $\frac{86}{99}$ and the answer is $\boxed{185}$.

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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