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

Revision as of 18:39, 4 July 2013

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 $m+n.$

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 $\boxed{185}$ .

Revision as of 22:29, 24 March 2011 (view source) Danielguo94 (talk \| contribs) (→‎Solution 2) ← Older edit		Revision as of 18:39, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
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	[[Category:Intermediate Geometry Problems]]		[[Category:Intermediate Geometry Problems]]
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