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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 [[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>
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The two [[square]]s shown share the same [[center]] <math>O_{}</math> and have sides of length 1. The length of <math>\overline{AB}</math> is <math>43/99</math> and the [[area]] of octagon <math>ABCDEFGH</math> is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are [[relatively prime]] [[positive]] [[integer]]s.  Find <math>m+n.</math>
  
 
[[Image:AIME_1999_Problem_4.png]]
 
[[Image:AIME_1999_Problem_4.png]]
 +
__TOC__
 
== Solution ==
 
== Solution ==
 +
=== Solution 1 ===
 
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\left(\frac{1}{2}xy\right) = 1 - 2xy</math>.  
 
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\left(\frac{1}{2}xy\right) = 1 - 2xy</math>.  
  
 
By the [[Pythagorean theorem]],
 
By the [[Pythagorean theorem]],
−
:<math>x^2 + y^2 = \left(\frac{43}{99}\right)^2</math>
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<cmath>x^2 + y^2 = \left(\frac{43}{99}\right)^2</cmath>
  
 
Also,
 
Also,
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:<math>x + y + \frac{43}{99} = 1</math>
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<cmath>\begin{align*}x + y + \frac{43}{99} &= 1\\
−
:<math>x^2 + 2xy + y^2 = \left(\frac{56}{99}\right)^2</math>
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x^2 + 2xy + y^2 &= \left(\frac{56}{99}\right)^2\end{align*}</cmath>
  
 
Substituting,
 
Substituting,
−
:<math>\left(\frac{43}{99}\right)^2 + 2xy = \left(\frac{56}{99}\right)^2</math>
+
<cmath>\begin{align*}\left(\frac{43}{99}\right)^2 + 2xy &= \left(\frac{56}{99}\right)^2 \\
−
:<math>2xy = \frac{(56 + 43)(56 - 43)}{99^2} = \frac{13}{99}</math>
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2xy = \frac{(56 + 43)(56 - 43)}{99^2} &= \frac{13}{99} \end{align*}</cmath>
  
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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>.
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Thus, the area of the octagon is <math>1 - \frac{13}{99} = \frac{86}{99}</math>, so <math>m + n = \boxed{185}</math>.
  
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== Solution 2 ==
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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)\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>.
−
Each of the triangle <math>AOB</math>, <math>BOC</math>, <math>COD</math>, etc. are congruent, and their areas are <math>((43/99)\cdot(1/2))/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 13:16, 26 April 2008

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 $((43/99)\cdot(1/2))/2$, 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
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All AIME Problems and Solutions
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