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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 squares 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 integers.  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>
  
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[[Image:AIME_1999_Problem_4.png]]
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<asy>
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== Solution ==
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//code taken from thread for problem
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real alpha = 25;
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pair W=dir(225), X=dir(315), Y=dir(45), Z=dir(135), O=origin;
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pair w=dir(alpha)*W, x=dir(alpha)*X, y=dir(alpha)*Y, z=dir(alpha)*Z;
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draw(W--X--Y--Z--cycle^^w--x--y--z--cycle);
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pair A=intersectionpoint(Y--Z, y--z),
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C=intersectionpoint(Y--X, y--x),
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E=intersectionpoint(W--X, w--x),
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G=intersectionpoint(W--Z, w--z),
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B=intersectionpoint(Y--Z, y--x),
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D=intersectionpoint(Y--X, w--x),
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F=intersectionpoint(W--X, w--z),
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H=intersectionpoint(W--Z, y--z);
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dot(O);
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label("$O$", O, SE);
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label("$A$", A, dir(O--A));
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label("$B$", B, dir(O--B));
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label("$C$", C, dir(O--C));
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label("$D$", D, dir(O--D));
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label("$E$", E, dir(O--E));
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label("$F$", F, dir(O--F));
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label("$G$", G, dir(O--G));
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label("$H$", H, dir(O--H));</asy>
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__TOC__
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== Solution 1 ==
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Triangles <math>AOB</math>, <math>BOC</math>, <math>COD</math>, etc. are congruent by symmetry (you can prove it rigorously by using the power of a point to argue that exactly two chords of length <math>1</math> in the circumcircle of the squares pass through <math>B</math>, etc.), and each area is <math>\frac{\frac{43}{99}\cdot\frac{1}{2}}{2}</math>. Since the area of a triangle is <math>bh/2</math>, 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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== Solution 2 ==
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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]],
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<cmath>x^2 + y^2 = \left(\frac{43}{99}\right)^2</cmath>
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 +
Also,
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<cmath>\begin{align*}x + y + \frac{43}{99} &= 1\\
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x^2 + 2xy + y^2 &= \left(\frac{56}{99}\right)^2\end{align*}</cmath>
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 +
Substituting,
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<cmath>\begin{align*}\left(\frac{43}{99}\right)^2 + 2xy &= \left(\frac{56}{99}\right)^2 \\
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2xy = \frac{(56 + 43)(56 - 43)}{99^2} &= \frac{13}{99} \end{align*}</cmath>
 +
 
 +
Thus, the area of the octagon is <math>1 - \frac{13}{99} = \frac{86}{99}</math>, so <math>m + n = \boxed{185}</math>.
  
 
== See also ==
 
== See also ==
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* [[1999_AIME_Problems/Problem_3|Previous Problem]]
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{{AIME box|year=1999|num-b=3|num-a=5}}
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* [[1999_AIME_Problems/Problem_5|Next Problem]]
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−
* [[1999 AIME Problems]]
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[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 12:27, 12 December 2020

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 $m+n.$

[asy] //code taken from thread for problem real alpha = 25; pair W=dir(225), X=dir(315), Y=dir(45), Z=dir(135), O=origin; pair w=dir(alpha)*W, x=dir(alpha)*X, y=dir(alpha)*Y, z=dir(alpha)*Z; draw(W--X--Y--Z--cycle^^w--x--y--z--cycle); pair A=intersectionpoint(Y--Z, y--z), C=intersectionpoint(Y--X, y--x), E=intersectionpoint(W--X, w--x), G=intersectionpoint(W--Z, w--z), B=intersectionpoint(Y--Z, y--x), D=intersectionpoint(Y--X, w--x), F=intersectionpoint(W--X, w--z), H=intersectionpoint(W--Z, y--z); dot(O); label("$O$", O, SE); label("$A$", A, dir(O--A)); label("$B$", B, dir(O--B)); label("$C$", C, dir(O--C)); label("$D$", D, dir(O--D)); label("$E$", E, dir(O--E)); label("$F$", F, dir(O--F)); label("$G$", G, dir(O--G)); label("$H$", H, dir(O--H));[/asy]

Solution 1

Triangles $AOB$, $BOC$, $COD$, etc. are congruent by symmetry (you can prove it rigorously by using the power of a point to argue that exactly two chords of length $1$ in the circumcircle of the squares pass through $B$, etc.), and each area is $\frac{\frac{43}{99}\cdot\frac{1}{2}}{2}$. Since the area of a triangle is $bh/2$, the area of all $8$ of them is $\frac{86}{99}$ and the answer is $\boxed{185}$.

Solution 2

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 $m + n = \boxed{185}$.

See also

1999 AIME (Problems • Answer Key • Resources)
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