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Difference between revisions of "2001 AIME I Problems/Problem 13"

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== Problem ==
 
== Problem ==
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In a certain circle, the chord of a <math>d</math>-degree arc is 22 centimeters long, and the chord of a <math>2d</math>-degree arc is 20 centimeters longer than the chord of a <math>3d</math>-degree arc, where <math>d < 120.</math>  The length of the chord of a <math>3d</math>-degree arc is <math>- m + \sqrt {n}</math> centimeters, where <math>m</math> and <math>n</math> are positive integers.  Find <math>m + n.</math>
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In a certain [[circle]], the [[chord]] of a <math>d</math>-degree arc is <math>22</math> centimeters long, and the chord of a <math>2d</math>-degree arc is <math>20</math> centimeters longer than the chord of a <math>3d</math>-degree arc, where <math>d < 120.</math>  The length of the chord of a <math>3d</math>-degree arc is <math>- m + \sqrt {n}</math> centimeters, where <math>m</math> and <math>n</math> are positive integers.  Find <math>m + n.</math>
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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{{image}}
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We let our chord of degree <math>d</math> be <math>\overline{AB}</math>, of degree <math>2d</math> be <math>\overline{AC}</math>, and of degree <math>3d</math> be <math>\overline{AD}</math>. We are given that <math>AC = AD + 20</math>. Since <math>AB = BC = CD = 22</math>, quadrilateral <math>ABCD</math> is a [[cyclic quadrilateral|cyclic]] [[isosceles trapezoid]], and so <math>BD = AC = AD + 20</math>. By [[Ptolemy's Theorem]], we have
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<cmath>\begin{align*}
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AB \cdot CD + AD \cdot BD &= AC \cdot BD\\
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22^2 + 22x = (x+20)^2 &\Longrightarrow x^2 + 18x - 84 = 0\\
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x = \frac{-18 + \sqrt{18^2 + 4\cdot 84}}{2} &= -9 + \sqrt{165}\end{align*}</cmath>
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Therefore, the answer is <math>m+n = \boxed{174}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2001|n=I|num-b=12|num-a=14}}
 
{{AIME box|year=2001|n=I|num-b=12|num-a=14}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 21:19, 13 June 2008

Problem

In a certain circle, the chord of a $d$-degree arc is $22$ centimeters long, and the chord of a $2d$-degree arc is $20$ centimeters longer than the chord of a $3d$-degree arc, where $d < 120.$ The length of the chord of a $3d$-degree arc is $- m + \sqrt {n}$ centimeters, where $m$ and $n$ are positive integers. Find $m + n.$

Solution

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We let our chord of degree $d$ be $\overline{AB}$, of degree $2d$ be $\overline{AC}$, and of degree $3d$ be $\overline{AD}$. We are given that $AC = AD + 20$. Since $AB = BC = CD = 22$, quadrilateral $ABCD$ is a cyclic isosceles trapezoid, and so $BD = AC = AD + 20$. By Ptolemy's Theorem, we have \begin{align*} AB \cdot CD + AD \cdot BD &= AC \cdot BD\\ 22^2 + 22x = (x+20)^2 &\Longrightarrow x^2 + 18x - 84 = 0\\ x = \frac{-18 + \sqrt{18^2 + 4\cdot 84}}{2} &= -9 + \sqrt{165}\end{align*} Therefore, the answer is $m+n = \boxed{174}$.

See also

2001 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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