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

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== Problem ==
 
== Problem ==
 
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>
 
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>
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 +
== Solution ==
 +
Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from the three chords of the three <math>d</math>-degree arcs and the chord of the <math>3d</math>-degree arc. The diagonals of this trapezoid turn out to be the two chords of the <math>2d</math>-degree arcs. Let <math>x</math> be equal to the chord of the <math>3d</math>-degree arc. Hence, the length of the chords of the <math>2d</math>-degree arcs can be represented as <math>x + 20</math>, as given in the problem.
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 +
Using Ptolemy's theorem,
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 +
<cmath>22(22) + 22x = (x + 20)^2</cmath>
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<cmath>484 + 22x = x^2 + 40x + 400</cmath>
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<cmath>0 = x^2 + 18x - 84</cmath>
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 +
We can then apply the quadratic formula to find the positive root to this equation since polygons obviously cannot have sides of negative length.
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<cmath>x = \[\frac{-18 + \sqrt{18^2 + 4(84)}}{2}}</cmath>
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<cmath>x = \[\frac{-18 + \sqrt{660}}{2}}</cmath>
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<math>x</math> simplifies to <math>\frac{-18 + 2\sqrt{165}}{2},</math> which equals <math>-9 + \sqrt{165}.</math> Thus, the answer is <math>9 + 165 = \boxed{174}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 00:26, 20 May 2014

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

Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from the three chords of the three $d$-degree arcs and the chord of the $3d$-degree arc. The diagonals of this trapezoid turn out to be the two chords of the $2d$-degree arcs. Let $x$ be equal to the chord of the $3d$-degree arc. Hence, the length of the chords of the $2d$-degree arcs can be represented as $x + 20$, as given in the problem.

Using Ptolemy's theorem,

\[22(22) + 22x = (x + 20)^2\] \[484 + 22x = x^2 + 40x + 400\] \[0 = x^2 + 18x - 84\]

We can then apply the quadratic formula to find the positive root to this equation since polygons obviously cannot have sides of negative length. 

x = \[\frac{-18 + \sqrt{18^2 + 4(84)}}{2}} (Error compiling LaTeX. Unknown error_msg)
x = \[\frac{-18 + \sqrt{660}}{2}} (Error compiling LaTeX. Unknown error_msg)

$x$ simplifies to $\frac{-18 + 2\sqrt{165}}{2},$ which equals $-9 + \sqrt{165}.$ Thus, the answer is $9 + 165 = \boxed{174}$.

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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