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

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Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from three chords of three <math>d</math>-degree arcs and one chord of one <math>3d</math>-degree arc. The diagonals of this trapezoid turn out to be two chords of two <math>2d</math>-degree arcs. Let <math>AB</math>, <math>AC</math>, and <math>AD</math> be the chords of the <math>d</math>-degree arcs, and let <math>CD</math> be the chord of the <math>3d</math>-degree arc. Also let <math>x</math> be equal to the chord length of the <math>3d</math>-degree arc. Hence, the length of the chords, <math>AD</math> and <math>BC</math>, of the <math>2d</math>-degree arcs can be represented as <math>x + 20</math>, as given in the problem.  
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Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from three chords of three <math>d</math>-degree arcs and one chord of one <math>3d</math>-degree arc. The diagonals of this trapezoid turn out to be two chords of two <math>2d</math>-degree arcs. Let <math>AB</math>, <math>AC</math>, and <math>BD</math> be the chords of the <math>d</math>-degree arcs, and let <math>CD</math> be the chord of the <math>3d</math>-degree arc. Also let <math>x</math> be equal to the chord length of the <math>3d</math>-degree arc. Hence, the length of the chords, <math>AD</math> and <math>BC</math>, of the <math>2d</math>-degree arcs can be represented as <math>x + 20</math>, as given in the problem.  
  
 
Using Ptolemy's theorem,
 
Using Ptolemy's theorem,

Revision as of 19:22, 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

2001AIME13.png

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

Using Ptolemy's theorem,

\[AB(CD) + AC(BD) = AD(BC)\] \[22x + 22(22) = (x + 20)^2\] \[22x + 484 = x^2 + 40x + 400\] \[0 = x^2 + 18x - 84\]

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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