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 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. | + | [[File:2001AIME13.jpg]] |
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| + | 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>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 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. | ||
Using Ptolemy's theorem, | Using Ptolemy's theorem, | ||
| + | <cmath>AB(CD) + AC(BD) = AD(BC)</cmath> | ||
<cmath>22(22) + 22x = (x + 20)^2</cmath> | <cmath>22(22) + 22x = (x + 20)^2</cmath> | ||
<cmath>484 + 22x = x^2 + 40x + 400</cmath> | <cmath>484 + 22x = x^2 + 40x + 400</cmath> | ||
Revision as of 03:23, 20 May 2014
Problem
In a certain circle, the chord of a 
-degree arc is 
centimeters long, and the chord of a 
-degree arc is 
centimeters longer than the chord of a 
-degree arc, where 
The length of the chord of a -degree arc is 
centimeters, where 
and 
are positive integers. Find 
Solution
Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from the three chords of the three -degree arcs and the chord of the -degree arc. The diagonals of this trapezoid turn out to be the two chords of the -degree arcs. Let 
, 
, and 
be the chords of the -degree arcs, and let 
be the chord of the -degree arc. Also let 
be equal to the chord of the -degree arc. Hence, the length of the chords of the -degree arcs can be represented as 
, as given in the problem.
Using Ptolemy's theorem,
![\[AB(CD) + AC(BD) = AD(BC)\]](http://latex.artofproblemsolving.com/e/4/9/e49713edcd6de6361f8d5ecf5df57120b91b00e0.png)
![\[22(22) + 22x = (x + 20)^2\]](http://latex.artofproblemsolving.com/0/8/7/087337c424f79b56d34a2243708e7a2c12c3e657.png)
![\[484 + 22x = x^2 + 40x + 400\]](http://latex.artofproblemsolving.com/f/1/2/f12b51222b3897dbd9bab052e0061dc17465bd65.png)
![\[0 = x^2 + 18x - 84\]](http://latex.artofproblemsolving.com/f/1/5/f15dd545e67e4393fd29d2a684400d699c39d06d.png)
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)
simplifies to 
which equals 
Thus, the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
