Difference between revisions of "2001 AIME I Problems/Problem 13"
(→Solution) |
m (→Solution) |
||
| Line 15: | Line 15: | ||
We can then apply the quadratic formula to find the positive root to this equation since polygons obviously cannot have sides of negative length. | We can then apply the quadratic formula to find the positive root to this equation since polygons obviously cannot have sides of negative length. | ||
| − | <cmath>x = | + | <cmath>x = \frac{-18 + \sqrt{18^2 + 4(84)}}{2}</cmath> |
| − | <cmath>x = | + | <cmath>x = \frac{-18 + \sqrt{660}}{2}</cmath> |
<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>. | <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>. | ||
Revision as of 10:28, 12 March 2015
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 three chords of three -degree arcs and one chord of one -degree arc. The diagonals of this trapezoid turn out to be two chords of two -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 length of the -degree arc. Hence, the length of the chords, 
and 
, 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)
![\[22x + 22(22) = (x + 20)^2\]](http://latex.artofproblemsolving.com/7/3/d/73d80bfed17d4f79cb1eb5e69a10849cd802677f.png)
![\[22x + 484 = x^2 + 40x + 400\]](http://latex.artofproblemsolving.com/e/2/3/e231d9b344d55300e625575317f4da19463a6f1a.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.
![\[x = \frac{-18 + \sqrt{18^2 + 4(84)}}{2}\]](http://latex.artofproblemsolving.com/b/9/5/b95cc316dc3871bbea07f30f421c37e37867f7a8.png)
![\[x = \frac{-18 + \sqrt{660}}{2}\]](http://latex.artofproblemsolving.com/7/1/b/71b7033bbcda8f696de0b52090b9fa78467c4937.png)
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. 
