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 | + | 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 a <math>3d</math>-degree arc. The diagonals of this trapezoid turn out to be 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 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 03:32, 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 three chords of three -degree arcs and one chord of a
-degree arc. The diagonals of this trapezoid turn out to be 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 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,
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.