Difference between revisions of "2001 AIME I Problems/Problem 13"
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Let <math>z=\frac{d}{2}</math>, <math>R</math> be the circumradius, and <math>a</math> be the length of 3d degree chord. Using the extended sine law, we obtain: | Let <math>z=\frac{d}{2}</math>, <math>R</math> be the circumradius, and <math>a</math> be the length of 3d degree chord. Using the extended sine law, we obtain: | ||
− | <cmath>22= | + | <cmath>22=2R\sin(z)</cmath> |
− | <cmath>20+a= | + | <cmath>20+a=2R\sin(2z)</cmath> |
− | <cmath>a= | + | <cmath>a=2R\sin(3z)</cmath> |
− | Dividing the second from the first we get <math>cos(z)=\frac{20+a}{44}</math> | + | Dividing the second from the first we get <math>\cos(z)=\frac{20+a}{44}</math> |
− | By the triple angle formula we can manipulate the third equation as follows: <math>a=2R\times sin(3z)=\frac{22}{sin(z)} \times ( | + | By the triple angle formula we can manipulate the third equation as follows: <math>a=2R\times \sin(3z)=\frac{22}{\sin(z)} \times (3\sin(z)-4\sin^3(z)) = 22(3-4\sin^2(z))=22(4\cos^2(z)-1)=\frac{(20+a)^2}{22}-22</math> |
Solving the quadratic equation gives the answer. | Solving the quadratic equation gives the answer. | ||
Revision as of 22:32, 25 October 2021
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
Solution 1
![2001AIME13.png](https://wiki-images.artofproblemsolving.com//0/01/2001AIME13.png)
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,
We can then apply the quadratic formula to find the positive root to this equation since polygons obviously cannot have sides of negative length.
simplifies to
which equals
Thus, the answer is
.
Solution 2
Let and
be the circumradius. From the given information,
Dividing the latter by the former,
We want to find
From
this is equivalent to
Using the quadratic formula, we find that the desired length is equal to
so our answer is
Solution 3
Let ,
be the circumradius, and
be the length of 3d degree chord. Using the extended sine law, we obtain:
Dividing the second from the first we get
By the triple angle formula we can manipulate the third equation as follows:
Solving the quadratic equation gives the answer.
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.