Difference between revisions of "2001 AIME I Problems/Problem 13"
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===Solution 3=== | ===Solution 3=== | ||
− | Let <math>z=\frac{d}{2} | + | Let <math>z=\frac{d}{2}</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=2Rsin(z)</cmath> | <cmath>22=2Rsin(z)</cmath> | ||
<cmath>20+a=2Rsin(2z)</cmath> | <cmath>20+a=2Rsin(2z)</cmath> | ||
<cmath>a=2Rsin(3z)</cmath> | <cmath>a=2Rsin(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: | + | By the triple angle formula we can manipulate the third equation as follows: </math>a=2R\times sin(3z)=\frac{22}{sin(z)} \times (3sin(z)-4sin^3(z)) = 22(3-4sin^2(z))=22(4cos^2(z))=\frac{(20+a)^2}{22}-22$ |
− | </math>a=2R\times sin(3z)=\frac{22}{sin(z)} \times (3sin(z)-4sin^3(z)) = 22(3-4sin^2(z))=22(4cos^2(z))$ | + | Solving the quadratic equation gives the answer. |
− | Solving | ||
== See also == | == See also == |
Revision as of 05:07, 18 June 2020
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
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 , Racos(z)=\frac{20+a}{44}a=2R\times sin(3z)=\frac{22}{sin(z)} \times (3sin(z)-4sin^3(z)) = 22(3-4sin^2(z))=22(4cos^2(z))=\frac{(20+a)^2}{22}-22$ 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 |
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