Difference between revisions of "2011 AIME II Problems/Problem 10"
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We are given that <math>EF</math> has length 12, so, using the [[Law of Cosines]] with <math>\triangle EPF</math>: | We are given that <math>EF</math> has length 12, so, using the [[Law of Cosines]] with <math>\triangle EPF</math>: | ||
− | <math>12^2 = a^2 + b^2 - 2ab \cos (\angle EPF) = a^2 + b^2 - 2ab \cos (\angle EPO | + | <math>12^2 = a^2 + b^2 - 2ab \cos (\angle EPF) = a^2 + b^2 - 2ab \cos (\angle EPO - \angle FPO)</math> |
Substituting for <math>a</math> and <math>b</math>, and applying the Cosine of Sum formula: | Substituting for <math>a</math> and <math>b</math>, and applying the Cosine of Sum formula: |
Revision as of 17:50, 25 March 2014
Contents
[hide]Problem 10
A circle with center has radius 25. Chord of length 30 and chord of length 14 intersect at point . The distance between the midpoints of the two chords is 12. The quantity can be represented as , where and are relatively prime positive integers. Find the remainder when is divided by 1000.
Solution 1
Let and be the midpoints of and , respectively, such that intersects .
Since and are midpoints, and .
and are located on the circumference of the circle, so .
The line through the midpoint of a chord of a circle and the center of that circle is perpendicular to that chord, so and are right triangles (with and being the right angles). By the Pythagorean Theorem, , and .
Let , , and be lengths , , and , respectively. OEP and OFP are also right triangles, so , and
We are given that has length 12, so, using the Law of Cosines with :
Substituting for and , and applying the Cosine of Sum formula:
and are acute angles in right triangles, so substitute opposite/hypotenuse for sines and adjacent/hypotenuse for cosines:
Combine terms and multiply both sides by : $144 x^2 = 2 x^4 - 976 x^2 - 2 (x^2 - 400) (x^2 - 576) + 960 \sqrt{(x^2 - 400)(x^2 - 576)$ (Error compiling LaTeX. Unknown error_msg)
Combine terms again, and divide both sides by 64:
Square both sides:
This reduces to ; .
Solution 2
We begin as in the first solution. Once we see that has side lengths 12,24, and 25, we can compute its area with Heron's formula:
.
So its circumradius is . Since is cyclic with diameter , we have , so and the answer is .
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.