Difference between revisions of "2013 AMC 12A Problems/Problem 19"
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Let <math>CX=x, BX=y</math>. Let the circle intersect <math>AC</math> at <math>D</math> and the diameter including <math>AD</math> intersect the circle again at <math>E</math>. | Let <math>CX=x, BX=y</math>. Let the circle intersect <math>AC</math> at <math>D</math> and the diameter including <math>AD</math> intersect the circle again at <math>E</math>. | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Number Theory]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:05, 27 October 2020
Contents
Problem
In , , and . A circle with center and radius intersects at points and . Moreover and have integer lengths. What is ?
Solution
Solution 1 (Number theoretic power of a point)
Let . Let the circle intersect at and the diameter including intersect the circle again at . Use power of a point on point C to the circle centered at A.
So .
Obviously so we have three solution pairs for . By the Triangle Inequality, only yields a possible length of .
Therefore, the answer is D) 61.
Solution 2
Let , , and meet the circle at and , with on . Then . Using the Power of a Point, we get that . We know that , and that by the triangle inequality on . Thus, we get that
Solution 3
Let represent , and let represent . Since the circle goes through and , . Then by Stewart's Theorem,
(Since cannot be equal to , dividing both sides of the equation by is allowed.)
The prime factors of are , , and . Obviously, . In addition, by the Triangle Inequality, , so . Therefore, must equal , and must equal
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2013amc12a/357
~dolphin7
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.