Difference between revisions of "2013 AMC 12A Problems/Problem 19"
Icematrix2 (talk | contribs) |
Sugar rush (talk | contribs) |
||
Line 49: | Line 49: | ||
~dolphin7 | ~dolphin7 | ||
+ | ==Video Solution== | ||
+ | https://youtu.be/zxW3uvCQFls | ||
+ | |||
+ | ~sugar_rush | ||
== See also == | == See also == |
Latest revision as of 00:36, 25 November 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
Video Solution
~sugar_rush
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.