2013 AMC 10A Problems/Problem 23
- The following problem is from both the 2013 AMC 12A #19 and 2013 AMC 10A #23, so both problems redirect to this page.
Contents
Problem
In , , and . A circle with center and radius intersects at points and . Moreover and have integer lengths. What is ?
Solutions
Solution 1 (Number Theoretic Power of a Point)
Let , , and meets the circle at and , with on . Then . Using the Power of a Point (Secant-Secant Power Theorem), we get that . We know that , so is either , , or . We also know that by the triangle inequality on . Thus, is so we get that .
Solution 2 (Stewart's Theorem)
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 .
Solution 3
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 .
Solution 4
We first draw the height of isosceles triangle and get two equations by the Pythagorean Theorem. First, . Second, . Subtracting these two equations, we get . We then add to both sides to get . We then complete the square to get . Because and are both integers, we get that is a square number. Simple guess and check reveals that . Because equals , therefore . We want , so we get that .
Solution 5
Let be the foot of the altitude from to Since is isosceles and the answer is by the Pythagorean Theorem. Only is a factor of such that
~dolphin7
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=f1nxu8MWWKc
Video Solution by OmegaLearn
https://youtu.be/NsQbhYfGh1Q?t=2692
~ pi_is_3.14
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 10 Problems and Solutions |
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.