Difference between revisions of "2013 AMC 12A Problems/Problem 19"
(→See also) |
(→See also) |
||
Line 56: | Line 56: | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 16:01, 3 July 2013
Problem
In , , and . A circle with center and radius intersects at points and . Moreover and have integer lengths. What is ?
Solution
Solution 1
Let CX=x, BX=y. Let the circle intersect AC at D and the diameter including AD intersect the circle again at E. Use power of a point on point C to the circle centered at A.
So CX*CB=CD*CE x(x+y)=(97-86)(97+86) x(x+y)=3*11*61.
Obviously x+y>x so we have three solution pairs for (x,x+y)=(1,2013),(3,671),(11,183),(33,61). By the Triangle Inequality, only x+y=61 yields a possible length of BX+CX=BC.
Therefore, the answer is D) 61.
Solution 2
Let be the perpendicular from to , , , then by Pythagorean Theorem,
Subtracting the two equations, we get ,
then the rest is similar to the above solution by power of points.
Solution 3
Let represent , and let represent . Since the circle goes through and , = = 86. Then by Stewart's Theorem,
(Since cannot be equal to 0, dividing both sides of the equation by is allowed.)
The prime factors of 2013 are 3, 11, and 61. Obviously, . In addition, by the Triangle Inequality, , so . Therefore, must equal 33, and must equal 61.
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.