2013 AMC 12A Problems/Problem 19
Problem
In 
, 
, and 
. A circle with center 
and radius 
intersects 
at points 
and 
. Moreover 
and 
have integer lengths. What is 
?
Solution
Solution 1
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 
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 | |
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