2013 AMC 12A Problems/Problem 19

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.

-Solution 3 by fowlmaster