2013 AMC 12A Problems/Problem 19: Difference between revisions
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Therefore, the answer is '''D) 61.''' | Therefore, the answer is '''D) 61.''' | ||
===Solution 2=== | |||
Revision as of 17:51, 8 February 2013
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.
