2013 AMC 12A Problem 19

Attempt Problem 19 of the 2013 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 12A solutions, or check the answer key.

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19.

In ABC,\triangle ABC, AB=86,AB = 86, and AC=97.AC = 97. A circle with center AA and radius ABAB intersects BC\overline{BC} at points BB and X.X. Moreover BX\overline{BX} and CX\overline{CX} have integer lengths. What is BC?BC?

1111

2828

3333

6161

7272

Answer: D
Concepts:power of a pointprime factorization
Difficulty rating: 2200
Solution:

By the Power of a Point Theorem, BCCX=AC2AB2BC\cdot CX = AC^2 - AB^2 where ABAB is the radius. Thus BCCX=972862=2013.BC\cdot CX = 97^2 - 86^2 = 2013.

Since BC=BX+CXBC = BX + CX and CXCX are integers, they are complementary factors of 2013=31161.2013 = 3\cdot 11\cdot 61. As CX<BC<AB+AC=183,CX \lt BC \lt AB + AC = 183, the only possibility is CX=33CX = 33 and BC=61.BC = 61.

Thus, the correct answer is D.

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