2013 AMC 10A Problem 23

Attempt Problem 23 of the 2013 AMC 10A 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 10A solutions, or check the answer key.

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23.

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 factorizationtriangle inequality
Small hint:
Use power of a point from CC
Big hint:
Factor AC2−AB2AC^2-AB^2 and use the triangle inequality to choose the valid factor pair
Solution:
By power of a point from CC, CB⋅CXCB\cdot CX =AC2−AB2=AC^2-AB^2 =972−862=97^2-86^2.
This equals (97−86)(97+86)(97-86)(97+86) =11⋅183=11\cdot183 =2013=2013 =3⋅11⋅61=3\cdot11\cdot61.
Both CXCX and BXBX are integers, so BC=BX+CXBC=BX+CX is an integer factor paired with CXCX. Also CX<BC<86+97=183CX<BC<86+97=183, so the only possible pair is CX=33CX=33, BC=61BC=61.
Thus, D is the correct answer.
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