2013 AMC 10A Problem 23

Below is the professionally curated solution for Problem 23 of the 2013 AMC 10A, 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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Concepts:power of a point prime factorization triangle inequality

Difficulty rating: 2010

23.

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

$11$

$28$

$33$

$61$

$72$

Solution:

By power of a point from $C$ , $CB\cdot CX=AC^2-AB^2=97^2-86^2$ .

This equals $(97-86)(97+86)=11\cdot183=2013=3\cdot11\cdot61$ .

Both $CX$ and $BX$ are integers, so $BC=BX+CX$ is an integer factor paired with $CX$ . Also $CX<BC<86+97=183$ , so the only possible pair is $CX=33$ , $BC=61$ .

Thus, D is the correct answer.

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