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

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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

19.

\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

Power of a Point from

C

gives

BC\cdot CX = AC^2 - AB^2

AC^2 - AB^2 = 2013 = 3\cdot 11\cdot 61;

factor with

CX \lt BC \lt AB + AC = 183

By the Power of a Point Theorem,

BC\cdot CX = AC^2 - AB^2

where

AB

is the radius. Thus

BC\cdot CX = 97^2 - 86^2 = 2013.

Since

BC = BX + CX

and

CX

are integers, they are complementary factors of

2013 = 3\cdot 11\cdot 61.

CX \lt BC \lt AB + AC = 183,

the only possibility is

CX = 33

and

BC = 61.

Thus, the correct answer is D.