2013 AMC 10A Problem 23

Problem 23 of 25HarderGeometry Number Theory

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?

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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.

More practice

Concepts: power of a point · prime factorization · triangle inequality

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.