Question

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 $BC?$

$11$

$28$

$33$

$61$

$72$

Answer: D

Concepts:power of a point prime factorization triangle inequality

Difficulty rating: 2010

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.

2013 AMC 10A Problem 23

Answer: D

Difficulty rating: 2010

Solution:

Problem 23 in Other Years