# Difference between revisions of "2013 AMC 12A Problems/Problem 19"

## Problem

In $\bigtriangleup ABC$, $AB = 86$, 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 $BC$? $\textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ 61 \qquad \textbf{(E)} \ 72$

## Solution

### Solution 1

Let CX=x, BX=y. Let the circle intersect AC at D and the diameter including AD intersect the circle again at E. Use power of a point on point C to the circle centered at A.

So CX*CB=CD*CE x(x+y)=(97-86)(97+86) x(x+y)=3*11*61.

Obviously x+y>x so we have three solution pairs for (x,x+y)=(1,2013),(3,671),(11,183),(33,61). By the Triangle Inequality, only x+y=61 yields a possible length of BX+CX=BC.

Therefore, the answer is D) 61.

### Solution 2

Let $h$ be the perpendicular from $B$ to $AC$, $AX=x$, $XC=y$, then by Pythagorean Theorem, $h^2 + (x/2)^2 = 86^2$ $h^2 + (x/2 + y)^2 = 97^2$

Subtracting the two equations, we get $(x+y)y = (97-86)(97+86)$,

then the rest is similar to the above solution by power of points.

### Solution 3

Let $x$ represent $CX$, and let $y$ represent $BX$. Since the circle goes through $B$ and $X$, $AB$ = $AX$ = 86. Then by Stewart's Theorem, $xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.$ $x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x$ $x^2 + xy + 86^2 = 97^2$

(Since $y$ cannot be equal to 0, dividing both sides of the equation by $y$ is allowed.) $x(x+y) = (97+86)(97-86)$ $x(x+y) = 2013$

The prime factors of 2013 are 3, 11, and 61. Obviously, $x < x+y$. In addition, by the Triangle Inequality, $BC < AB + AC$, so $x+y < 183$. Therefore, $x$ must equal 33, and $x+y$ must equal 61. $\boxed{D}$

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