2008 AMC 12B Problems/Problem 17

Let the coordinates of $A$ be $(m, m^2)$ and the coordinates of $C$ be $(n, n^2)$ . Since the line $AB$ is parallel to the $x$ -axis, the coordinates of $B$ must be $(-m, m^2)$ . Then the slope of line $AC$ is $\frac{m^2-n^2}{m-n}=\frac{(m+n)(m-n)}{m-n}=m+n$ . The slope of line $BC$ is $\frac{m^2-n^2}{-m-n}=-\frac{(m+n)(m-n)}{m+n}=-(m-n)$ .

Supposing $\angle A=90^\circ$ , $AC$ is perpendicular to $AB$ and, it follows, to the $x$ -axis, making $AB$ a segment of the line x=m. But that would mean that the coordinates of $C$ $(m, m^2)$ , contradicting the given that points $A$ and $C$ are distinct. So $\angle A$ is not $90^\circ$ . By a similar logic, neither is $\angle B$ .

This means that $\angle C=90^\circ$ and $AC$ is perpendicular to $BC$ . So the slope of $BC$ is the negative reciprocal of the slope of $AC$ , yielding $m+n=\frac{1}{m-n}$ $\Rightarrow$ $m^2-n^2=1$ .

Because $m^2-n^2$ is the length of the altitude of triangle $ABC$ from $AB$ , and $2m$ is the length of $AB$ , the area of $\triangle ABC=m(m^2-n^2)=2008$ . Since $m^2-n^2=1$ , $m=2008$ . Substituting, $2008^2-n^2=1$ $\Rightarrow$ $n^2=2008^2-1=4032063$ , whose digits sum to $18 \Rightarrow \textbf{(C)}$ .

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2008 AMC 12B Problems/Problem 17