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Difference between revisions of "2008 AMC 12B Problems/Problem 17"
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<math> \textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20 </math>
 
<math> \textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20 </math>
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Revision as of 17:05, 29 December 2009

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 $\frac{m^2-n^2}{-m-n}=-\frac{(m+n)(m-n)}{m+n}=-(m-n)$.


$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20$


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 $C$ are $(m, m^2)$, 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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