2008 AMC 12B Problems/Problem 17

Problem

Let $A$ , $B$ and $C$ be three distinct points on the graph of $y=x^2$ such that line $AB$ is parallel to the $x$ -axis and $\triangle ABC$ is a right triangle with area $2008$ . What is the sum of the digits of the $y$ -coordinate of $C$ ?

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

Solution

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 $BC$ . Let C be the point $(n, n^2)$ . 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, $2008^2-n^2=1$ $\Rightarrow$ $n^2=2008^2-1=(2000+8)^2-1=4000000+32000+64-1=4032063$ , whose digits sum to $18 \Rightarrow \textbf{(C)}$ .

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 12 Problems and Solutions

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2008 AMC 12B Problems/Problem 17

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