2011 AIME I Problems/Problem 3

Problem

Let $L$ be the line with slope $\frac{5}{12}$ that contains the point $A=(24,-1)$ , and let $M$ be the line perpendicular to line $L$ that contains the point $B=(5,6)$ . The original coordinate axes are erased, and line $L$ is made the $x$ -axis and line $M$ the $y$ -axis. In the new coordinate system, point $A$ is on the positive $x$ -axis, and point $B$ is on the positive $y$ -axis. The point $P$ with coordinates $(-14,27)$ in the original system has coordinates $(\alpha,\beta)$ in the new coordinate system. Find $\alpha+\beta$ .

Solution

Given that $L$ has slope $\frac{5}{12}$ and contains the point $A=(24,-1)$ , we may write the point-slope equation for $L$ as $y+1=\frac{5}{12}(x-24)$ . Since $M$ is perpendicular to $L$ and contains the point $B=(5,6)$ , we have that the slope of $M$ is $-\frac{12}{5}$ , and consequently that the point-slope equation for $M$ is $y-6=-\frac{12}{5}(x-5)$ .

Converting both equations to the form $0=Ax+By+C$ , we have that $L$ has the equation $0=12x+5y-90$ and that $M$ has the equation $0=5x-12y-132$ . Applying the point-to-line distance formula, $\frac{\abs{Ax+By+C}}{\sqrt{A^2+B^2}}$ (Error compiling LaTeX. Unknown error_msg), to point $P$ and lines $L$ and $M$ , we find that the distance from $P$ to $L$ and $M$ are $\frac{123}{13}$ and $\frac{526}{13}$ , respectively.

Since $A$ and $B$ lie on the positive axes of the shifted coordinate plane, we may show by graphing the given system that point P will lie in the second quadrant in the new coordinate system. Therefore, the abscissa of $P$ is negative, and is therefore $-\frac{123}{13}$ ; similarly, the ordinate of $P$ is positive and is therefore $\frac{526}{13}$ .

Thus, we have that $\alpha=-\frac{123}{13}$ and that $\beta=\frac{526}{13}$ . It follows that $\alpha+\beta=-\frac{123}{13}+\frac{526}{13}=\frac{403}{13}=\boxed{031}$ .

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2011 AIME I Problems/Problem 3

Problem

Solution