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2011 AIME I Problems/Problem 3

Revision as of 15:21, 27 December 2022 by Jd9 (talk | contribs) (Terrible alternate solution.)

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 $0=5x-12y-132$ and that $M$ has the equation $0=12x+5y-90$. Applying the point-to-line distance formula, $\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$, to point $P$ and lines $L$ and $M$, we find that the distance from $P$ to $L$ and $M$ are $\frac{526}{13}$ and $\frac{123}{13}$, 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 $x$-coordinate of $P$ is negative, and is therefore $-\frac{123}{13}$; similarly, the $y$-coordinate 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 $\alpha+\beta=-\frac{123}{13}+\frac{526}{13}=\frac{403}{13}=\boxed{031}$.

Solution 2 (alternate bash)

The equations for the axes are $\frac{5}{12} (x-24) = y+1$ and $-\frac{12}{5}(x-5) = y - 6$. We can solve the system to find that they intersect at the point $\left( \frac{1740}{169},\frac{-1134}{169} \right)$

The unit basis vectors of our new axes are $\begin{pmatrix} 12/13 \\ 5/13 \end{pmatrix}$ and $\begin{pmatrix} -5/13 \\ 12/13 \end{pmatrix}$ for the $x$ and $y$ axes respectively (taking into account which direction is positive).

Then, we solve the following system for $\alpha$ and $\beta$ :

\[\alpha \begin{pmatrix} 12/13 \\ 5/13 \end{pmatrix} + \beta \begin{pmatrix} -5/13 \\ 12/13 \end{pmatrix} + \begin{pmatrix} 1740/169 \\ -1134/169 \end{pmatrix} = \begin{pmatrix} -146 \\ 27 \end{pmatrix}\]

Painful bashing gives $\alpha = -\frac{123}{13}$ and $\beta = \frac{526}{13}$. Adding gives $\alpha + \beta = \frac{403}{13} = \boxed{031}$

~jd9

Video Solution

https://www.youtube.com/watch?v=_znugFEst6E&t=919s

~Shreyas S

See also

2011 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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