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Difference between revisions of "2011 AIME I Problems/Problem 3"
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Thus, we have that <math>\alpha=-\frac{123}{13}</math> and that <math>\beta=\frac{526}{13}</math>. It follows that <math>\alpha+\beta=-\frac{123}{13}+\frac{526}{13}=\frac{403}{13}=\boxed{031}</math>.
 
Thus, we have that <math>\alpha=-\frac{123}{13}</math> and that <math>\beta=\frac{526}{13}</math>. It follows that <math>\alpha+\beta=-\frac{123}{13}+\frac{526}{13}=\frac{403}{13}=\boxed{031}</math>.
 +
== Note ==
 +
Since AIME only accepts nonnegative integer solutions up to 999, once we find the distances, since the sum of the absolute values of the abscissa and ordinate is not divisible by <math>13</math> and therefore cannot be a valid solution, the answer must be the difference instead.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 23:57, 11 July 2022

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 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 $\alpha+\beta=-\frac{123}{13}+\frac{526}{13}=\frac{403}{13}=\boxed{031}$.

Note

Since AIME only accepts nonnegative integer solutions up to 999, once we find the distances, since the sum of the absolute values of the abscissa and ordinate is not divisible by $13$ and therefore cannot be a valid solution, the answer must be the difference instead.

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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