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>. | ||
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+ | ==Video Solution== | ||
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+ | https://www.youtube.com/watch?v=_znugFEst6E&t=919s | ||
== See also == | == See also == | ||
{{AIME box|year=2011|n=I|num-b=2|num-a=4}} | {{AIME box|year=2011|n=I|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:06, 30 June 2020
Contents
Problem
Let be the line with slope that contains the point , and let be the line perpendicular to line that contains the point . The original coordinate axes are erased, and line is made the -axis and line the -axis. In the new coordinate system, point is on the positive -axis, and point is on the positive -axis. The point with coordinates in the original system has coordinates in the new coordinate system. Find .
Solution
Given that has slope and contains the point , we may write the point-slope equation for as . Since is perpendicular to and contains the point , we have that the slope of is , and consequently that the point-slope equation for is .
Converting both equations to the form , we have that has the equation and that has the equation . Applying the point-to-line distance formula, , to point and lines and , we find that the distance from to and are and , respectively.
Since and 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 is negative, and is therefore ; similarly, the ordinate of is positive and is therefore .
Thus, we have that and that . It follows that .
Video Solution
https://www.youtube.com/watch?v=_znugFEst6E&t=919s
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.