Difference between revisions of "2011 AIME I Problems/Problem 3"
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== Problem == | == Problem == | ||
Let <math>L</math> be the line with slope <math>\frac{5}{12}</math> that contains the point <math>A=(24,-1)</math>, and let <math>M</math> be the line perpendicular to line <math>L</math> that contains the point <math>B=(5,6)</math>. The original coordinate axes are erased, and line <math>L</math> is made the <math>x</math>-axis and line <math>M</math> the <math>y</math>-axis. In the new coordinate system, point <math>A</math> is on the positive <math>x</math>-axis, and point <math>B</math> is on the positive <math>y</math>-axis. The point <math>P</math> with coordinates <math>(-14,27)</math> in the original system has coordinates <math>(\alpha,\beta)</math> in the new coordinate system. Find <math>\alpha+\beta</math>. | Let <math>L</math> be the line with slope <math>\frac{5}{12}</math> that contains the point <math>A=(24,-1)</math>, and let <math>M</math> be the line perpendicular to line <math>L</math> that contains the point <math>B=(5,6)</math>. The original coordinate axes are erased, and line <math>L</math> is made the <math>x</math>-axis and line <math>M</math> the <math>y</math>-axis. In the new coordinate system, point <math>A</math> is on the positive <math>x</math>-axis, and point <math>B</math> is on the positive <math>y</math>-axis. The point <math>P</math> with coordinates <math>(-14,27)</math> in the original system has coordinates <math>(\alpha,\beta)</math> in the new coordinate system. Find <math>\alpha+\beta</math>. |
Revision as of 14:44, 9 August 2018
Contents
Problem
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 .
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.