2011 AIME I Problems/Problem 3
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 -coordinate of is negative, and is therefore ; similarly, the -coordinate of is positive and is therefore .
Thus, we have that and that . It follows that .
Solution 2 (alternate bash)
The equations for the axes are and . We can solve the system to find that they intersect at the point
The unit basis vectors of our new axes are and for the and axes respectively (taking into account which direction is positive).
Then, we solve the following system for and :
Painful bashing gives and . Adding gives
We can also attempt to manipulate the system of equations to solve for , and avoid solving for and separately.
~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 | |
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