Difference between revisions of "2011 AIME I Problems/Problem 3"
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== See also ==
== See also ==
Latest revision as of 20:07, 30 June 2020
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 .
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 .
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