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== | ==Video Solution== |
Revision as of 10:38, 18 August 2022
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
-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
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.