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>. | ||
− | == | + | == Solution 2 (alternate bash) == |
− | + | ||
+ | The equations for the axes are <math>\frac{5}{12} (x-24) = y+1</math> and <math>-\frac{12}{5}(x-5) = y - 6</math>. We can solve the system to find that they intersect at the point <math>\left( \frac{1740}{169},\frac{-1134}{169} \right)</math> | ||
+ | |||
+ | The unit basis vectors of our new axes are <math>\begin{pmatrix} 12/13 \\ 5/13 \end{pmatrix}</math> and <math>\begin{pmatrix} -5/13 \\ 12/13 \end{pmatrix}</math> for the <math>x</math> and <math>y</math> axes respectively (taking into account which direction is positive). | ||
+ | |||
+ | Then, we solve the following system for <math>\alpha</math> and <math>\beta</math> : | ||
+ | |||
+ | <cmath>\alpha \begin{pmatrix} 12/13 \\ 5/13 \end{pmatrix} + \beta \begin{pmatrix} -5/13 \\ 12/13 \end{pmatrix} + \begin{pmatrix} 1740/169 \\ -1134/169 \end{pmatrix} = \begin{pmatrix} -146 \\ 27 \end{pmatrix} </cmath> | ||
+ | |||
+ | Painful bashing gives <math>\alpha = -\frac{123}{13}</math> and <math>\beta = \frac{526}{13}</math>. Adding gives <math>\alpha + \beta = \frac{403}{13} = \boxed{031}</math> | ||
+ | |||
+ | We can also attempt to manipulate the system of equations to solve for <math>\alpha + \beta</math>, and avoid solving for <math>\alpha</math> and <math>\beta</math> separately. | ||
+ | |||
+ | ~jd9 | ||
+ | |||
+ | == Possibly a solution 3 == | ||
+ | |||
+ | First, find where the new positive x and y axis are by comparing the points A and B to the line it is not on (higher or lower). | ||
+ | Next, just find the new quadrant the point is in from the x and y axis. Finally, to find the distance from the point to the axis, we can just use both slopes to find both projections of the line onto the axis. The intersection of the projection line and the axis's previous line would give us information on the new point's coordinate points. | ||
+ | |||
+ | ~#EmilyQ | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 16:01, 13 April 2024
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
.
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
Possibly a solution 3
First, find where the new positive x and y axis are by comparing the points A and B to the line it is not on (higher or lower). Next, just find the new quadrant the point is in from the x and y axis. Finally, to find the distance from the point to the axis, we can just use both slopes to find both projections of the line onto the axis. The intersection of the projection line and the axis's previous line would give us information on the new point's coordinate points.
~#EmilyQ
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.