Difference between revisions of "1988 AIME Problems/Problem 14"
(→Solution 3) |
m (Replaced x and y with x' and y' in some places to make the solution more proper) |
||
Line 9: | Line 9: | ||
Given a point <math>P (x,y)</math> on <math>C</math>, we look to find a formula for <math>P' (x', y')</math> on <math>C^*</math>. Both points lie on a line that is [[perpendicular]] to <math>y=2x</math>, so the slope of <math>\overline{PP'}</math> is <math>\frac{-1}{2}</math>. Thus <math>\frac{y' - y}{x' - x} = \frac{-1}{2} \Longrightarrow x' + 2y' = x + 2y</math>. Also, the midpoint of <math>\overline{PP'}</math>, <math>\left(\frac{x + x'}{2}, \frac{y + y'}{2}\right)</math>, lies on the line <math>y = 2x</math>. Therefore <math>\frac{y + y'}{2} = x + x' \Longrightarrow 2x' - y' = y - 2x</math>. | Given a point <math>P (x,y)</math> on <math>C</math>, we look to find a formula for <math>P' (x', y')</math> on <math>C^*</math>. Both points lie on a line that is [[perpendicular]] to <math>y=2x</math>, so the slope of <math>\overline{PP'}</math> is <math>\frac{-1}{2}</math>. Thus <math>\frac{y' - y}{x' - x} = \frac{-1}{2} \Longrightarrow x' + 2y' = x + 2y</math>. Also, the midpoint of <math>\overline{PP'}</math>, <math>\left(\frac{x + x'}{2}, \frac{y + y'}{2}\right)</math>, lies on the line <math>y = 2x</math>. Therefore <math>\frac{y + y'}{2} = x + x' \Longrightarrow 2x' - y' = y - 2x</math>. | ||
− | Solving these two equations, we find <math>x | + | Solving these two equations, we find <math>x = \frac{-3x' + 4y'}{5}</math> and <math>y = \frac{4x' + 3y'}{5}</math>. Substituting these points into the equation of <math>C</math>, we get <math>\frac{(-3x'+4y')(4x'+3y')}{25}=1</math>, which when expanded becomes <math>12x'^2-7x'y'-12y'^2+25=0</math>. |
Thus, <math>bc=(-7)(-12)=\boxed{084}</math>. | Thus, <math>bc=(-7)(-12)=\boxed{084}</math>. |
Revision as of 17:03, 30 March 2020
Contents
[hide]Problem
Let be the graph of
, and denote by
the reflection of
in the line
. Let the equation of
be written in the form
Find the product .
Solution 1
Given a point on
, we look to find a formula for
on
. Both points lie on a line that is perpendicular to
, so the slope of
is
. Thus
. Also, the midpoint of
,
, lies on the line
. Therefore
.
Solving these two equations, we find and
. Substituting these points into the equation of
, we get
, which when expanded becomes
.
Thus, .
Solution 2
The asymptotes of are given by
and
. Now if we represent the line
by the complex number
, then we find the direction of the reflection of the asymptote
by multiplying this by
, getting
. Therefore, the asymptotes of
are given by
and
.
Now to find the equation of the hyperbola, we multiply the two expressions together to get one side of the equation: . At this point, the right hand side of the equation will be determined by plugging the point
, which is unchanged by the reflection, into the expression. But this is not necessary. We see that
,
, so
.
Solution 3
The matrix for a reflection about the polar line is:
This is not hard to derive using a basic knowledge of linear transformations. You can refer here for more information: https://en.wikipedia.org/wiki/Orthogonal_matrix
Let . Note that the line of reflection,
, is the polar line
. Then
, so
and
.
Therefore, if is mapped to
under the reflection, then
and
. Since the transformation matrix represents a reflection, it must be its own inverse; therefore,
and
.
The original coordinates must satisfy
. Therefore,
Thus,
and
, so
. The answer is
.
See also
1988 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.