Difference between revisions of "1988 AIME Problems/Problem 14"
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== Problem == | == Problem == | ||
+ | Let <math>C</math> be the [[graph]] of <math>xy = 1</math>, and denote by <math>C^*</math> the [[reflection]] of <math>C</math> in the line <math>y = 2x</math>. Let the [[equation]] of <math>C^*</math> be written in the form | ||
− | == Solution == | + | <cmath>12x^2 + bxy + cy^2 + d = 0.</cmath> |
+ | |||
+ | Find the product <math>bc</math>. | ||
+ | |||
+ | == Solution 1 == | ||
+ | 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 = \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>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | The [[asymptote]]s of <math>C</math> are given by <math>x=0</math> and <math>y=0</math>. Now if we represent the line <math>y=2x</math> by the complex number <math>1+2i</math>, then we find the direction of the reflection of the asymptote <math>x=0</math> by multiplying this by <math>2-i</math>, getting <math>4+3i</math>. Therefore, the asymptotes of <math>C^*</math> are given by <math>4y-3x=0</math> and <math>3y+4x=0</math>. | ||
+ | |||
+ | Now to find the equation of the hyperbola, we multiply the two expressions together to get one side of the equation: <math>(3x-4y)(4x+3y)=12x^2-7xy-12y^2</math>. At this point, the right hand side of the equation will be determined by plugging the point <math>(\frac{\sqrt{2}}{2},\sqrt{2})</math>, which is unchanged by the reflection, into the expression. But this is not necessary. We see that <math>b=-7</math>, <math>c=-12</math>, so <math>bc=\boxed{084}</math>. | ||
+ | |||
+ | |||
+ | == Solution 3 == | ||
+ | The matrix for a reflection about the polar line <math>\theta = \alpha, \alpha+\pi</math> is: | ||
+ | <cmath>\[ \left[ \begin{array}{ccc} | ||
+ | \cos(2\alpha) & \sin(2\alpha) \\ | ||
+ | \sin(2\alpha) & -\cos(2\alpha) | ||
+ | \end{array} \right] \]</cmath> | ||
+ | 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 <math>\alpha = \arctan 2</math>. Note that the line of reflection, <math>y = 2x</math>, is the polar line <math>\theta = \alpha, \alpha+\pi</math>. Then <math>2\alpha = \arctan\left(-\frac{4}{3}\right)</math>, so <math>\cos(2\alpha) = -\frac{3}{5}</math> and <math>\sin(2\alpha) = \frac{4}{5}</math>. | ||
+ | |||
+ | Therefore, if <math>(x', y')</math> is mapped to <math>(x, y)</math> under the reflection, then <math>x = -\frac{3}{5}x'+\frac{4}{5}y'</math> and <math>y = \frac{4}{5}x'+\frac{3}{5}y'</math>. Since the transformation matrix represents a reflection, it must be its own inverse; therefore, <math>x' = -\frac{3}{5}x+\frac{4}{5}y</math> and <math>y' = \frac{4}{5}x+\frac{3}{5}y</math>. | ||
+ | |||
+ | The original coordinates <math>(x', y')</math> must satisfy <math>x'y' = 1</math>. Therefore, | ||
+ | <cmath>\left(-\frac{3}{5}x+\frac{4}{5}y\right)\left(\frac{4}{5}x+\frac{5}{5}y\right) = 1</cmath> | ||
+ | <cmath>-\frac{12}{25}x^2 + \frac{7}{25}xy + \frac{12}{25}y^2 - 1 = 0</cmath> | ||
+ | <cmath>12x^2 - 7xy - 12y^2 + 25 = 0</cmath> | ||
+ | Thus, <math>b = -7</math> and <math>c = -12</math>, so <math>bc = 84</math>. The answer is <math>\boxed{084}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | Find some simple points on the graph of C, reflect them and note down the new coordinates, and plug 'em into the given equation. After some plugging and chugging and solving the system of equations that follow, we get that <math>b = -7</math> and <math>c = -12</math>, so <math>bc = 84</math>. The answer is <math>\boxed{084}</math>. | ||
+ | |||
+ | pi_is_3.141 | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1988|num-b=13|num-a=15}} | |
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 17:21, 3 May 2021
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
.
Solution 4
Find some simple points on the graph of C, reflect them and note down the new coordinates, and plug 'em into the given equation. After some plugging and chugging and solving the system of equations that follow, we get that and
, so
. The answer is
.
pi_is_3.141
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.