1988 AIME Problems/Problem 14


Let $C$ be the graph of $xy = 1$, and denote by $C^*$ the reflection of $C$ in the line $y = 2x$. Let the equation of $C^*$ be written in the form

\[12x^2 + bxy + cy^2 + d = 0.\]

Find the product $bc$.

Solution 1

Given a point $P (x,y)$ on $C$, we look to find a formula for $P' (x', y')$ on $C^*$. Both points lie on a line that is perpendicular to $y=2x$, so the slope of $\overline{PP'}$ is $\frac{-1}{2}$. Thus $\frac{y' - y}{x' - x} = \frac{-1}{2} \Longrightarrow x' + 2y' = x + 2y$. Also, the midpoint of $\overline{PP'}$, $\left(\frac{x + x'}{2}, \frac{y + y'}{2}\right)$, lies on the line $y = 2x$. Therefore $\frac{y + y'}{2} = x + x' \Longrightarrow 2x' - y' = y - 2x$.

Solving these two equations, we find $x = \frac{-3x' + 4y'}{5}$ and $y = \frac{4x' + 3y'}{5}$. Substituting these points into the equation of $C$, we get $\frac{(-3x'+4y')(4x'+3y')}{25}=1$, which when expanded becomes $12x'^2-7x'y'-12y'^2+25=0$.

Thus, $bc=(-7)(-12)=\boxed{084}$.

Solution 2

The asymptotes of $C$ are given by $x=0$ and $y=0$. Now if we represent the line $y=2x$ by the complex number $1+2i$, then we find the direction of the reflection of the asymptote $x=0$ by multiplying this by $2-i$, getting $4+3i$. Therefore, the asymptotes of $C^*$ are given by $4y-3x=0$ and $3y+4x=0$.

Now to find the equation of the hyperbola, we multiply the two expressions together to get one side of the equation: $(3x-4y)(4x+3y)=12x^2-7xy-12y^2$. At this point, the right hand side of the equation will be determined by plugging the point $(\frac{\sqrt{2}}{2},\sqrt{2})$, which is unchanged by the reflection, into the expression. But this is not necessary. We see that $b=-7$, $c=-12$, so $bc=\boxed{084}$.

Solution 3

The matrix for a reflection about the polar line $\theta = \alpha, \alpha+\pi$ is: \[ \left[ \begin{array}{ccc} \cos(2\alpha) & \sin(2\alpha) \\ \sin(2\alpha) & -\cos(2\alpha) \end{array} \right] \] 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 $\alpha = \arctan 2$. Note that the line of reflection, $y = 2x$, is the polar line $\theta = \alpha, \alpha+\pi$. Then $2\alpha = \arctan\left(-\frac{4}{3}\right)$, so $\cos(2\alpha) = -\frac{3}{5}$ and $\sin(2\alpha) = \frac{4}{5}$.

Therefore, if $(x', y')$ is mapped to $(x, y)$ under the reflection, then $x = -\frac{3}{5}x'+\frac{4}{5}y'$ and $y = \frac{4}{5}x'+\frac{3}{5}y'$. Since the transformation matrix represents a reflection, it must be its own inverse; therefore, $x' = -\frac{3}{5}x+\frac{4}{5}y$ and $y' = \frac{4}{5}x+\frac{3}{5}y$.

The original coordinates $(x', y')$ must satisfy $x'y' = 1$. Therefore, \[\left(-\frac{3}{5}x+\frac{4}{5}y\right)\left(\frac{4}{5}x+\frac{5}{5}y\right) = 1\] \[-\frac{12}{25}x^2 + \frac{7}{25}xy + \frac{12}{25}y^2 - 1 = 0\] \[12x^2 - 7xy - 12y^2 + 25 = 0\] Thus, $b = -7$ and $c = -12$, so $bc = 84$. The answer is $\boxed{084}$.

Solution 4(the best solution)

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 $b = -7$ and $c = -12$, so $bc = 84$. The answer is $\boxed{084}$.


See also

1988 AIME (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS