1988 AIME Problems/Problem 14

Revision as of 17:03, 30 March 2020 by Jjmath123 (talk | contribs) (Replaced x and y with x' and y' in some places to make the solution more proper)


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 $\boxed{084}$.

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

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