1988 AIME Problems/Problem 14

Revision as of 01:22, 16 February 2014 by PythagorasofSamos (talk | contribs) (Solution 2)

Problem

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 $12x^2-7xy-12y^2+25=0$.

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 $bc=\boxed{084}$.

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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