Difference between revisions of "1988 AIME Problems/Problem 14"

Revision as of 02:37, 26 November 2015

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 $bc=\boxed{084}$ .

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 $\alpha = \arctan 2$ . 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 $y' = \frac{4}{5}x+\frac{5}{5}y$ .

The new 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 $084$ .

@@ Line 21: / Line 21: @@
 == Solution 3 ==
 The matrix for a reflection about the polar line <math>\theta = \alpha, \alpha+\pi</math> is:
-<cmath>\[ \left( \begin{array}{ccc}
+<cmath>\[ \left[ \begin{array}{ccc}
 \cos(2\alpha) & \sin(2\alpha) \\
 \sin(2\alpha) & -\cos(2\alpha)
-\end{array} \right)\]</cmath>
+\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

1988 AIME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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