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

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Find the product <math>bc</math>.
 
Find the product <math>bc</math>.
  
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== Solution ==
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== 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>.
 
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>.
  
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Thus, <math>bc=(-7)(-12)=\boxed{084}</math>.
 
Thus, <math>bc=(-7)(-12)=\boxed{084}</math>.
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== Solution 2 ==
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The asymptotes 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>.
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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>.
  
 
== See also ==
 
== See also ==

Revision as of 17:47, 9 June 2013

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 (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
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