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

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Solving these two equations, we find <math>x' = \frac{-3x + 4y}{5}</math> and <math>y' = \frac{4x + 3y}{5}</math>. Substituting these points into the equation of <math>C</math>, we get <math>\frac{(-3x+4y)(4x+3y)}{25}=1</math>, which when expanded becomes <math>12x^2-7xy-12y^2+25=0</math>.
 
Solving these two equations, we find <math>x' = \frac{-3x + 4y}{5}</math> and <math>y' = \frac{4x + 3y}{5}</math>. Substituting these points into the equation of <math>C</math>, we get <math>\frac{(-3x+4y)(4x+3y)}{25}=1</math>, which when expanded becomes <math>12x^2-7xy-12y^2+25=0</math>.
  
Thus, <math>bc=(-7)(-12)=\boxed{84}</math>.
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Thus, <math>bc=(-7)(-12)=\boxed{084}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 18:43, 10 July 2009

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

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

See also

1988 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AIME Problems and Solutions
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