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

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== Solution ==
 
== Solution ==
<math>bc=84</math>
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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>.
  
We need to find a formula for <math>P^*</math> if <math>P=(x,y)</math>.
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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>.
  
With some algebra, we find this to be:
+
Thus, <math>bc=(-7)(-12)=\boxed{84}</math>.
 
 
<math>P^*=(\frac{-3x+4y}{5},\frac{4x+3y}{5})</math>.
 
 
 
Substituting in, we get:
 
 
 
<math>\frac{(-3x+4y)(4x+3y)}{25}=1</math>
 
 
 
Expanding,
 
 
 
<math>12x^2-7xy-12y^2+25=0</math>.
 
 
 
Thus, <math>bc=(-7)(-12)=84.</math>
 
  
 
== See also ==
 
== See also ==

Revision as of 13:29, 29 December 2007

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, $bc=(-7)(-12)=\boxed{84}$.

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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