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

Revision as of 17:43, 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, $bc=(-7)(-12)=\boxed{084}$ .

@@ Line 11: / Line 11: @@
 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>.
+Thus, <math>bc=(-7)(-12)=\boxed{084}</math>.
 == 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
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Difference between revisions of "1988 AIME Problems/Problem 14"

Revision as of 17:43, 10 July 2009

Problem

Solution

See also