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

m (Solution)
(Solution)
Line 7: Line 7:
  
 
== Solution ==
 
== Solution ==
{{solution}}
+
<math>bc=84</math>
 +
 
 +
We need to find a formula for <math>P^*</math> if <math>P=(x,y)</math>.
 +
 
 +
With some algebra, we find this to be:
 +
 
 +
<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 02:16, 17 November 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

$bc=84$

We need to find a formula for $P^*$ if $P=(x,y)$.

With some algebra, we find this to be:

$P^*=(\frac{-3x+4y}{5},\frac{4x+3y}{5})$.

Substituting in, we get:

$\frac{(-3x+4y)(4x+3y)}{25}=1$

Expanding,

$12x^2-7xy-12y^2+25=0$.

Thus, $bc=(-7)(-12)=84.$

See also

1988 AIME (ProblemsAnswer KeyResources)
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