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{ | + | Thus, <math>bc=(-7)(-12)=\boxed{084}</math>. |
== See also == | == See also == |
Revision as of 18:43, 10 July 2009
Problem
Let be the graph of , and denote by the reflection of in the line . Let the equation of be written in the form
Find the product .
Solution
Given a point on , we look to find a formula for on . Both points lie on a line that is perpendicular to , so the slope of is . Thus . Also, the midpoint of , , lies on the line . Therefore .
Solving these two equations, we find and . Substituting these points into the equation of , we get , which when expanded becomes .
Thus, .
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 |