2007 AMC 12B Problems/Problem 20
Problem
The parallelogram bounded by the lines , , , and has area . The parallelogram bounded by the lines , , , and has area . Given that , , , and are positive integers, what is the smallest possible value of ?
Solution
Plotting the parallelogram on the coordinate plane, the 4 corners are at . Because , we have that or that , which gives . From the 4 corners of the parallelogram, we have that . Substituting ,
$\begin{eqnarray*} 2d\left(\frac{3bd-ad}{a-b}\right) &=& 18 \\ d\left(\frac{bd-ad}{a-b}+\frac{2bd}{a-b}\right) &=& 9 \\ d\left(-d+\frac{2bd}{a-b}\right) &=& 9 \\ -d^2 + \frac{2b}{a-b}d^2 &=& 9 \\ \frac{2b}{a-b}d^2 &=& 9 + d^2 \end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)
Taking , we have that . This gives us that . Taking the minimum values of we have that
See also
2007 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |