2007 AMC 12B Problems/Problem 20
Contents
[hide]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
Template:Incomplete Plotting the parallelogram on the coordinate plane, the 4 corners are at . Because , we have that or that , which gives (consider a homothety, or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by , it follows that the stretch along the diagonal is ). The area of triangular half of the parallelogram on the right side of the y-axis is given by , so substituting :
\[\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) &= 9 \quad \Longrightarrow \quad 2d^2 &= 9(a-b)\] (Error compiling LaTeX. Unknown error_msg)
Thus , and we verify that , will give us a minimum value for . Then .
Solution 2
The key to this solution is that area is invariant under translation. By suitably shifting the plane, the problem is mapped to the lines and . Now, the area of the parallelogram contained by is the former is equal to the area of a rectangle with sides and , , and the area contained by the latter is . Thus, and must be even if the former quantity is to equal . so is a multiple of . Putting this all together, the minimal solution is
See also
2007 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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All AMC 12 Problems and Solutions |