2007 AMC 12B Problems/Problem 20
Contents
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
Template:Incomplete
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 19 |
Followed by Problem 21 |
| 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 | |
