Difference between revisions of "2007 AMC 12B Problems/Problem 20"
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Revision as of 10:52, 4 July 2013
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 for
, so the sum 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 |
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