Physics..

Wouldn't 'by a factor of n' mean you multiply (A-B) by n to get (A+B)?

That is correct.

Code:

|[B]A[/B]-[B]B[/B]|*n = |[B]A[/B]+[B]B[/B]|
|[B]A[/B]| = |[B]B[/B]|
a*[B][I]i[/I][/B] + b*[B][I]j[/I][/B] = [B]A[/B]
x*[B][I]i[/I][/B] + y*[B][I]j[/I][/B] = [B]B[/B]

I'm thinking rhombi and triangles.

If n = 1, then theta = 90 degrees. That much I can tell you immediately.

 Workspace

Code:

Law of Cosines: C^2 = A^2 + B^2 - 2ABcos(c), where C is the side of a triangle opposite of angle c, and A, B are the other sides of the triangle.

Assume C1 = |[B]A[/B]-[B]B[/B]| and C2 = |[B]A[/B]+[B]B[/B]|.  Using substitution, we get...
|[B]A[/B]-[B]B[/B]|*n = |[B]A[/B]+[B]B[/B]|
C1*n = C2
[|[B]A[/B]|^2 + |[B]B[/B]|^2 - 2|[B]A[/B]||[B]B[/B]|cos(θ)]n^2 = |[B]A[/B]|^2 + |[B]B[/B]|^2 - 2|[B]A[/B]||[B]B[/B]|cos(180-θ)
The reason why I put 180-θ for the 2nd angle for C2 is due to the property of parallelograms in general.

Further simplification...
(n^2 - 1) * [|[B]A[/B]|^2 - |[B]A[/B]|^2 + |[B]B[/B]|^2 - |[B]B[/B]|^2] - 2|[B]A[/B]||[B]B[/B]|cos(θ)*n^2 + 2|[B]A[/B]||[B]B[/B]|cos(180-θ) = 0
- 2|[B]A[/B]||[B]B[/B]|cos(θ)*n^2 + 2|[B]A[/B]||[B]B[/B]|cos(180-θ) = 0
2|[B]A[/B]||[B]B[/B]|cos(180-θ) = 2|[B]A[/B]||[B]B[/B]|cos(θ)*n^2
cos(180-θ) = cos(θ)*n^2