TITLE: How to prove that it is possible to make rhombuses with any number of interior points?
QUESTION [18 upvotes]: I was given some square dot paper which can be found on this link: http://lrt.ednet.ns.ca/PD/BLM/pdf_files/dot_paper/sq_dot_1cm.pdf  and was told to draw a few rhombuses with the vertices on the dots but no other dots on the edges.
Here is an example:
This is not a rhombus it just shows how the shape must be lay out. 

                   A = 6

The number of interior dots would be represented by A. I was then told that it is possible to draw rhombuses with any number of interior points. All I had to do was prove this.
I started by trying to draw these shapes and came up with some of the below:

A = 2

A = 4

A = 9  
However, I am still unsure of how you can prove something like this since there may be numerous ways of drawing these rhombuses.
I would appreciate support.
Thank you :)

REPLY [2 votes]: The proof is straight-forward. Lets use the property that rhombus is a quadrilateral whose diagonals are perpendicular bisectors. We'll accommodate all the interior points we want along a line. 

Odd number of interior points

Say we want $p_o$ interior points; consider $p_o$ consecutive points along a line (notice the square paper has two symmetries - straight and diagonal, so we can choose either a straight or a diagonal line. Lets choose straight line). Now take one point each on either side of these $p_o$ points; they shall serve as two of the vertices of our desired rhombus. Consistent with our definition, choose the point which bisects the distance between these $p_o+2$ points; from the point of bisection, draw a perpendicular line. Immediate points on each side that this line intersects are the other two vertices of our rhombus. Join the four vertices and we have our desired rhombus. 

Even number of interior points

We follow the same procedure, but since the line perpendicularly bisecting the distance between two $p_e+2$ points will never intersect an other point on the square paper, we make use of its diagonal symmetry. So, we choose our $p_e$ points along a diagonal instead and follow the same procedure.
So, the proof follows from Rhombus's diagonals being perpendicular bisectors.