4? one for each quadrant...

I tried using 45-45-90 but the vertex coordinate would be (5sqrt(2), 5sqrt(2)) not integers..

I tried using 30-60-90 but the vertex coordinate would be (5,5sqrt(3)) not integers...

Unless two sides of the square are the x and y axis...hmm....

******deleted****see my post below...

My approach is bit different.

Try to find the possible vertices,

1. Set1 - 4 squares, one in each quadrant. Vertices- (0,0) (0,10) (0,-10) (10,0) (-10,0)

2 Set2 - Rotate the set 1 of the squares in clockwise so that the the vertex of (10,0) becomes (8,6). There will be 4 squares with this orientation.

3 Set3 -Rotate the set1 squares further so that the the vertex of (10,0) becomes (6,8).There will be 4 squares with this orientation.

Hence there will be 4+4+4 = 12 squares.
It's hard to imagine that way without putting it diagrammatically.
Attached is the diagram. It;s not accurate but will certainly help you visualizing the squares orientation.

(E) for me ... Is it really a GMAT question

First of all, the square pocesses 4 sides of 10.

Let set:
o Point 1 : (x1,y1)
o Point 2 : (x2,y2) : diagonally opposed to the vertice on 0(0,0)
o Point 3 : (x3,y3)

Then, we have:
o x1^2+y1^2 = 100
o x3^2+y3^2 = 100
o x2^2+y2^2 = 200

Also, notice that y1/x1, y2/x2 and y3/x3 are the slopes of the equation passing by 0(0,0) and respectively the point 1, 2 and 3.

By the property of a square, we know that the sides 01 and 03 are perpendicular. Thus, we have:
o y1/x1 * y3/x3 = -1

Also, by properties of vectors, notice that:
o x2 = x1+x3
o y2 = y1+y3

But how could we obtain 100 with 2 integers?
o 8^2 + 6^2 = 100
o 10^2 + 0^2 = 100

Also, there are symetries that recreate similar squares by coordonates of integers in other quadrans.

To solve this issue, we should consider only 1 cadran and we should not exchange values of x1 and y1 : x1 to be equal to y1 and y1 to be equal x1 Indeed, it recreates only another point 3.

Arbitrarly, I choose the cadran 1.

How many square are done from x1 > 0 and y1 > 0 (Cadran 1)?
o If x1 = 8 and y1 = 6, then:
> x3=-6 and y3=8 and x2 = 2 and y2 = 14
or
> x3=6 and y3=-8 and x2 = 14 and y2 = -2

>>>> 2

Then, how many square possesses 2 sides : one on X and one on Y axes?
>>>> 4 : 1 by cadran.

Finally,

we have : 4*2 + 4 = 12

Such a square must have a point P(x,y) that is sqrt(200) from the origin
i.e. x^2+y^2=200 where x and y are integers

|x|=sqrt(200-y^2)

|y| could be 2,10,14

So, I guess there will be 12 such squares

Let's confirm:

Let's look at the squares in an organized way. Let P(x,y) be the vertex of the side OP we would encounter if we started at the y axis and looked clockwise. How many such pairs (x,y) are there? |x|=sqrt(100-y^2), where x and y are both integers, so we can make a chart:

|y| ..... |x|
0..........10
6...........8
8...........6
10.........0

So P(x,y) must be one of {(0,10),(6,8),(8,6),(10,0), (8,-6), (6,-8),(0,-10), (-6,-8),(-8,-6),(-10,0),(-8,6),(-6,-8)} Thus there are 12 squares, confirming the simplier method above

indeed its E. Thanks!

wow!! nice Fig!!!

Thanks!

I think GMAT doesn't ask for the whole solution but want you to find the trick.

Trick is to find points on circle (origin 0,0 and radius 10).
Equation of circle
x^2 + y^2 = 10

how many times both x and y will be integers?
only when |x| and |y| (6,8) and (0,10)
so there are 12 points
(0,10), (10,0) (0,-10) (-10,0)
(6,8) (8,6) (8,-6) (6,-8) (-6,-8) (-8,-6) (-8,6) (-6,8)

Ans : 12