How many squares?

[levocap]

A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an area of 100. If all coordinates of the vertices must be integers, how many different ways can this square be drawn?

OE is 12. How do you conceptualize this? The answer given was not very explanatory and hoping for another way to look at the problem.

[rajeev.bajpai]

I saw the answer to this problem in some forums and they rely mostly on symmetry even though without really mentioning it. The problem with those answers is that they dont really verify that all the vertices have integer x and y values, specifically the fourth one.

Here is the solution:
Each side of this square would br sqrt(100) = 10 units.

There are four obvious sqaures having two sides on X and Y axis, one in each quadrant. The (x,y) vertices of all these squares are out of (0,0), (10,0),(0,10),(0,-10), (-10,0),(10,10),(-10,10),(10,-10) and (-10,-10).

To calculate the number of squares that do not have any of the vertices on X or Y axes except the fixed one on (0,0), we can try to find the squares in the first quadrant only and multiply the number by four (symmetry suggests that each quadrant would have similar squares).

Realize that all such squares can be drawn by rotating the already known square ((0,0),(10,0),(0,10),(10,10) around the center. Two of the vertices would travel on the circle x^2 + Y^2 = 100 and one on x^2 + Y^2 = 200 (think how).

Now, in the first quadrant:

The X cordinate with integer values not on the two axes can be 1,2,3,4,5,6,7,8,9 and trying these values in the circle x^2 + Y^2 = 100 gives the only integer values of 8 and 6 corresponding to X = 6 and 8. So we find that the vertices can be (8,6) and (6,8). Though you also find that they can be (-8,6) and (-6,8) but they are not in First quadrant and infact make the other vertex of these suqares.

The square known in first quadrant are (8,6) and (-6,8) and (6,8) and (-8,6) - (parts go to the second quadrant).

So, there can be two squares in the first quadrant as long as we verify that the fourth vertex also has integer values. This can be trickey but easily done if solved for the line passing thru origin and the mid point of (-6,8) and (8,6) and intersecting the circle x^2 + Y^2 = 200. This would give you x,y as (2,14) and (-2,-14) that are integers. Now note that (2,14) belongs to first quadrant and (-2,-14) would belong to one such square in third quadrant. We can similarly verify that the fourth vertex corresponding to (6,8) and (-8,6) is also integer.

So knowing that there are two such squares in first quadrant, we can deduce that there would be 8 such squares, 2 in each quadrant.

So Total number of such squares = 4 + 8 =12.


GMAT problem? I dont think so for all the calculations involved.

Thanks.

[Ben Ku]

It might be helpful to search for the question before posting it. Here are several links to explanations for this question.

http://www.manhattangmat.com/forums/square-coordinate-plane-question-from-online-exam-t3940.html
http://www.manhattangmat.com/forums/cat-1-how-many-squares-t3554.html
http://www.manhattangmat.com/forums/manhattan-gmat-cat-test-maths-question-t2401.html
http://www.manhattangmat.com/forums/a-certain-square-is-to-be-drawn-on-a-coordinate-plane-t1016.html

Let me know if you have additional questions on this problem.