This week, we will further build up on what we have discussed in the past two weeks. You will need to sum up everything we discussed last week in a few seconds and arrive at a conclusion and then, move on and solve the question on the basis of that conclusion. We will take you through the ‘summing up’ and ‘getting a feel for it’ process step by step so that it’s intuitive to you next time you come across this concept.

**Question**: A certain square is to be drawn on a coordinate plane such that all the coordinates of its vertices are integers. One of the vertices must be at the origin, and the area of the square must be 25. How many different squares can be drawn?

(A) 4

(B) 6

(C) 8

(D) 10

(E) 12

**Solution**:

Since the question tells us that there are many different ways to draw the square, let’s draw it in one particular way. We will have something with which to proceed.

One vertex must be at (0, 0). Area must be 25 which means the side of the square must be 5. The easiest way to draw such a square would be the blue square shown in the figure.

How will we get many different squares? By turning the square around (0, 0) as shown (Push one side of the square – notice the vertex with the green dot at (0, 5) moving clockwise). Now the problem is that we need the coordinates of all the vertices to be integers. How do we assure that? One vertex is at (0, 0) and will always stay at (0, 0) so we don’t need to worry about that. What about the rest of the three vertices? When you turn the square, at any one position, they may or may not have integer coordinates – and we have three of them to worry about!

This is where our last two posts come in. First let’s establish that we only need to focus on one vertex – the other two will follow. Look at the blue square. Each one of its vertices is at integer coordinates: (0, 0), (0, 5) – the green dot vertex, (5, 5) and (5, 0). Let’s say, when you push the side joining the center and the green dot, the green dot moves by 1 unit down and 3 units to the right. What happens to the side perpendicular to this side (the one joining the green and the yellow dots)? Let’s think about it.

Pay attention to the diagram. When we turn the square, the coordinates of all three vertices change. The change is similar in all three vertices.

Hope it’s intuitive that if one vertex takes integral steps, all vertices will move in a similar fashion. The diagram shown above is just to reinforce this point. So if our first square has all integral coordinates and we move one vertex such that its coordinates remain integers, all the other vertices will follow suit.

Let’s catch hold of the green dot vertex with coordinates (0, 5). We want the x and y coordinates to stay integers such that

x^2 + y^2 = 5^2 (since the length of the side must stay 5)

As discussed above, x = 0 and y = 5 satisfies this. When x = 1, will y be integral? No.

What about when x = 2? No.

When x = 3? Yes, y will be 4.

When x = 4? Yes, y will be 3.

When x = 5? Yes, y will be 0.

So you have two pairs of numbers 0, 5 and 3, 4 (and their variations) which satisfy this equation. (Hope it reminded you of Pythagorean theorem and you jumped to this conclusion right away!)

Notice that coordinates can be negative too so 0, -5 and -3, -4 will also work. So how many total squares do we get?

So the green dot can take any of the following coordinates:

(0, 5), (0, -5)

(5, 0), (-5, 0)

(3, 4), (3, -4)

(4, 3), (4, -3)

(-3, 4), (-3, -4)

(-4, 3), (-4, -3)

A total of 12 values.

Alternatively, you have 3 different squares keeping the green dot in the first quadrant and when rotated, they will give you one square in each of the three other quadrants too.

Total 12 such squares.

**The correct answer is (E).**

If it seems difficult, that is because it is – 750 level.

But keep practicing. Most of it will become intuitive once you get the hang of it.

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

How you determined that, when the green dot moves by 1 unit down, then 3 units to the right?

And is the sum of squared x axis and y axis of any vertics of square, always euql to the area of spuare?

Thanks.

You don’t really need to figure it out. I took that case as an example to show that when one coordinate moves appropriately, so do the rest of the two.

You figure out the various ways in which you can move one coordinate appropriately when you start with x = 0 and y = 5 and go to x = 1, x = 2 etc later on.

Also, if you do want to figure out how to get x = 3 at the initial stage, note that the length of the line from the center to the green dot is 5. When you turn it and reduce the y coordinate by 1 (in second figure), the hypotenuse is 5 and one side is 4 so the other side must be 3 for integral values.

Area of the square is given by the square of the length of any side. If the side lies on the x or y axis, its length is simply its y or x coordinate. If it doesn’t, then the square of its length if the sum of the squares of its x and y coordinates.

Thank-you.