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A certain square is to be drawn on a coordinate plane. One [#permalink]
27 Mar 2007, 18:54

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?
4
6
8
10
12

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.

4 like you all say plus another because
when the side of the square which willbe drawn diagonally such that co-ordinate x = 6 and y=8, forming a right triangle.

Plus another 4 when x & y interchange values
ie
when x=8 & y=10,
the side of the square will be 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

Last edited by Fig on 29 Mar 2007, 00:28, edited 1 time in total.

Re: how many squares in a plane [#permalink]
28 Mar 2007, 14:55

amd08 wrote:

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? 4 6 8 10 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:

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

I have represented the point that I epressed from 1 (x1,y1) in the cadran 1 and I have added the special square with 1 and 3 on X & Y axes to give a point 2 in cadran 1.

As u can see, by mirroring with X and Y axes of the figure, we obtain the other squares in cadran II, III and IV.

And also, to avoid confusion, we should not flip X1 > Y1 and Y1 > X1 because we recreate a point 3 and so a square already considered.

Attachments

Fig1_Sqaure for (x1,y1) in cardan 1.gif [ 9.29 KiB | Viewed 1467 times ]

Last edited by Fig on 29 Mar 2007, 03:17, edited 1 time in total.

Fig's explanation is good. There is also a visual way to do this:

First, if one vertex is on integer coordinates, then all of them will be. You can draw a picture and use geometric similarities to see this. Thus, the problem reduces to rotating a vertex adjacent to the origin in a circle, and counting the times it passes through an integer coordinate. This then, will give (going counterclockwise) (10,0), (8,6), (6,8), (0,10), (-6,8), ... (8,-6) for a total of 12.

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)

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

Small correction,

Equation of Circle:

x^2 + y^2 = r^2 = 10^2 = 100 (If center of the circle is at origin)