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Re: A certain square is to be drawn on a coordinate plane. One [#permalink]
27 Mar 2013, 22:06

Perimeter of a circle of radius 10 centered at the origin will coincide with 12 points where where both the x and y values are integers. (10, 0); (8, 6); (6, 8); (0, 10); (-8, 6), (-6, 8); (-10, 0); (-8, -6); (-6, -8); (0, -10); (8, -6); (6, -8)

Re: A certain square is to be drawn on a coordinate plane. One [#permalink]
28 Mar 2013, 02:04

Expert's post

mun23 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? (A)4 (B)6 (C)8 (D)10 (E)12 Need help

Re: A certain square is to be drawn on a coordinate plane [#permalink]
28 Apr 2013, 08:55

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?

Doesn't one of the vertices must be on the origin mean one vertices of the square has always to be (0,0) hence only 4 possibilities ??? Please clarify !!

Re: A certain square is to be drawn on a coordinate plane [#permalink]
28 Apr 2013, 09:39

1

This post received KUDOS

Since area of the square is 100, each side = 10 One of the vertices of the square = (0,0) Let the co-ordinates of another vertex of the square be (x,y) Using the formula \(d^2 = x^2 + y^2\) (d = Distance from the origin to any point in the co-ordinate) So, \(100 = x^2 + y^2\) As the vertices, must be integers, solve for different values for x and y When x = 0, y = 10 Also, x = 0, y = -10 x=10, y = 0 x = -10, y = 0 Also x = 6 , y = 8 (as \(100 = 6^2+8^2\)) x = 6, y = -8 x = -6, y= 8 x = -6, y = -8 Similarly, x = 8, y = 6 x = 8, y = -6 x = -8, y = 6 x = -8, y = -6

that is 12 possible values _________________

Consider giving +1 Kudo when my post helps you. Also, Good Questions deserve Kudos..!

Re: A certain square is to be drawn on a coordinate plane [#permalink]
28 Apr 2013, 19:50

karishmatandon wrote:

Since area of the square is 100, each side = 10 One of the vertices of the square = (0,0) Let the co-ordinates of another vertex of the square be (x,y) Using the formula \(d^2 = x^2 + y^2\) (d = Distance from the origin to any point in the co-ordinate) So, \(100 = x^2 + y^2\) As the vertices, must be integers, solve for different values for x and y When x = 0, y = 10 Also, x = 0, y = -10 x=10, y = 0 x = -10, y = 0 Also x = 6 , y = 8 (as \(100 = 6^2+8^2\)) x = 6, y = -8 x = -6, y= 8 x = -6, y = -8 Similarly, x = 8, y = 6 x = 8, y = -6 x = -8, y = 6 x = -8, y = -6

that is 12 possible values

This diagram posted earlier in the forum explains everything

Re: A certain square is to be drawn on a coordinate plane [#permalink]
29 Dec 2013, 01:16

Got this question wrong on the mgmat cat also.

What I still don't understand how you confirm that a square that has a hypotenuse as an edge running from 0,0 to 6,8, would also have vertices at integer co-ordinates 8,6, and 14,2 and not fractional co-ordinates. How do you reach that conclusion?

Re: A certain square is to be drawn on a coordinate plane [#permalink]
10 Jan 2014, 06:40

1

This post received KUDOS

jpr200012 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?

I'll post the official explanation, but it doesn't make sense to me

Each side of the square must have a length of 10. If each side were to be 6, 7, 8, or most other numbers, there could only be four possible squares drawn, because each side, in order to have integer coordinates, would have to be drawn on the x- or y-axis. What makes a length of 10 different is that it could be the hypotenuse of a Pythagorean triple, meaning the vertices could have integer coordinates without lying on the x- or y-axis.

For example, a square could be drawn with the coordinates (0,0), (6,8), (-2, 14) and (-8, 6). (It is tedious and unnecessary to figure out all four coordinates for each square).

If we label the square abcd, with a at the origin and the letters representing points in a clockwise direction, we can get the number of possible squares by figuring out the number of unique ways ab can be drawn.

a has coordinates (0,0) and b could have the following coordinates, as shown in the picture:

Re: MGMAT How many squares? [#permalink]
10 May 2014, 12:11

Bunuel wrote:

jpr200012 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?

(A) 4 (B) 6 (C) 8 (D) 10 (E) 12

This question becomes much easier if you visualize/draw it.

Let the origin be O and one of the vertices be A. Now, we are told that length of OA must be 10 (area to be 100). So if the coordinates of A is (x, y) then we would have \(x^2+y^2=100\) (distance from the origin to the point A(x, y) can be found by the formula \(d^2=x^2+y^2\))

Now, \(x^2+y^2=100\) has several integer solutions for \(x\) and \(y\), so several positions of vertex A, note that when vertex A has integer coordinates other vertices also have integer coordinates. For example imagine the case when square rests on X-axis to the right of Y-axis, then the vertices are: A(10,0), (10,10), (0,10) and (0,0).

Also you can notice that 100=6^2+8^2 and 100=0^2+10^2, so \(x\) can tale 7 values: -10, -8, -6, 0, 6, 8, 10. For \(x=-10\) and \(x=10\), \(y\) can take only 1 value 0, but for other values of \(x\), \(y\) can take two values positive or negative. For example: when \(x=6\) then \(y=8\) or \(y=-8\). This gives us 1+1+5*2=12 coordinates of point A:

\(x=10\) and \(y=0\), imagine this one to be the square which rests on X-axis and to get the other options rotate OA anticlockwise to get all possible cases; \(x=8\) and \(y=6\); \(x=6\) and \(y=8\); \(x=0\) and \(y=10\); \(x=-6\) and \(y=8\); \(x=-8\) and \(y=6\); \(x=-10\) and \(y=0\); \(x=-8\) and \(y=-6\); \(x=-6\) and \(y=-8\); \(x=0\) and \(y=-10\); \(x=6\) and \(y=-8\); \(x=8\) and \(y=-6\).

Answer: E.

Hi Bunuel,

How did you come up with 8 & 6 being the viable options? I can obviously see it once you point it out but how did you come up with that in the first place?

Additionally, if the square had an area of 50 and we still had to maintain integer lengths, then our answer would be 4, correct?

Re: MGMAT How many squares? [#permalink]
11 May 2014, 20:56

Bunuel wrote:

sandeep800 wrote:

hey Bunuel Thanx,but can u please come out with a Image,only 2 or 3 coordinate value drawn in it....i am sooooo confused....

All 12 squares.

Image posted on our forum by GMATGuruNY:

Attachment:

square.PNG

Bunuel I have a doubt in the figure. The question says that one of the vertices must be origin but in the figure it shows the centre of the square at origin. Isn't this a fallacy ?

In other words one of the vertex of the circle will always have to be (0,0) . Now rotating along this point and considering any one quadrant at a time , we can say distance of any adjacent vertex ( x,y) must be 10 units.

So x^2 + y^2 = 100. Given the constraint of co-ordinates being integers , we see that 8,6 and 6,8 satisfy this . So considering quadrant one only two vertex are possible i.e. (6,8) and (8,6) . Thus 2 squares are possible in quad 1. For four quadrants the possibilities are 4* 2 = 8. Now squares can be also be formed along the x-y axis . They would be 4 in number i.e. 1 in each quadrant with two adjacents sides as x/y axis.

This makes the total as 8+4 = 12.

So although the same answer is coming but the figure in question is confusing.

Is this the correct approach ?

Last edited by himanshujovi on 12 May 2014, 02:33, edited 1 time in total.

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