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# A certain square is to be drawn on a coordinate plane. One

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Re: need help counting prob with geometry [#permalink]

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21 Aug 2010, 13:30
If area is 100 than to calculate length of the side of area:10
Co-ordinates Square1 (0,0),(10,0)(10,10)(0,10)
Co-ordinates Square2 (0,0),(-10,0)(-10,-10)(0,-10)
Co-ordinates Square3 (0,0),(-10,0)(-10,10)(0,10)
Co-ordinates Square4 (0,0),(10,0)(10,-10)(0,-10)

Let me know If you didn't get it
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Re: MGMAT How many squares? [#permalink]

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05 Feb 2011, 04:55
4 apparent squares; with one of the vertices as: (0,10),(10,0),(-10,0)(0,-10).

Then,
I took the square in the first quadrant and slowly moved anticlockwise.

I could guess that finding integral co-ordinates for just one of the other vertices in the square( other than 0,0) would suffice and I did find all the required points, {(8,6),(6,8)(-6,8)(-8,6)(-8,-6)(-6,-8)(6,-8)(8,-6)} using pythagoras theorem.

However, I still couldn't formulate how the remaining two points will be integers as well.

By what principle of Geometry can we conclude that, in an xy plane; if two points of a square are integers, such as (0,0) and (8,6); the other two vertices will be all integers as well without explicitly solving for specific values of those vertices.
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Re: MGMAT How many squares? [#permalink]

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01 Mar 2011, 20:21
fluke
Principle of the triplets. 3:4:5 or 6:8:10. Karishma's assumptions are correct. She is the best.

8 squares are obvious and NOT tilted. With one vertex on the origin.
4 are tilted with diagonals on the x and y axis. With one vertex on the origin.

Hope that helps ! Btw if this question comes in real exam, I will recall that answer is 12.

fluke wrote:
By what principle of Geometry can we conclude that, in an xy plane; if two points of a square are integers, such as (0,0) and (8,6); the other two vertices will be all integers as well without explicitly solving for specific values of those vertices.
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Re: MGMAT How many squares? [#permalink]

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01 Mar 2011, 22:20
This is doable in a minute or so. 4 on the vertices should come fairly quickly. Then, once you realize that 10 is a multiple of a 3-4-5 triagle, you should recognize more are possible. I drew just one, by arbitrarily picking one to draw. My choice was (-6, -8) from the origin. Recognize that You can do the same shape in the other four quadrants, and can flip to (-8, -6). 12 total.
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Re: Tuesday Q1 - # of Squares [#permalink]

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13 May 2011, 03:10
at 90deg to x, at 45 deg to x and at 60 deg to x.

4*3=12 total.
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Re: Tuesday Q1 - # of Squares [#permalink]

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13 May 2011, 13:19
vannu wrote:
IMO, OA is 12

We can draw 12 squares with one vertix at the origin.
The regular 4 squares with all vertices on the x and y coordinates - 4 squares
Tilt the side of a square to 45 degrees - 4 squares
Tilt the side of the square to 60 degrees - 4 squares

Can you post a diagram and explain.

With 45 degree the diagonal will lie ong the X aix.

The co-ordinates of diagonal on X aixs will be (0,0) and (10\sqrt{2} , 0)
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20 Jul 2011, 21:07
i think E

12 ways....

possibilities are of type -
6,8
8,6
10,0

4 quadrants... so 4*3 =12 ways
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21 Jul 2011, 05:02
You're correct. I wouldn't come up with that in 90 seconds.

agdimple333 wrote:
i think E

12 ways....

possibilities are of type -
6,8
8,6
10,0

4 quadrants... so 4*3 =12 ways
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Re: MGMAT How many squares? [#permalink]

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31 Jul 2011, 07:57
Can anybody explain how can we apply combination concepts in this question?
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Re: how many squares ? [#permalink]

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19 Aug 2011, 15:04
I am totally confused, shouldn't x and y be the same as it is a square and there are only 4 possibilities.
Isn't it (+/- 10, 10) and (10, +/- 10)? Help Please!
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Re: how many squares ? [#permalink]

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23 Aug 2011, 20:31
Hi Gyanone .. Can you please describe your solution in detail ?

I really want to know hw you reached the solution ...

Thanks
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Re: how many squares ? [#permalink]

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24 Aug 2011, 06:32
For the square to have an area of 100, the length of any one side of the square must be 10.

For this to be true, if one vertex of the square is given to be the origin (0,0) then we can take one vertex adjacent to this vertex as (x,y)

For the length of this vertex to be 10, using the distance formula,
sqrt [(x-0)^2 + (y-0)^2] = 10
=> x^2 + y^2 = 100

Therefore the possible coordinates of the adjacent vertex are the values of x and y which satisfy this equation. There are 12 such values (as detailed above), so there are 12 possible squares.

Note that (10,10) is not a solution for a vertex adjacent to (0,0) as its distance from (0,0) is not 10.
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Re: how many squares ? [#permalink]

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24 Aug 2011, 16:51
Thank you very much Gyanone. Yes, (6,8) and (0,0) don't make the diagonal but one of the sides of the square. I got it now. Thank you again for the explanation.
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Re: how many squares ? [#permalink]

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12 Sep 2011, 02:20
GyanOne wrote:
Let the other vertex by (x,y)

Then sqrt (x^2 + y^2) = 10
=> x^2 + y^2 = 100
This has 12 solutions as the pairs can be (0,+/- 10); (+/- 10,0); (+/- 6, +/- 8); (+/- 8, +/- 6)

Option (E)

Thanks Gyanone.. MGMAT CAT solution is much more complicated than yours..

Is there a possibility of picking decimals in this question(if the question did not say that they have to be integers)
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23 Sep 2011, 06:11
shahideh wrote:
scthakur wrote:
vr4indian wrote:
Hey dude I am not getting how you come with (6,8) and (8,6) as valid vertices point.

As per my knowledge... if we have one vertices to be (0,0) then one of vertices will lie on X axis (10,0) and other one must lie on Y axis (0,10) ..

So we are left with only one choice for fourth vertices... (10,10)

This is for 1 quadrants , as we have total 4 quadrants answer should be 4 .

Hope i have not done anything wrong here and i had understand question.

Please correct me if i am wrong

Thanks
Vishal Shah

Distance between (0,0) and (6,8) or (8,6) will be 10.

And this distance will make the side of square. So do I think that the answer is 12 (3 square in 4 areas)

I still don't understand.
The distance between (0,0) and (6,8) equals to 10 - you're right, but area of the square is not 100, but 48.
Or where am I mistaken?
4 is correct IMO.
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23 Sep 2011, 06:14
I think i got it:)
It isn't said that square cannot cross any axis
Probably yes, 12 - is correct
Re: Zumit PS 023   [#permalink] 23 Sep 2011, 06:14

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