A certain square is to be drawn on a coordinate plane. One : GMAT Problem Solving (PS)
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# A certain square is to be drawn on a coordinate plane. One

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

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09 Feb 2007, 00:01
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-square-is-to-be-drawn-on-a-coordinate-plane-127018.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Mar 2012, 23:21, edited 1 time in total.
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21 Aug 2010, 20:37
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Here are the conditions:
Each side of every square must have a length of 10.
Each square must have a vertex at the origin.
The coordinates of the other vertices must be integers.

A length of 10 always should make us think of a 6-8-10 triangle. To satisfy the conditions above, we have two options:

Make each side of every square horizontal or vertical.
Make each side of every square the hypotenuse of a 6-8-10 triangle.

Please see the attached file for a picture of the 12 squares that can be drawn.

Hope this helps!

Mitch Hunt
GMAT tutor and instructor
New York, NY
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integer squares.pdf [47.68 KiB]

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Re: how many squares ? [#permalink]

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19 Aug 2011, 03:14
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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)
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19 Aug 2008, 12:30
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lexis 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

(E)

One of the vertices of the square is fixed. Think of it as a pivot. One of the points that comes easily is (0,10) since 10 should be the length of one side ( Area being 100). Now let the point (0,0 ) act as a hinge and sweep the other parallel vertex across the quadrands.
Now, since one side is 10 (use that as hypotenuse) and find the other sides of the right triangle - 8 and 6. Now you can find the combination of the points as follows:
(0,10), (0,-10), (10,0), (-10, 0),
(8,6), (-8,6), (-8,-6), (8,-6),
(6,8), (-6,8), (-6,-8), (6,-8)

Total is 12.
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19 Aug 2008, 13:10
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lexis 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

E

You can think of this problem as a right triangle problem.

With the hypotenuse of the right triangle going out to (x,y) from the origin and one of the sides perpendicular to one of the axes. In these cases, x is the length of the base of the triangle and y is the height.

So the question can be rephrased what are all possible integers x and y such that x^2 + y^2 = 10^2.

(10,0), (-10,0), (0,10), (0,-10), (8,6), (8,-6), (-8,6), (-8,-6), (6,8), (6,-8), (-6,8), (-6, -8)
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20 Aug 2008, 00:23
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My solution: 100 = 10x10 = 2x5x2x5. There are 3 ways to arrange 2x5x2x5 into two integers. Then, multiply by 4, because the square can be in any of the 4 quadrants.

Also, this is a good question for guessing. The answer must be divisible by 4, and answers A and C assume only 1 and 2 ways, respectively, in which the square can be drawn in a given quadrant. So E is left.
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17 Jun 2009, 19:09
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Number of squares in the xy-coordinate plane with the following criteria.
1. Total area of the square = 100 i.e. each side = 10
2. One of the vertices must be on the origin
3. All coordinates of the vertices must be integers

The first four are quite obvious. Four squares, one in each quadrant, with x and y axis as the two sides with the following vertices:

1. (0,0) (10,0) (0,10) (10,10)
2 (0,0) (-10,0) (0,10) (-10,10)
3. (0,0) (-10,0) (0,-10) (-10,-10)
4. (0,0) (10,0) (0,-10) (10,-10)

Now lets see which the remaining 8 are.

As each side is 10, if you rotate the side of the first square that falls on (0,0) and (10,0) in the counter clockwise direction with the vertex (0,0) fixed, the whole square will be rotated in the same direction.

Now if we think of this side as the hypotenuse of the right angle triangle that can be created by drawing a perpendicular from the rotating vertex to the x-axis then there are two points in the first quadrant where the rotating vertex has xy co-ordinates that are integers and they can be calculated using the Pythagorean formula.

$$x^2+y^2=10^2$$

the two values for x and y where x,y are integers and $$x^2+y^2=10^2$$ are (8,6) and (6,8)

$$8^2+6^2=10^2$$

$$6^2+8^2=10^2$$

Two additional squares can be created with the vertices for the bases as (0,0) (8,6) & (0,0) (6,8) in the first quadrant.

Similarly if you continue rotating the base counter-clockwise you can create two additional squares in quadrant II [bases (0,0) (-6,8) and (0,0) (-8,6)], two more in quadrant III [bases (0,0) (-6,-8) and (0,0) (-8,-6)] and the last two in quadrant IV [bases (0,0) (6,-8) and (0,0) (8,-6)] for a total of 8 additional squares.

Total squares = 4 + 8 = 12
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Re: Tuesday Q1 - # of Squares [#permalink]

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13 Oct 2009, 06:54
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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
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14 Aug 2010, 08:33
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I saw this answer somewhere. Basically the diagonal is length 10$$sqrt{2}$$, so x^2+y^2=200, If you imagine a stick rotating around (0,0) then the places where the other end of the stick will encounter integers and still satisfy this condition are (14,2), (2,14), (10,10), there are 3more quadrants = 4x3=12
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22 Sep 2011, 14:04
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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)
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Re: MGMAT PS- How many ways can the square be drawn [#permalink]

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25 Sep 2011, 09:04
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given that origin is one of the vertex of the square and also area of the square is 100.

=> In other words its saying length of side of square has to be 10.

let (x,y) be the point that is at a distance of 10 from origin => x^2+y^2 = 100

36,64
0,100 are such square combinations that are possible here.

Number of possible points in first quadrant = (6,8),(8,6)
Number of possible points in all four quadrants = 2*4 = 8

there are also 4 points on the co-ordinate axis = (10,0) (-10,0)(0,10)(0,-10) = 4

=> total possible points = 12

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09 Feb 2007, 00:18
I can count 8 by a simple raw method... not sure if this is correct thougth...
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Re: PS - Number of Squares [#permalink]

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09 Feb 2007, 00:25
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VirtualM wrote:
From MGMAT. I couldn't understand the explanation...

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

1. 0,0; 10,0; 10,10; 0,10.
2. 0,0; 8, 6; 2,14; -6,8
3. 0,0; 6, 8; -2,14; -8,6
4. 0,0; 0, 10; -10,10; -10,0

this way every quadrant has 3 squars. so, there should be 12 possibilities.

indeed a very good question.....
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Re: PS - Number of Squares [#permalink]

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09 Feb 2007, 00:32
VirtualM wrote:
From MGMAT. I couldn't understand the explanation...

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

Ok giving it more thougt here is what I think are the options:

A(0,0) B(10,0) C(10,10) D(0,10) -1st quadrant

A(0,0) B(-10,0) C(-10,10) D(0,10) -2nd quadrant

A(0,0) B(-10,0) C(-10,-10) D(0,-10) -3rd quadrant

A(0,0) B(10,0) C(10,-10) D(0,-10) -4th quadrant
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09 Feb 2007, 00:53
The OA is E. The explanation says that if the square's length were 6, 7, 8, or almost any number, you could only have 4 squares, but a side of 10 is a special case. Do you guys know why?
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09 Feb 2007, 00:59
VirtualM wrote:
The OA is E. The explanation says that if the square's length were 6, 7, 8, or almost any number, you could only have 4 squares, but a side of 10 is a special case. Do you guys know why?

No idea for me with 10 also I get only 4 options
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09 Feb 2007, 01:04
Posting...
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11 Feb 2007, 05:00
The issue here is that all coordinates must be integers.

For every quadrant, there is one evident possibility:

(0,0) (10,0) (10,10) (0,10)

But since the lenght of the side is 10, there is one combination of coordinates that generate the same segment: the segment joining points (0,0) and (8,6) since 8^2+6^2=10^2. In a similar way you can build a new square from the segment (0,0) and (6,8).

The reason why with a square with side of 7 or 8 there would only be 4 possibilities is that you cannot have a segment of the same lenght with integers other than (0,7) or (7,0)
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11 Feb 2007, 07:14
VirtualM wrote:
Posting...

Now I get this problem. There can be squares formed at an angle... so imagine A(0,0) and point B(6,8) AB = 10...

So I think the answer is 12
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