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

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Re: Geometry Question - need help [#permalink] New post 04 Feb 2009, 08:12
srpradhan wrote:
I've been having a hard time understanding the solution to this problem. Can anyone offer a simple explanation? Thanks in advance.

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?

***The Answer is 12 but I don't understand why.


http://gmatclub.com/forum/7-p509129?t=69095&hilit=A+certain+square+is+to+be+drawn+on+a+coordinate+plane.#p509129
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Re: square is to be drawn [#permalink] New post 30 Jul 2009, 12:43
I know this answer b/c I reviewed it a few times from MGMAT so I won't go into detail and let others try. I got it wrong during the actual test.
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Re: square is to be drawn [#permalink] New post 30 Jul 2009, 13:08
i would like to know too... I tried to draw it out besides the 4 and i could draw some diamonds on the 4 directions but to get a side of 10 on the square in the 1-1-sqrt(2) format the x and y need to be 5*sqrt(2) which isnt integer... my guess would be B...
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Re: square is to be drawn [#permalink] New post 30 Jul 2009, 15:05
I am neonate in coordinate geometry.

I think A is the answer.

Area = 100
each side=10
as one vertex is at the origin, one vertex is a x intercept and another y intercept.the fourth can be on the four coordinates.
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Re: square is to be drawn [#permalink] New post 30 Jul 2009, 17:06
i say 10 because i can draw 8 shapes and there's prob 2 special ones...
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Re: square is to be drawn [#permalink] New post 31 Jul 2009, 01:33
noboru wrote:
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


first + 1 for posting a good question !

IMO It should be e, We can 4 simple squares with 2 sides on the X and Y coordinates as shown ( 1 in each Quadrant )
And now Lets consider one side of the square as OA and try moving it.
OA = 10, Lets consider OA as hypotenuse of a right angled triangle with one side X axis as one coordinate, A can have integers as it coordinates only when either x= 8 or 6 and Y as 6 or 8 .. as OA^2 = X^2 + Y^2 ( 10 x10 = 8x8 + 6x6) for any other combination A will no be having its coordinates as integers for eg if X =9 and then y = sqrt 19.
So , This way we can have 2 squares in 1st quadrant and similarly total 8 in 4 quadrants.

Hence total no. of squares will 8 + 4 =12.

Let me know if this is correct. !
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Re: square is to be drawn [#permalink] New post 31 Jul 2009, 02:47
nitishmahajan wrote:
noboru wrote:
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


first + 1 for posting a good question !

IMO It should be e, We can 4 simple squares with 2 sides on the X and Y coordinates as shown ( 1 in each Quadrant )
And now Lets consider one side of the square as OA and try moving it.
OA = 10, Lets consider OA as hypotenuse of a right angled triangle with one side X axis as one coordinate, A can have integers as it coordinates only when either x= 8 or 6 and Y as 6 or 8 .. as OA^2 = X^2 + Y^2 ( 10 x10 = 8x8 + 6x6) for any other combination A will no be having its coordinates as integers for eg if X =9 and then y = sqrt 19.
So , This way we can have 2 squares in 1st quadrant and similarly total 8 in 4 quadrants.

Hence total no. of squares will 8 + 4 =12.

Let me know if this is correct. !
Cheers



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:

(-10,0)
(-8,6)
(-6,8)
(0,10)
(6,8)
(8,6)
(10,0)
(8, -6)
(6, -8)
(0, 10)
(-6, -8)
(-8, -6)

There are 12 different ways to draw ab, and so there are 12 ways to draw abcd.

The correct answer is E.
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Re: coordinate geometry [#permalink] New post 03 Aug 2009, 14:31
Jozu 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


i think the question boiles down to the following (a,b) is a coordinate pair that must satisfy the following criteria

intigers as well as a^2+b^2 = 100

(10,10),(-10,10),(10,-10),(-10,-10) (8,6),(6,8),(-8,-6),(-6,-8),(-8,6),(6,-8),(-6,8),(8,-6) total number = 12
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Re: coordinate geometry [#permalink] New post 03 Aug 2009, 16:08
E, so sides must be 10, with either x, y to be a +-10 and 0 pair, or x^2+y^2 = 10^2, so x could be +-6, +-8, and y could be +- 8, +- 6 so that's 12 different starting lines, 12 different squares
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Re: Tuesday Q1 - # of Squares [#permalink] New post 13 Oct 2009, 06:24
hogann 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?
1. 4
2. 6
3. 8
4. 10
5. 12

OA to come this afternoon


Well I think they are only 4 ways so the answer is A (or 1. in this case)

it says that there must be one vertice on the origin so you can draw a square in each of the cuadrants having 4 in total.
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Re: Tuesday Q1 - # of Squares [#permalink] New post 13 Oct 2009, 07:02
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


Hmm... thanks. I didn't think of tilting the squares.. +1
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Re: Tuesday Q1 - # of Squares [#permalink] New post 13 Oct 2009, 07:11
I will go with 12

We can draw 12 squares with one vertix at the origin.

Since \sqrt{100} = \sqrt{36 + 64}

we can have 2 points with distance 10 from origin. (8,6) and (6,8). Also not to forget (10,0). So in all 3 valid points in first quadrant. As we rotate this square in each quadrant we will have
3*4 = 12

So the answer is 12.
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Re: Tuesday Q1 - # of Squares [#permalink] New post 13 Oct 2009, 10:29
RJAGmat wrote:
I will go with 12

We can draw 12 squares with one vertix at the origin.

Since \sqrt{100} = \sqrt{36 + 64}

we can have 2 points with distance 10 from origin. (8,6) and (6,8). Also not to forget (10,0). So in all 3 valid points in first quadrant. As we rotate this square in each quadrant we will have
3*4 = 12

So the answer is 12.


Perfect explanation above - OA is 12
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Re: Squares [#permalink] New post 09 Dec 2009, 12:31
maybe this is to simple, but if its a square than all sides have to be the same. And if there are 4 vertices of a square and you cant change the size of the square than wouldnt there be four options? Drawing a square in each of the 4 quadrants?
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Re: Squares [#permalink] New post 09 Dec 2009, 12:45
Gmatter111 wrote:
Is there a quick way to solve this:


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


In short, this question is asking how many combinations of integers x & y that form the relationship (x)squared * (y)squared = 100
they are (0,10),(6,8) and (8,6)
Ignoring the other vertices, this shows the 3 options of one vertix in a single quadrant.
Now multiply this by the number of quadrants i.e. 4
You get the number of ways of drawing the square will be 3*4 = 12!
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Re: Squares [#permalink] New post 20 Jan 2010, 05:01
yes its 12

3 in each quadrant and total = 12

x^2 +y^2 = 10

only comes when x,y = 0,10 ; 6,8 ; 8,6

note: x:y 0,10 will have 10,0 on another vertex so we will take it once only
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Re: Squares [#permalink] New post 20 Jan 2010, 05:31
It should be 4 simply 4. I fail to understand the concept of three in each quadrant, 6,8, 8,6 etc. Dear gurpreetsingh, please elaborate.

Thank you
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Re: Squares [#permalink] New post 20 Jan 2010, 07:58
Consider quadrant I

area of square = 100

side of square = 10
now x^2 + y^2 must be equal to 10.

only 10,0 ; 0,10 ; 6,8; 8,6 satisfy this.
If you take x,y = 10,0.....x,y =0,10 has been already satisfied on other vertex.

thus you are left with 3 options only

Since this can be same for all the 4 quadrants, total possibilities are 3*4 = 12
I hope you got the concept now.
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Re: Geometry Question - need help [#permalink] New post 11 Mar 2010, 05:14
srpradhan wrote:
I've been having a hard time understanding the solution to this problem. Can anyone offer a simple explanation? Thanks in advance.

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?

***The Answer is 12 but I don't understand why.


one of the vertex is origin so (0,0)
now area is 100
to get 100 the length of the side can be 10, the set of coordinates on x and y plane dat can have hypotenuse as 10 from origin
(0,10), (10,0), (0,-10), (-10, 0)
(6,8),(-6,-8), (-6,8), (6,-8),
(8,6), (-8,6), (8,-6), (-8,-6),


hence 12 ways...
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Re: Geometry Question - need help [#permalink] New post 11 Mar 2010, 05:23
djveed wrote:
Hi all. Taking the GMATs tomorrow. Needed some help for the day. Geometry:


stmt1: QR = RS, => <RQS = <RSQ not suff to find angle x
stmt2: ST = TU, => <TSU = <TUS not suff to find angle x

if we use both
in quad PQSU,
=> <p+<q+<s+<u = 360
=> 90+180 - <RQS+x+180 - <SUT = 360
=>90+x-<RSQ-<TSU = 0 since <RQS = <RSQ and <TSU = <TUS
but x+<RSQ+TSU = 180 => <RSU+<TSU = 180 - x
replace this in above equation
90+x-180+x = 0 => 2x-90=0 => x is 45

so both the stmts are required.
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Re: Geometry Question - need help   [#permalink] 11 Mar 2010, 05:23
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