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

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03 Aug 2009, 15:31
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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03 Aug 2009, 17:08
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]

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13 Oct 2009, 07:24
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]

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13 Oct 2009, 07:54
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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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Re: Tuesday Q1 - # of Squares [#permalink]

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13 Oct 2009, 08:02
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]

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13 Oct 2009, 08:11
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

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Re: Tuesday Q1 - # of Squares [#permalink]

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13 Oct 2009, 11:29
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

Perfect explanation above - OA is 12
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09 Dec 2009, 13:31
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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09 Dec 2009, 13:45
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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20 Jan 2010, 06:01
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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20 Jan 2010, 06:31
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
-Sunny.
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20 Jan 2010, 08:58

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]

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11 Mar 2010, 06:14
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]

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11 Mar 2010, 06:23
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
=> <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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14 Aug 2010, 09:33
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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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Last edited by mainhoon on 15 Aug 2010, 15:51, edited 1 time in total.
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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: need help counting prob with geometry [#permalink]

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21 Aug 2010, 21:37
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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!

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Attachments

integer squares.pdf [47.68 KiB]

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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.
Re: MGMAT How many squares?   [#permalink] 01 Mar 2011, 22:20

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