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A certain square is to be drawn on a coordinate plane. One [#permalink]
 09 Feb 2007, 01:01
Question Stats:
45% (01:28) correct
55% (00:51) wrong based on 5 sessions
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
Last edited by Bunuel on 27 Mar 2012, 00:21, edited 1 time in total.
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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]
09 Feb 2007, 01:25
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]
09 Feb 2007, 01:32
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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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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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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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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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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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
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I get 8 as well. By drawing 1 big square from 4 little squares resting on the axis. Then I get another big square that has the diagonals of each square resting on an axis.
I have no idea if this is correct though.
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beckee529 wrote: ashkrs 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 get 8 (C). Two different ways in each of the four regions. oh shoot.. I think this is a trick question. I change my answer to A (4) because if the set of four squares drawn diagonally cannot generate non-integer coordinates (hence (0, 10sqrt2), etc) we cannot consider those!
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Should be 8.
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Nice drawing exactly what I was saying. However, I I agree w/ Kumar now. We can't have a vertice of (0, 10sqrt2)
Nice catch.
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GMATBLACKBELT wrote: Nice drawing exactly what I was saying. However, I I agree w/ Kumar now. We can't have a vertice of (0, 10sqrt2)
Nice catch.
yes, that's exactly what i was thinking too.. but do you mean you agree with my revised answer of 4 or kumar's answer of 8??
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should be 'E'
There will be 12 squares. See the link below:
http://www.gmatclub.com/forum/t43813
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vshaunak@gmail.com wrote: should be 'E' There will be 12 squares. See the link below: http://www.gmatclub.com/forum/t43813Yes OA is E .
And I am to get in terms with this question.
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vshaunak@gmail.com wrote: should be 'E' There will be 12 squares. See the link below: http://www.gmatclub.com/forum/t43813thanks vshaunak.. completely forgot about the 6-8-10 right triangle.. tricky!!
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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?
4
6
8
10
12
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I think its A, just because it is a square with area 100, so each side must be 10, it cannot be any other shape than a 10x10 square. so if this is the case, then it can only be drawn four ways, one in each quadrant bc the origin (0,0) has to be one of the vertices. unless I am missing something...
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close
