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# Square ABCD is to be drawn in the xy-plane such that the origin

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Square ABCD is to be drawn in the xy-plane such that the origin  [#permalink]

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09 Sep 2018, 22:06
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Difficulty:

95% (hard)

Question Stats:

30% (02:09) correct 70% (01:59) wrong based on 105 sessions

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Square ABCD is to be drawn in the xy-plane such that the origin is one of vertices and the perimeter is 20. If all the rest three vertices have integer coordinates (x, y), how many different squares can be drawn?

1. 12
2. 10
3. 8
4. 6
5. 4
Math Expert
Joined: 02 Aug 2009
Posts: 8341
Square ABCD is to be drawn in the xy-plane such that the origin  [#permalink]

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10 Sep 2018, 01:45
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rahul16singh28 wrote:
Square ABCD is to be drawn in the xy-plane such that the origin is one of vertices and the perimeter is 20. If all the rest three vertices have integer coordinates (x, y), how many different squares can be drawn?

1. 12
2. 10
3. 8
4. 6
5. 4

so one vertex is fixed and we rotate this square. In a way we are rotating the square around its diagonal.
thus it is a circle with radius as diagonal of the square..
perimeter = 20...side = 20/4=5
diagonal = $$5\sqrt{2}$$
thus $$x^2+y^2=r^2=(5\sqrt{2})^2=50$$
now $$50 = 1+49=1^2+7^2$$ , so x and y can take values 1 or -1 and 7 or -7
In each quadrant - 2 values

next $$50=5^2+5^2$$ so 5 or -5
in each quadrant = 1 value

total $$(2+1)*4=12$$

A
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Untitled1212.png [ 12.51 KiB | Viewed 377 times ]

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Re: Square ABCD is to be drawn in the xy-plane such that the origin  [#permalink]

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09 Dec 2019, 05:21
1
devavrat wrote:
Hi can you please explain this problem in detail

Thanks

If the perimeter of the square is 20, then each side of the square has to be equal to 20/4 = 5.

Let us consider a square ABCD of side 5 units with A on the origin. There are two ways to draw such a square.

1) Please B on one of the axes and proceeding counter-clockwise. 4 such squares are possible, where B is on:

- positive x axis A(0,0), B(5,0), C(5,5), D(0,5)
- positive y axis A(0,0), B(0,5), C(-5,5), D(-5,0)
- negative x axis A(0,0), B(-5,0), C(-5,-5), D(0,-5)
- negative y axis A(0,0), B(0,-5), C(5,-5), D(5,0)

2) Place B at a point not on the x or y axis.

Any line segment from the origin on the co-ordinate plane (such as our side AB) can be considered to be the hypotenuse of a right angled triangle with one of the other 2 sides being parallel to (including coinciding with) x axis and other parallel to (including coinciding with) y axis.

Let this triangle be ABX. For B (which is the other vertex) to have integer co-ordinates, the lengths of AX and BX must be integers.

Pythagorus theorem tells us that with a hypotenuse AB of 5 units, the AX and BX must be 3,4 units (either one can be 3, the other 4). We can find 8 such locations for the point X [such that (AX = 3, BX = 4) or (AX = 4, BX = 3)]

- positive x axis [X(3,0), B(3,4)] and [X(4,0), B(4,3)]
- positive y axis [X(0,3), B(-4,3)] and [X(0,4), B(-3,4)]
- negative x axis [X(-3,0), B(-3,-4)] and [X(-4,0), B(-4,-3)]
- negative y axis [X(0,-3), B(4,-3)] and [X(0,-4), B(3,-4)]

Rules of reflection of points on the co-ordinate place imply that for these co-ordinates of A(0,0) and B(as above), C and D must also have integer co-ordinates.

Hope this clarifies.
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Re: Square ABCD is to be drawn in the xy-plane such that the origin  [#permalink]

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11 Sep 2018, 22:02
Bunuel could you please explain this
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Re: Square ABCD is to be drawn in the xy-plane such that the origin  [#permalink]

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09 Dec 2019, 00:21
chetan2u
Hi can you please explain this problem in detail

Thanks
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Posts: 8341
Re: Square ABCD is to be drawn in the xy-plane such that the origin  [#permalink]

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12 Dec 2019, 06:10
devavrat wrote:
chetan2u
Hi can you please explain this problem in detail

Thanks

Please see the attached figure and connect with my solution above.
Attachments

Untitled1212.png [ 12.51 KiB | Viewed 377 times ]

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Re: Square ABCD is to be drawn in the xy-plane such that the origin   [#permalink] 12 Dec 2019, 06:10
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