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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?

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?

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:

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?

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
_________________

"Live as if you were to die tomorrow. Learn as if you were to live forever." - Mahatma Gandhi

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)

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?

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.

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!

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

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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