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coordinate plane - Manhattan CAT

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coordinate plane - Manhattan CAT

by [email protected] » Sat Mar 31, 2012 6:02 am
Although they gave an explanation, but i didnt quite understand how to approach such problem, Using some common sense, I could imagine 8 squares but thats not the correct answer, Anybody??

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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by rijul007 » Sat Mar 31, 2012 6:27 am
why 8?

in my opinion it should be 4..
a coordinate plane has 4 quadrants.. 10x10 square can be placed in any of the four quadrants
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by [email protected] » Sat Mar 31, 2012 6:40 am
rijul007 wrote:why 8?

in my opinion it should be 4..
a coordinate plane has 4 quadrants.. 10x10 square can be placed in any of the four quadrants
Thanks for your reply
4 considering the edges along the x and y axis, another 4 when the edges are within the quadrants
but the correct answer as per the test is 12.
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by Pharo » Sat Mar 31, 2012 6:56 am
The answer should be E, 12.

Here is my explanation:

The fact that the area of the square is 100 means that the sides should be 10. There are three ways to get this on the coordinate plane and have vertices as integers.

1. You set the coordinates as (10,0) etc.. Meaning you should always have 2 sides of the square on the axis at all times; this way you will get 4 distinct squares.

2. Having one of the vertices attached to the origin, imagine as if you are turning this square 360 degrees around the origin and you want to only stop at such angles where x and y coordinates are integers. That means you will get:
a. 6-8-10 triangles (where x = 6, y = 8 for one of the cases)
b. 8-6-10 triangles (where x = 8, y = 6 for one of the cases)
And you will have 4 repetition of the above two cases; meaning 8 new squares.

Total = 12 squares.

Let me know if you need me to post a picture :)
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by GMATGuruNY » Sat Mar 31, 2012 2:09 pm
I posted an illustration here:

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