coordinate plane

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coordinate plane

by cuty » Fri Mar 04, 2011 9:07 am
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


got it from MGMAT practice test !!
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by Night reader » Fri Mar 04, 2011 11:05 am
Hi cuty, to approach this problem decisively :) we need to look at the diagonal of one of our possible squares with one vertice in the origin at (x=0, y=0) and another in point (x=10, y=10). We are ready to derive the equation for line (diagonal) in our square --> y=x and we know that y*x=100 OR x^2=100. So x and y can be integers only when x=10, -10 {|10|} and y=10, -10 {|10|}, hence answer is A) 4 ways.
cuty 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


got it from MGMAT practice test !!
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by cuty » Fri Mar 04, 2011 11:10 am
@night reader ...thanks but answer is E :|

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by Anurag@Gurome » Fri Mar 04, 2011 11:19 am
cuty 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
One simple way to construct the square is to select the sides along the coordinate axes, which results in 4 possible squares.

But now the square can be slanted.
In which case the co-ordinates of the other vertices of the sides of the square originating from the origin should be at a distance of 10 from (0,0).

Say the those vertices are at (x,y).
Using the distance formula, x² + y² = 10²

Since we know both x and y are integers, the only possible values for (x, y) in 1st quadrant are (0,10), (10,0), (8,6), and (6,8). Out of these (0, 10) and (10, 0) are on the coordinate axes, and we have included that case.

Hence, two new solution in 1st quadrant. Similarly each of the other quadrants will also have two solutions.

Hence, total number of possible squares = 4 + 2*4 = 12

The correct answer is E.
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by vineeshp » Fri Mar 04, 2011 11:23 am
Night reader wrote:Hi cuty, to approach this problem decisively :) we need to look at the diagonal of one of our possible squares with one vertice in the origin at (x=0, y=0) and another in point (x=10, y=10). We are ready to derive the equation for line (diagonal) in our square --> y=x and we know that y*x=100 OR x^2=100. So x and y can be integers only when x=10, -10 {|10|} and y=10, -10 {|10|}, hence answer is A) 4 ways.
cuty 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


got it from MGMAT practice test !!
You have not taken into consideration the Pythagoras theorem and hence points 6,8 and 8,6 come into play.
So answer should be 12.
Vineesh,
Just telling you what I know and think. I am not the expert. :)

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by vineeshp » Fri Mar 04, 2011 11:24 am
Anurag@Gurome wrote:
cuty 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
One simple way to construct the square is to select the sides along the coordinate axes, which results in 4 possible squares.

But now the square can be slanted.
In which case the co-ordinates of the other vertices of the sides of the square originating from the origin should be at a distance of 10 from (0,0).

Say the those vertices are at (x,y).
Using the distance formula, x² + y² = 10²

Since we know both x and y are integers, the only possible values for (x, y) in 1st quadrant are (0,10), (10,0), (8,6), and (6,8). Out of these (0, 10) and (10, 0) are on the coordinate axes, and we have included that case.

Hence, two new solution in 1st quadrant. Similarly each of the other quadrants will also have two solutions.

Hence, total number of possible squares = 4 + 2*4 = 12

The correct answer is E.
You seem to be solving the problems faster than the aspirants :D
Vineesh,
Just telling you what I know and think. I am not the expert. :)

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by Night reader » Fri Mar 04, 2011 11:38 am
Please explain selected?
also (0,10) and (10,0) are the same we are looking for relative changes and not change in the order, I guess?
Anurag@Gurome wrote:
cuty 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
One simple way to construct the square is to select the sides along the coordinate axes, which results in 4 possible squares.

But now the square can be slanted.
In which case the co-ordinates of the other vertices of the sides of the square originating from the origin should be at a distance of 10 from (0,0).

Say the those vertices are at (x,y).
Using the distance formula, x² + y² = 10²

Since we know both x and y are integers, the only possible values for (x, y) in 1st quadrant are (0,10), (10,0), (8,6), and (6,8). Out of these (0, 10) and (10, 0) are on the coordinate axes, and we have included that case.

Hence, two new solution in 1st quadrant. Similarly each of the other quadrants will also have two solutions.

Hence, total number of possible squares = 4 + 2*4 = 12

The correct answer is E.
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by Night reader » Fri Mar 04, 2011 11:39 am
@vineeshp, please consider my question to the solution by Anurag
vineeshp wrote:
You have not taken into consideration the Pythagoras theorem and hence points 6,8 and 8,6 come into play.
So answer should be 12.
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by Anurag@Gurome » Fri Mar 04, 2011 11:47 am
Image

Refer to the figure above.
I have drawn it for the point (8, 6) in the first quadrant.
Night reader wrote:... also (0,10) and (10,0) are the same we are looking for relative changes and not change in the order, I guess?
I don't understand the phrase "relative changes and not change in the order". Can you please express your doubt using a little more words?
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by Night reader » Fri Mar 04, 2011 11:59 am
cleared, I figured all 12 squares :) thanks
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by cuty » Fri Mar 04, 2011 10:14 pm
thnks anurag :)

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by theCEO » Sat Jun 30, 2012 3:10 pm
Alternative Approach:

Draw a circle with radius of 10 with center at the origin.
In quadrant 1 the points crosses the axis at 10 and crosses the x axis at 10.

At any point along the arc, y = sin (angle).
There are 5 angles that you should memorize:
Sin 0 = 0
Sin 30 = 0.5
Sin 45 = Sqrt 2 / 2
Sin 60 = Sqrt 3 / 2
Sin 90 = 1

If we multiply 10 times each of the angles above, we see that only 3 gives integers
We just looked at the 1st quadrant and there are 4 quadrants on the graph.

Therefore we have to multiple 3 x 4 = 12