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Hard MGMAT 700-800 level question. Co-ordinate plane

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khurram Master | Next Rank: 500 Posts Default Avatar
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Hard MGMAT 700-800 level question. Co-ordinate plane

Post Mon Apr 14, 2008 6:37 pm
Thanks
I had 4 as answer. Can someone please elaborate on the answer explanation.

Khurram


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
Each side of the square must have a length of 10. If each side were to be 6, 7, 8, or most other numbers, there could only be four possible squares drawn, because each side, in order to have integer coordinates, would have to be drawn on the x- or y-axis. What makes a length of 10 different is that it could be the hypotenuse of a Pythagorean triple, meaning the vertices could have integer coordinates without lying on the x- or y-axis.

For example, a square could be drawn with the coordinates (0,0), (6,8), (-2, 14) and (-8, 6). (It is tedious and unnecessary to figure out all four coordinates for each square).

If we label the square abcd, with a at the origin and the letters representing points in a clockwise direction, we can get the number of possible squares by figuring out the number of unique ways ab can be drawn.

a has coordinates (0,0) and b could have the following coordinates, as shown in the picture:


(-10,0)
(-8,6)
(-6,8)
(0,10)
(6,8)
(8,6)
(10,0)
(8, -6)
(6, -8)
(0, 10)
(-6, -8)
(-8, -6)

There are 12 different ways to draw ab, and so there are 12 ways to draw abcd.

The correct answer is E.

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jerryragland Senior | Next Rank: 100 Posts Default Avatar
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Post Mon May 02, 2011 8:30 am
unbelievable.. this is a 2 minute question?? comon.. MGMAT

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gmatapril Senior | Next Rank: 100 Posts Default Avatar
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Post Mon Feb 28, 2011 4:53 pm
why we have to take points 6, 8 .( we know that side of square is 10 )please correct me



bluementor wrote:
mjjking wrote:
can somebody explain how and why we find 8,6 and 6,8?
I think its best to use a diagram to illustrate this.

We know that each of the sides of the square is 10. If you drew a line from the origin to point A(10, 0), you have a line of length 10. You can form a (red) square using this line as the basis.

Now, move point A anti-clock wise (with the other end of the line fixed to the origin).Notice the right-angled triangle that is formed with OA' and the x-axis. With the hypoteneus equal to 10, the only possible integer combinations for the height and base of this triangle is 6 and 8 (remember the 6-8-10 triangle). With OA', you have height = 6, base =8, and you will be able to form a (blue) square.

You can also move point A' to A'' and construct a triangle with OA'' and the x-axis. The other combination for the height and base is 8 and 6, respectively. And with this you have another (green) square.

If you repeat this for the other quadrants, you will have a total of 3x4 squares = 12 squares.

-BM-

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aas550 Newbie | Next Rank: 10 Posts Default Avatar
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Post Thu Apr 17, 2008 11:11 am
Can someone explain this me in a different way?
What does smiley stand for?

Thanks

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gabriel Legendary Member
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Post Thu Apr 17, 2008 11:25 am
aas550 wrote:
Can someone explain this me in a different way?
What does smiley stand for?

Thanks
The smiley stands for 8.

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Stuart Kovinsky GMAT Instructor
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Post Thu Apr 17, 2008 11:27 am
aas550 wrote:
Can someone explain this me in a different way?
What does smiley stand for?

Thanks
If you don't have smileys disabled (box you can click under the message that you type), if you ever type 8 followed by a closed bracket you'll get 8)

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netigen Legendary Member Default Avatar
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Post Thu Apr 17, 2008 11:30 am
An alternative way to think about this:

1. There are 4 quadrants we need to draw the squares in so if you find the number of square vertices possible in 1 quadrants , you can just multiple that with 4 to get the answer

2. The area of the sq = 100 so if a side is s then s^2 = 100 and for this to be possible s =10

3. Rephrasing the problem:

to get s = 10 the co-ordinates of each side of the square should be at a distance of 10 from (0,0)

lets say the vertices of sq are at (x,y), using the distance formula

x^2 + y^2 = 10^2

Since we know both x,y are integers the only possible values for x,y in quadrants I are

(0,10), (10,0) (8,6) (6,8)

Out of these possibilities (0,10) and (10,0) lie on the same sq so it means we can have 3 unique possibilities

so total = 4*3 = 12

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mberkowitz Senior | Next Rank: 100 Posts Default Avatar
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Post Tue Sep 09, 2008 1:44 pm
very helpful thanks very much

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Montreal06 Senior | Next Rank: 100 Posts Default Avatar
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Post Wed Oct 01, 2008 2:15 am
Hi,

I'm a bit confused about this part:

to get s = 10 the co-ordinates of each side of the square should be at a distance of 10 from (0,0)

lets say the vertices of sq are at (x,y), using the distance formula

x^2 + y^2 = 10^2

Since we know both x,y are integers the only possible values for x,y in quadrants I are

(0,10), (10,0) (8,6) (6,8 )


If the coordinate needs to be 10 units away from (0,0), how does the vertice (8,6) or (6,8 ) come into play? i.e. 8^2*6^2 = 10^2

but how does the pythagorean principle come into play if we're just talking sides of the square?

thanks,
r

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mjjking Master | Next Rank: 500 Posts Default Avatar
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Post Sat Feb 28, 2009 11:39 pm
can somebody explain how and why we find 8,6 and 6,8?

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bluementor Master | Next Rank: 500 Posts Default Avatar
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Post Tue Mar 03, 2009 5:24 am
mjjking wrote:
can somebody explain how and why we find 8,6 and 6,8?
I think its best to use a diagram to illustrate this.

We know that each of the sides of the square is 10. If you drew a line from the origin to point A(10, 0), you have a line of length 10. You can form a (red) square using this line as the basis.

Now, move point A anti-clock wise (with the other end of the line fixed to the origin).Notice the right-angled triangle that is formed with OA' and the x-axis. With the hypoteneus equal to 10, the only possible integer combinations for the height and base of this triangle is 6 and 8 (remember the 6-8-10 triangle). With OA', you have height = 6, base =8, and you will be able to form a (blue) square.

You can also move point A' to A'' and construct a triangle with OA'' and the x-axis. The other combination for the height and base is 8 and 6, respectively. And with this you have another (green) square.

If you repeat this for the other quadrants, you will have a total of 3x4 squares = 12 squares.

-BM-
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Whitney Garner GMAT Instructor
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Post Mon May 02, 2011 8:56 am
khurram 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?
4
6
8
10
12
A couple of notes on this one. I think everyone is comfortable with the fact that the square must have sides of length 10, so I will ignore that.

Good Guessing Strategy
- I can quickly sketch the 4 "easy" squares that sit on the axes, so 4 feels too easy - eliminate A.
- I can also see that anything I can draw in one coordinate will be mirrored in the other 3, so the answer must be a multiple of 4 - eliminate B and D.
- This just leaves C and E - not a bad guess if stuck (and that would only have taken a minute or so to do).

Actually Solving
The problem gives us a square in a coordinate plane so step 1 should be to sketch something. I would draw at least one of the "easy" squares (the ones that sit on the axes), but then start to think about rotating it. The side length 10 becomes a distance from the origin = This should be sending off a beacon for you to use Pythagorean Theorem rules. Add to that the fact that the vertices have to be integers = which right triangles have integer side lengths?? AH thats right, the Pythagorean triples.

We should have committed the first few iterations of the 3-4-5 triangle to memory, so hypotenuse 10 would be the 6-8-10 triangle. We just need to use the coordinates (6,8) or (8,6) to see if our hunch is correct. Draw one (see image attached).



Now, I chose to make the x-axis length 8 (using the coordinate (8,6) as the vertices), but I could have just as easily picked (6,8). So that means there are 2 of these "weird" squares in each quadrant - so 8 of those when I mirror them around. And then the 4 "easy" squares. Total = 12.

Smile
Whit

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Math is a lot like love - a simple idea that can easily get complicated Smile

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Post Mon May 02, 2011 2:12 pm
Quote:
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
Step 1: If area = 100, side = 10.
Step 2: Recognize that the hypotenuse of a 6-8-10 triangle is 10.
Step 3: Plot coordinate pairs using every possible combination of (±6,±8), (±8,±6),(0,±10) and (±10,0).
Step 4: Using the plotted points, draw sets of squares centered about the origin. The following sets are possible:



Number of possible squares = 12.

The correct answer is E.

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sarang_gmt11 Newbie | Next Rank: 10 Posts Default Avatar
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Post Mon Sep 26, 2011 8:35 pm
GMATGuruNY wrote:
Quote:
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
Step 1: If area = 100, side = 10.
Step 2: Recognize that the hypotenuse of a 6-8-10 triangle is 10.
Step 3: Plot coordinate pairs using every possible combination of (±6,±8), (±8,±6),(0,±10) and (±10,0).
Step 4: Using the plotted points, draw sets of squares centered about the origin. The following sets are possible:



Number of possible squares = 12.

The correct answer is E.
Hi ,
In the above diagram you have shown the centre of the square on the origin but in the problem it is mentioned that one of the vertices must be on the origin ?

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imhimanshu Senior | Next Rank: 100 Posts Default Avatar
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Post Tue Nov 29, 2011 5:13 am
Sorry to bring up the old post..but somehow I am not able to get the solution.. request you to please clarify my below doubts -

1- I understand that we are employing Pythagorean triplets, but I am not been able to follow how that triplets will help me in solving/understanding that all other points will be integer.
Please explain

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