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by adi_800 » Wed Aug 18, 2010 7:18 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
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by adi_800 » Wed Aug 18, 2010 7:24 am
What i did was..
we can have 4 squares along the axis..
1. one side along +ve X axis
2. one side along -ve X axis
3. one side along +ve Y axis
4. one side along -ve Y axis...

Now i drew one line along 45 degrees in the first quadrant...now this can be side for one square..then draw a perpendicular to this side and we have one more square...Take a reflection of this square along the line y = x...So we have 2 squares..
So, in 4 quadrants, we can have 8 squares...
so total of 12 squares..
is my approach fine?

Mgmat explanatn is weird...
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by beatthegmatinsept » Wed Aug 18, 2010 7:30 am
adi_800 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
I think only 4 squares with area 100 are possible. When you make the additional 4 squares by drawing a line at 45 degree angle, you change the side of the square from 10 to sqrt 200, which will not give you the area 100 (which is what the question is asking).

I might be wrong in understanding what the question is asking here... Whats the OA?
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by kvcpk » Wed Aug 18, 2010 7:38 am
adi_800 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
square will have an area of 100, if the side length is 10

Which means distance from origin needs to be 10

Possible points in first quadrant are:
(10,0)
(6,8)
(8,6)

For each of these points, The line connecting them to the origin ca have 2 squares on the either side.
Hence 6 squares are possible.

Similarly, 6 squares can be drawn using the 3rd quadrant.
Hence total of 12 squares are possible.

Hope this helps!!
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by gmatmachoman » Wed Aug 18, 2010 7:44 am
coordinates are :

Following are the starting point(base) of the squares that can be formed.

(10,0)
(-10,0)
(0,10)
(0,-10)

(8,6)
(-8,6)
(8,-6)
(-8,-6)

(6,8)
(6,-8)
(-6,8)
(-6,-8)

So 12 squares can be formed....
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by beatthegmatinsept » Wed Aug 18, 2010 7:51 am
gmatmachoman wrote:coordinates are :

Following are the starting point(base) of the squares that can be formed.

(10,0)
(-10,0)
(0,10)
(0,-10)

(8,6)
(-8,6)
(8,-6)
(-8,-6)

(6,8)
(6,-8)
(-6,8)
(-6,-8)

So 12 squares can be formed....
"One of the vertices must be on the origin" - I understand this as a requirement that one of the corners of the square has to be on 0. But the points (8,6) for example, doesnt meet that requirement.
What am I missing here?
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by GMATGuruNY » Wed Aug 18, 2010 8:37 pm
beatthegmatinsept wrote:
gmatmachoman wrote:coordinates are :

Following are the starting point(base) of the squares that can be formed.

(10,0)
(-10,0)
(0,10)
(0,-10)

(8,6)
(-8,6)
(8,-6)
(-8,-6)

(6,8)
(6,-8)
(-6,8)
(-6,-8)

So 12 squares can be formed....
"One of the vertices must be on the origin" - I understand this as a requirement that one of the corners of the square has to be on 0. But the points (8,6) for example, doesnt meet that requirement.
What am I missing here?
The attached .pdf shows the 12 squares that can be formed. Each square satisfies the conditions of the problem:

-- one vertex is at the origin
-- the area of each square is 100
-- the coordinates of each vertex are integers

Hope this helps!
Attachments
integer squares.pdf
(47.68 KiB) Downloaded 149 times
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by praveen_gmat » Sat Aug 21, 2010 9:20 pm
GMATGuruNY wrote:
beatthegmatinsept wrote:
gmatmachoman wrote:coordinates are :

Following are the starting point(base) of the squares that can be formed.

(10,0)
(-10,0)
(0,10)
(0,-10)

(8,6)
(-8,6)
(8,-6)
(-8,-6)

(6,8)
(6,-8)
(-6,8)
(-6,-8)

So 12 squares can be formed....
"One of the vertices must be on the origin" - I understand this as a requirement that one of the corners of the square has to be on 0. But the points (8,6) for example, doesnt meet that requirement.
What am I missing here?
The attached .pdf shows the 12 squares that can be formed. Each square satisfies the conditions of the problem:

-- one vertex is at the origin
-- the area of each square is 100
-- the coordinates of each vertex are integers

Hope this helps!

Hey GMATGuruNY,

Thanks for the answer... But how do I approach the problem? How to solve it quickly?
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by GMATGuruNY » Sun Aug 22, 2010 2:50 am
praveen_gmat wrote:
GMATGuruNY wrote:
beatthegmatinsept wrote:
gmatmachoman wrote:coordinates are :

Following are the starting point(base) of the squares that can be formed.

(10,0)
(-10,0)
(0,10)
(0,-10)

(8,6)
(-8,6)
(8,-6)
(-8,-6)

(6,8)
(6,-8)
(-6,8)
(-6,-8)

So 12 squares can be formed....
"One of the vertices must be on the origin" - I understand this as a requirement that one of the corners of the square has to be on 0. But the points (8,6) for example, doesnt meet that requirement.
What am I missing here?
The attached .pdf shows the 12 squares that can be formed. Each square satisfies the conditions of the problem:

-- one vertex is at the origin
-- the area of each square is 100
-- the coordinates of each vertex are integers

Hope this helps!

Hey GMATGuruNY,

Thanks for the answer... But how do I approach the problem? How to solve it quickly?
Here are the conditions:
Each side of every square must have a length of 10.
Each square must have a vertex at the origin.
The coordinates of the other vertices must be integers.

A length of 10 always should make us think of a 6-8-10 triangle. To satisfy the conditions above, we can:

Make each side of every square horizontal or vertical.
Building off (10,0), we can draw 4 squares.
Make each side of every square the hypotenuse of a 6-8-10 triangle.
Building off (6,8), we can draw 4 squares
Building off (8,6) we can draw 4 squares.

12 squares total, as shown in the attached file.
Attachments
integer squares.pdf
(47.68 KiB) Downloaded 102 times
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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