A certain square is to be drawn on a coordinate plane

This question becomes much easier if you visualize/draw it.

Let the origin be O and one of the vertices be A. Now, we are told that length of OA must be 10 (area to be 100). So if the coordinates of A is (x, y) then we would have \(x^2+y^2=100\) (distance from the origin to the point A(x, y) can be found by the formula \(d^2=x^2+y^2\))

Now, \(x^2+y^2=100\) has several integer solutions for \(x\) and \(y\), so several positions of vertex A, note that when vertex A has integer coordinates other vertices also have integer coordinates. For example imagine the case when square rests on X-axis to the right of Y-axis, then the vertices are: A(10,0), (10,10), (0,10) and (0,0).

Also you can notice that 100=6^2+8^2 and 100=0^2+10^2, so \(x\) can tale 7 values: -10, -8, -6, 0, 6, 8, 10. For \(x=-10\) and \(x=10\), \(y\) can take only 1 value 0, but for other values of \(x\), \(y\) can take two values positive or negative. For example: when \(x=6\) then \(y=8\) or \(y=-8\). This gives us 1+1+5*2=12 coordinates of point A:

\(x=10\) and \(y=0\), imagine this one to be the square which rests on X-axis and to get the other options rotate OA anticlockwise to get all possible cases;
\(x=8\) and \(y=6\);
\(x=6\) and \(y=8\);
\(x=0\) and \(y=10\);
\(x=-6\) and \(y=8\);
\(x=-8\) and \(y=6\);
\(x=-10\) and \(y=0\);
\(x=-8\) and \(y=-6\);
\(x=-6\) and \(y=-8\);
\(x=0\) and \(y=-10\);
\(x=6\) and \(y=-8\);
\(x=8\) and \(y=-6\).

Answer: E.
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All 12 squares.

Image posted on our forum by GMATGuruNY:
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Best way is to find for one quadrant and multiply by 4.

6,8 satisfy the point for the vertex of the square.

=> 8,6 will also satisfy => 2 squares per quadrant ---> if you are confused why this is true then draw the x-y axis and try to visualize what happens when x is replaced with y.
=> 4*2 = 8 squares

Now 10,0 also satisfy the point or the vertex.
but when we will replace x with y the same square is generated
=> 10,0 and 0,10 are part of same squares.
=> 1 per quadrant
=> 4*1 = 4 squares

total = 4+8 = 12 hence E

hey Bunuel Thanx,but can u please come out with a Image,only 2 or 3 coordinate value drawn in it....i am sooooo confused....

Merging similar topics.



As for your question I doubt that this is a realistic GMAT question. Though if you find that # of squares should be multiple of 4 you'll be left with A, C and E choices right away. Next, you can also rule out A as at least 2 squares per quadrant can be easily found and then make an educated guess for E thus "solving" in less than 2 minutes. Refer for complete solution to the posts above.
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The reason the other vertex will be integral is that square is a symmetrical figure. I have explained this in the following post:
https://gmatclub.com/forum/coordinate-plane-90772.html#p807400

So you don't need to find the 4th vertex and hence don't need to spend that time. You just need to figure out the integral values of x and y such that x^2 + y^2 = 100 which is quite straight forward.
Take x = 0. y = 10 satisfies.
Now check for x = 1/2/3 etc which will take just a few secs each. You will see that x = 6 and y = 8 satisfies.

x can be 0/10/6/8, y will be 10/0/8/6 or -10/-8/-6.
Taking negative sign of x, you will get: x = -10/-6/-8 and y will be 0/8/6 or -8/-6.

Total 12 such squares.

And yes, it is one of the tougher questions, definitely above 700.

To construct a square, we think of one of its side which is a line from (0,0) to some (x,y).
Since the area of the square is 100, its side will have a side = 10.
We can use the distance formula: \(d^2 = x^2 + y^2\) Thus, \(100 = x^2 + y^2\)

Let's think of combinations of perfectly squared x and y that adds up to 100.
{0,10} and {6,8} The best way to think of these combinations is to list the perfect squares and experiment on the combinations that adds up to 100.

Now {0,10}, Both numbers could be x,y or reversed and 10 can be negative or positive. Thus, we already have \(2*2 = 4\) points.
Now {6,8}, Both numbers could be x,y or reversed and 6 and 8 could be negative or positive. Thus, we have \(2*2*2 = 8\)points

And your possible points that form a distance of 10 from the (0,0)... \(4+8 = 12\)

Answer: 12

Since area of the square is 100, each side = 10
One of the vertices of the square = (0,0)
Let the co-ordinates of another vertex of the square be (x,y)
Using the formula \(d^2 = x^2 + y^2\) (d = Distance from the origin to any point in the co-ordinate)
So, \(100 = x^2 + y^2\)
As the vertices, must be integers, solve for different values for x and y
When x = 0, y = 10
Also, x = 0, y = -10
x=10, y = 0
x = -10, y = 0
Also x = 6 , y = 8 (as \(100 = 6^2+8^2\))
x = 6, y = -8
x = -6, y= 8
x = -6, y = -8
Similarly, x = 8, y = 6
x = 8, y = -6
x = -8, y = 6
x = -8, y = -6

that is 12 possible values

Got this question wrong on the mgmat cat also.

What I still don't understand how you confirm that a square that has a hypotenuse as an edge running from 0,0 to 6,8, would also have vertices at integer co-ordinates 8,6, and 14,2 and not fractional co-ordinates. How do you reach that conclusion?

Think of this as a circle with radius 10 ok?

10 will be the hypothenuse of the triangle.

Now x^2 + y^2 = 100 since it is based in the origin as per the question.

Now since x and y must be integers we only have 2 sets of numbers that satisfy this

0 and 10 obviously and 6 and 8 (Think of pythagorean triples 3-4-5 only doubled)

Now, since we have 4 quadrants we are going to have a bunch of different combinations between (x,y) order and signs but shouldn't be too hard

10 and 0, since 0 does not have a sign then we have 0,10, 10,0, -10,0 and 0,-10 easy to count them out total of 4

Then, since 6 and 8 will have positive and negative signs then its better to use a combinatorics approach instead of counting

Two slots _ _

We have 2 options for the first (6 and 8), 1 option for the first (either 6 or 8). And then for each of them we have 2 possible signs (+ or -)

Then (2)(2)*(2)(1) = 2^3 = 8

Now add em up 8 + 4 = 12

E is the correct answer

Hope it helps
Gimme Kudos

Cheers
J

Hi Bunuel,

Could you help to explain why we can assure that whenever vertex A has integer coordinates other vertices also have integer coordinates? I mean do we have some theorem about this one or do we have some way to justify it?

Thanks for your response

Hi thuyduong91vnu,

If we only knew that we were drawing a square with one vertice at the Origin, then the other 3 vertices COULD be on non-integer co-ordinates. However, the original prompt STATES that all 3 vertices are on integer co-ordinates, so we have to use the 'restrictions' that the question places on us.

GMAT assassins aren't born, they're made,
Rich

Hi Rich,

Thanks for your response. I understand your explanation, but that is not my point though

My question is, let's say we have to calculate the number of squares OABC, whose each side has to equal 10 and the coordinates of all 4 vertices have to be integers. By determining possible combinations of x and y-coordinates of A vertice, we could find the questioned number, right? But, assume that we have found these combinations of x and y-coordinators of A vertice (like (8,6) or (-8,-6)..), then how can we assure that the remaining vertices B and C will also have integer coordinates?

Thanks for helping

Hi,

Yes,
and as you ask why?

we are not taking ONLY one set of integers as one vertice..
One is already existing as ORIGIN and the other we have taken as (8,6)..
the line joining origin and (8,6) and the ORIGIN and one diagonally opposite to (8,6) are at 90 degree or perpendicular..
their slope are in ratio -1..
1) let me show you with an example slope of line joining 0,0 and 8,6 ---\(m= \frac{(8-0)}{(6-0)}= \frac{8}{6}\)...
the line perpendicular to it will have -\(\frac{1}{m}\)..
so slope = \(-\frac{6}{8}=\frac{(x-0)}{(y-0)}\)..
thus x= -6 and y=8
so third point is also integer and similarly 4th vertice will also be integer
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Hi chetan2u,

Thanks for your clarification. I think I got it now But, one more question, as I found a new concept here: "diagonally opposite". As you explained, after figuring out the slope of perpendicular line (which is \(\frac{-6}{8}\)), we could use such slope to get the coordinates of diagonally opposite point (which are x= -6 and y=8) by substituing \(\frac{x-0}{y-0}\) for \(\frac{-6}{8}\), right? But to do this, we should not reduce the fractional value of the slope, I mean we should not reduce \(\frac{−6}{8}\) to \(\frac{-3}{4}\)? It is the way to find out coordinates of any diagonally opposite point?

Thanks for your help

hi,
we did not reduce the ratio because the length of the line is the same..
Had it been a rectangle, the ratio could have changed..
Even if we reduce the ratio, we will still get the same answer

even for this example

x/y = -6/8=-3/4..
let the common ratio be a..
so x= -3a and y =4a...

the length of each side is 10..
so \(\sqrt{(-3a)^2+(4a)^2}\) = 10..
\(9a^2+16a^2 = 100\)..
\(a^2 = \frac{100}{25}=4\)..
a= 2, -2..
depending on which Quad the point is we can calculate the coord..
x=-3*2; y=4*2..
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Since the square's area is 100, each side must have length 10.

If all coordinates of the vertices must be integers, we might immediately see that these 4 squares all meet the given conditions:


Any others?
You bet.

At this point, we're looking for something called Pythagorean Triplets. These are sets of three INTEGER values that could be the 3 sides of a right triangle.
The two most common ones to remember for the GMAT are 3-4-5 and 5-12-13.
We should also look out for MULTIPLES of these, such as 6-8-10 and 50-120-130
For this question, the 6-8-10 triplet comes into play. So, the hypotenuse (with length 10) will be one side of the square and the 6 and 8 will be the coordinated of the vertex.
Here are 2 such squares in Quadrant I:


As you might guess, there will be 2 of these types of squares possible in each of the 4 quadrants for a total of 8 squares.

So, the TOTAL number of squares = 4 + 8
= 12

Answer: E


Cheers,
Brent

Please allow me to respond.

Let's take four coordinates that we found: (0, 0), (6, 8), (x, y), (-8,6). Here, (x, y) is the missing coordinate and you are saying that you want to confirm that this is an integer coordinate.

I'm sure there are many ways to do this. We can simply use midpoint formula. Using the 2nd and 4th coordinates the its mid-point is (\(\frac{6-8}{2}\), \(\frac{8+6}{2}\)) which should equal the mid-point of the remaining coordinates, (\(\frac{x+0}{2}\), \(\frac{y+0}{2}\)).

So we have, \(\frac{-2}{2}\) = \(\frac{x}{2}\) and \(\frac{14}{2}\)=\(\frac{y}{2}\), giving us the last coordinate (x, y) = (-2, 14).

Hope it helped.

Three points in 1st Quadrant

So total Cases = 3*4 = 12

Answer: Option E