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A certain square is to be drawn on a coordinate plane

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Re: A certain square is to be drawn on a coordinate plane  [#permalink]

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New post Updated on: 12 May 2014, 03:33
Bunuel wrote:
sandeep800 wrote:
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....


All 12 squares.

Image posted on our forum by GMATGuruNY:
Attachment:
square.PNG


Bunuel I have a doubt in the figure. The question says that one of the vertices must be origin but in the figure it shows the centre of the square at origin. Isn't this a fallacy ?

In other words one of the vertex of the circle will always have to be (0,0) . Now rotating along this point and considering any one quadrant at a time , we can say distance of any adjacent vertex ( x,y) must be 10 units.

So x^2 + y^2 = 100. Given the constraint of co-ordinates being integers , we see that 8,6 and 6,8 satisfy this . So considering quadrant one only two vertex are possible i.e. (6,8) and (8,6) . Thus 2 squares are possible in quad 1. For four quadrants the possibilities are 4* 2 = 8. Now squares can be also be formed along the x-y axis . They would be 4 in number i.e. 1 in each quadrant with two adjacents sides as x/y axis.

This makes the total as 8+4 = 12.

So although the same answer is coming but the figure in question is confusing.

Is this the correct approach ?

Originally posted by himanshujovi on 11 May 2014, 21:56.
Last edited by himanshujovi on 12 May 2014, 03:33, edited 1 time in total.
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Re: A certain square is to be drawn on a coordinate plane  [#permalink]

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New post 11 May 2014, 22:05
Bunuel wrote:
sandeep800 wrote:
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....


All 12 squares.

Image posted on our forum by GMATGuruNY:
Attachment:
square.PNG


The figure seems to indicate that the area is 20*20 = 400 sq units.
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Re: A certain square is to be drawn on a coordinate plane  [#permalink]

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New post 12 May 2014, 03:18
himanshujovi wrote:
Bunuel wrote:
sandeep800 wrote:
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....


All 12 squares.

Image posted on our forum by GMATGuruNY:
Attachment:
square.PNG


The figure seems to indicate that the area is 20*20 = 400 sq units.


Please read the whole thread: a-certain-square-is-to-be-drawn-on-a-coordinate-plane-127018.html#p782255

Each diagram shows 4 squares not 1, so if you take first diagram you'll see 4 squares and each has one vertex at the origin.
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Re: A certain square is to be drawn on a coordinate plane  [#permalink]

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New post 29 May 2015, 15:29
Hi pawanCEO,

This question includes a number of "restrictions" that you must follow:

1) You have to draw a SQUARE
2) One of the vertices MUST be at the ORIGIN (0, 0)
3) EVERY vertices MUST be an INTEGER
4) Since the area is 100, each side length MUST be 10

Given these restrictions, there are only 12 possible squares that can be drawn.

You mentioned that you think that there are MORE than 12 possibilities.....if so, then why do you think that? Do you have any examples?

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Re: A certain square is to be drawn on a coordinate plane  [#permalink]

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New post 07 Apr 2016, 17:09
Bunuel wrote:
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).


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 :)
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Re: A certain square is to be drawn on a coordinate plane  [#permalink]

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New post 07 Apr 2016, 19:41
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.

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Re: A certain square is to be drawn on a coordinate plane  [#permalink]

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New post 08 Apr 2016, 06:13
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 :)
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Re: A certain square is to be drawn on a coordinate plane  [#permalink]

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New post 08 Apr 2016, 06:42
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thuyduong91vnu wrote:
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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New post 08 Apr 2016, 15:58
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 :)
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New post 08 Apr 2016, 20:08
thuyduong91vnu wrote:
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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New post 14 Jun 2016, 07:45
Where can I find questions on Co-Ordinate Geometry?
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New post 14 Jun 2016, 07:55
vedantkabra wrote:
Where can I find questions on Co-Ordinate Geometry?
Bunuel bb walker


First of all you might find the following post useful ALL YOU NEED FOR QUANT ! ! !.

Also, check our questions bank: viewforumtags.php

Theory on Coordinate Geometry: math-coordinate-geometry-87652.html

All DS Coordinate Geometry Problems to practice: search.php?search_id=tag&tag_id=41
All PS Coordinate Geometry Problems to practice: search.php?search_id=tag&tag_id=62

Hope it helps.
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New post 17 May 2017, 16:19
Area of square = 100. Therefore, Side = 10 and Diagnol = 10sqrt2.

First, calculate possible squares per quadrant:
Let end point of Diagnol be point A (x,y).
So, OA^2 (fixed length of diagnol) = x^2 + y^2 = 200.
Now, calculate [Integers]^2 satisfying above equation:
Only two matching pairs are found: (2,14) & (10,10).
So, we have three coordinates possible for point A: (2,14) (14,2) & (10,10).
Thus, possible squares: per quadrant : 3 / for all four quadrants : 4 × 3 = 12.

Ans. E
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Re: A certain square is to be drawn on a coordinate plane  [#permalink]

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New post 22 May 2017, 18:47
jpr200012 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'll post the official explanation, but it doesn't make sense to me :)

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.


This problem is easy if you simplify the possibilities in one quadrant and extend to the other quadrants. So for instance, in this case, the two obvious choices for a given quadrant are vertices that extend along the x-axis and y-axis. Those are two possible triangles, so we know that there are at least 8 possibilities and we can eliminate answer choice A.

Since the length of the side of the square is 10 and we know that it's a right triangle, two lengths of triangles would satisfy the pythagorean triple: 6^2 + 8^2 + 10^2. So (6,8) and (8,6) are two possible points for the vertex of the square. As a result, there are 4 possible squares that can be drawn in the first quadrant and 12 total squares. It's not equal to 16 because then you would double count the squares that extend along the x- and y-axes.
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Re: A certain square is to be drawn on a coordinate plane  [#permalink]

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New post 09 Aug 2017, 08:49
Hello!

I had a different way to approach this...

Since we are looking for an area of 100 in a square, I factorised it and got 2,2,5,5

That's 4 factors and order matters.

Using permutations I get ---- 4 P 2 = 12.

Is this a correct way to look at this problem, or did I just get lucky?

Thanks for helping out !
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New post 27 Oct 2017, 04:12
Hi,
I am very confused about this question, I saw this on a Manhattan CAT test, the question might be hard, but other than that, I was thinking the vertices of a square are simply its corners, so in the question, I simply thought one of the corners should be in the origin. Just to clarify the concept, aren't the vertices of a square the corners?
Thanks.
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Re: A certain square is to be drawn on a coordinate plane  [#permalink]

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New post 13 Nov 2017, 07:36
Bunuel wrote:
jpr200012 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


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.


I did something exactly like you. But in the middle I got confused. Since we are only focusing on determining the coordinated of one vertex, how do we say for sure that the other two vertices will be necessarily integers?
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Re: A certain square is to be drawn on a coordinate plane  [#permalink]

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New post 30 Apr 2018, 07:08
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jpr200012 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



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:
Image

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:
Image

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


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Re: A certain square is to be drawn on a coordinate plane  [#permalink]

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New post 12 May 2018, 12:02
VeritasPrepKarishma wrote:
TheBirla wrote:
Great approach nookway, but in the 2nd part of your your solution, you are assuming the 4rth vertex is also an integer. And though your assumption is correct, i.e. the 4rth vertex is (14,2) , (2,14) and so on and so forth, I am not sure if this is a 2 minute problem and i got this in one of the mock CAT's that i was doing. Is this the level of problems one has to expect if you are aiming for a 750 + ? Thanks.


The reason the other vertex will be integral is that square is a symmetrical figure. I have explained this in the following post:
http://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.


Hi Karishma,
I went through your explanation in the link that you have provided but I did not understand it very much with respect to this question. Could you please explain again how to ascertain that the other two vertices will definitely be integral co-ordinates.

Could you also please provide a generalized explanation on how to use the technique in other similar questions?
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New post 03 Apr 2019, 14:14
actionj wrote:
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?


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.
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