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

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11 May 2014, 21:56

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 ?

Last edited by himanshujovi on 12 May 2014, 03:33, edited 1 time in total.

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

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29 May 2015, 12:34

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?

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?

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

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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?

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.

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

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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 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
_________________

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

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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 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..
_________________

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

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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.

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

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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?

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:

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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