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

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01 Nov 2012, 08:14

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 _________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

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

Merging similar topics. Please ask if anything remains unclear. _________________

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

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27 Mar 2013, 22:40

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 Need help

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 Need help

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

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28 Apr 2013, 09:55

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?

Doesn't one of the vertices must be on the origin mean one vertices of the square has always to be (0,0) hence only 4 possibilities ??? Please clarify !!

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

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28 Apr 2013, 20:50

karishmatandon wrote:

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

This diagram posted earlier in the forum explains everything

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

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29 Dec 2013, 02:16

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?

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

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10 May 2014, 13:11

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.

Hi Bunuel,

How did you come up with 8 & 6 being the viable options? I can obviously see it once you point it out but how did you come up with that in the first place?

Additionally, if the square had an area of 50 and we still had to maintain integer lengths, then our answer would be 4, correct?

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?

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.. _________________

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