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

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14 Sep 2010, 22:06

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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:

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 \(y=-6\).

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

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14 Oct 2010, 15:19

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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

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.

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.
_________________

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

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25 Jan 2013, 07:32

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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

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)... \(4+8 = 12\)

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

Question says "One of the vertices must be on the origin", then why center of square is at origin of co-ordinate system? Shouldn't one of the vertices (the corner) of the square be at the origin?

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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14 Oct 2010, 15:27

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I got this question on a MGMAT CAT as well, but I refuse to believe a question this hard can be on the real GMAT. It is not obvious at all how to solve this in a straight forward manner.
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Re: A certain square is to be drawn on a coordinate plane [#permalink]

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14 Oct 2010, 16:13

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shrouded1 wrote:

I got this question on a MGMAT CAT as well, but I refuse to believe a question this hard can be on the real GMAT. It is not obvious at all how to solve this in a straight forward manner.

I got this on my first Mgmat Cat as well. Most of the questions are time consuming in Mgmat cat's. Have you seen similar level of mgmat cat in Gmat?
_________________

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.

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.
_________________

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

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27 Mar 2013, 23:06

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Perimeter of a circle of radius 10 centered at the origin will coincide with 12 points where where both the x and y values are integers. (10, 0); (8, 6); (6, 8); (0, 10); (-8, 6), (-6, 8); (-10, 0); (-8, -6); (-6, -8); (0, -10); (8, -6); (6, -8)

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

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28 Apr 2013, 10:39

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

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10 Jan 2014, 07:40

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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?

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:

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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14 Sep 2010, 22:22

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.

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....

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

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22 Dec 2010, 16:03

Thanks a lot Bunuel and Karishma.

Karishma, took me a while to get my head around the solution (symmetry of squares), but once i did its just given me a different perspective for these sort of problems. Great approach ! And thanks once again.

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