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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?

(A) 4 (B) 6 (C) 8 (D) 10 (E) 12

i think the question boiles down to the following (a,b) is a coordinate pair that must satisfy the following criteria

intigers as well as a^2+b^2 = 100

(10,10),(-10,10),(10,-10),(-10,-10) (8,6),(6,8),(-8,-6),(-6,-8),(-8,6),(6,-8),(-6,8),(8,-6) total number = 12

E, so sides must be 10, with either x, y to be a +-10 and 0 pair, or x^2+y^2 = 10^2, so x could be +-6, +-8, and y could be +- 8, +- 6 so that's 12 different starting lines, 12 different squares

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? 1. 4 2. 6 3. 8 4. 10 5. 12

OA to come this afternoon

Well I think they are only 4 ways so the answer is A (or 1. in this case)

it says that there must be one vertice on the origin so you can draw a square in each of the cuadrants having 4 in total.

We can draw 12 squares with one vertix at the origin. The regular 4 squares with all vertices on the x and y coordinates - 4 squares Tilt the side of a square to 45 degrees - 4 squares Tilt the side of the square to 60 degrees - 4 squares _________________

Keep trying no matter how hard it seems, it will get easier.

We can draw 12 squares with one vertix at the origin. The regular 4 squares with all vertices on the x and y coordinates - 4 squares Tilt the side of a square to 45 degrees - 4 squares Tilt the side of the square to 60 degrees - 4 squares

Hmm... thanks. I didn't think of tilting the squares.. +1 _________________

We can draw 12 squares with one vertix at the origin.

Since \sqrt{100} = \sqrt{36 + 64}

we can have 2 points with distance 10 from origin. (8,6) and (6,8). Also not to forget (10,0). So in all 3 valid points in first quadrant. As we rotate this square in each quadrant we will have 3*4 = 12

So the answer is 12. _________________

Thanks, Just think differently, there is a easier solution:)

We can draw 12 squares with one vertix at the origin.

Since \sqrt{100} = \sqrt{36 + 64}

we can have 2 points with distance 10 from origin. (8,6) and (6,8). Also not to forget (10,0). So in all 3 valid points in first quadrant. As we rotate this square in each quadrant we will have 3*4 = 12

maybe this is to simple, but if its a square than all sides have to be the same. And if there are 4 vertices of a square and you cant change the size of the square than wouldnt there be four options? Drawing a square in each of the 4 quadrants?

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? 4 6 8 10 12

In short, this question is asking how many combinations of integers x & y that form the relationship (x)squared * (y)squared = 100 they are (0,10),(6,8) and (8,6) Ignoring the other vertices, this shows the 3 options of one vertix in a single quadrant. Now multiply this by the number of quadrants i.e. 4 You get the number of ways of drawing the square will be 3*4 = 12!

I've been having a hard time understanding the solution to this problem. Can anyone offer a simple explanation? Thanks in advance.

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?

***The Answer is 12 but I don't understand why.

one of the vertex is origin so (0,0) now area is 100 to get 100 the length of the side can be 10, the set of coordinates on x and y plane dat can have hypotenuse as 10 from origin (0,10), (10,0), (0,-10), (-10, 0) (6,8),(-6,-8), (-6,8), (6,-8), (8,6), (-8,6), (8,-6), (-8,-6),

Hi all. Taking the GMATs tomorrow. Needed some help for the day. Geometry:

stmt1: QR = RS, => <RQS = <RSQ not suff to find angle x stmt2: ST = TU, => <TSU = <TUS not suff to find angle x

if we use both in quad PQSU, => <p+<q+<s+<u = 360 => 90+180 - <RQS+x+180 - <SUT = 360 =>90+x-<RSQ-<TSU = 0 since <RQS = <RSQ and <TSU = <TUS but x+<RSQ+TSU = 180 => <RSU+<TSU = 180 - x replace this in above equation 90+x-180+x = 0 => 2x-90=0 => x is 45

I saw this answer somewhere. Basically the diagonal is length 10\(sqrt{2}\), so x^2+y^2=200, If you imagine a stick rotating around (0,0) then the places where the other end of the stick will encounter integers and still satisfy this condition are (14,2), (2,14), (10,10), there are 3more quadrants = 4x3=12 _________________

Consider kudos, they are good for health

Last edited by mainhoon on 15 Aug 2010, 15:51, edited 1 time in total.

Re: need help counting prob with geometry [#permalink]

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21 Aug 2010, 13:30

Answer is 4 If area is 100 than to calculate length of the side of area:10 Co-ordinates Square1 (0,0),(10,0)(10,10)(0,10) Co-ordinates Square2 (0,0),(-10,0)(-10,-10)(0,-10) Co-ordinates Square3 (0,0),(-10,0)(-10,10)(0,10) Co-ordinates Square4 (0,0),(10,0)(10,-10)(0,-10)

Let me know If you didn't get it _________________

I will give a Fight till the End

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed." - Bernard Edmonds

A person who is afraid of Failure can never succeed -- Amneet Padda

Re: need help counting prob with geometry [#permalink]

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21 Aug 2010, 21:37

2

This post received KUDOS

Here are the conditions: Each side of every square must have a length of 10. Each square must have a vertex at the origin. The coordinates of the other vertices must be integers.

A length of 10 always should make us think of a 6-8-10 triangle. To satisfy the conditions above, we have two options:

Make each side of every square horizontal or vertical. Make each side of every square the hypotenuse of a 6-8-10 triangle.

Please see the attached file for a picture of the 12 squares that can be drawn.

4 apparent squares; with one of the vertices as: (0,10),(10,0),(-10,0)(0,-10).

Then, I took the square in the first quadrant and slowly moved anticlockwise.

I could guess that finding integral co-ordinates for just one of the other vertices in the square( other than 0,0) would suffice and I did find all the required points, {(8,6),(6,8)(-6,8)(-8,6)(-8,-6)(-6,-8)(6,-8)(8,-6)} using pythagoras theorem.

However, I still couldn't formulate how the remaining two points will be integers as well.

By what principle of Geometry can we conclude that, in an xy plane; if two points of a square are integers, such as (0,0) and (8,6); the other two vertices will be all integers as well without explicitly solving for specific values of those vertices. _________________

fluke Principle of the triplets. 3:4:5 or 6:8:10. Karishma's assumptions are correct. She is the best.

8 squares are obvious and NOT tilted. With one vertex on the origin. 4 are tilted with diagonals on the x and y axis. With one vertex on the origin.

Hope that helps ! Btw if this question comes in real exam, I will recall that answer is 12.

fluke wrote:

By what principle of Geometry can we conclude that, in an xy plane; if two points of a square are integers, such as (0,0) and (8,6); the other two vertices will be all integers as well without explicitly solving for specific values of those vertices.

This is doable in a minute or so. 4 on the vertices should come fairly quickly. Then, once you realize that 10 is a multiple of a 3-4-5 triagle, you should recognize more are possible. I drew just one, by arbitrarily picking one to draw. My choice was (-6, -8) from the origin. Recognize that You can do the same shape in the other four quadrants, and can flip to (-8, -6). 12 total.

gmatclubot

Re: MGMAT How many squares?
[#permalink]
01 Mar 2011, 22:20

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