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by PAB2706 » Fri Jun 05, 2009 7:15 am
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?
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by deagez » Fri Jun 05, 2009 7:37 am
I would have to say that the answer is 40.

There should be 10 different squares per quadrant.

for example in Q2 there should be a square with a vertice at (0,10),(-1.9).(-2,8)..........(-9,1) and the same pattern per other quadrant.
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by ssmiles08 » Fri Jun 05, 2009 8:00 am
wow this one took up a lot of time. I got 12. Not sure if it is right but here is my explanation:

automatically you know there are 4 squares with origin (0,0)

(0,0) (10,0) (10,10) (0,10)
(0,0) (10,0) (10,-10) (0,-10)
(0,0) (-10,0) (-10,10) (0,10)
(0,0) (-10,0) (-10,-10) (0,-10)

To have a distance of 10 and vertex of of (0,0)

(0,0) (x,y) = 100

plug in distance formula and we get x^2+y^2 = 100

The only squares that I could think of are 36 and 64 which is (6,8) and (8,6)

(0,0) (6,8) (0,16) (-6,8)----> one more alternative possibility
(0,0) (6,8) (12,0) (6,-8)----> one more alternative possibility

to get (0,16) and (12,0) we have to set up (6,8) (0,y) find the distance : y^2 - 16y = 0; y = 16

(6,8) (x,0) x^2+12x = 0; x = 12

There are 4 more possibilites for (8,6)

(0,0) (8,6) (16,0) (8,-6)---->3 other possibilities


So in total we have (10,0), (6,8), (8,6) four possibilities each: 12 total

What is the OA?
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by mike22629 » Fri Jun 05, 2009 8:56 am
IMO 36

What is OA?
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by mike22629 » Fri Jun 05, 2009 9:03 am
Ah, forget my answer, just reread it. Forgot they have to be squares.
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by PAB2706 » Fri Jun 05, 2009 9:23 am
guys this was the 4th problem i got on my mgmat practice test....i ended up doing miserable in the test cos till 25th prob i got almost all 700-800 level questions..wasted a lot of time as some prob were very very very conceptual and ended up randomly guessing last 15 problems.... :cry:

btw one of you has given the right answer and i am sorry for avoiding to post the OA..i need to see some more ways to solve this problem.
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by dtweah » Fri Jun 05, 2009 9:40 am
There are only 4 possibilities in which the sqaure completely lies within one quadrant. The vertices of the square in this case are:

QI: (0, 0) (10, 0), (10, 10) (0,10)
QII (0 0) (-10, 0), (-10,10) (0,10)
....
Vertices of QIII and QIV can be constructed similarly. IN this case u don't even bother about vertices, just draw four square boxes in each quardrant and you are done.

Find possibilities in which the square overlaps between 2 adjacent quadrants. For example for the overlap b/w Q1 and Q2, only two squares can be drawn. Now since the side of the sqaure will be at an angle, it boils down to finding the distance between the origin and a point such that the distance is 10. To find the distance between the origin and any point we just square the x and y coordinates of the point. The only possibilities are 36 and 64. The vertices of the the overlapping square can be

(0, 0)( 8, 6) (-8 6 ) (0,12)
(0, 0) 6, 8 (6-,8) (0, 16)

All such adjacent combinations will produce 2 squares each. Since there are 4 Quandrants and 2 squares when we combine adjacent quadrants (note I and III are not possible) there are 4 x 2 =8 sqaures that can overlap between adjacent quadrants.

It is not possible to have the square overlap in 3 or 4 quadrants. So
total possibilities are:

4 +8 =12
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by ghacker » Fri Jun 05, 2009 9:56 am
Actually it wont take a lot of time .

It is given that one vertices must be (0,0) and all the vertices must be integers hence they should be of the form (10,0) or (8,6)

For (10,0) we have 4 possible out comes = (10,0),(-10,0),(0,10) and (0-10)

But for (8,6) we have 8

altogether 8+4 =12 squares

(8,6) group = 2ways for 8 and 6 and 4 ways for the sign = 8 ways
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by PAB2706 » Fri Jun 05, 2009 10:15 am
ok..the answer is 12....and yes it wont take a lot of time if concepts are crystal clear.....

as stated by everyone visualising 4 squares is an easy task...

now by distance formula we get 36 and 64 satisfying the given condition...the possible(|6|,|8|) and (|8|,|6|)....we get these two points in all the quadrants..so 4x2=8

@ssmile you really are filled with potential energy....you just need to gather some kinetic energy and you will conquer gmatland.

@dtweah and ghacker..nice thinking.
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by ssmiles08 » Fri Jun 05, 2009 10:24 am
PAB2706 wrote:
@ssmile you really are filled with potential energy....you just need to gather some kinetic energy and you will conquer gmatland.
.
HAHAh thanks man...thats what I am trying o so hard on!! :lol:
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by Osirus@VeritasPrep » Sat Jun 06, 2009 5:14 pm
Can someone explain how 8,6 is an option? If one of the vertices is at (8,6) how can one of the other vertices be at (0,0) and the square still have an area of 100?
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by dtweah » Sun Jun 07, 2009 1:21 am
osirus0830 wrote:Can someone explain how 8,6 is an option? If one of the vertices is at (8,6) how can one of the other vertices be at (0,0) and the square still have an area of 100?
The question states that one of the veritices MUST be (0,0). The side of the square is 10. Find the distance between (0,0) and (8,6). Distance between (0,0) and (x, y) is

D=(x^2 +y^2)^.5= (8^2 +6^2)^.5=100^.5=10
That is why (8 6) and all of its negative variation ( -6 8) are used. Also from the origin, draw a line to the points 8 6 and -8, 6. or (6 8) ( -6 8) You will see two right triangles of the form 8 6 10 or 6 8 10 . The sides of the square are the hypotenuses of both right triangles. If you turn the bottom rays upside down you will have the top part of the square. The points (0 12) and (0 16) do for the top rays what (0 0) does for the bottom rays. Hope this helps.
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