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A certain square is to be drawn on a coordinate plane [#permalink]
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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'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 yaxis. 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 yaxis. 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: (10,0) (8,6) (6,8) (0,10) (6,8) (8,6) (10,0) (8, 6) (6, 8) (0, 10) (6, 8) (8, 6) There are 12 different ways to draw ab, and so there are 12 ways to draw abcd. The correct answer is E.
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Originally posted by jpr200012 on 14 Sep 2010, 22:06.
Last edited by Bunuel on 23 Mar 2016, 10:57, edited 2 times in total.
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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 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 Xaxis to the right of Yaxis, 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 Xaxis 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.
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Re: A certain square is to be drawn on a coordinate plane [#permalink]
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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 Xaxis to the right of Yaxis, 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 Xaxis 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....



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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 xy 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 total = 4+8 = 12 hence E
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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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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?
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Merging similar topics. TheBirla wrote: 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.
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TheBirla wrote: 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. The reason the other vertex will be integral is that square is a symmetrical figure. I have explained this in the following post: http://gmatclub.com/forum/coordinateplane90772.html#p807400So 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.
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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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how can 8 and 6 be X and Y.. when we multiply both we r geting 48 ..bt ans should be 100.. m not geting how can 6 8 and 0 be the x and y value
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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)... \(4+8 = 12\) Answer: 12
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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)
E. 12



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



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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 coordinates 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 coordinate) 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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karishmatandon wrote: Since area of the square is 100, each side = 10 One of the vertices of the square = (0,0) Let the coordinates 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 coordinate) 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
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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 coordinates 8,6, and 14,2 and not fractional coordinates. How do you reach that conclusion?



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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 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 yaxis. 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 yaxis. 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: (10,0) (8,6) (6,8) (0,10) (6,8) (8,6) (10,0) (8, 6) (6, 8) (0, 10) (6, 8) (8, 6) There are 12 different ways to draw ab, and so there are 12 ways to draw abcd. The correct answer is E. Think of this as a circle with radius 10 ok? 10 will be the hypothenuse of the triangle. Now x^2 + y^2 = 100 since it is based in the origin as per the question. Now since x and y must be integers we only have 2 sets of numbers that satisfy this 0 and 10 obviously and 6 and 8 (Think of pythagorean triples 345 only doubled) Now, since we have 4 quadrants we are going to have a bunch of different combinations between (x,y) order and signs but shouldn't be too hard 10 and 0, since 0 does not have a sign then we have 0,10, 10,0, 10,0 and 0,10 easy to count them out total of 4 Then, since 6 and 8 will have positive and negative signs then its better to use a combinatorics approach instead of counting Two slots _ _ We have 2 options for the first (6 and 8), 1 option for the first (either 6 or 8). And then for each of them we have 2 possible signs (+ or ) Then (2)(2)*(2)(1) = 2^3 = 8 Now add em up 8 + 4 = 12 E is the correct answer Hope it helps Gimme Kudos Cheers J



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Re: A certain square is to be drawn on a coordinate plane [#permalink]
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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 Xaxis to the right of Yaxis, 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 Xaxis 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?




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