Mitch's illustration is perfect. Please see the diagrams carefully. The smaller squares only are the matter of our concern here, which are having one of its vertices at the origin all the time.sarang_gmt11 wrote:GMATGuruNY wrote:Step 1: If area = 100, side = 10.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
Step 2: Recognize that the hypotenuse of a 6-8-10 triangle is 10.
Step 3: Plot coordinate pairs using every possible combination of (±6,±8), (±8,±6),(0,±10) and (±10,0).
Step 4: Using the plotted points, draw sets of squares centered about the origin. The following sets are possible:
Number of possible squares = 12.
The correct answer is E.
Hi ,
In the above diagram you have shown the centre of the square on the origin but in the problem it is mentioned that one of the vertices must be on the origin ?
Hard MGMAT 700-800 level question. Co-ordinate plane
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Last edited by sanju09 on Sat Mar 24, 2012 2:07 am, edited 1 time in total.
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The favorite Pythagorean triplets on GMAT include integers only. Suitable rotation about the origin and/or reflection through origin or axes would ensure that the coordinates of all the vertices would remain integers with the triplet in hand.imhimanshu wrote:Sorry to bring up the old post..but somehow I am not able to get the solution.. request you to please clarify my below doubts -
1- I understand that we are employing Pythagorean triplets, but I am not been able to follow how that triplets will help me in solving/understanding that all other points will be integer.
Please explain
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Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
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How did you figure out the 4th vertex has coordinates with the variation of 2 and 14? I get the vertices with 6 and 8 by Pythagorean Theorem. Alot of solutions here don't explain that last part. It was the reason I chose 4 instead of 8 or 12.GMATGuruNY wrote:Step 1: If area = 100, side = 10.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
Step 2: Recognize that the hypotenuse of a 6-8-10 triangle is 10.
Step 3: Plot coordinate pairs using every possible combination of (±6,±8), (±8,±6),(0,±10) and (±10,0).
Step 4: Using the plotted points, draw sets of squares centered about the origin. The following sets are possible:
Number of possible squares = 12.
The correct answer is E.
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Solution:
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 the basic (simplest) case, there are 4 squares, and these 4 all have sides that are parallel to the coordinate axes. However, there are 8 other configurations that satisfy the stated requirements, and in these, the sides of the squares are not parallel to the coordinate axes.
If one of the vertices of the square is at the origin and all the coordinates of the vertices must be integers, then a second vertex of the square must be on the circle with radius 10 centered at the origin. Such a circle has the equation x^2 + y^2 = 100. Notice that the radius of the circle is 10 and it could be a side of the square so that the area of the area is 10^2 = 100. Since there are 12 points (see note below) on this circle that have integer coordinates, 12 squares can be drawn.
(Note: The 12 points are (±6, ±8), (±8, ±6), (0, ±10), and (±10, 0).)
Answer: E
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