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

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09 Feb 2007, 01:01

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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?

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

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21 Aug 2010, 21:37

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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.

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

(E)

One of the vertices of the square is fixed. Think of it as a pivot. One of the points that comes easily is (0,10) since 10 should be the length of one side ( Area being 100). Now let the point (0,0 ) act as a hinge and sweep the other parallel vertex across the quadrands. Now, since one side is 10 (use that as hypotenuse) and find the other sides of the right triangle - 8 and 6. Now you can find the combination of the points as follows: (0,10), (0,-10), (10,0), (-10, 0), (8,6), (-8,6), (-8,-6), (8,-6), (6,8), (-6,8), (-6,-8), (6,-8)

Total is 12. _________________

To find what you seek in the road of life, the best proverb of all is that which says: "Leave no stone unturned." -Edward Bulwer Lytton

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

E

You can think of this problem as a right triangle problem.

With the hypotenuse of the right triangle going out to (x,y) from the origin and one of the sides perpendicular to one of the axes. In these cases, x is the length of the base of the triangle and y is the height.

So the question can be rephrased what are all possible integers x and y such that x^2 + y^2 = 10^2.

My solution: 100 = 10x10 = 2x5x2x5. There are 3 ways to arrange 2x5x2x5 into two integers. Then, multiply by 4, because the square can be in any of the 4 quadrants.

Also, this is a good question for guessing. The answer must be divisible by 4, and answers A and C assume only 1 and 2 ways, respectively, in which the square can be drawn in a given quadrant. So E is left. _________________

Number of squares in the xy-coordinate plane with the following criteria. 1. Total area of the square = 100 i.e. each side = 10 2. One of the vertices must be on the origin 3. All coordinates of the vertices must be integers

The first four are quite obvious. Four squares, one in each quadrant, with x and y axis as the two sides with the following vertices:

As each side is 10, if you rotate the side of the first square that falls on (0,0) and (10,0) in the counter clockwise direction with the vertex (0,0) fixed, the whole square will be rotated in the same direction.

Now if we think of this side as the hypotenuse of the right angle triangle that can be created by drawing a perpendicular from the rotating vertex to the x-axis then there are two points in the first quadrant where the rotating vertex has xy co-ordinates that are integers and they can be calculated using the Pythagorean formula.

\(x^2+y^2=10^2\)

the two values for x and y where x,y are integers and \(x^2+y^2=10^2\) are (8,6) and (6,8)

\(8^2+6^2=10^2\)

\(6^2+8^2=10^2\)

Two additional squares can be created with the vertices for the bases as (0,0) (8,6) & (0,0) (6,8) in the first quadrant.

Similarly if you continue rotating the base counter-clockwise you can create two additional squares in quadrant II [bases (0,0) (-6,8) and (0,0) (-8,6)], two more in quadrant III [bases (0,0) (-6,-8) and (0,0) (-8,-6)] and the last two in quadrant IV [bases (0,0) (6,-8) and (0,0) (8,-6)] for a total of 8 additional squares.

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.

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: Looking for modern floor lamps........ [#permalink]

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From MGMAT. I couldn't understand the explanation...

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?

From MGMAT. I couldn't understand the explanation...

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

Ok giving it more thougt here is what I think are the options:

The OA is E. The explanation says that if the square's length were 6, 7, 8, or almost any number, you could only have 4 squares, but a side of 10 is a special case. Do you guys know why?

The OA is E. The explanation says that if the square's length were 6, 7, 8, or almost any number, you could only have 4 squares, but a side of 10 is a special case. Do you guys know why?

No idea for me with 10 also I get only 4 options _________________

"Live as if you were to die tomorrow. Learn as if you were to live forever." - Mahatma Gandhi

The issue here is that all coordinates must be integers.

For every quadrant, there is one evident possibility:

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

But since the lenght of the side is 10, there is one combination of coordinates that generate the same segment: the segment joining points (0,0) and (8,6) since 8^2+6^2=10^2. In a similar way you can build a new square from the segment (0,0) and (6,8).

The reason why with a square with side of 7 or 8 there would only be 4 possibilities is that you cannot have a segment of the same lenght with integers other than (0,7) or (7,0)

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