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FJ24
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A certain square is to be drawn on a coordinate plane 03 Jul 2019, 15:31
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So i have question about we use the x^2 + Y^2 = 10^2 concept to solve this problem. These other values that you mentioned 6^2 + 8^2 may solve to 100 but isn't Area of a square = l*w. In this case, 6*8 isn't 100 so how are we justifying this? I've looked at the drawn out images as well, and I know I'm missing something but don't know what it is. Can you please clarify? Thanks!

Bunuel
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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

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 X-axis to the right of Y-axis, 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 X-axis 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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A certain square is to be drawn on a coordinate plane 04 Jul 2019, 08:43
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Hi FJ24,

The reason why we're using the Pythagorean Theorem is because we want to define where the sides of the square COULD be. For example, draw a graph with a line from (0,0) to (6,8). That line would have a length of 10 (you can use the Pythagorean Theorem to prove it) and would create one side of a 10x10 square. If you draw the rest of that square, you'll find that all of the vertices are INTEGERS (which is one of the 'restrictions' that the question places on us). By drawing similar diagonal lines, you can find the other squares that fit these descriptions.

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A certain square is to be drawn on a coordinate plane 17 Jul 2020, 07:45
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jpr200012
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


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Solution:

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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A certain square is to be drawn on a coordinate plane 26 Feb 2023, 03:48
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Bunuel
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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....

All 12 squares.

Image posted on our forum by GMATGuruNY:
Attachment:
square.PNG
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How is the condition of 1 vertex on origin satisfied in these squares shown in the diagram??
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A certain square is to be drawn on a coordinate plane 26 Feb 2023, 03:50
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NikTek
Bunuel
sandeep800

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....

All 12 squares.

Image posted on our forum by GMATGuruNY:
Attachment:
square.PNG
How is the condition of 1 vertex on origin satisfied in these squares shown in the diagram??
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Each image has four squares and one of the vertices of each square is at the origin.
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A certain square is to be drawn on a coordinate plane 14 Sep 2024, 04:29
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When 2 vertices of the square are integer coordinates, rest should also be integers ?
So we need to establish a relationship between origin and one vertex only?

If correct, does this extend to 1 vertex is integer. Then all other 3 vertices are integer coordinates?

Oh I see Scott has already made that statement.
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A certain square is to be drawn on a coordinate plane 20 Mar 2025, 13:42
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Bunuel
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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

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 X-axis to the right of Y-axis, 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 X-axis 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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Hi Bunuel, as per the soln we have two vertices of the square, the origin (0,0) and 12 cases for point A, I have two queries:

How are we sure that the subsequent other vertices will be integers, and not some fractional values?
Also how are we sure that there is no repetition of squares when we use any of these 12 vertices, as we don't know their corresponding other vertices?
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