Square ABCD is to be drawn in the xy-plane such that the origin

chetan2u
Hi can you please explain this problem in detail

Thanks

If the perimeter of the square is 20, then each side of the square has to be equal to 20/4 = 5.

Let us consider a square ABCD of side 5 units with A on the origin. There are two ways to draw such a square.

1) Please B on one of the axes and proceeding counter-clockwise. 4 such squares are possible, where B is on:

- positive x axis A(0,0), B(5,0), C(5,5), D(0,5)
- positive y axis A(0,0), B(0,5), C(-5,5), D(-5,0)
- negative x axis A(0,0), B(-5,0), C(-5,-5), D(0,-5)
- negative y axis A(0,0), B(0,-5), C(5,-5), D(5,0)

2) Place B at a point not on the x or y axis.

Any line segment from the origin on the co-ordinate plane (such as our side AB) can be considered to be the hypotenuse of a right angled triangle with one of the other 2 sides being parallel to (including coinciding with) x axis and other parallel to (including coinciding with) y axis.

Let this triangle be ABX. For B (which is the other vertex) to have integer co-ordinates, the lengths of AX and BX must be integers.

Pythagorus theorem tells us that with a hypotenuse AB of 5 units, the AX and BX must be 3,4 units (either one can be 3, the other 4). We can find 8 such locations for the point X [such that (AX = 3, BX = 4) or (AX = 4, BX = 3)]

- positive x axis [X(3,0), B(3,4)] and [X(4,0), B(4,3)]
- positive y axis [X(0,3), B(-4,3)] and [X(0,4), B(-3,4)]
- negative x axis [X(-3,0), B(-3,-4)] and [X(-4,0), B(-4,-3)]
- negative y axis [X(0,-3), B(4,-3)] and [X(0,-4), B(3,-4)]

Rules of reflection of points on the co-ordinate place imply that for these co-ordinates of A(0,0) and B(as above), C and D must also have integer co-ordinates.

Hope this clarifies.

Please see the attached figure and connect with my solution above.
If any query please revert.

Marking some co-oridnates will help a lot! chetan2u.... what I'm confused with is that a pair of lines of a right angled triangle that measure 1 and 7 don't make a square.

I don’t know if this will help at all, but I can show the way I thought about it.

Knowing that the distance formula is just utilizing the Pythagorean Formula, when I saw 5 I immediately thought of the Pythagorean triples 3-4-5.


If one vertex starts at the origin

(1st) we can move 4 units to the right, and 3 units up. (4 , 3)

Connect this to the Origin

The next vertex will be the perpendicular line with the same length as this line we created

Slope of this line = + 3/4

Slope of perpendicular line = - 4/3

Which means from the vertex adjacent to the left of (4 , 3) —— we would have to move 4 units down, and 3 units to the right to get to point (4 , 3)

So we can just back this up and instead do:

Starting from (4 , 3) ———> move 3 to the left, and then 4 up

This will put us at vertex (1 , 7)

Then, to find the final integer vertex, since the opposite side must be parallel, we can make the same moves from the origin (0 , 0)

3 to the left, and 4 up —— gives us vertex (-3, 4)

So we have one square with side length 5:

(0 , 0)
(4 , 3)
(1 , 7)
(-3 , 4)


Now, because when you rotate each vertex 90 degrees about the origin it will maintain a symmetrical movement, we should be able fo make 3 90 degree rotations of this square about the origin. They will have integer vertices.

That is 4 squares


(2nd)

From the origin, we could use the same logic and move 3 units to the right and 4 units up

(3 , 4)

This gives a slope = + 4/3

The perpendicular side will have slope = - 3/4

Move 4 to the left, and 3 up from point (3, 4)

Point (-1 , 7)

Then we make the same move from the origin

(-4 , 3)

So now we have a different square with integer vertices:

(0 , 0)
(3 , 4)
(-1 , 7)
(-4 , 3)

Again, we can make 3 90 degree rotations about the origin to get 3 more square like this one with integer coordinates.

That is 4 more squares.

We now have 8 unique square and the last 4 square are the ones drawn on the X and Y Axis

Total of 12 squares can be made

I don’t think it explains chetan’s method, but maybe it’s a different way to think about it? It helps to figure out how to determine the vertices of a square.


I hope something was helpful.



Posted from my mobile device

Diagonal x^2 +y^2 = sq( 5*rt2)

1^2 +7^2 =50
5^2 +5^2 = 50

With all possibilities allianged with we get 12 different squares

Therefore IMO A