A certain square is to be drawn on a coordinate plane.

[tim]

When you say "this", what exactly do you mean? Your question is too ambiguous for us to help with, unfortunately. There are a lot of things that apply to all Pythagorean triples, and a lot of things that apply only selectively.

[rishijmehta]

Sorry for the lack of clarity. Hopefully this makes my question more clear:

What if the problem stated, "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 169. If all coordinates of the vertices must be integers, how many different ways can this square be drawn?"

This would mean the length of AB would be 13 and thus is a Pythagorean triplet (5, 12, 13). So does the same logic hold as used to solve the problem when the length of AB was 10?

Thanks

Rishi

[RonPurewal]

Sorry for the lack of clarity. Hopefully this makes my question more clear:

What if the problem stated, "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 169. If all coordinates of the vertices must be integers, how many different ways can this square be drawn?"

This would mean the length of AB would be 13 and thus is a Pythagorean triplet (5, 12, 13). So does the same logic hold as used to solve the problem when the length of AB was 10?

Thanks

Rishi

ya, sure.
although you do have to be careful -- if the numbers start getting bigger (which they almost certainly won't, on a non-calculator test), then there may be multiple pythagorean triples with the same hypotenuse.
for instance, if you are given a square with a side length of 25, then you'd have three different kinds of possibilities:
1/
horizontal and vertical sides of length 25 units each
2/
diagonal sides according to the right triangle 15-20-25 (part of the 3-4-5 family)
3/
diagonal sides according to the right triangle 7-24-25

but, unless you encounter one of these types of cases (which, again, you almost certainly won't), you'd be ok here.

[machengcheng]

Hi, i'm confused, why not the result can be countless? If the square requires a fixed point and a length of 10 each, then we can draw a circle from the center with a 10 radius, then the number of possible square can be countless.

What's wrong with my process? Can anyone give a hint? Thanks!

[RonPurewal]

Read the question again, this time more carefully.

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?

[FannieD726]

Hello,

One question regarding this problem: with the method proposed above we make sure that both a and b have integers as coordinate, but how do we know that the other two vertices of the square are also integers?
This is quite clear in the case of a (0;0) and b (0,10) -we can see that the other two vertices will also have integer coordinates by drawing the square-, but not so much in the case of b (6,4) for example.
How can we say that finding the number of ways to trace ab to ensure both a and b have integer coordinates is enough, when we don't know the coordinates of the other two vertices?

Thank you!

[Sage Pearce-Higgins]

Good question. It works like this: if one of the sides of the square is the hypotenuse of a 6-8-10 triangle, then the other sides will also be hypotenuses of other 6-8-10 triangles. Just as an exercise, try drawing out one of these squares in the co-ordinate plane (e.g. start with vertices (0,0), and (8,6) and figure the other vertices out). You'll see that all the vertices lie on integer co-ordinates. The reason for this is that the interior angles of a square are all 90 degrees, meaning that we can effectively rotate the 6-8-10 triangles 90 degrees so that they will line up with other integer co-ordinates. Hopefully you'll see this when you try it out.

Of course, the next question should be: how can I see this in a test situation? And that's not an easy one to answer.