## Kainui one year ago In a square with integer side lengths, is it possible to find a point that's a rational number length away from all 4 corners?

1. Kainui

I believe that it's not possible.

2. Kainui

Give an example, because this is an unproven problem in number theory. :)

3. freckles

|dw:1438879607827:dw| $h=\frac{1}{\sqrt{2}}a \cancel{ \in} \mathbb{Q}$

4. Kainui

|dw:1438879604541:dw| I was kinda thinking about trying to show that we can't shift the origin to any point and end up with these 4 vectors equal to rational numbers. I don't really know how to approach this though, just my first guess.

5. freckles

for some reason i thought we had to choose a point where all corners had the same rational length from this point we are trying to find but i think i misunderstood it like I took "a rational number length" to mean each corner had length b from point P where b is a rational number. that is not right right? like say the square is called ABCD where A,B,C,D are the vertices. We are just wanting to choose point P such that the following are all rational lengths not necessarily the same: AP BP CP DP

6. Kainui

Yeah, so like specifically for anyone else that might be confused, |dw:1438880210962:dw| So in my picture, a,b,c,d are all rational lengths. Is this possible?

7. freckles

|dw:1438880394272:dw| I found a point which made two of those rational but the other two aren't :p

8. freckles

i found one where 3 are rational one isn't though choosing one of the vertices to be our P

9. ganeshie8

since scaling the square by the LCM of denominators gives a square with all integer lengths, I think the problem is equivalent to looking for a point inside a square which has integer distances from all the four corners

10. Kainui

Ohhh that's a cool observation I was thinking the other way around and looking for a point inside a unit square.

11. ganeshie8

its a trade off, we can choose either an unit square with rational distances from corners or some random square with integer distances..

12. Kainui

When you say it that way, that means something about stacking pythagorean triples together doesn't it?

13. Kainui

|dw:1438880869802:dw| All pythagorean triples. I remember a while back we proved something like all pythagorean triples have like they have to have at least one even numbered leg?

14. Kainui

Nevermind, that was a specific case I think you asked (not sure but I think): "If the length of one leg of a right triangle is 16 and the length of the other leg is odd, what are the lengths of all the sides of the triangle?"

15. Kainui

|dw:1438881465435:dw| Equivalently (?) can we stick 4 pythagorean triples together like this to form a quadrilateral?

16. ganeshie8

$$a^2+b^2=c^2$$ if $$(a,b,c)$$ is primitive pythagorean triple, then $$a\not \equiv b \pmod{2}$$

17. ganeshie8

Let $$s$$ be the length of side of square. |dw:1438881952775:dw| I think we want to show that there exist $$(x,y)$$ such that all below expressions are perfect squares : $$x^2+y^2$$ $$(x-s)^2+y^2$$ $$x^2+(y-s)^2$$ $$(x-s)^2+(y-s)^2$$ and $$0\le x,y\le s$$ all are integers

18. Kainui

Hmmmm I don't know anything about quadratic residues, do you know if they might be possibly useful here?

19. ganeshie8

yeah that seem to work, im able to find few squares with 3 integer distances https://i.gyazo.com/a2ddaf3248da9fb3781cdc5aa56d75fd.png

20. ganeshie8

I'll need to review, but im pretty sure @oldrin.bataku has these on his finger tips!

21. Kainui

Hahaha true, I think the problem is getting that fourth one. That's the toughie to make them all line up hmmm.

22. freckles

http://mathworld.wolfram.com/RationalDistanceProblem.html yeah this conway and guy dude were only able to find an infinite number of solutions such that three of the distances were rational

23. freckles

well integers

24. ganeshie8

Here is a fiddle (dumb, fast script) if you want to play with http://jsfiddle.net/ganeshie8/pdcgprc9/

25. Kainui

Ok thanks I'll definitely play with this and see what I can do.

26. thomas5267

Is this equivalent to asking is it possible to draw four circles of integer radius with the center located at the four corners such that they intersect at one point?

27. Kainui

Possibly, could you maybe draw a rough picture to give me a better idea, I'm having some trouble visualizing it.