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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?

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  1. Kainui
    • one year ago
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    I believe that it's not possible.

  2. Kainui
    • one year ago
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    Give an example, because this is an unproven problem in number theory. :)

  3. freckles
    • one year ago
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    |dw:1438879607827:dw| \[h=\frac{1}{\sqrt{2}}a \cancel{ \in} \mathbb{Q}\]

  4. Kainui
    • one year ago
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    |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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    |dw:1438880394272:dw| I found a point which made two of those rational but the other two aren't :p

  8. freckles
    • one year ago
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    i found one where 3 are rational one isn't though choosing one of the vertices to be our P

  9. ganeshie8
    • one year ago
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    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
    • one year ago
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    Ohhh that's a cool observation I was thinking the other way around and looking for a point inside a unit square.

  11. ganeshie8
    • one year ago
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    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
    • one year ago
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    When you say it that way, that means something about stacking pythagorean triples together doesn't it?

  13. Kainui
    • one year ago
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    |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
    • one year ago
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    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
    • one year ago
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    |dw:1438881465435:dw| Equivalently (?) can we stick 4 pythagorean triples together like this to form a quadrilateral?

  16. ganeshie8
    • one year ago
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    \(a^2+b^2=c^2\) if \((a,b,c)\) is primitive pythagorean triple, then \(a\not \equiv b \pmod{2}\)

  17. ganeshie8
    • one year ago
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    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
    • one year ago
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    Hmmmm I don't know anything about quadratic residues, do you know if they might be possibly useful here?

  19. ganeshie8
    • one year ago
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    yeah that seem to work, im able to find few squares with 3 integer distances https://i.gyazo.com/a2ddaf3248da9fb3781cdc5aa56d75fd.png

  20. ganeshie8
    • one year ago
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    I'll need to review, but im pretty sure @oldrin.bataku has these on his finger tips!

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

  22. freckles
    • one year ago
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    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
    • one year ago
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    well integers

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

  25. Kainui
    • one year ago
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    Ok thanks I'll definitely play with this and see what I can do.

  26. thomas5267
    • one year ago
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    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
    • one year ago
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    Possibly, could you maybe draw a rough picture to give me a better idea, I'm having some trouble visualizing it.

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