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Rational points on a circle

  1. Mar 24, 2007 #1
    Does anyone have an idea how to prove the following (or prove that it is not true):

    For any positive integer k, you can find k points on a circle such that each point is a rational distance from every other point.
     
  2. jcsd
  3. Mar 24, 2007 #2

    matt grime

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    Not immediately, but you shouldn't call them rational points - that has a strictly different meaning, i.e points on the circle with rational coordinates.
     
  4. Mar 24, 2007 #3
    True, I actually know that but didn't think of it. Sorry.
     
  5. Mar 24, 2007 #4
    I can see a complication.

    Suppose the conjecture is true.

    Take a circle with with unit radius, then take the k points so that each point is a rational distance from each other.

    Let x be an irrational number

    Scale the circle by x, now the distances between the points are irrational.

    So it seems the position of the points must be radius dependent.
     
  6. Mar 24, 2007 #5
    The exact position will be radius dependent, sure. But I'm only looking for a proof of possibility, not a constructive proof. So you can just as well assume the unit circle.
     
  7. Mar 24, 2007 #6
    What I am saying is that proving it for a unit circle will not necessarily prove it for all circles.

    Suppose k=2

    One a unit circle two points at either end of a diameter would be two such points but these would not work if the circle had a diameter of irrational length.

    But I see what you mean about a non constructive proof

    for k=2

    Let A and B be two points on the circumference of a circle radius r. The chord AB subtending an angle x at the centre. For 0<= x <= pi let f:x-->length AB

    f is a continuous function 0<= f(x) <=2r

    there exists a positive integer n such that 2^(-n) <2r and so there is a chord of rational length.

    Bit rough and ready but would I think give a proof for k=2

    For k=3

    Three points A, B, C

    Fix A, arrange for AB to be rational, fixing B
    Could arrange for C so that BC is rational or BC + CA is rational

    Is there a way of forcing both BC and CA or BC and BC+CA to be rational?

    Cannot think of a way at present.
     
  8. Mar 24, 2007 #7
    Proving it for a unit circle is good enough for what I need, whether or not it works for any circle (though I suspect it would anyway).
     
  9. Mar 24, 2007 #8
    "2^(-n) <2r" This is kind of unnecessary to say... in any continuous closed or open interval of real numbers, there are infinity many rational numbers. BC + AC could be equal to a rational number without BC or AC being rational themselves.
     
  10. Mar 24, 2007 #9
    The case k = 3 is possible. We have:

    [tex]C^{2} = A^{2} + B^{2} + 2ABcos {\theta}[/tex]

    Where A and B are known to be rational. We have the conditions:

    [tex]0 < \theta < 180 [/tex].

    [tex] A, B, C < 2r [/tex]

    We set [tex]\theta = 90[/tex] and we simply choose Pythagoras triples with C under or equal to 2r. For example, we could choose a suitable natural n such as

    [tex]A,B < C = \frac{5}{n} < 2r [/tex]

    to get the distances

    [tex]\frac{5}{n}, \frac{4}{n}, \frac{3}{n}[/tex]
     
    Last edited: Mar 24, 2007
  11. Mar 24, 2007 #10
    Werg, if A and B are fixed, then if their endpoints lie on the unit circle, C can have at most 2 distinct values, so I don't think your argument works.
     
  12. Mar 24, 2007 #11
    Hummm. You're right. Discard what I said.
     
  13. Mar 24, 2007 #12
    The statement is true!
    Think of inscribed poligones (Hint:triangles ).You should arrive at set of a very known diophantine equations :smile:
     
  14. Mar 24, 2007 #13
    I don't see the answer. And I don't see how triangles alone help since you need each point to be a rational distance from EVERY other point.

    But if you have something clever, PLEASE explain!
     
  15. Mar 25, 2007 #14
    In general triangles may or may not help. However for k=3 you only have to consider a triangle and if it proves difficult in this case then what does this say for k>3 when joining all points produces numerous triangles?
     
  16. Mar 25, 2007 #15
    Take the case k=3.

    OK you can draw a triangle with sides of rational length and you can draw the circumscribed circle to this triangle. However Gonzo wants this true for all circles or at least for a unit circle.

    So can you draw a triangle with rational sides such that the circumscribed circle has unit radius?
     
    Last edited: Mar 25, 2007
  17. Mar 25, 2007 #16
    I'd settle for an arbitrary circle. I just chose the unit circle since i figured if it would be true for any circle it should be true for the unit circle. However, it is entirely possible that it is only true for circles with irrational radii or something like that, which is fine too. I don't want to limit that part too much.
     
  18. Mar 25, 2007 #17
    So is the question now

    'given k, a positive integer, is it possible to find a circle with k points on the circumference of the circle such that the distance between any two of these points is rational?' ?
     
  19. Mar 25, 2007 #18
    Yes, that is correct. That was what I asked from the beginning, but I thought it would be easier to limit it to a unit circle when someone pointed out that it could depend on the radius, but I realize that might be too much of a limit.
     
  20. Mar 25, 2007 #19
    True for both irrational and rational radii circles...:smile:

    That's the easiest case .
    Points on circle you can choose to satisfy http://www.karlscalculus.org/pythtrip.html" [Broken]
     
    Last edited by a moderator: May 2, 2017
  21. Mar 25, 2007 #20
    Okay, what's the answer then?
     
  22. Mar 25, 2007 #21
    "dense" set modulo q

    Similarily you can generate triangles with rational side lenghts.
    See also "Unit circle relationship" section in wikipedia:
    http://en.wikipedia.org/wiki/Pythagorean_triple

    and take triangle with notation in pic there as B=2b and other two sides =c.

    Think why scale of circle radius or even its' rationality doesn't matter to conclude there are infinitely many rational side triangles.
    From there ,working out conclusion for poligons is also possible ,but longer becouse it requires reccurence relation for Q field (every poligon can be subdivided in triangles).
     
    Last edited: Mar 25, 2007
  23. Mar 25, 2007 #22
    I understand that triangles are easy, I've never had a problem with that part, and that polygons can be divided into triangles. But I still don't see how that helps.

    It's not enough to divide polygon into distinct triangles, you would have draw every possible triangle, all overlapping, and they are all dependent on each other in some very complex way. I just don't see how this will solve the problem.

    Can you be more specific?
     
  24. Mar 25, 2007 #23

    matt grime

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    Gonzo is correct here, tehno. At best your idea finds k points x_1 to x_k with each distance |x_i - x_{i+1}| rational, and that tells you nothing about |x_i - x_j| in general (apart from it being algebraic).
     
  25. Mar 25, 2007 #24
    k=4 is possible

    draw a rectangle with sides 3 and 4, hence diagonals 5. All vertices lie on circle radius 2.5. Do not see how this helps in general though.
     
  26. Mar 25, 2007 #25
    links were provided to answer this part ^^ of jing's question (obviously he asked for it).
    Without proof I answered OP's question.And my answer is correct.


    My idea ?What idea?I didn't give the algorithm how to find rational side distances from disposition of infinitelly many rational points on circle (Probably there isn't such).But,trust me there are always infinitelly many such distances.]

    What kind of a curve is x2+y2=c2 (x2+y2=1 namely)?
    What we know about arithmetic of elliptic curves?
    Mordell-Weil Tm. states there can be only finitely many rational points on elliptic curve.
    x2+y2=c2 is equation of a central circle,which belongs to the familiy of conics.A small differece but that makes it can have only 0 or infinitely many rational points.But regardless of rationality ,there can be choosen infinitely many consequtive triplet vertices determining rational distances.
     
    Last edited: Mar 26, 2007
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