## Rational points on a circle

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.
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 Recognitions: Homework Help Science Advisor 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.
 True, I actually know that but didn't think of it. Sorry.

## Rational points on a circle

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.
 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.
 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.
 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).
 "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.
 The case k = 3 is possible. We have: $$C^{2} = A^{2} + B^{2} + 2ABcos {\theta}$$ Where A and B are known to be rational. We have the conditions: $$0 < \theta < 180$$. $$A, B, C < 2r$$ We set $$\theta = 90$$ and we simply choose Pythagoras triples with C under or equal to 2r. For example, we could choose a suitable natural n such as $$A,B < C = \frac{5}{n} < 2r$$ to get the distances $$\frac{5}{n}, \frac{4}{n}, \frac{3}{n}$$
 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.
 Hummm. You're right. Discard what I said.

 Quote by gonzo 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.
The statement is true!
Think of inscribed poligones (Hint:triangles ).You should arrive at set of a very known diophantine equations

 Quote by tehno The statement is true! Think of inscribed poligones (Hint:triangles ).You should arrive at set of a very known diophantine equations
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!
 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?

 Quote by tehno The statement is true! Think of inscribed poligones (Hint:triangles ).You should arrive at set of a very known diophantine equations
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?
 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.
 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?' ?

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