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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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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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matt grime

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True, I actually know that but didn't think of it. Sorry.

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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.

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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.

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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]

[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]

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Hummm. You're right. Discard what I said.

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The statement is 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.

Think of inscribed poligones (Hint:triangles ).You should arrive at set of a very known diophantine equations

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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.The statement is true!

Think of inscribed poligones (Hint:triangles ).You should arrive at set of a very known diophantine equations

But if you have something clever, PLEASE explain!

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Take the case k=3.The statement is true!

Think of inscribed poligones (Hint:triangles ).You should arrive at set of a very known diophantine equations

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?

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'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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True for both irrational and rational radii circles...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.

That's the easiest case .So can you draw a triangle with rational sides such that the circumscribed circle has unit radius?

Points on circle you can choose to satisfy http://www.karlscalculus.org/pythtrip.html" [Broken]

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Okay, what's the answer then?

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Similarily you can generate triangles with rational side lenghts.Okay, what's the answer then?

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).

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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?

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matt grime

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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.

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links were provided to answer this part ^^ of jing's question (obviously he asked for it).So can you draw a triangle with rational sides such that the circumscribed circle has unit radius?

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 x

What we know about arithmetic of elliptic curves?

Mordell-Weil Tm. states there can be only

x

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