# 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.

matt grime
Homework Helper
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.

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

$$\frac{5}{n}, \frac{4}{n}, \frac{3}{n}$$

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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.

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

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?

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?

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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?' ?

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.

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

So can you draw a triangle with rational sides such that the circumscribed circle has unit radius?
That's the easiest case .
Points on circle you can choose to satisfy http://www.karlscalculus.org/pythtrip.html" [Broken]

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"dense" set modulo q

Similarily you can generate triangles with rational side lenghts.
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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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?

matt grime
Homework Helper
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).

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.

So can you draw a triangle with rational sides such that the circumscribed circle has unit radius?

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).
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.

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matt grime
Homework Helper
What we know about modular arithmetic on elliptic curves?

not a lot since it seems to be a phrase you just invented. Elliptic curves. Yes. Modular arithmetic, yes. The two? No. Modular *forms* and elliptic curves, perhaps.

Gonzo - At first I was not entrirely sure what it was you wanted to prove (or find a counter example for). Following our discussions I am now clear on that. Like you however I am not clear how Techno's statements help.

Techno - Thank you for answering my question re: unit circles and triangles, I follow this and am now clear on this point. I am not at all clear how you extend this to polygons meeting Gonzo's requirement. In fact I am still not clear how they meet the partial requirement of the vertices being on a circle.

Perhaps you would be good enough to explain the process for say k=4 and k=5?

not a lot since it seems to be a phrase you just invented.
Thanks for the phraseology check.I edited it.

Perhaps you would be good enough to explain the process for say k=4 and k=5?
For k=5 see related much stronger requirement dealing with http://www.geocities.com/teufel_pi/papers/maa_2005.pdf".
Gauß (I think) who worked both on regular polygons and non-regular showed that there are infinitely many inscribed non-regular polygons with property of their sides being rational.But not for every arbitrary large k.Note this doesn't even include rational diagonals assumption.I thought the OP's question asks merely this..
Now rereading it I see that OP's question asks most likely about complete triangulation of polygons where all diagonals and sides are rational for all k>5 polygons inscribed in unit circle!?
If so,I think ,very probably,such conjecture isn't true.
Don't know how to disprove it though.

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matt grime
Homework Helper
What we know about arithmetic of elliptic curves?
Mordell-Weil Tm. states there can be only finitely many rational points on elliptic curve.

what has this to do with anything?

x2+y2=c2 is equation of a central circle,

so is that or is that not an elliptic curve? If it isn't then what did invoking the mordell-weil theorem do?

which belongs to the familiy of conics.A small differece but that makes it can have only 0 or infinitely many rational points.

why? you invoked M-W, but didn't say that was some kind of iff statement (and when did 0 stop being a finite number?)

there can be choosen infinitely many consequtive triplet vertices determining rational distances.

that doesn't appear to follow at all from anything you have written before which appears to be a random citing of various high powered theorems without explanation.

In case anyone is interested, I've now seen two proofs of this, both of which are in principle the same. It actually turns out to be pretty simple. And it's fine to work with the unit circle.

The basic idea is to make triangles with the center of the circle and two adjacent points. Then you just need to see that the distance between those two points being rational is dependent on the sine of the half the angle being rational.

Then you can see that the the angles between any two non-adjacent points is the sum of angles of adjacent points, and the sine of the sum of angles can always be broken down into the sum of the product of sines and cosines of the base angles.

So the basic idea is just to choose rational points on the unit circle $$e^{i \theta}$$ such that all points $$e^{2i \theta}$$ are distinct.