Rational points on a circle

gonzo

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.

tehno

True for both irrational and rational radii circles...

That's the easiest case .
Points on circle you can choose to satisfy Rational Pythagorean Triplets

gonzo

Okay, what's the answer then?

tehno

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).

gonzo

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

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).

jing

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.

tehno

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.

matt grime

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.

jing

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?

tehno

Thanks for the phraseology check.I edited it.

For k=5 see related much stronger requirement dealing with Cyclic Rational Area Pentagons.
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.

matt grime

what has this to do with anything?


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

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?)


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.

gonzo

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 [tex] e^{i \theta}[/tex] such that all points [tex]e^{2i \theta}[/tex] are distinct.

tehno

Google is your friend.What do you think (eq (2) and (3))?

Elliptic curves I reffered to as an example.One power degree higher than conics but can have only finitely many rational points.
Circles and other conics ,in Cartesian plane have:
a) 0 rational points
or
b) infinitely many rational points.

For central circles case (a) or (b) depends exclusively on c.
For instance x²+y²=3 has no rational points while
x²+y²=1;x²+y²=2 have infinitely many of them.
Your only reasonable question ,as far as I can see, would be something like this:"What on earth existance of circle rational points has to do with rational distances among them?".
As I already emphasized not much ,if the interest is only in rational distances.
But OP is to be blamed for the reason I brought this up.

He changed his mind on the formulation of the problem several times during the thread.
Reducing considerations to the unit circle and other things suggested to me he could be actually interested in situations where both points and distances are rational.That's interesting,but it doesn't turn to be case (unfortunately).
Again ,his concluding sentence prolongs that confusion,in the light of what he wrote before:
So,your basic idea is to choose:
-rational points on the unit circle
or
-points at the rational distance
or
-rational points at rational distance
?
Rational points means to me points with coordinates [tex](p,q)\in\mathbb{Q}[/tex]

gonzo

Yes, I mean points with rational coordinates on the unit circle. It isn't hard to prove that if you pick two points with rational coordinates on the unit circle the distance between the points with double the angles is also rational.

tehno

Can't you (dis)prove it for x²+y²=3 ?

gonzo

The proof I have for the unit circle (and it's not my proof, I can't take credit for it) doesn't directly extend to circles with non-perfect square radii. However, there is another proof I've seen that might.

Why do you ask?