Proving the Existence of Rational Points on a Circle

[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.

 

[matt grime]

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.

 

[gonzo]

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

 

[jing]

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.

 

[gonzo]

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.

 

[jing]

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.

 

[gonzo]

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).

 

[Werg22]

"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.

 

[Werg22]

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 &lt; \theta &lt; 180.

A, B, C &lt; 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 &lt; C = \frac{5}{n} &lt; 2r

to get the distances

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

 

[Moo Of Doom]

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.

 

[Werg22]

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

 

[tehno]

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

 

[gonzo]

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!

 

[jing]

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?

 

[jing]

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?

 

[gonzo]

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.

 

[jing]

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?' ?

 

[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]

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"

 

[gonzo]

Okay, what's the answer then?

 

[tehno]

"dense" set modulo q

Okay, what's the answer then?

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]

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

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.

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.

 

[matt grime]

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.

 

[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]

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.

 

[matt grime]

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.

 

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

 

[tehno]

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

so is that or is that not an elliptic curve?

http://mathworld.wolfram.com/EllipticCurve.html" .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 existence 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 the basic idea is just to choose rational points on the unit circle...

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 (p,q)\in\mathbb{Q}

 

[gonzo]

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 (p,q)\in\mathbb{Q}

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]

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.

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?

 

[matt grime]

I know what elliptic curves are, thank you very much. I also know what conics are, and the M-W theorem. The question is not about my knowledge but about your attempts at writing mathematics, tehno. Elliptic curves are abelian varieties - they have an addition of points. Do conics? Is there a Mordell Weil theorem (which is a statement about the finite generation of rational points for ab vars) for conics? None of what you wrote joins together well. And your rebuttal with a link to eq 2 and 3 is nonsense - they both point to

y^2=cubic in x (and you can't set the leading coefficient to 0).

Your only reasonable question ,as far as I can see, would be something like this:"What on Earth existence 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 that is the only question i asked, and one you utterly failed to answer until now.

But OP is to be blamed for the reason I brought this up.

No, they are not.

He changed his mind on the formulation of the problem several times during the thread.

at no point did the OP require rationality of the points only of the distances between them.

 

[jing]

Techno - I agree with Matt Grime Gonzo never changed his mind about what he wanted proof of. However there was a lack of clarity (for me) about the statement.

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.

I took this to mean

For any positive integer k, you can find k points on any circle such that each point is a rational distance from every other point.

There followed a discussion with Gonzo about unit circles and such that led to an agreed restatement

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?' ?

which meant that I should have read the OP as

For any positive integer k, you can find k points on some circle such that each point is a rational distance from every other point.


Reading posts in general on the site has led me to conclude that:

Communicating in posts can be difficult as we know what we mean and so expect others to know exactly what we mean when we write the posts but often do not write with sufficient clarity to put over our exact meaning.

So extra care must be taken to read all posts carefully, to ask for clarification when needed and to give clarification when required.

Gonzo - I am sure everyone's posts have been helpful in clarifying your thoughts and in finding the proof you required, which I am pleased you have done.

 

[gonzo]

Thanks jing. But like I said, I can't take credit for it. Wish I could since it was really obvious and simple when I saw it. But I can't.

 

[tehno]

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?

Probably becouse I came up with solution for all circles.
OTOH,I'm not satisfied with it.
I'm almost certain there must be more elegant ,shorter,and
elementar proof of it.
It would be intereting to see other proof for c=1 case you saw.
Can you post it?Maybe it can be extended to the all cases.
Here I'll present an "ugly" proof.
Firstly ,about terminology and some definitions I will use.

For collection of points C \subset \mathbb{R}^2 I say to be at r-distance if the distance between every pair of points is rational.Let's call such collection of points
a rational distance set.Note that rationals form a rational
distance subset of the reals.Particularly,if C is a line
in \mathbb{R}^2,we say C is a dense
set of points at r-distance.
Prior to the proof two known lemmas I will reffer to (without proof).

Lm1.

A triple (x,y,z) of naturals is a primitive pythagorean triple
if x^2+y^2=z^2 holds and if x,y,z are coprime.
All primitive pythagorean triplets where y is even are given by:
x=m^2-n^2,y=2mn,z=m^2+n^2;m&gt;n
m,n are relatively prime and one of them is even.

Lm2.

Affine-rational solutions to \alpha^2 =\beta^2 + 1

are parameterized by:

\alpha=\frac{m^2+1}{2m},\beta=\frac{m^2-1}{2m};m\in\mathbb{Q}
.

_____________________________________________________________
With terminology given above your proposition can be formulated now in this way:

Any circle contains a dense set of points at r-distance.
_____________________________________________________________
Proof:

Let's identify the complex plane,as is often done,with \mathbb{R}^2.
Two points in the complex plane z_{1},z_{2} we can always choose that
they are at r-distance and have rational lenght.Then,becouse of the identity:
\frac{||z_{1}-z_{2}||}{||z_{1}||\cdot ||z_{2}||}=||\frac{1}{z_{1}}-\frac{1}{z_{2}}||

Points 1/z_{1}[/tex] and 1/z_{2} are at r-distance too.<br /> <br /> Consider a vertical line <b>I</b> in the complex plane.<br /> By Lm1 &amp; Lm2 ascertain that we can always parameterize all points on <b>I</b> whose<br /> lenghts and imaginary parts are rational.This set of points is dense in <b>I</b>.<br /> Furthery ,this set points is mapped, as long as <b>I</b> isn&#039;t imaginary axis,<br /> by the complex mapf(z)=z^{-1} to a dense ,rational distance subset of a<br /> circle.Since <b>I</b> can be translated at will through all vertical lines,we can obtain<br /> circless of all radii.<br /> <b>QED</b><br /> <br /> That&#039;s a cumbersome proof of something geometrically obvious (to me).<br /> Ugly and quite probably overkill to the problem.<br /> But it&#039;s a proof.<br /> [EDIT: That <b>I</b> can be parameterized in described manner make sure like this<br /> Vertical line is x=a.We seek points (a,y) on it such that y is rational and also<br /> (a^2+y^2)^(1/2}=z is rational.We get pythagorean equation a^2+y^2=z^2 or 1+(y/a)^2=(z/a)^2.By Lm2 we have y= a[(m^2-1)/(2m)],for m\in\mathbb{Q} .The parametrization occurs ]<br /> If need be I&#039;ll post the proof for lemma 2.It&#039;s not long.<br /> But that&#039;s is only thing what I&#039;m willing to do more ( lack of time and nerves for a chit-chat on this subject).<br /> <br /> <blockquote data-attributes="" data-quote="matt grimme" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> matt grimme said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> but that is the only question i asked, and one you utterly failed to answer until now. </div> </div> </blockquote>Ok.How about posting your solution instead?<br /> <br /> <blockquote data-attributes="" data-quote="jing" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> jing said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Communicating in posts can be difficult as we know what we mean and so expect others to know exactly what we mean when we write the posts but often do not write with sufficient clarity to put over our exact meaning.<br /> <br /> So extra care must be taken to read all posts carefully, to ask for clarification when needed and to give clarification when required. </div> </div> </blockquote>Well said.<br /> Quite sometimes,I read to quickly.Somewhere around half of this thread I saw the title<br /> &quot;Rational points on the circle&quot; .Didn&#039;t care before that about rational points,and I don&#039;t know why I thought he was interested in both distances and points being rational.<br /> he wanted both.

 

[gonzo]

I already posted the basis of my proof, the rest is trivial.

Pick two rational points (points with rational coordinates) on the unit circle:

e^{i \theta_1} and e^{i \theta_2}

Then it is really easy to show that the distance between the points with double the angles e^{2i \theta_1} and e^{2i \theta_2} is rational.

Since there are infinite rational points, you can always find what you need.

 

[matt grime]

l
Ok.How about posting your solution instead?

did I say I had one? I merely have very strong objections to your misuse of mathematics (M-W theorems, and invoking elliptic curves: x^2+y^2=-1 is a perfectly good conic (over C - birational, isomorphic as a curve over C, in fact, to x^2+y^2=1) and has precisely one Q-rational point contradicting your previous assertions, for example).

 

[tehno]

I pleadge gulity as charged for invoking elliptic curves (and for misreading problem conditions as the thread developed as I already said).
I reffered to elliptics only in comparative purposes to conics,but that was completely unnecessary complication to the problem I agree. I accept your criticism as concerns that matter.
I never said x²+y²=-1 has infinitely many rational points. x²+y²=1 in Cartesian plane has that property.

 

[matt grime]

I never said x²+y²=-1 has infinitely many rational points. x²+y²=1 in Cartesian plane has that property.

I meant x^2+y^2=0, not -1, a degnerate case - but what differentiates it from x^2-y^2=0? Or for that matter, what about x^4=3y^4? That has 1 Q-rational point, and is a quadratic, so what's special about conics? Or x^2-1=sqrt(3)y^2? Another curve with only one Q-rational point, although it isn't defined over Q.