You can even find infinitely many such points on the unit circle: Let $\mathscr S$ be the set of all points on the unit circle such that $\tan \left(\frac {\theta}4\right)\in \mathbb Q$. If $(\cos(\alpha),\sin(\alpha))$ and $(\cos(\beta),\sin(\beta))$ are two points on the circle then a little ...
Suppose two points on the circle have angle $\phi$ between them. Bisect the chord connecting them to get two right triangles, each with central angle $\frac {\phi}2$. The length of the side opposite that central angle is $\sin \frac {\phi}2$. I then use the addition formula for $\sin$ applied to the angle $\frac {\alpha - \beta}2$.
From the distance formula, I see that I need sine and cosine of both half-angles to be rational. But the z=tan x/2 substitution tells us that that we're good if the quarter angles have rational tangent, so I assume that from the start.
Specifically: if tan(x/4 ) is rational then sin(x/2) and cos(x/2) are rational (from the substitution rules I wrote out.). Thus if tan(x/2) and tan(y/2) are both rational then the distance between the points (cos(x),sin(x)) and (cos(y),sin(y)) is rational (from the formula I gave).
I see, so in other words, The two points on the unit circle should have a specific angle between each others.. And by moving them we get an infinite set..
No...My set S consists of all the points P(x)=(cos(x),sin(x)) such that tan(x/4) is rational. All I need to prove if that, for P(x),P(y) in S, the distance between them is rational. But the distance between then is given by that expression I gave, which depends on sine and cosine of the half angles. Each of those terms, in turn, is rational because of the identities I wrote out.
To me, this all comes down to the famous tan(x/2) substitution. This comes up all the time in looking for rational properties of points on the unit circle.
Well, if I want to choose any two points on the unit circle to test if they have a rational distance between each others.. Just give me three or more points to test them..
Like, point one has 30 degrees.. point two has 44 degrees..
Points that have a rational distance between each others..
Exactly. It is generally true that, with a few clear exceptions, you can't ask for an angle to be reasonable (e.g. rational) and for its trig functions to also be reasonable. Here I want the trig functions to be nice and friendly, so I have to put up with ugly angles.
Not a problem. It's a sign of how difficult this is that, unless there's something new out there, nobody knows if you can do this with a dense set on the plane. Personally I'd vote no on that....but who knows? It's very hard to get a good handle on it.
The other answer on my question shows that it's possible on aplane.. He used the fact that there are an infinte numbers of Pythagorean triples.. Take a look on that answer and the comment.. It's so simple
No, of course you can do it with a discrete set. the open problem is if you can do it with a dense set. That is, I want a set such that every little open set on the plane contains infinitely many members of the set. My example is dense on the circle, for example.
What you meant by a dense set is that like a set between, for example, 1 - 100 on the x axis.. Not like the example on that answer whose points on the x axis are smaller than one?
the integers are not dense on the line...the open interval (.5,.75) contains no integers. The rationals are dense on the line, every open interval contains a rational. My set S is dense on the circle...that's pretty good! My set also has what I think is the nice property that no three points on it lie in a single line.
Discussion between Ahmed Amir and lulu
Discussion between Ahmed Amir and lulu