Hmm... I think the answer is yes and this will go by induction. Assume we have such a structure for $n$, and consider a chord [a,b] on the circle. Then if there is a point with rational distance to a and to b, by ptolemy's theorem it has rational distance to every other point in the structure. (since the quadrilateral is cyclic) But now we have to prove that there is a chord for which there are an infinity of points having rational distance to both points... Wlog assume one point is $(1,0)$, and represent a second point as $[\cos(t),\sin(t)]$. Then we find that $\sqrt{2-2\cos(t)}=2\sin\left(\frac t2\right)$ has to be rational, so there are an infinity of points at rational distance of that one point, more specifically all points with rational y-ordinate... but then I miss the rigorous argument to complete the proof... anyone able to finish?

Start with the two points A=(-1,0) and B=(1,0). Another point (cos t, sin t) has rational distance to both if both cos(t/2) and sin(t/2) are rational. An infinite number of these points can easily be found by either using the formula for Pythagorean triples or intersecting a line through (-1,0) with rational slope with the circle. Ptomely's theorem ensures that any two points X,Y among them have also rational distance as XY is the only unknown distance in ABXY.

Since there are infinitely many primitive Pythagorean triples, there are infinitely many angles $\theta$ s.t. $sin\theta ,cos\theta$ are both rational. Call such angles good. By angle-sum formulas, if $a,b$ are good, then $a+b,a-b$ are also good. For points $A,B$ on the circle $\omega$, let $\angle AB$ be the angle subtended by $AB$. Now inductively construct points on $\omega$ s.t. all angles formed by them are good; for 1,2 take any good angle. If there are $n$ points chosen, pick a good angle $\theta$ and a marked point $A$ s.t. the point $B$ on $\omega$ with $\angle AB=\theta$ is distinct from the $n$ points. Since there are infinitely many good angles but finitely many marked points, such $\theta$ exists. For a previously marked point $P$ we have $\angle BP=\pm \angle AP\pm \angle AB$ for suitable choices for the two $\pm$. Since $\angle AP ,\angle AB$ are both good, it follows that $\angle BP$ is good, which finishes induction by adding $B$. Observe that these points for $n=1975$ work: since $AB=2sin\angle AB$ for $A,B$ on the circle, it follows that $AB$ is rational, and so we're done.

Here are 2 very nice solutions: 1) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2016134#p2016134 2) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2182412#p2182412

kunterbunt wrote: Ptomely's theorem ensures that any two points X,Y among them have also rational distance as XY is the only unknown distance in ABXY. Your proof is very nice but maybe it is 'Ptolemy' and not 'Ptomely'?