We can prove that we can draw a point such that its distance is unbounded (i.e. can be greater than any constant \(c\)). We prove the following lemma: Let \(XYZ\) be an equilateral triangle with circumcenter \(O\). I claim we can construct the equilateral triangle with a side on \(YZ\) and the third vertex being \(O\). In other words, if the midpoint of \(YZ\) is \(M\), we take a homothety centered at \(M\) with scale factor 1/3. This may be done by constructing the circumcenter of \(OYZ\), which is the midpoint \(N\) of arc \(YZ\) not containing \(X\) of the circumcircle of \(XYZ\), then constructing the circumcenters of \(NOY\) and \(NOZ\), which are the two vertices of the new equilateral triangle that is not \(O\). Let \(D\) be the midpoint of \(BC\). We can use this lemma to take homotheties centered at \(D\) with negligible scale factors, obtaining points \(P\) on \(BC\) negligibly close to \(D\). Finally, we can draw \(BPC\)'s circumcenter such that the antipode of \(P\), or \(P'\) is arbitrarily far away from \(P\) (as by similar triangles, \(PC^2=PD\cdot PP'\)), implying that the circumcenter can also be arbitrarily far away from \(P\), as desired.

Does this work? Let the circumcenter of $ABC$ be $O$. It is easy to see that the circumcenter of $AOB$(and others) is the reflection of $O$ over $AB$(and others), so due to symmetry over $AB$, the triangles formed by circumcenter of $AOB$(Call it $O_{AB}$), $O$ and $A$ and $B$ separately(ie- the triangles $O_{AB}OA$ and $O_{AB}OB$) have circumcentres that trisect $AB$. Call them $X$ and $Y$. Hence, $OXY$ is a triangle that is one-thrice as small as $ABC$. So the circumcenter of $OXY$ is closer to $AB$ than $O$, hence the circumcenter of $O_{XY}AB$($O_{XY}$ being the circumcenter of $OXY$) is farther away from $C$ than $O_{AB}$. Now iter starting from $OXY$. So, the new circumcenters can be made as far away from $C$ as we want. This solves both parts.

Can we get exactly 7 and/or 2019?

Hexagrammum16 wrote: Can we get exactly 7 and/or 2019? Bump?

This problem is amenable to a so-called "proof without words." After a bit of pondering, it should be clear how the $n$th equilateral triangle is constructed.

Does circumcenter of three collinear points count as infinite distance?

niksic wrote: Does circumcenter of three collinear points count as infinite distance? I'm assuming they say "triangle" as "non-degenerate triangle".