So I've been thinking of this problem for the past two years, but could't get it.
You are given a circle of center O and arbitrary radius and a segment AB of arbitrary length. Both are arbitray placed in the infinite 2 D plane.
Using only the circle, a sharp pen and a ruler of arbitrary length but without markings on (this is important, because it excludes a solution which won't consider using the circle at all), construct two points inside the segment AB, call them M and N, so that
$$\frac{AM}{7} = \frac{MN}{5} = \frac{NB}{6} $$
I've ran accross the Apollonius circles (one circle is the locus in the plane of all points - including one inside the segment AB - whose distances to two fixed points A and B are in a particular ratio). But here I have to buid two points, not just one.
So any brilliant geometer among you?
Since we know where the center $O$ of the circle $\Gamma$ lies, it is simple to construct perpendiculars.
If $E\neq F$ lie on $\Gamma$ and $E'$ is the antipode of $E$ in $\Gamma$, $EF\perp FE'$.
If we know how to construct perpendiculars, we also know how to construct parallels.
If we construct a segment with a vertex at $A$ that is divided in $18$ equal parts, we may also divide $AB$ in $18$ equal parts through parallels. Then $M$ and $N$ just lie at $\frac{7}{18}$ and $\frac{2}{3}$ of $AB$.
This is an improved solution. Take a point $C$ anywhere outside the $AB$ line. Draw the parallels to the triangle sides through $A,B,C$, finding the anticomplementary triangle $A'B'C'$ and the centroid $G$. The parallel to $BC$ through $G$ meets $AB$ at $N$. Let $H=AA'\cap NB'$ and let $J$ be the intersection between $CH$ and the parallel to $AA'$ through $N$. $GJ$ meets $AB$ at $M$.
Addendum - segment replication through parallels: