Given a line segment $A B$, construct a segment half as long as $A B$ using only a compass. Construct a segment one-third as long as $A B$ using only a compass.
First, we provide (part of) an algorithm for circular inversion. Suppose we are given point $O$ and a circle centered at $O$ of some radius $r$. If $P$ is a point outside the circle, we wish to construct point $Q$ on ray $O P$, satisfying $O P \cdot O Q=r^{2}$. Let the circle centered at $P$, with radius $O P$, intersect the given circle centered at $O$ at points $X$ and $Y$. Then let the circle with center $X$,
radius $r$ (which is constructible since $r=O X$ ), and the circle with center $Y$, radius $r$, intersect at $O$ and $Q$. We claim this point $Q$ is what we want. Indeed, it is clear from symmetry that $O P$ is the perpendicular bisector of $X Y$. Since $X Q=r=Y Q, Q$ lies on this bisector as well - that is, on line $O P$. Moreover, let $H$ denote the intersection of $X Y$ with $O P$; then $P H<P X=P O \Rightarrow H$ lies on ray $O P$, and, since $Q$ is the reflection of $O$ across $X Y, Q$ will lie on the opposite side of $H$ from $O$. This shows that $Q$ lies on ray $O P$, not just on line $O P$. Finally, observe that $P X=P O$ and $X O=X Q \Rightarrow \angle O X P=\angle P O X=\angle Q O X=\angle X Q O$, so, by equal angles, $\triangle P O X \sim \triangle X Q O$. Thus, $O P / O X=Q X / Q O \Rightarrow O P \cdot O Q=O X \cdot Q X=r^{2}$, as needed.

Now that this is done, we return to the original problem. By scaling, assume $A B=1$. By drawing circles centered at $A$ and $B$ of radius 1 , and letting $C$ be one of their intersection points, we obtain an equilateral $\triangle A B C$. Similarly, we successively construct equilateral triangles $B C D, B D E$ (with $D \neq A, E \neq C)$. We have $\angle A B E=\angle A B C+\angle C B D+\angle D B E=3(\pi / 3)=\pi$, so $A, B, E$ are collinear, and $A E=A B+B E=2$. Then, since we have drawn the circle with center $A$ and radius 1 , we can invert $E$ across it according to the paragraph above, obtaining $F$ such that $A F=1 / 2$. In fact, $F$ lies on the given segment $A B$, so the segment $A F$ is fully drawn.

Similarly, to construct a segment of length $1 / 3$, let $A B$ be given; extend $A B$ to $E$ as above so that $B E=1$, and then repeat the process, extending $B E$ to $G$ so that $E G=1$. Then $A G=3$, and inverting $G$ across our circle (center $A$, radius 1 ) will give what we need.