In high school geometry, one of the most useful ruler and compass constructions is bisecting a line segment. This beautiful construction can be done with only two circles and one line.
Euclid's very first proposition contains the essence of this construction, though he does not explicitly bisect a segment until the tenth proposition.
But what if we wanted to trisect the line segment? Given a line with segment \(AB\), construct a point \(F\) on the segment so that \(AF = (1/3) AB\), using the classical straightedge and compass.
Trisecting with Two Circles and Four Lines
Scott Coble found a clever construction, reprinted in the wonderful book Proofs without Words . (Here is a proof why it works.)
This construction uses two circles and four additional lines. Certainly two is the fewest number of circles that is possible, since one needs two circles to construct a point not on the given line through \(AB\).
Can we do better than two circles and four lines?
Trisecting with Two Circles and Three Lines
This elegant construction takes two circles and only three additional lines. (Here is a proof why it works.)
Trisecting with Three Circles and Two Lines
What if one allows a third circle?
Hartshorne in his wonderful textbook Companion to Euclid  says the average good geometer can trisect a segment with a "par" of 5 (average of five steps). Here is a version of the classical five step construction.
We have used three circles and two additional lines, which is Hartshorne's "par 5." (Here is a proof.)
Trisecting with Four Circles
If we wanted to use circles and no additional lines, how many would it take?
Only four circles, one under "par"!
Here is a proof. This construction can be generalized to construct a segment of length \(1/n\) by replacing the circle of radius 3 by one of radius \(n\).
Four circles is in fact the best possible, since three circles or three circles and one line cannot suffice (details). Nor are two circles and two additional lines enough to trisect the segment (details).