Straightedge-only constructions

Question

I know Poncelet-Steiner tells us that given a circle and its center, straightedge alone is equivalent to straightedge and compass. My question is, what can we construct with purely straightedge? We certainly can't construct any square roots in a finite number of steps. Given a segment of unit length, is it possible to construct any rational number?

Thanks in advance. I wanted to know because I wanted to show that you can construct any square root with straightedge alone in an infinite number of steps.

EDIT: What would you need to construct every rational? Would some manner of constructing parallel lines suffice? Would a segment of length 2 in addition to the unit segment suffice?

Answer 1

It is known that, given just a circle, using straightedge alone it is not possible to construct the center of the circle. Using straightedge with compass, it is easy, draw the perpendicular bisectors of two chords. These are radii and meet in the center. If they are identical then take a third chord.

Answer 2

I'll describe the idea I have for a straightedge-only construction. We are working in the projective plane for simplicity. You are allowed to

1. Connect two points with a line.
2. Find the intersection of two lines.
3. Mark an arbitrary point lying on/not lying on some already constructed lines.

A construction's result should be independent of arbitrary points. Imagine as if they're supplied by an "evil goblin" and you want your result to be independent of his malice.

Let $f$ be any collineation of the projective plane. Then if your arbitrary points in a certain were $A_1, A_2, \dots A_n$ and the resulting point/line of the construction was $B$ then the "evil goblin" could have given you the arbitrary points $f(A_1), f(A_2), \dots f(A_n)$ and these result would have been $f(B)$.

Therefore you can only construct projective invariants of the initially given points. You cannot, then, halve an interval, as that is not a projective invariant of the two endpoints. This is because collineations act 3-transitively on a line.

Having three points with given distances on a line (one may be at infinity, this is equivalent to the ability to construct parallel lines) is enough for the projective invariance to go away. We can use this construction: https://en.wikipedia.org/wiki/Cross-ratio#/media/File:Pappusharmonic.svg to get the harmonic conjugate point on the line and from there we can construct any rational distance. With duality this extends to angles, resulting in right angles, so we pretty much regain Euclidean geometry.

This is considered folklore among Hungarian students, mostly thanks to Lajos Pósa.