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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?

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Given straightedge alone, what are you given to start with? 2 points? If that is the case, all you can construct is a line, since that's the only thing you can build over those two points. –  Patrick Da Silva Jan 3 '12 at 20:08
With only a straightedge, you cannot make copies of your unit length segment, so it is rather useless :) –  Mariano Suárez-Alvarez Jan 3 '12 at 20:08
The only thing that a straightedge lets you do is connect two points to get a line (and by extension, find intersections of two lines). I think that, given a segment of length 1 and a straightedge, you can't even produce a segment of length 2, let alone the other rational numbers. (Did you mean a ruler, instead of a straightedge?) –  Arturo Magidin Jan 3 '12 at 20:09
(If one can construct a segment of length 2, then I think one can construct all rationals, no?) –  Mariano Suárez-Alvarez Jan 3 '12 at 20:13
@MarianoSuárez-Alvarez I think you also need to solve the problem of drawing parallel lines through given points.... –  N. S. Jan 3 '12 at 20:15
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2 Answers 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: http://en.wikipedia.org/wiki/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.

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I would think that applying any linear transform to all of the given points would cause any constructable points to be transformed likewise; that would imply that to construct a right angle one would need three basis points that either formed an isosceles right triangle, or else formed a specific known triangle from which one could be constructed. If one was told three points formed such a triangle but they didn't, one's constructions would be skewed by whatever transform would have turned such a triangle into the thing one actually had. –  supercat Feb 24 at 1:14
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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.

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