geometric-construction
I'm trying to draw a line through point P, parallel to given line l, with only a blank ruler (a straightedge of a certain length). I know this is easy with a compass, but I don't know how to do it without anything else. Is this even possible?
Thanks!
If you have a segment $AB$, place the unit length segment on the line where $AB$ lies, starting with $A$ and in the direction opposite to $B$; let $C$ be the other point of the segment. Now draw a semicircle with diameter $BC$ and the perpendicular to $A$; this line crosses the semicircle in a point $D$. Now $AD$ is the square root of $AB$.
$\triangle BCD$ is a right triangle, like $\triangle ACD$ and $\triangle ABD$; all of these are similar, so you find out that $AC/AD = AD/AB$. But $AC=1$, so $AD = \sqrt{AB}$.
See the drawing below:
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
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.
Best Answer
Whether or not its possible depends on what you have available to you. In general, with only a straightedge, it is not possible to make a line through an arbitrary point that is parallel to an arbitrary line. There are special conditions that do make it possible though...
Under the Poncelet-Steiner theorem,
Variants on this theme also exist that restrict (or generalize?) the above even further. The center of the circle may be substituted for some other sufficient information. For example, instead of the circle center given to you, you can have:
From any of these scenarios the center of any or all of the circles can be constructed and the problem reduces to the aforementioned Poncelet-Steiner straightedge-only construction.
Whats more, any of the above can be modified further by eliminating a portion of the circle itself. As it turns out, any full circle is equivalent to any portion of the circle.
Lets eliminate the circle entirely now.
If the line you wish to make a parallel of has three points on it, A,M,B, where M is the midpoint between A and B, then you can create a parallel of it.
If you have two parallel lines already, you can create a third parallel to them through any arbitrary point.
If you have an arbitrary parallelogram anywhere on the plane, you can also create a parallel to any arbitrary line through any arbitrary point.
There may indeed be other tricks and conditions, but these are the ones I am aware of. They are all pretty fun constructions.
The above are all restricted Euclidean constructions, obviously. I emphasize that fact because you mentioned "rulers" rather than just sticking to the traditional straightedges.
If you are broadening the scope to physical objects and tools... Rulers tend to provide two parallels and two perpendiculars right off the bat, plus the capacity to measure length. All of this is immensely powerful and I wont even bother getting into the various options you have.
I am embedding animated GIF files below to demonstrate the constructions of parallels...
If youre given three points on a line, one of which is the midpoint of the other two:
If, however, your line just happens to pass through the center of a circle, the translation into three points is a trivial property of the circle. The parallel is finished by the previous construction:
If your line does not go through the center of a circle then you must construct your three points. This is done by choosing an arbitrary line through the circle center and a parallel is constructed from that. Ultimately the previous two constructions are both used.
But if instead of a circle you are given two parallels and wish to construct a third:
Or if instead of a circle you are given a parallelogram (square in this case). Use the square to construct a second parallel then use your two parallels in the previous construction to get your desired third.