Some thoughts that I had while trying a bunch of problems.
This is an excellent problem.
One thing it reminded me of is that I still don’t understand planar transformations that look like involutions from one or more perspectives. A very basic question is
Q. Classify hexagons $ABCDEF$ where it looks like an involution from any other point $P$.
We know from Dual Desargues (DDIT) that this is true for complete quadrilaterals, but what about other things?
Something that you want to abuse is probably that you can set $P$ to actually be on $AB$, and this should wreck havoc on your hopes of it being an involution (e.g. then $P$ is also forced to be on $DE$, which is clearly not true). But wait:
Q. Why doesn’t this argument break the complete quadrilateral case?
Turns out you can have “degenerate involutions”, where you pair up one ray with all other rays. It’s still valid in some sense, and you can get this from a limiting case.
So indeed, the only such hexagons are complete quadrilaterals, which is somewhat saddening to hear. It also implies that we don’t have general plane transformations that look like an involution from every perspective.
If you restrict the perspective to certain curves, the prospects become slightly better:
UQ. Characterize the involutions on lines, conics (circles), maybe even cubics?
We’ll come back to this in a while.
Side note from the above: isogonal conjugates (w.r.t a triangle) of lines are conics passing through $A,B,C$. The easiest way to do it is just to use the cross-ratio condition for conics.
I was reading up about the polynomial method of moving points (1, 2). The underlying idea is quite interesting: we can turn any geometry problem into essentially “finite” case checking, with the upper bound on the number of cases being the degree of the condition in terms of the free variable $t$.
One realization I had while reading this is that it seems like functions that may seem pretty gnarly to do in coordinate are often tame (polynomials!) over projective coordinates. We can define, say:
A polynomial transformation is a map $[x:y:z]$ to $[P(x,y,z):Q(x,y,z):R(x,y,z)]$ for homogeneous polynomials $P,Q,R$.
UQ. Efficient ways of showing things are polynomial transformations?
It seems like once we know that things are polynomial transformations, we can just study the image of lines under transformations. With a very sloppy degree argument, we can resonably guess that
Q. Conversely: if you have quadratic $P,Q,R$, must lines get sent to conics?
And of course, isogonal conjugation is really interesting because applying it twice should get us the same thing (UQ. except…?)
It’s kinda strange, but I’m imagining it’s something like $[x:y:z] \mapsto [yz:zx:xy]$-type of thing. More investigation soon?
Also a cool idea: we know that the circumcircle <-> line at infinity, so maybe this gets us $R(x,y,z) = (\text{circle})$?
Q. Suppose a quadrilateral isogonal conjugation (on a isoptic cubic) matches a triangular isogonal conjugation. Are they the same map?
They should be, since this means the vertices of the triangle also lie on the cubic, and I’m hoping that the conjugation on the cubic will be a conjugation from the perspective of any point on the cubic… actually that’s true!
UQ. More generally: what are the family of isogonal cubics that can be embedded into such maps (probably those passing through $A,B,C$ is)? How do we understand this?
UQ. Do isogonal cubics pass through the circle points? Please tell me :<
UQ. Is the loci of general conjugation for isogonal $ABCDEF$ a cubic? What about just coharmonic? What about in general?
UQ. Let’s think about a specific version of this: (a) for isogonal conjugation, are there any other points from which this looks like an involution (other than $A,B,C$, of course)? Discrete points or curve (we know it’s not the whole plane)? (b) for a transform that looks like an involution from $A,B$, must there be a $C$ from which it also looks like an involution? Where is it?