0

I'm talking about the real projective plane. Let's say we have a triangle, and we do a projective transformation, taking the vertices of the triangle to $(1,0,0), (0,1,0)$ and $(0,0,1)$, does this transformation take centroid of our original triangle to the centroid of our new triangle?

Also, how does this transformation affect the incircle? Does it turn into a general conic? Or does it still pass through $(1,i,0)$ and $(1,-i,0)$?

  • 1
    Projective transformations preserve nothing of interest about a triangle. If you have a triangle $ABC$ and two points $P$ and $Q$ in general position, then you can find a projective transformation of the projective plane that fixes $A$, $B$ and $C$ and sends $P$ to $Q$. (This follows by an easy "degrees of freedom" calculation, or more explicitly by encoding projective transformations as $3 \times 3$-matrices in homogeneous coordinates.) – darij grinberg Jul 26 '19 at 13:44
  • @darijgrinberg yes we can send 4 points to another 4 points. What if we let the 4th point we send be the centroid of our triangle to centroid of our new triangle? Actually I have seen through examples in my book how powerful projective transformations can be to prove that some points are collinear, and was hoping I can use it in problems that involve lines passing through certain centers. –  Jul 26 '19 at 15:40
  • 2
    Yes, projective transformations are powerful -- not because they preserve remarkable points, but precisely because they can take them (almost) anywhere. – darij grinberg Jul 26 '19 at 15:41
  • 1
    Better yet: If $A, B, C, D, E, A', B', C', D', E'$ are $10$ points in general position, then you can find a projective transformation sending $A, B, C, D, E$ to $A', B', C', D', E'$, respectively. Thus you can take any two points to any two sufficiently general points (e.g., to the circumcenter and the orthocenter of your triangle). – darij grinberg Jul 26 '19 at 15:45
  • @darijgrinberg one last question, is inversion a type of of projective transformation? If yes, is their a matrix representation for inversion? –  Jul 26 '19 at 16:57
  • No, inversion is not projective. (You can tell it by the fact that it isn't defined everywhere on the projective plane. To make it defined everywhere, you have to extend the plane by a point at infinity, not by a line at infinity.) – darij grinberg Jul 26 '19 at 18:23
  • As a general rule, if you want to prove things about centers of triangles you have to restrict yourself to affine transformations. – amd Jul 26 '19 at 22:25

1 Answers1

1

If you assign vectors $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ to corners of a triangle, there are essentially two ways of picturing this.

Either you are changing coordinate system but leaving the triangle as is. In this case you likely intend to switch to barycentric or trilinear coordinates, which are two specific forms of homogeneous coordinates so a lot of projective geometry machinery will work for them. Triangle centers often are expressed in trilinear or barycentric coordinates, so this is a natural coordinate system here.

See also a past answer of mine discussing the connection between barycentric or trilinear coordinates and projective transformations. Following that you can see that if you assign the three unit vectors to the three corners during a change of coordinate system, that does not fully specify the coordinate system you are using. One convenient way to make the specification unique is by prescribing the coordinates of the line at infinity, too. Or you can pick arbitrary coordinates (not on one of the triangle edges) for any single triangle center, since a change of projective basis is algebraically equivalent to a projective transformation and as such is defined by the images of four points.

Or you apply a projective transformation to your triangle while keeping the projective plane in its conventional $z=1$ embedding. In that case, two of your triangle corners are now at infinity, which likely will break the very definition of a huge number of triangle centers. Concepts like incircle have no clear definition in this situation, and even a limit process will likely converge on the line at infinity as an infinite circle, which has no well-defined center.

You can of course apply a projective transformation to the original triangle and all its associated constructs like triangle centers, incircle and so on. This will turn the incircle into a parabola (a conic section that touches the line at infinity which is one of your triangle edges) and map the original incenter to a point finite which has no special relationship to the new triangle. In general the image of the incenter under a transformation that is not a similarity will not be the incenter of the image of the triangle, and the same is true for other triangle centers.

MvG
  • 42,596