Does this line preserving, continuous bijection on an equilateral triangle exist?

Question

I'm trying to define a continuous bijection on the points (x/y coordinates) of an equilateral triangle.

The vertices and the midpoints of each edge need to remain fixed. The (6) points on the edges that are roughly 19.1% of the way from one vertex to another need to be mapped to the corresponding points 25% of the way from one vertex to another.

I think the above is possible with many different maps, however, I'd like a map that preserves linearity. In other words, line segments get mapped to line segments. This is where I'm stuck.

I don't even know where to start. Is this related to anything famous that I should have learned in topology and now could google?

For ease of notation, I'm assuming unit side lengths with vertices at (0,0) (1,0) and (${1\over2}$, $\cos 60$).

Thanks in advance for any help!

PS - I don't think it is relevant, but the exact value of the "roughly 19.1%" from above is ${1\over2\phi^2}$ or ${1\over3+\sqrt5}$ or $ {3-\sqrt5 \over4}$ or ...

Edit: Here are some links to pictures that may help:

• Here are some point of my original triangle. The red point is one of the 6 that are roughly 19.1% of the way from vertex to vertex.

• Here are those same points under a map that I created. You can see the points get "bunched up" in the middle. The red point is the image of the red point in the original triangle and is 1/4 of the way from the bottom left vertex to the bottom right one (which is good). This map fixes the vertices and midpoints (also good), but does not preserve linearity (not good) as you can see with the arbitrary line that I included.

Answer 1

This is not possible.

A continuous transformation of (a nicely behaved region of) the plane where the image of a straight line is always a straight line is a projective transformation.

A projective transformation is completely determined by its values on four points of which no three are collinear. This is sometimes known as the first fundamental theorem of projective geometry.

Since you want a map that fixes both the corners and the midpoints of the sides, it will also have to fix the centroid (where the lines from a corner to its opposite midpoints cross). This is enough to see that the projective transformation has to be the identity, and therefore it cannot move your 19%-points anywhere.