Post #3179044
2025-09-03 14:30 UTC
@gregeganSF@mathstodon.xyz What's not necessarily so obvious is that this linear transformation preserves probability distributions. I mean ... it does in this case, but it's not a gimme.
For example, if you said - "Pick a random line that intersects the triangle" then it no longer works. But why not? You have to get into the nature of linear transformations and what it means to pick a "random line" ...
Replies (1)
-
@gregeganSF@mathstodon.xyz 2025-09-03 14:42
@isaackuo@spacey.space Sure. In more detail, the full configuration space for this problem is the set of all pairs of points in the interior of the triangle T, which is a 4-dimensional set, T x T, which can be split into 3 subsets consisting of pairs of points such that the line through them intersects the 3 different pairs of sides of the triangle. Any linear transformation that acts on the plane to map T into an equilateral triangle E also acts on R^4 to map T x T into E x E, while preserving all the ratios between the 4-volumes of the subsets of T x T associated with the 3 different outcomes.