Timeline for partition of a convex set into squares

11 events

when toggle format	what		by	license	comment
Jun 25, 2015 at 14:57	vote	accept	student
Jun 24, 2015 at 9:29	comment	added	Wolfgang		Your constant of 6 in the rectangle case isn't sharp, is it? I don't think there are rectangles which need $C>2$.
Jun 24, 2015 at 8:54	answer	added	Fedor Petrov		timeline score: 1
Jun 24, 2015 at 4:42	comment	added	Fedor Petrov		@student how do you do it for, say, right isosceles triangle? I try to remove a square, then I get two similar triangles with the same total perimeter. Such process leads to divergent sum of perimeters.
Jun 22, 2015 at 22:23	history	edited	student	CC BY-SA 3.0	added 1186 characters in body
Jun 22, 2015 at 21:56	comment	added	student		@Fedor Petrov. Now I proved the case for the triangle, by first dividing it into countable union of rectangles, and the convergence of geometric progression.
Jun 22, 2015 at 19:53	comment	added	Dylan Thurston		Can you say some more about where it comes up? It does seem intriguing.
Jun 22, 2015 at 19:09	comment	added	student		@Fedor Petrov, yes, I consider the open polygon. I'm frustrated because I could not either prove for triangles in $\mathbb{R}^2$ or for rectangles in $\mathbb{R}^3$. The thing is, I've no idea how to disprove it.
Jun 22, 2015 at 18:57	comment	added	Fedor Petrov		I am afraid that triangle can not be partitioned into squares. Or do you consider open polygon?
Jun 22, 2015 at 18:54	history	edited	student		edited tags
Jun 22, 2015 at 18:48	history	asked	student	CC BY-SA 3.0