Your solution seems to assume that all cuts need to be directed towards a single point, but doesn't it seem likely that an even more optimal solution increases h (depth of cut target) as the cuts move outward? Or did I miss a reason that's not the case?
"So, the best depth for an onion with ten layers would be somewhere between 0 and 0.5573066. I have not investigated this in depth, but this seems like a fun next step."
You are suggesting something even more advanced. :)
If you're still checking, I have a semi-related question:
You're solving the problem for a circle in a plane (actually, a semicircle in a plane), and the reduction in dimensions is related to something that has bothered me.
I can easily segment a circle into a bunch of identical arcs (say, by making each arc 3 degrees long and getting 120 identical copies). Polar coordinates are great for this.
But spherical coordinates are terrible for accomplishing the same thing on a sphere, and my understanding is that the analogous effect - tiling the surface of a sphere with a single shape - can't be achieved?
What motivated me to thinking about this was the idea of a coordinate system that would allow every "square" on a map to be the same as the other squares, regardless of how much distortion there might be between the shape of the region on the spherical surface and the shape of the same region as a square on this fancy map. But it also seems relevant to the question of how well your two-dimensional analogue to the onion problem answers the original three-dimensional question. (I'm writing this comment in the middle of reading your article, so I don't know if the 3D solution is ultimately addressed.)
I'd be happy for any comments you might have related to this.
I have slides that detail the problem setup and the mathematics, as well as a consideration of three-dimensional onions, here: https://drspoulsen.github.io/Onion_Marp/index.html
I have submitted a formal write-up of the details of the problem and the solution to a recreational mathematics journal.
I'm also happy to answer any questions about this!