"Since the internal diagonal of a unit cube has length √(3), that meant y was equal to √(3)−(1/75)1/6, or about 1.245. Finally, this “optimal” ratio x/y was roughly 0.3911. Right?"
"Wrong!"
This suggests solving for the density of the iceberg at which the wrong answer would become the right answer. Therefore,
(1-ρ) = x³√3/2 ≈ 0.866025x³ ===> x = [2(1-ρ)/√3]^⅓ and y = √3 - x for the vertex-high case so the center of mass is a distance (y-x)/2 = (√3 - 2x)/2 = ½√3 - [2(1-ρ)/√3]^⅓ below the surface.
For the floating-horizontal case, x = (1-ρ) and y = ρ, so (y-x)/2 = (2ρ-1)/2 = ρ - ½
So, to solve for the critical value of ρ, I set these two CM depths equal to obtain
½√3 - [2(1-ρ)/√3]^⅓ = ρ - ½
The solution of interest is ρ = 0.9264336, at which cubes in either orientation would have their centers of mass a distance of 0.4264336 below the surface and therefore be equally likely to float in either orientation. Not a huge difference from 0.90.
Another question of interest is whether a cube with the equilibrium density calculated above would be bistable in one or the other of the two orientations considered, or would it prefer yet another orientation.
"Since the internal diagonal of a unit cube has length √(3), that meant y was equal to √(3)−(1/75)1/6, or about 1.245. Finally, this “optimal” ratio x/y was roughly 0.3911. Right?"
"Wrong!"
This suggests solving for the density of the iceberg at which the wrong answer would become the right answer. Therefore,
(1-ρ) = x³√3/2 ≈ 0.866025x³ ===> x = [2(1-ρ)/√3]^⅓ and y = √3 - x for the vertex-high case so the center of mass is a distance (y-x)/2 = (√3 - 2x)/2 = ½√3 - [2(1-ρ)/√3]^⅓ below the surface.
For the floating-horizontal case, x = (1-ρ) and y = ρ, so (y-x)/2 = (2ρ-1)/2 = ρ - ½
So, to solve for the critical value of ρ, I set these two CM depths equal to obtain
½√3 - [2(1-ρ)/√3]^⅓ = ρ - ½
The solution of interest is ρ = 0.9264336, at which cubes in either orientation would have their centers of mass a distance of 0.4264336 below the surface and therefore be equally likely to float in either orientation. Not a huge difference from 0.90.
https://i.servimg.com/u/f89/19/74/46/40/densit10.jpg
Another question of interest is whether a cube with the equilibrium density calculated above would be bistable in one or the other of the two orientations considered, or would it prefer yet another orientation.
Is the answer to the main Fiddler as obvious as I think it is?
Man I wish I'd submitted my pyramid figure made in Geogebra.
I guess up to isomorphism Round = Red.