"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.
I have considered the case of the equilibrium density ≈ 0.9264336.
If the cube floats horizontally, it is stable, but not if it floats in the symmetrically vertex-high orientation.
If the cube floats in the vertex-high orientation, a small rotation shifts the center of mass further down.
In other words, the symmetrically vertex-high orientation is neither a global optimum nor a local optimum; a slight tilt will result in a local optimum for the cube, with the center of mass 0.4273997 away from the surface.
However, the global optimum is taken in a different orientation.
If one edge of the cube is horizontal and higher than the water surface, the distance of the cube's center of mass from the water surface is 0.4358756, which is the deepest.
Edit:
0.4273997 is not a local optimum.
This is the time when one of the edges of the cube appears completely above the water surface.
I'm not sure the problem is even sound. The person you are giving the word to doesn't know the selection of words you were given to choose between. What if the options had been "round, red, square"?
According to later posts in the linked twitter thread, the person you are giving the word to DOES know the selection of words you were given to choose from. But this is not stated in the original post.
"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.
I have considered the case of the equilibrium density ≈ 0.9264336.
If the cube floats horizontally, it is stable, but not if it floats in the symmetrically vertex-high orientation.
If the cube floats in the vertex-high orientation, a small rotation shifts the center of mass further down.
In other words, the symmetrically vertex-high orientation is neither a global optimum nor a local optimum; a slight tilt will result in a local optimum for the cube, with the center of mass 0.4273997 away from the surface.
However, the global optimum is taken in a different orientation.
If one edge of the cube is horizontal and higher than the water surface, the distance of the cube's center of mass from the water surface is 0.4358756, which is the deepest.
Edit:
0.4273997 is not a local optimum.
This is the time when one of the edges of the cube appears completely above the water surface.
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.
I'm not sure the problem is even sound. The person you are giving the word to doesn't know the selection of words you were given to choose between. What if the options had been "round, red, square"?
According to later posts in the linked twitter thread, the person you are giving the word to DOES know the selection of words you were given to choose from. But this is not stated in the original post.