Christmas Streak 15/88: This Is A Bit Repulsive

Geometry Level 3

Tom has a set of particles that repel each other with a force of $F(r)$ whose strength is proportional to $\frac{1}{r^2},$ where $r$ is the distance of separation between any pair of particles.

If Tom puts 8 of these particles into a rigid, frictionless sphere and allows them to come to rest, they'll align to the vertices of a certain $n$ -faced polyhedron.

What is $n?$

For example , if Tom puts 4 of the particles into the sphere, they'll come to rest at the vertices of a regular tetrahedron, as shown.

4 5 6 7 8 9 10 11

2 solutions

Boi (보이)
Oct 13, 2017

This problem is well-known as the Thomson problem .

If there are multiple repulsive balls in a spherical container, the balls will try to be far from each other as much as possible, with minimal deviation between the distances.

The outcome is as below, contrary to the popular assumption that they would form a cube:

The shape above is called a "square antiprism."

This is because the deviation between the distances is minimal if the two squares on the top and the bottom are not aligned; notice that the sides of a square antiprism are equilateral triangles where the sides of a cube are square. (compare the side length and the diagonal length)

A square antiprism has $\boxed{10}$ sides.

This (and the other solution) don't really offer any justification as to why the square antiprism is the answer. Sure it's better than a cube, but where is the proof that there is not a better option still? As it stands this seems like a test of having seen it before, rather than a problem that readers could have solved having not seen it before.

Matt McNabb - 3 years, 7 months ago

Well

If there's eight particles, then intuitively one can assume that they need to split into four-four in both hemispheres.

And from that the majority of people can think of a cube.

But this question asks if people can think a bit farther to twist the cube to make it a better solution.

This is not about computationally exacting the particles' location from calculations, it's about intuition and a bit of careful thinking .

Boi (보이) - 3 years, 7 months ago

This only addresses the comparison of the cube and square antiprism. It doesn’t address other possibilities, such as a trigonal dodecahedron or a bicapped trigonal antiprism with the caps over the top and bottom faces. Intuitively, I think these would be better. The part of the sphere above the two four-sided faces of the square antiprism is in a sense wasted space that could be used to move the other vertices farther apart. Both the DD and the BTAP have only triangular faces. I’m not good enough at solid geometry to prove it, however.

Eric Lucas - 3 years, 7 months ago

For a trigonal antiprism with the caps over the top and bottom faces, their lengths vary too much

And for a trigonal dodecahedron, the whole volume is irregular -- that is, there's no symmetry in the shape, and thus the particles will collapse right after they form it (if they ever get to do).

Boi (보이) - 3 years, 7 months ago

Thanks, I got it wrong so appreciate your solution.

Nelson Bibby - 3 years, 7 months ago

Andy Hayes
Oct 19, 2017

As per @H.M. 유 's solution, the shape is a square antiprism with 10 faces. Below is a cube and a square antiprism inscribed in the same sphere.

As you can see, the vertices of the square antiprism are ever-so-slightly further apart than the vertices of the cube.

That is surreal!

James Wilson - 3 years, 7 months ago

Christmas Streak 15/88: This Is A Bit Repulsive

2 solutions

0 pending reports