# The 2 million, er, 20 dollar problem

Alfred is a guy who really likes solid shapes. He also really likes to keep his money in his wallet.

What makes him the happiest? Getting a boxful of shapes at the Cheap Solids Store!

The store has a new promotion: Get a sphere-shaped container for $20 and fill it up with tetrahedrons from the Tetrahedron Bin™. Each tetrahedron must have a corner on the center of the sphere. As long as nothing is sticking out, you may take the whole sphere home. The Tetrahedron Bin™ in the corner is full of these funny things: Each tetrahedron in the bin has side length 1, and the sphere has radius 1. As Alfred is a very stingy guy, he wants to know in advance how many tetrahedrons he can get for his$20.

TL;DR: Given a sphere of radius 1 and tetrahedrons of side length 1, what is the maximum number of tetrahedrons you can pack into the sphere, such that each tetrahedron has a vertex on the center?

Note: Notice there is no lateral-thinking tag on this one. That means you are not allowed to deform the tetrahedrons or the sphere, or feed them into a shrinking machine, or give them to an alien. Everything is subject to the normal laws of physics.

Edit: I am aware that this is an open problem, with no proof found as of now, hence the title, "The two million dollar question," given by my math professor. I just thought the puzzlng community might want to see this and think about it over lunch.

• How about 12? Seems too easy to be unsolved, so there must be a trick. – Ian MacDonald Jun 28 '15 at 16:46
• @IanMacDonald How did you get 12? Also, there's really no trick. You just need to find a proof that the number you got is the maximum. – mmking Jun 28 '15 at 18:11
• @2012rcampion From the link in the answer below, this one is unsolved too. – mmking Jun 28 '15 at 19:12
• there must be the High bound of 35 by calculating volumes (of course it is not possible) ... – Tomer W Jun 28 '15 at 19:26
• I hear proof by intimidation is very effective for these types of problems.. – Conor O'Brien Jun 30 '15 at 0:47

This question is not about packing tetrahedra, it is about how many tetrahedra can share a common vertex.

The constraints of this question are that the side length of the tetrahedra is equal to the radius of the sphere and all tetrahedra share a common vertex at the center of the sphere. Given these constraints, all vertices are either at the center of the sphere or on the surface of the sphere. Thus, the sphere is not relevant; the question becomes "How many regular, identical tetrahedra can share a single vertex?"

The solid angle at a vertex of a regular tetrahedron is $\arccos (23/27) \approx 0.55$ steradians $\approx 4\pi / 22.8$. So we have an upper limit of 22.

A regular icosahedron has triangular sides with edges that are $\approx 1.05$ times the radius of a circumscribing sphere. Thus, we could arrange a 20 tetrahedra such that each outer face is aligned with that of an icosahedron (same icosahedron for all faces, if that wasn't clear).

Thus, we have a lower limit of 20 and an upper limit of 22. This is a much simpler problem than finding an efficient packing algorithm.

Proving that 20 is the maximum is probably possible. Given that the symmetries make packing very efficient (edges aligned with edges), I do not believe it could be possible to squeeze any more in. However, proving it formally would require 3D modeling software, which I don't have.

For reference, this image (Wikipedia) shows how little space there is left for fudging:

I can't believe it would be possible to rearrange the tetrahedra to fit in more than 20.

• From the article linked by Ivo Beckers; "To rule out 21 seems like a nonlinear optimization problem in some 63-dimensional space." The article note the problem described in OP as a simplification of tetrahedra packing, one that is still open. – Taemyr Jun 30 '15 at 13:35

According to https://matthewkahle.wordpress.com/2010/11/05/packing-tetrahedra/ this is a open problem so there is no solution yet

• I was aware of that, hence the title, "The two million dollar question," given by my math professor. I believe I had found a proof, but it is so trivial I think I must have messed up somewhere terribly. I will post it later, after anyone else who wishes to gets a chance to post their solution. – mmking Jun 28 '15 at 15:55

Another way to look at it is to project the base of each tetrahedron onto the sphere.

The length of each segment is $\pi/3$. By using the law of cosines, each angle is $arcsec(3)$ and the area of one of the spherical triangles is $3\ arcsec(3)-\pi$. The area of the sphere is $4\pi$. Therefore, the max number of spherical triangles we can theoretically fit on the surface of the sphere is $\frac{4\pi}{3\ arcsec(3)-\pi}\approx22.7947$.

Doesn't improve the upper bound at all but it takes it from packing tetrahedrons in 3D to packing triangles on the surface of a sphere.

• Did you find that picture on the internet, or did you draw it yourself? I would like to draw some similar pictures. – mmking Jun 30 '15 at 16:30
• I got it from the wolfram article on spherical triangles. – Kyle G Jun 30 '15 at 16:32
• The problem with "packaging problems" which are famous for having no solution, is that the non trivial ways of packing tend to be more efficient. Though this may get you a number for how many you can pack this way, sometimes just randomly tossing them in the sphere with no particular order can get you more. – Albert Renshaw Nov 5 '15 at 17:15
• For example, here is the most efficient packing method for a particular problem, and it's not algorithmic. These problems are famously unsolved: i.stack.imgur.com/O3eXC.png – Albert Renshaw Nov 5 '15 at 17:16
• I completely agree, this is not a simple thing to solve. It's just a different way of looking at the original problem from a different angle by embedding the tetrahedrons onto the surface of the sphere. For me at least, I find it far easier to visualize and reason about triangles on a surface than tetrahedrons in a sphere. – Kyle G Nov 5 '15 at 17:59

Once I weaved a 3d framework of tetrahedra with pen-refills and thread. It ended at a regular Icosahedron (20 triangular surface); whose each triangle contained a tetrahedron underneath. And all of them had a common vertex at the centremost point of the icosahedron. So the ans should be 20.