The Vertices of a Cube

I recently got a couple of tiny wooden cubes – and by a couple, I mean about 250. They’re wonderful for visualizing all sorts of interesting problems about cubes and discrete 3-space. So, with a bunch of cubes in hand, I set about imagining some problems. I came up with three problems, all about the vertices of cubes, and here they are.

Problem 1

In my initial “problem searching” wandering, I drew the numbers 1 through 6 on the cube, with pairs adding up to seven on opposite sides like a standard die. Then I noticed that each of the eight vertices had three adjacent numbers, and I started to think about the eight different sets of numbers represented by the vertices. What if I could choose a number for each face, but each of those eight sets had to have the same sum? What would my choices be limited to?

This problem proved to be quite simple. Two adjacent vertices, as shown in the diagram, share two elements of their three element set. 

The vertex on the left has the sum A + B + D, and the vertex on the right has the sum A + B + C. If these sums had to be the same, then A + B + D = A + B + C, so D = C. Now, because of symmetry, we know that all opposite faces must have the same value. We didn’t assume anything; we just started with two arbitrary vertices, and you can verify that if we pick two vertices in a different adjacent direction, it leads to proving that the two faces oriented in that direction are equal. And it turns out, that’s it! Every vertex is a combination of a x-oriented, y-oriented, and z-oriented face, and if faces with the same orientation have the same value, then every vertex is the sum of those three numbers! The only restriction on this cube is that opposite faces must have the same value!

Problem 2

Then, of course, I asked myself the reverse question. What if each of the vertex sums had to have a different value? Moreover, what is the optimal cube that satisfies this property? My intuition told me that there were probably many cubes that fit the description, so I needed to find a “best” one. Maybe one whose sums were all consecutive numbers, or one whose total sum was the lowest possible. So I decided to limit myself to positive integers.

I started by lazily finding an upper bound. I like powers of two, so I started with the good old 1, 2, 4, 8, 16, 32. The sum of any three of them can’t equal the sum of any other three, because if you write the sum in binary, you get a number with three ones and three zeroes, with the ones like flags pointing towards which numbers were in the sum. Not all of these sums would be present, because of the geometry of cubes, but it would easily pass the condition as long as none of the sets of three had the same faces, which of course they didn’t. So the lazy upper bound for total sum was 63. Didn’t have a great way to quantify an upper bound for how close the sums were, so I ignored it for now.

Then, I made myself a little adjacency map of a cube’s faces and vertices, so it would be easier than turning a cube around in my hands and losing track of what was where. This went through a few iterations before I settled on this one – it’s simple, somewhat symmetric, and keeps most of the properties of the three-dimensional model. The areas are the faces, and the other areas they touch are are the faces they’re adjacent to. If only the earth was a cube – maps would be way more fun.

Then, I outlined what the optimal cube would look like. The smallest possible sum of three positive integers was three, and there were eight vertices, so the best possible sums would be 3, 4, 5, 6, 7, 8, 9, and 10. Well, I thought, let’s jump right in and see if that’s possible.

I started with the three ones, necessary to have a sum of three.

I would also need a sum of 4, so I added a 2 on another face. I can pick any face without losing generality, because the cube is still symmetric and all remaining blank faces have identical properties.

Now, I need a 5. You can only get a 3 with 1+1+1, and you can only get a 4 with 1+1+2, but you can get a 5 with either 1+2+2 or 1+1+3. Anyone familiar with Kakuro should know the score by now. Luckily, adding a 2 to either blank space ends in another 1+1+2 situation, which breaks our only rule. We can’t have two different vertices adding up to 4, after all. So if this mythical perfect cube exists, it must have a 3 in one of these two remaining faces. And the map of the cube itself shows you that the two remaining faces have symmetry, so we can pick whichever we like to put the 3 on.

As you can see, that 3 neatly gets us the 6, too. The next number we need is 7, and we have two ones bordering the last face, so if we’re to get a 7, it must be 5. If the perfect cube exists, this face must be a 5, and we have to hope that the rest of the vertices work out to be different somehow.

And amazingly, they work out perfectly. The perfect cube with all unique vertex sums does exist, and it’s not only definitely the cube with the lowest sum, but all of its sums are adjacent! I thought I was going to find a contradiction and have to trudge through a slew of messy cases to find the optimal cube, but there was no need.

Problem 3

The final question of the three came from a completely different angle, almost literally. One of the cubes had an imperfection, a little chip off of a vertex. As I studied it, I thought about the triangle formed when slicing off a vertex of a cube. As I imagined all the ways you could cut a vertex, it occurred to me that there was something strange. None of these triangles were obtuse, and only the most extreme edge cases were right triangles! But there wasn’t much of a limitation – I tried to imagine a good way to phrase the problem without the context of a cube, and this is what I came up with. If you have a triangle with one point on the x-axis, one point on the y-axis, and the final point on the z-axis, can this triangle be obtuse?

I haven’t proven this one yet. My intuition tells me it’s not possible, without offering any theoretical basis for that claim. I can fix two of the points and see what the angles look like when the third point goes to infinity, and it always seems to approach a right triangle or a line, but that’s not exactly a rigorous proof.