I noticed that David Akers' Chapter 8 paper discusses an interesting collection of phenomena regarding slices perpendicular to the longest diagonal. He and I have been discussing this aspects of this in planning applets for a discussion of polytope slicing for the polytope group project.

David, barring what I figure are some typos in his coordinates (it should be {(1,0,0), (0,1,0), and (0,0,1)} that define the largest triangle before the onset of truncation), has the right idea about the plane that contains that triangle being 1/3 of the way along the longest diagonal of the cube. That is not true for the hypercube, however. I want to show that the last tetrahedron before truncation is 1/4 of the way along the longest diagonal of the hypercube (that is, 1/2-unit along the 2-unit longest diagonal of the hypercube).

If we define the unit cube as based at the origin and consisting of the other points (x,y,z) where those three coordinates represent all combinations of zero and one, then the three vertices closest to (and not including) the origin are (1,0,0), (0,1,0), and (0,0,1). These three points define the largest triangle in a plane perpendicular to the longest diagonal of the cube. We can use coordinate geometry to show that David is right about that plane being 1/3 of the way along the longest diagonal:

The plane can be defined by a fairly simple formula:

The plane defined by the three corners of the unit cube closest to but not including the origin: a. x + y + z = 1

We can test this quickly by checking all three of those points--they (trivially) fulfill the conditions of this formula for a plane. Now we need to find that plane's intersection with the diagonal of the cube. First we define the diagonal:

The diagonal of the unit cube: b. x = y = z

This should be no surprise--the longest diagonal of the cube is the line along which the values along all three axes are the same.

But now we can solve for the intersection of these two formulae:

Given formula b, substitute x for y and z in formula a. x + y + z = 1 x + x + x = 1 3x = 1 c. x = 1/3 Since a is true, then also: d. x = y = z = 1/3

So we know that the plane including those three points intersects the diagonal at a point one-third of the way along each of the three axes, which should convince most of you that it is one-third along the diagonal, but just in case you aren't convinced:

Let I = the point (1/3,1/3,1/3). The distance from I to the origin equals: SQR(1/9 + 1/9 + 1/9) SQR[1/3] (SQR[3])/3 The diagonal's length is SQR[3].

Therefore the distance from I to the origin is the diagonal's length divided by three.

Now we can step up a dimension and perform the same mathematics to find a different answer.

First we must define the hyperplane that contains the four vertices of the unit hypercube closest to but not including the origin.

The vertices: (1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1) The hyperplane including all four: a': x + y + u + v = 1

We also need a formula for the line that makes the diagonal, and we can look above:

b': x = y = u = v

By solving using a' and b', we can see that:

x + y + u + v = 1 x + x + x + x = 1 4x = 1 c': x = 1/4 Since a' is true, then also: d': x = y = u = v = 1/4

Since this point is 1/4 in the direction parallel to each axis, it is 1/4 along the diagonal, but we can do another algebraic proof, like above, for the skeptics:

Let I' = the point (1/4,1/4,1/4,1/4). The distance from I' to the origin equals: SQR[1/16 + 1/16 + 1/16 + 1/16] SQR[1/4] = 1/2 The diagonal's length is 2.

Therefore the distance from I' to the origin is the diagonal's length divided by four (not three, which is the three-space equivalent).

Having done three and four dimensions, we can extend this to *n* dimensions. First we need a formula for an (n-1)-plane that includes the *n* vertices of the n-cube closest to but not including the origin:

a'': x + y +... (nth coordinate) = 1

We also need a formula for the diagonal of the n-cube:

b'': x = y = ...If we solve these two, we find that:

x + y + ... (nth coordinate) = 1 x + x + ... (n x's) = 1 (n)(x) = 1 c'': x = 1/n

and:

d'': x = y = ... = 1/n

And the proof for skeptics that this is 1/n along the diagonal: We let I'' be the point defined by the intersection of the longest diagonal of the n-cube with the plane containing the n vertices of the n-cube closest to but not including the origin.

I'' = (1/n,1/n,...) [n terms in the tuple]

The distance from I'' to the origin is:

|I''| = SQR[1/(n^2) + 1/(n^2) + ...]

There are n identical terms inside the square root, so:

|I''| = SQR[n/(n^2)] = SQR[1/n] = 1/SQR[n] = SQR[n]/n

The length of the long diagonal D of an n-cube is:

|D| = SQR[n]

So the ratio of the distance from I'' to the origin over the length of D is:

|I''| / |D| (SQR[n]/n) / SQR[n] 1/n

In other words, the segment from the origin to I'' is 1/n times the length of the longest diagonal of the n-cube.

I just noticed that |I''| as defined above goes to zero as n goes to infinity, while |D| as defined above goes to infinity as n goes to infinity, which means that though the diagonal of the n-cube gets larger as we add to n, the distance to the plane intercepting these closest points gets slowly shorter and shorter!

*Jeremy Kahn* x6753