Dime Time

Everyone knows the shapes for one and two dimensions, far fewer know the shapes for three and four dimensions, let alone five, six and seven. And what shapes are those? The shapes that answer this question:

• How many equidistant points are possible in 1d, 2d, 3d, 4d…?

In one dimension it’s obvious that the answer is 2. In other words, you can get only two equidistant points, (a,b), on a straight line. Point a must be as far from point b as point b is from point a. You can’t add a third point, c, such that (a,b,c) are equidistant. Not on a straight line in 1d. But suppose you bend the line into a circle, so that you’re working in two dimensions. It’s easy to place three equidistant points, (a,b,c), on a circle.

Three equidistant points around a circle forming the vertices of an equilateral triangle

And it’s also easy to see that the three points will form the vertices of an equilateral triangle. Now try adding a fourth point, d. If you place it in the center of the triangle, it will be equidistant from (a,b,c), but it will be nearer to (a,b,c) than they are to each other. So you can have only 2 equidistant points in 1d and 3 equidistant points in 2d.

But what are the co-ordinates of the equidistant points in 1d and 2d? Suppose (a,b) in 1d are given the co-ordinates (0) and (1), so that a is 1 unit distant from b. When you move to 2d and add point c, the co-ordinates for (a,b) become (0,0) and (1,0). They’re still 1 unit distant from each other. But what are the co-ordinates for c? Start by placing c exactly midway between a and b, so that it has the co-ordinates (0.5,0) and is 0.5 units distant from both a and b. Now, if you move c in the first dimension, it will become nearer either to a or b: (0.49,0) or (0.51,0) or (0.48,0) or (0.52,0)…

But if you move c in the second dimension, it will always be equidistant from a and b, because (a,b) stay in the first dimension, as it were, and c moves equally away from both into the second dimension. So where in 2d will c be 1 unit distant from both a and b just as a and b are 1 unit distant from each other in 1d? You can see the answer here:

Height of an equilateral triangle

The co-ordinates for c are (0.5,√3/2) or (0.5,0.8660254…), because the second co-ordinate satisfies the Pythagorean equation 1^2 = 0.5^2 + (√3/2)^2 = 0.25 + 0.75. That is, to find the second co-ordinate of c for 2d, you find the answer to √(1 – 0.5^2) = √(1-0.25) = √0.75 = √(3/4) = √3/2 = 0.8660254….

But you can’t add a fourth point, d, in 2d such that (a,b,c,d) are equidistant. So let’s move to 3d for the points (a,b,c,d). Begin with point d in the center of the triangle formed by (a,b,c), where it will have the co-ordinates (0.5,√3/6,0) = (0.5,0.28867513…,0) and will be equidistant from (a,b,c). But d will be nearer to (a,b,c) than they are to each other. However, if you move d in the third dimension, it will be moving equally away from (a,b,c). So where in 3d will d be 1 unit from (a,b,c)? By analogy with 2d, the third co-ordinate for d will satisfy the generalized Pythagorean equation √(1 – 0.5^2 – (√3/6)^2). And √6/3 = √(1 – 0.5^2 – (√3/6)^2) = 0.81649658… So point d will have the co-ordinates (0.5,√3/6,√6/3) = (0.5, 0.288675135…, 0.816496581…).

And the four points (a,b,c,d) will be the vertices of a three-dimensional shape called the tetrahedron:

Rotating tetrahedron

But you can’t add a fifth point, e, in 3d such that (a,b,c,d,e) are equidistant. So let’s move to 4d, the fourth dimension, for the points (a,b,c,d,e). Begin with point e in the center of the tetrahedron formed by (a,b,c,d), where it will have the co-ordinates (0.5,√3/6,√6/12,0) = (0.5,0.28867513…, 0.2041241…, 0) and will be equidistant from (a,b,c,d). But e will be nearer to (a,b,c,d) than they are to each other. However, if you move e in the fourth dimension, it will be moving equally away from (a,b,c,e). So where in 4d will e be 1 unit from (a,b,c,d)? By analogy with 2d and 3d, the co-ordinate for 4d will satisfy the equation √(1 – 0.5^2 – (√3/6)^2 – (√6/12)^2). And √10/4 = √(1 – 0.5^2 – (√3/6)^2 – (√6/12)^2) = 0.79056941… So point e will have the co-ordinates (0.5,√3/6,√6/3,√10/4) = (0.5, 0.288675135…, 0.816496581…, 0.79056941…).

And the five points (a,b,c,d,e) will be the vertices of a four-dimensional shape called variously the hyperpyramid, the 5-cell, the pentachoron, the 4-simplex, the pentatope, the pentahedroid and the tetrahedral pyramid. It’s impossible for 3d creatures like human beings (at present) to visualize the hyperpyramid, but we can see its 3d shadow, as it were. And here is the 3d shadow of a rotating hyperpyramid:

Rotating hyperpyramid

N.B. Wikipedia reveals the mathematically beautiful fact that the “simplest set of coordinates [for a hyperpyramid] is: (2,0,0,0), (0,2,0,0), (0,0,2,0), (0,0,0,2), (φ,φ,φ,φ), with edge length 2√2, where φ is the golden ratio.”

So that’s the hyperpyramid, with 5 points in 4d. But you can’t add a sixth point, f, in 4d such that (a,b,c,d,e,f) are equidistant. You have to move to 5d. And it should be clear by now that in any dimension nd, the maximum possible number of equidistant points, p, in that dimension will be p = n+1. And here are the co-ordinates for p in dimensions 1 to 10 (the co-ordinates are given in full for 1d to 4d, then for 5d to 10d only the co-ordinates of the additional point are given):

d1: (0), (1)
d2: (0,0), (1,0), (0.5,0.866025404)
d3: (0,0,0), (1,0,0), (0.5,0.866025404,0), (0.5,0.288675135,0.816496581)
d4: 0.5, 0.288675135, 0.204124145, 0.790569415
d5: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.774596669
d6: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.763762616
d7: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.109108945, 0.755928946
d8: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.109108945, 0.0944911183, 0.75
d9: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.109108945, 0.0944911183, 0.0833333333, 0.745355992
d10: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.109108945, 0.0944911183, 0.0833333333, 0.0745355992, 0.741619849

In each dimension d, the final co-ordinate, cd+1, of the additional point satisfies the generalized Pythagorean equation cd+1 = √(1 – c1^2 – c2^2 – … cd^2).

Readers’ advisory: I am not a mathematician and the discussion above cannot be trusted to be free of errors, whether major or minor.