Abstract. The diagonals of  hypercubes may sometimes
intersect at right-angles, but only in even-dimensional
spaces. They are never at right-angles in odd
dimensional spaces.

INTRODUCTION

You can not measure the angle between two diagonals in space because, at best, we can only have a distorted projection onto a piece of paper and the higher the dimension, the greater the distortion. Although the word diagonals has a general meaning, as used here, , it was found necessary to introduce new terminology when working with hypercubes to specify that diagonals will be restricted to squares and to use “triagonals” for cubes; “quadragonals” for tesseracts, and, in general, “n-agonals” when dealing with the higher dimensional magic hypercubes.

Sin(φ/2) = (1/2)/(√3/2)                …(1)

  φ = 2.arcsin(1/√3)                 …(2)

 φ = 2arcsin( 0.577350269189625764509148780501957)

φ = 70.5287793655093086307540006600376 degrees

 or, approximately 70 degrees. This is for adjacent triagonals. By adjacent triagonals, I mean those which pass through an adjacent corners of a face. By opposite triagonals, I mean those passing through opposite corners of a face. In higher spaces, there may be many in between.

We have:

φ = 2.arcsin(√2/√3)                …(3)

 φ =109.471220634490691369245999339962 degrees    

 or,   φ =  70.528779365509308630754000660038 degrees

 which is the same as for adjacent triagonals, (only reversed).
 Can you prove it analytically?

THE TESSERACT & HYPERCUBES

Starting at a corner of a unit hypercube, and proceeding in each perpendicular direction a unit distance, you will end up in the opposite corner. An n-agonal is a straight-line joining opposite corners of a hypercube. The length of a diagonal turns out to be √2; the length of a triagonal turns out to be √3; and, by continuing this very same process into higher dimensional spaces, the length of an n-agonal is √n. There are 2ⁿ corners. There are 2^n-1 n-agonals which can be paired. Fewer than n moves are required on a unit cube to go to any other corner to reach a different diagonal. 

In Figure 4, one sees the isosceles triangle defined by O, B₁ and B₂. Since OB_1­ is in 4-space and half the length of a quadragonal, its length will be 1 and the triangle becomes equilateral, so that adjacent quadragonals intersect at 60 degrees. (or, 120 degrees)

For non-adjacent quadragonals such as the ones passing through A₁ and A₃, which are two corners apart, the base of the quadragonal triangle would be of length √2 and the isosceles sides would be 1. Hence,

                             φ = 2arcsin[(√2/2)/(1)] = 90 degrees                    …(4)

Finally, for quadragonals of a tesseract passing through corners which are 3 removed one obtains:

                             φ = 2arcsin[(√3/2)/1] = 120 degrees                 …(5)

IN GENERAL

  The angle of intersection of two n-agonals of a hypercube in n-Dimensional space is given as a function of the dimension “n” and of proximity “p", of the two corners of the bounding hyperplane through which they pass. One designates the next, or adjacent corner, 2 means “two away.” The proximity “p” is determined by how many (minimum) coordinates (x₁, x₂, …, x_n) have to be altered to obtain the coordinates of the other corner, or its opposite corner. The equation simplifies to:

φ = 2.arcsin(√p/√n)        p<n          …(6)

Equation (6)  holds the key. If p is a half of n, then the two n-agonals are at right angles. This can only happen in an even dimensional space. Also, it is seen that a 60 degree angle occurs with p=n/4.

  Table of Angles of Intersection

Last updated Thursday March 22, 2007

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