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        riddles >> medium >> Volume of n-Simplex
        
(Message started by: THUDandBLUNDER on Nov 30^th, 2006, 5:03pm)

Title: Volume of n-Simplex
Post by THUDandBLUNDER on Nov 30^th, 2006, 5:03pm

An n-simplex is an n-dimensional equivalent of the regular tetrahedron. That is, it has n+1 vertices, all of which are equidistant from each other. Thus a 1-simplex is a line segment, a 2-simplex is an equilateral triangle (area = sqrt(3)/4), and a 3-simplex is a regular tetrahedron (volume = sqrt(2)/12), etc.

For which values of n is the hypervolume of an n-simplex with unit side length rational?

(Jock would like this one but he seems to have been absent for a while.)

Title: Re: Volume of n-Simplex
Post by Barukh on Dec 1^st, 2006, 8:51am

[hide]n = 1[/hide]?

;)

Title: Re: Volume of n-Simplex
Post by THUDandBLUNDER on Dec 2^nd, 2006, 6:01am

In fact, there are more.

Title: Re: Volume of n-Simplex
Post by Eigenray on Dec 3^rd, 2006, 12:52am

Let the vertices of an (n-1)-simplex S_n-1 in R^n-1 be x₁,...,x_n, and suppose x₁+...+x_n=0. By symmetry, say |x_i|=r for all i, and <x_i,x_j> = s whenever i != j. Then from

0 = |x₁+...+x_n|² = [sum]_i,j <x_i,x_j> = n*r² + n(n-1)*s

we get s = -r²/(n-1), and then

1 = |x_i - x_j|² = 2r² - 2s = 2r²(1+1/(n-1))

gives r² = (n-1)/(2n). Let d = sqrt(1-r²) = sqrt((n+1)/(2n)). Then the origin, together with the points {(x_i, d)} in Rⁿ, are the vertices of an n-simplex S_n with unit side lengths.

For t in [0, d], the intersection of S_n with the hyperplane { (x,t) : x in R^n-1 } is a copy of S_n-1, scaled by a factor of (t/d), so has (n-1)-volume (t/d)^n-1V_n-1, where V_n-1 is the volume of the original (n-1)-simplex. Hence the n-volume of S_n is given by

V_n = [int]₀^d (t/d)^n-1 V_n-1 dt
= V_n-1*d/n = V_n-1*sqrt[(n+1)/(2n)]/n,

and since V₁ = 1, we find inductively

V_n = sqrt[(n+1)/2ⁿ]/n!.

This is rational iff (n+1)2ⁿ is a perfect square. If n is even, then we need n+1 to be a square, and if n is odd, then we need n+1 to be twice a square. Hence n is of the form (2k+1)²-1 or 2k²-1. That is,

n=1, 7, 8, 17, 24, 31, 48, 49, 71, 80, 97, ....

Title: Re: Volume of n-Simplex
Post by Barukh on Dec 3^rd, 2006, 4:09am

Yes. Here is a slightly different approach.

The simplest way to put an n-simplex into coordinate system, is to put it into an (n+1)-space, like in this thread (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_putnam;action=display;num=1163875917). Then, the i-th vertex will have all coordinates 0 except the i-th, which will equal 1. The center of n-simplex is then at (1/(n+1), …, 1/(n+1)), the side length is 2^1/2, and the circum-radius is R_n = [n/(n+1)]^1/2. Therefore, the circum-radius of the unit-side simplex is R_n = [n/2(n+1)]^1/2

Next, the in-radius r_n satisfies the identity r_n² = R_n² - R_n-1², and therefore equals [1/2n(n+1)]^1/2.

Finally, we have the hyper-volume V_n = r_nV_n-1(n+1)/n = V_n-1[(n+1)/2]^1/2/n, arriving at Eigenray’s formula.

on 12/03/06 at 00:52:34, Eigenray wrote:

n=1, 7, 8, 17, 24, 31, 48, 49, 71, 80, 97, ....

BTW, this sequence is not in Sloane’s database. Do you think it’s worth to file an addition form?