Is the centroid the same as the average of boundary points?

Higher-dimensional geometry (previously "Polyshapes").

Is the centroid the same as the average of boundary points?

Postby quickfur » Sat Jan 23, 2021 1:18 am

Given a convex set P in Rn, is it true that the centroid of P (the average of all the points in P) is equal to the average of the points on the boundary of P? Intuitively it would appear to be so, but just wanted to make sure. :P
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Re: Is the centroid the same as the average of boundary poin

Postby mr_e_man » Mon Jan 25, 2021 3:03 am

No.

An isosceles triangle, with vertices at (±b, 0) and (0, h), has centroid (0, h/3). Its edges have lengths 2b and √(b² + h²), and centroids (0, 0) and (±b/2, h/2). So the triangle's boundary has centroid

( 2b(0,0) + √(b² + h²)(b/2,h/2) + √(b² + h²)(-b/2,h/2) ) / ( 2b + √(b² + h²) + √(b² + h²) )

= √(b² + h²)(0, h) / ( 2b + 2√(b² + h²) )

= (0, h√(b² + h²) / (2b + 2√(b² + h²)) )

≠ (0, h/3).

(If you try to set them equal, you get h² = 3b², indicating a regular triangle.)

Now let's try a trapezoid, with vertices at (±(b + a), 0) and (±b, h). It can be split into two right triangles and a rectangle; this gives the centroid

( ah/2 (b + a/3, h/3) + ah/2 (-(b + a/3), h/3) + 2bh (0, h/2) ) / ( ah/2 + ah/2 + 2bh )

= (0, ah²/3 + bh²) / (ah + 2bh)

= (0, h/3 (a + 3b)/(a + 2b) ).

But the centroid of its vertices is (0, h/2).

So we conclude that the centroids of the various k-dimensional skeletons (or frames, or whatever they're called) are generally unrelated. It really is a coincidence that, for an n-dimensional simplex, the n-frame and the 0-frame (vertices) have the same centroid.
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Re: Is the centroid the same as the average of boundary poin

Postby quickfur » Mon Jan 25, 2021 4:05 pm

Interesting. So basically, the average of vertices, the average of edges, ... the average of n-dimensional elements, are in general all different?

Edit: nevermind, that's exactly what you said. :D
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Re: Is the centroid the same as the average of boundary poin

Postby mr_e_man » Tue Mar 16, 2021 1:49 am

But we might guess that, for an n-dimensional simplex, the k-frame and the (n - k)-frame have the same centroid, for any k.

This is obviously true in 1D or 2D. So let's try a 3D simplex, with not too much symmetry: a triangular pyramid. We already know that the 0-frame and the 3-frame have the same centroid; we want to know whether the 1-frame (edges) and the 2-frame (faces) have the same centroid.

Say the three base edges have length a, and the three rising edges have length b. The height (in its own plane) of the base triangle is √3/2 a. The height (in its own plane) of a lateral face is l = √(b² - (1/2 a)²) = √(b² - 1/4 a²). The 3D height of the pyramid is h = √(l² - (√3/6 a)²) = √(b² - 1/3 a²); notice that this requires b > 1/√3 a. The area of the base is √3/4 a², and the area of a lateral face is 1/2 a l.

We can put the vertices at coordinates (√3/3 a, 0, 0), (-√3/6 a, ±1/2 a, 0), and (0, 0, h). Then the 1-frame's centroid is (0, 0, (3b)/(3a + 3b) 1/2 h), while the 2-frame's centroid is (0, 0, (3/2 a l)/(√3/4 a² + 3/2 a l) 1/3 h). And of course the 0-frame and 3-frame have centroid (0, 0, 1/4 h).

These are not equal in general. But let's equate them and see what happens:

b/(a + b) 1/2 h = l/(√3/2 a + 3 l) h

b (√3/2 a + 3l) = 2(a + b) l

√3/2 ab = (2a - b) l

The left side is positive, while the right side is positive only if b < 2a. Continuing, with x = b/a to simplify:

3/4 a²b² = (4a² - 4ab + b²) (b² - 1/4 a²)

3x² = (4 - 4x + x²) (4x² - 1)

3x² = 4x⁴ - 16x³ + 16x² - x² + 4x - 4

0 = 4x⁴ - 16x³ + 12x² + 4x - 4

Obviously x = 1 is a solution, corresponding to a regular tetrahedron, so we can factor that out:

0 = 4 (x - 1) (x³ - 3x² + 1)

This cubic polynomial has one solution in the relevant interval 1/√3 < x < 2, which happens to be exactly

x = 1 - 2 cos(4π/9) = 1/(2 cos(2π/9)) ≈ 0.6527

Very interesting! Somehow this equal-centroid pyramid is related to the regular enneagon. :o_o: :?:

triangularPyramidEqualCentroids.png
triangularPyramidEqualCentroids.png (41.58 KiB) Viewed 5118 times
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Re: Is the centroid the same as the average of boundary poin

Postby quickfur » Tue Mar 16, 2021 4:57 am

Hmm. This is indeed an extremely interesting coincidence. I wonder if there may be some kind of construction involving this equal-centroid pyramid that results in enneagonal symmetry? Perhaps some kind of tiling of space with 11-gonal symmetry, that contains this pyramid as one of the tile shapes?
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Re: Is the centroid the same as the average of boundary poin

Postby mr_e_man » Wed May 12, 2021 10:40 pm

Due to crystallographic restriction, there is no tiling of 3D space with 9-fold or 11-fold rotation symmetry.

Indeed, in any dimension, a tiling can't have simple 9-fold rotation symmetry. But a compound rotation is possible! The general condition of crystallographic restriction is that the characteristic polynomial of the transformation matrix needs to have integer coefficients.

Let's consider a triple rotation in 6D, with angles θ₁,θ₂,θ₃ in three orthogonal 2D planes. The rotation matrix has eigenvalues

z₁=exp(-iθ₁), z₂=exp(iθ₁),
z₃=exp(-iθ₂), z₄=exp(iθ₂),
z₅=exp(-iθ₃), z₆=exp(iθ₃),

and the characteristic polynomial is (t - z₁)(t - z₂)(t - z₃)(t - z₄)(t - z₅)(t - z₆). This paper by P. A. Damianou shows that, if the roots z of an integer-coefficient monic polynomial are all on the unit circle |z|=1, then they must be roots of unity; that is, each θ must be a rational fraction of a circle. Furthermore, the polynomial must be a product of cyclotomic polynomials (which are just the minimal polynomials of roots of unity).

(I don't know why Damianou mentions Newton's formulas; the coefficients of fk are not all simple sums of powers of z₁,...,z₆. What matters is that they are at least symmetric polynomials in z₁,...,z₆, and all symmetric polynomials are generated by the elementary symmetric polynomials; since the latter have integer values at (z₁,...,z₆) (being the coefficients of f itself), the former have integer values as well. Then fj=fk for some j≠k because there are finitely many such f.)

So we could have a characteristic polynomial t⁶+t³+1, for a triple rotation by 2π/9,4π/9,8π/9. But I don't know any specific example of a 6D tiling or symmetry group containing this rotation. Crystallographic restriction is a necessary condition, not a sufficient condition, for a tiling to exist with such a symmetry.
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