^{n}, 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.

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Given a convex set P in R^{n}, 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.

- quickfur
- Pentonian
**Posts:**2708**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

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.

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.

ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Trionian
**Posts:**187**Joined:**Tue Sep 18, 2018 4:10 am

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.

Edit: nevermind, that's exactly what you said.

- quickfur
- Pentonian
**Posts:**2708**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

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.

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.

ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Trionian
**Posts:**187**Joined:**Tue Sep 18, 2018 4:10 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?

- quickfur
- Pentonian
**Posts:**2708**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

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