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.
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