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> 2 cubes you can hit 8 by having the inter cube rotated so it's producing an edge from each of the 6 faces each just C2, then 2 volumes one from C1 and 1 from

by Steko 9y ago

> 2 cubes you can hit 8 by having the inter cube rotated so it's producing an edge from each of the 6 faces each just C2, then 2 volumes one from C1 and 1 from C1 & C2. Seems like you should be able to get at least 11 with 2 cubes. 8 triangular prisms and 3 "interior" sections corresponding to what you get with 2 spheres. I'm not entirely clear on what "not further subdivided" limits but ... for 3 cubes each face of the first cube can hold 6 volumes of the other cubes (2 overlapping triangular pyramids), then 8 sections of the vertices of the base cube and the interior which is at least 1. That's 45.

deleted 9y ago
[deleted]
Retric 9y ago
I suspect you can go higher than what I was thinking, but not necessarily with the same sized cubes and picturing 3 cubes without pencil an paper is very mind bending. Anyway, cubes have 8 corners, but I am having trouble visualizing them coming out of the six faces like your describing while sectioning the inner parts into 3 sections. Can you describe the relative size and orientation your thinking of? PS: It's easy to be thinking of an Octahedron not a cube inside like so: https://commons.wikimedia.org/wiki/File:Dual_Cube-Octahedron.svg https://commons.wikimedia.org/wiki/File:Dual_Cube-Octahedron...
- Steko 9y ago
  > cubes have 8 corners, but I am having trouble visualizing them coming out of the six faces like your describing Yeah that was some bad extrapolating. Uncle Google gave me a helpful picture [1] but I'll try and work through the math the right way: Starting with two cubes: two of the faces of the base cube will have prisms and four of the faces will have a tetrahedron. Adding a third cube you can either have the prism faces on the same side or not. The former gives two faces with 5 volumes (prisms intersecting) and 4 faces with 7 (tetrahedrons intersecting). The latter gives 4 faces with 5 volumes (tetrahedron intersecting prism) and 2 faces with 7 volumes (tetrahedrons intersecting). So go with prisms on the same side. This tells us there's 38 volumes completely outside C1 - 32 truly exterior pieces that are only part of C2 or C3 and 6 interior where C2 and C3 overlap on each face. Repeating this process for C2 and C3 will count all the truly exterior pieces twice 32 * 3 / 2 = 48 exterior pieces. There are 6 volumes for each combination of 2 cubes so 6 * C(3,2) = 18 intersections of two cubes. And there's one dodecahedron in the center where all 3 cubes overlap. 48 + 18 + 1 = 67 which appears to be the correct answer according to quora. [1] https://en.wikipedia.org/wiki/Compound_of_three_cubes https://en.wikipedia.org/wiki/Compound_of_three_cubes
  - Retric 9y ago
    Cool, I realized you could get 4 not just 2 faces to 7 with summitry. And you could increase the interior volumes if the three bases of the triangular pyramid extending to the edge of the cube. But, I had yet to work out you could get five volumes on the final two sides that way.

by Steko 9y ago

deleted 9y ago

[deleted]

Retric 9y ago

I suspect you can go higher than what I was thinking, but not necessarily with the same sized cubes and picturing 3 cubes without pencil an paper is very mind bending. Anyway, cubes have 8 corners, but I am having trouble visualizing them coming out of the six faces like your describing while sectioning the inner parts into 3 sections. Can you describe the relative size and orientation your thinking of? PS: It's easy to be thinking of an Octahedron not a cube inside like so: https://commons.wikimedia.org/wiki/File:Dual_Cube-Octahedron.svg https://commons.wikimedia.org/wiki/File:Dual_Cube-Octahedron...

Steko 9y ago
> cubes have 8 corners, but I am having trouble visualizing them coming out of the six faces like your describing Yeah that was some bad extrapolating. Uncle Google gave me a helpful picture [1] but I'll try and work through the math the right way: Starting with two cubes: two of the faces of the base cube will have prisms and four of the faces will have a tetrahedron. Adding a third cube you can either have the prism faces on the same side or not. The former gives two faces with 5 volumes (prisms intersecting) and 4 faces with 7 (tetrahedrons intersecting). The latter gives 4 faces with 5 volumes (tetrahedron intersecting prism) and 2 faces with 7 volumes (tetrahedrons intersecting). So go with prisms on the same side. This tells us there's 38 volumes completely outside C1 - 32 truly exterior pieces that are only part of C2 or C3 and 6 interior where C2 and C3 overlap on each face. Repeating this process for C2 and C3 will count all the truly exterior pieces twice 32 * 3 / 2 = 48 exterior pieces. There are 6 volumes for each combination of 2 cubes so 6 * C(3,2) = 18 intersections of two cubes. And there's one dodecahedron in the center where all 3 cubes overlap. 48 + 18 + 1 = 67 which appears to be the correct answer according to quora. [1] https://en.wikipedia.org/wiki/Compound_of_three_cubes https://en.wikipedia.org/wiki/Compound_of_three_cubes
- Retric 9y ago
  Cool, I realized you could get 4 not just 2 faces to 7 with summitry. And you could increase the interior volumes if the three bases of the triangular pyramid extending to the edge of the cube. But, I had yet to work out you could get five volumes on the final two sides that way.