4 ms·

Some of my favorite issues are the ones with ultra hard IQ tests, the LAIT, Mega Test, and Titan Test, which appeared in 1979, 1985, and 1990, always in April.

Some of my favorite issues are the ones with ultra hard IQ tests, the LAIT, Mega Test, and Titan Test, which appeared in 1979, 1985, and 1990, always in April. Some info on those tests is available at miyaguchi.4sigma.org (or simply 4sigma.org). Sample question from the Mega Test: What is the maximum number of completely bounded volumes that can be formed by three interpenetrating cubes, considering only the surfaces of the cubes as bounds and counting only volumes that are not further subdivided?

criddell 9y ago
42?
Retric 9y ago
I don't think that qualifies as an IQ test. But it's fun to visualize. 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. Adding a third offset cube to that geometry would give 3 protrusion per face (c2, c3, c2 & c3) * 6 = 18 Inside would have cube C1, C1 & C2, C1 & C3, C1 & C2 & C3 = 4 with the most obvious orientation. So minimum bound of 22. However, you can push that push that higher by keeping 2 of the exterior points aligned on C2/C3 and rotating C2 and C3. So, now 4 faces are still at 3, but 2 faces have 7. Making 26 just on the outside + C1 + C1 & C2 & C3 + 4 * C1 & C2, 4 * C1 & C3 = 10 inside. For a total of 36 assuming I am counting the insides correctly. I don't see anything obvious to improve this.
- 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.
deleted 9y ago
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criddell 9y ago

42?

Retric 9y ago

I don't think that qualifies as an IQ test. But it's fun to visualize. 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. Adding a third offset cube to that geometry would give 3 protrusion per face (c2, c3, c2 & c3) * 6 = 18 Inside would have cube C1, C1 & C2, C1 & C3, C1 & C2 & C3 = 4 with the most obvious orientation. So minimum bound of 22. However, you can push that push that higher by keeping 2 of the exterior points aligned on C2/C3 and rotating C2 and C3. So, now 4 faces are still at 3, but 2 faces have 7. Making 26 just on the outside + C1 + C1 & C2 & C3 + 4 * C1 & C2, 4 * C1 & C3 = 10 inside. For a total of 36 assuming I am counting the insides correctly. I don't see anything obvious to improve this.

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.

deleted 9y ago

[deleted]