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Do the cubes inscribed in a tetrahedron intercept at points collinear with the vertex of the tetrahedron?

People can answer the question by using a $3D$ Geometry Tool, but unfortunately, I do not have enough knowledge to construct the necessary shapes to verify the hypothesis.

Could people help me prove or disprove the hypothesis?

enter image description here

For more context about the question, please visit Sides of inscribed squares of a triangle meet at points collinear with a vertex of the triangle.

Larry
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    When you mentioned this as part of your other question, I first envisioned "inscribed cube" somewhat differently: a cube with a face contained in one face-plane of the tetrahedron, an edge (parallel to that first face) contained in another face-plane (in one of three ways), and then the remaining two vertices in the other two face-planes. This seemed more of a direct analogy to your inscribed squares. But now there's also your configuration above. There appear to be many possible generalizations to explore. :) – Blue Sep 12 '18 at 22:20
  • @Blue: Do you know where to find the source code for inscribed cubes in a tetrahedron? I am trying to visualize the problem by using Wolfram Mathematica. – Larry Sep 15 '18 at 13:01
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    I don't have anything that could help you. I suggest asking the Mathematica Stack Exchange]. – Blue Sep 15 '18 at 16:14
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    About the inscribed cubes in your figure ... The green cube has edges in the "front" and "bottom" planes of the tetrahedron; necessarily, these edges have to be parallel to the edge where "front" and "bottom" meet. Likewise, the edges in the other two planes must be parallel to opposite edge of the tetrahedron. Since the pairs of cube edges are orthogonal to each other, the tetrahedron edges must be, too. This must be true for each pair of opposite edges. So, that construction is limited to orthocentric tetrahedra. – Blue Sep 17 '18 at 09:23
  • Do you mean that the cubes can only be inscribed in such a way as appeared in the figure in the orthocentric tetrahedra particularly, but not in other tatrehedra? – Larry Sep 17 '18 at 10:18
  • Yes, that's what I mean. This is not the worst restriction in the world, however. Orthocentric tetrahedra are a fairly nice family. (I've tended to call them "perfect" in my own research.) – Blue Sep 17 '18 at 10:25
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    Maybe I should try to construct some non-orthocentric tetrahedra and see how inscribed cubes behave there. – Larry Sep 17 '18 at 10:30
  • Inscribing a cube in the way I described (cube-face in one tetrahedron-face, parallel cube-edge in another tet-face, a vertex in each remaining tet-face) seems to work for any tetrahedron, provided that "tetrahedron face" means "tetrahedron face plane", so that the resulting cube might not actually live in the tetrahedron's interior. (This is the same type of thing I described with the inscribed squares in your other question.) The collinearity property looks doubtful. I've done some coordinate-based investigation, which resulted in some very ugly expressions that I need to double-check. – Blue Sep 17 '18 at 10:36
  • I am doubtful about the collinearity, too. Someone provided me the code for the orthocentric tetrahedron in Mathematica, and I played around with it, but the property of collinearity didn’t seem promising. – Larry Sep 17 '18 at 12:12

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