Ponder This Challenge - February 2014 - Juxtaposing a 4D pyramids

Ponder This Challenge:

If you juxtapose a regular hexagon to an equilateral triangle (where all the edges of both polygons are of the same unit length) on a common edge - you get a polygon with five vertices; five edges; and one face.

If you juxtapose a four-dimensional pyramid whose base is a right triangular prism to a four-dimensional pyramid whose base is a tetrahedron (again: all the edges of both polychora are of the same unit length) on a common tetrahedron - how many vertices, edges, faces, and cells would the resulting polychoron have?

Update 2/12: Hint: you may use Euler's formula to check your answer.

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Solution

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  When designing this problem, I used four-dimensional space. However, Misha Lavrov gave us a much nicer answer using five-dimensional space, so we present his solution here:

  The union of the two polytopes has
  Eight vertices
  Sixteen edges
  Fourteen faces: eight triangles and six squares
  Six cells: two tetrahedra and four triangular prisms

  The resulting polytope is a polyhedral prism whose base is a tetrahedron. The prism is not right: if the two tetrahedra have vertices ABCD and A'B'C'D', then A'B'C'D' is a translation of ABCD placed so that A is at a unit distance from each of A', B', C', and D'.

  1. The convex hull of {A, B, C, D, B', C', D'} is a four-dimensional pyramid whose base is the triangular prism BCDB'C'D' and whose vertex is A. This is our first polytope.
  2. The convex hull of {A, A', B', C', D'} is a four-dimensional pyramid whose base is the tetrahedron A'B'C'D' and whose vertex is A. This is our second polytope.
  3. The two polytopes are joined along the tetrahedral cell AB'C'D'.

  We can give the points A, B, C, D, A', B', C', D' particularly nice coordinates by embedding the four dimensional polytope in five dimensional space while remaining on the hyperplane x1 + x2 + x3 + x4 + x5 = 1. Then we have:

  A = (1, 0, 0, 0, 0)
  B = (1, -1, 1, 0, 0)
  C = (1, -1, 0, 1, 0)
  D = (1, -1, 0, 0, 1)
  A' = (0, 1, 0, 0, 0)
  B' = (0, 0, 1, 0, 0)
  C' = (0, 0, 0, 1, 0)
  D' = (0, 0, 0, 0, 1)

Solvers

• Dan Dima (02/01/2014 09:27 AM EDT)
• Oleg Vlasii (02/02/2014 05:46 AM EDT)
• Igor Novak (02/03/2014 01:54 PM EDT)
• Liubing Yu (02/04/2014 03:24 AM EDT)
• Florian Fischer (02/07/2014 09:37 AM EDT)
• Kevin Bauer (02/09/2014 08:09 PM EDT)
• Christian Blatter (02/11/2014 03:54 AM EDT)
• Misha Lavrov (02/11/2014 03:52 PM EDT)
• Olivier Mercier (02/12/2014 03:47 PM EDT)
• Don Dodson (02/12/2014 03:54 PM EDT)
• Chuck Carroll (02/12/2014 04:15 PM EDT)
• Gilles-Philippe Paillé (02/12/2014 05:11 PM EDT)
• Alexandre Gilotte (02/12/2014 07:29 PM EDT)
• Antoine Comeau (02/12/2014 09:32 PM EDT)
• Andrea Andenna (02/13/2014 12:45 AM EDT)
• Todd Will (02/14/2014 11:05 PM EDT)
• Cynthia Beauchemin (02/15/2014 02:13 AM EDT)
• Motty Porat (02/15/2014 07:06 PM EDT)
• Masoud Alipour (02/15/2014 07:21 PM EDT)
• Michael Wilms (02/16/2014 06:15 AM EDT)
• Mathias Schenker (02/17/2014 05:31 PM EDT)
• Adam Daire (02/18/2014 12:09 PM EDT)
• Armin Krauss (02/19/2014 10:31 AM EDT)
• Philipp Seemann (02/19/2014 05:51 PM EDT)
• Radu-Alexandru Todor (02/24/2014 12:56 PM EDT)
• Daniel Bitin (02/26/2014 02:56 PM EDT)
• David Greer (02/27/2014 12:22 PM EDT)
• William Heller (02/28/2014 05:35 AM EDT)