Paperweight puzzle: Answers

The puzzle

This page attempts to answer the puzzle whose full statement can be summarised

A polyhedron is convex, face-transitive, and has 12 faces. What shape could its faces be?

The intended solution

The puzzle's author, Hubert Phillips, gives the answer "The paperweight is a perfectly symmetrical figure with twelve faces. Therefore it can only be a regular dodecahedron. Each of its faces is a regular pentagon." He mistakenly assumed that his definition "on whichever of its faces it stood, it presented exactly the same appearance" implies "perfectly symmetrical".

Further solutions

I do not accept that the regular dodecahedron is the only solution. I believe that "on whichever of its faces it stood, it presented exactly the same appearance" specifies only "face-transitive". A rhombic dodecahedron, for instance, clearly satisfies this condition. In the table below, I give details of other face-transitive polyhedra with twelve faces. I consider only those whose faces are planar and simply-connected, as I believe was Phillips' intention.

These solutions fall into three classes:

A continuum of pyritohedra from the cube (which is not a solution) to the rhombic dodecahedron, including the regular dodecahedron as a special case.
A continuum of triakis tetrahedra from the tetrahedron to a larger, dual, tetrahedron, with the cube as a special case. The two tetrahedra and the cube are not solutions. Each solution that appears in this continuum appears twice.
A two-dimensional continuum of skew trapezohedra from the hexagonal bipyramid to the hexagonal bipyramid, with the hexagonal trapezohedron as a special case. For each solution in this continuum, its mirror image is also present. (The second dimension of this continuum is the apical angle of the pyramids.)

Name	Type	Faces	Rotational Symmetry Group	Description of construction method	Coordinates of vertices
Cube	(not a solution)		(S4)	Take a cube. Create a new vertex at the midpoint of each edge. Join these new vertices in pairs by new edges, with each new edge bisecting a former face. Pull the new vertices away from the centre of the whole, all at the same rate.	All permutations of (±1,±1,±1) all permutations of (0,1+h,1-h²)	h = 0
Pyritohedron	A one-dimensional continuum of solutions	Pentagons with mirror symmetry, the mirror crossing the longest side	A4			0 < h < (√5-1)/2
Regular dodecahedron	One solution (the one intended by the author of the puzzle)	regular pentagons	A5			h = (√5-1)/2
Pyritohedron	A one-dimensional continuum of solutions	Pentagons with mirror symmetry, the mirror crossing the shortest side	A4			(√5-1)/2 < h < 1
Rhombic dodecahedron	One solution	Rhombi	A4			h = 1
Tetrahedron	(not a solution)		A4	Take a tetrahedron. Add a point in the centre of each face. Pull these points away from the centre of the whole, all at the same rate, while taking the convex hull of the original tetrahedron with the four new points.	(1,1,1) all permutations of (1,-1,-1) all permutations of (h,h,-h) (-h,-h,-h)	h = ⅓
Triakis tetrahedron	A one-dimensional continuum of solutions	Obtuse-angled isosceles triangles	A4			⅓ < h < 1
Cube	(not a solution)		(S4)			h = 1
Triakis tetrahedron	The same one-dimensional continuum of solutions as given above	Obtuse-angled isosceles triangles	A4			1 < h < 3
Tetrahedron	(not a solution)		A4			h = 3
Bipyramid	A one-dimensional continuum of solutions	Acute-angled isosceles triangles	D12	Place two hexagonal pyramids base-to-base.	Using one complex and one real dimension: (0,h) all 6 values of (1^⅙,0) all 6 values of (1^⅙(cos(θ)+i.sin(θ)),-d) (0,-h-d) h measures the heights of the pyramids. θ is the angle of twist. d = h . ( cos(θ) + sin(θ)/√3 - 1) is the separation given to the pyramid bases.	h > 0, θ = 0, d = 0
Skew trapezohedron	A two-dimensional continuum of solutions	Irregular quadrilaterals	D12 chiral	Place two hexagonal pyramids base-to-base, rotate one of them clockwise through less than π/6, and move them apart just far enough that their convex hull has four-sided faces.		h > 0, 0 < θ < π/6
Trapezohedron	A one-dimensional continuum of solutions	Kites	D12	Place two hexagonal pyramids base-to-base, rotate one of them through π/6, and move them apart as above.		h > 0, θ = π/6, d = h.(2/√3 - 1)
Skew trapezohedron	A two-dimensional continuum of solutions, the same as above but of opposite chirality	Irregular quadrilaterals	D12 chiral	Place two hexagonal pyramids base-to-base, rotate one of them clockwise through more than π/6 but less than π/3, and move them apart as above.		h > 0, π/6 < θ < π/3
Bipyramid	Identical to the bipyramid above	Acute-angled isosceles triangles	D12			h = π/3, θ=π/3, d = 0

This page answers the puzzle given at Paperweight puzzle.
It relates to What do we mean by "Regular" for Orientable Regular Maps?
Index to other pages on regular maps.