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 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".
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:
|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)
|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|
|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.
|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:
h measures the heights of the pyramids.
|h > 0,
θ = 0,
d = 0
|Skew trapezohedron||A two-dimensional continuum of solutions||Irregular quadrilaterals||D12
|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,
|Skew trapezohedron||A two-dimensional continuum of solutions, the same as above but of opposite chirality||Irregular quadrilaterals||D12
|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,
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.
Copyright N.S.Wedd 2012