This page attempts to answer the puzzle whose full statement can be summarised
A polyhedron is convex, facetransitive, 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 "facetransitive". A rhombic dodecahedron, for instance, clearly satisfies this condition. In the table below, I give details of other facetransitive polyhedra with twelve faces. I consider only those whose faces are planar and simplyconnected, as I believe was Phillips' intention.
These solutions fall into three classes:
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)  h = 0  
Pyritohedron  A onedimensional continuum of solutions  Pentagons with mirror symmetry, the mirror crossing the longest side  A4  0 < h < (√51)/2  
Regular dodecahedron  One solution (the one intended by the author of the puzzle)  regular pentagons  A5  h = (√51)/2  
Pyritohedron  A onedimensional 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. 
(1,1,1) 
h = ⅓  
Triakis tetrahedron  A onedimensional continuum of solutions  Obtuseangled isosceles triangles  A4  ⅓ < h < 1  
Cube  (not a solution)  (S4)  h = 1  
Triakis tetrahedron  The same onedimensional continuum of solutions as given above  Obtuseangled isosceles triangles  A4  1 < h < 3  
Tetrahedron  (not a solution)  A4  h = 3  
Bipyramid  A onedimensional continuum of solutions  Acuteangled isosceles triangles  D12  Place two hexagonal pyramids basetobase.  Using one complex and one real dimension: (0,h) h measures the heights of the pyramids. 
h > 0, θ = 0, d = 0 
Skew trapezohedron  A twodimensional continuum of solutions  Irregular quadrilaterals  D12 chiral 
Place two hexagonal pyramids basetobase, rotate one of them clockwise through less than π/6, and move them apart just far enough that their convex hull has foursided faces.  h > 0, 0 < θ < π/6 

Trapezohedron  A onedimensional continuum of solutions  Kites  D12  Place two hexagonal pyramids basetobase, rotate one of them through π/6, and move them apart as above.  h > 0, θ = π/6, 

Skew trapezohedron  A twodimensional continuum of solutions, the same as above but of opposite chirality  Irregular quadrilaterals  D12 chiral 
Place two hexagonal pyramids basetobase, 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  Acuteangled 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.
Copyright N.S.Wedd 2012