The four unit vectors v1…v4 and the closing vector v5 = −(v1+v2+v3+v4) sum to zero, so by Minkowski's theorem there is a convex polyhedron whose five faces have these vectors as outward normals, with face i of area |vi|.
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Parameters (multiples of π)
Parameter e is computed from a, b, c and d so that |v5| = 1, and its slider is locked. The shaded band on the d slider marks the values of d for which some e works.
Vectors
| Vector | x | y | z | Area |
|---|
Polyhedron
v1 = (0, 0, −1), v2 = (−sin a, 0, cos a), v3 = (sin b cos c, sin b sin c, cos b), v4 = (sin d cos e, sin d sin e, cos d). The sliders give each angle as a multiple of π, with a, b, d ∈ [0, π] and c, e ∈ [0, 2π].
The polyhedron is found by maximizing volume over support numbers hi with Σ |vi| hi fixed; at the maximum each face area is proportional to |vi|. It is unique up to translation and is drawn with its centre of mass at the origin; each face normal is drawn from the face's centre of area. A generic configuration gives a triangular prism; a square pyramid appears where four side planes meet in one point, and parallel normals merge into one face. When the normals become nearly coplanar the polyhedron grows very long and is hidden.