Three squares meet at a cube vertex, three triangles at a regular tetrahedron vertex. Are those corners equal because the face counts match? This additional lesson separates angles within faces from angles between faces and explains why there are only five convex regular solids.

Prerequisites: spatial angles, sections, spatial coordinates, cubes and prisms.

Several faces form a corner

A polyhedral angle at a convex polyhedron vertex is the region between the incident faces, including its boundary. The common vertex starts its edge rays; consecutive rays bound plane angles within faces. Three faces give a trihedral angle; its three rays are not coplanar.

A cube vertex has three plane angles of 90°, totaling 270°. A regular tetrahedron has three of 60°, totaling 180°. These sums are not one dihedral angle. A dihedral angle measures how two faces open across their shared edge, as in G026.

Each plane angle of a nondegenerate convex trihedral angle is smaller than the sum of the other two; all three sum to less than 360°. A convex vertex with more faces also needs a sum below 360°: a flat full turn does not make a spatial convex corner.

Three faces: different plane angles

3 × 90°3 × 60°

Tetrahedron, cube, octahedron: p and q

p=3q=3p=4q=3p=3q=4

Two different spatial loci

ABx=0x²+y²=1z
Original parallel projections: screen u=x−y/2, v=0.3x+0.45y−0.85z, each object with its own uniform scale. Spatial angles are not preserved: 90°/60° describe original models, not screen triangles. Tetrahedron vertices are (1,1,1),(1,−1,−1),(−1,1,−1),(−1,−1,1); cube (±1,±1,±1); octahedron the signed unit coordinate vectors. Frames are transparent; all edges are dashed equally, without visibility classification. Bottom left is a window in infinite plane x=0 between A(−2,0,0), B(2,0,0); right a fragment of the infinite radius-1 cylindrical surface. End circles bound the displayed fragment, not lids.

Why five regular solids?

Here a convex regular polyhedron has congruent regular polygon faces and the same arrangement at every vertex. Let each face have p sides and q faces meet at each vertex, p,q≥3. Each face angle is 180°(p−2)/p, hence:

q(180(p − 2))/(p) < 360, (p − 2)(q − 2) < 4.

Only these integer pairs remain:

SolidpqVertex angle sum
Regular tetrahedron33180°
Cube43270°
Regular octahedron34240°
Regular dodecahedron53324°
Regular icosahedron35300°

The inequality eliminates other possibilities: it is necessary. Separate constructions establish that these five exist; the inequality alone does not build them. This classification concerns convex regular polyhedra in three-dimensional Euclidean space, not star polyhedra or arbitrary meshes.

Symmetry preserves the entire solid

A cube is preserved by a 90° rotation around opposite face centers and a 120° rotation around a body diagonal. For vertices (±1,±1,±1), the maps (x,y,z)→(−y,x,z) and (x,y,z)→(y,z,x) send vertices to vertices and preserve distances and edges.

A matching photographic silhouette is insufficient. Spatial symmetry preserves the whole point set, including the hidden side; projection can conceal differences.

Loci in space

A locus contains all and only points satisfying a property. In space the answer may be a plane or surface:

  • Points equidistant from distinct A,B form the perpendicular bisector plane of AB. For A(−2,0,0), B(2,0,0), equality of squared distances simplifies to x=0, proving both directions.
  • Points equidistant from three noncollinear points form the line through their triangle’s circumcenter perpendicular to its plane: the intersection of two perpendicular bisector planes.
  • Points at distance a>0 from z=0 form two planes z=±a; for a=0 only one remains.
  • Points equidistant from parallel planes z=0 and z=4 form z=2, since |z|=|z−4|.
  • Points equidistant from intersecting planes x=0 and y=0 form two bisector planes x=y and x=−y, since |x|=|y|.
  • Points equidistant from three noncollinear points form the line through their triangle’s circumcenter perpendicular to its plane: the intersection of two perpendicular bisector planes.
  • Points at distance a>0 from the z-axis form the infinite cylindrical surface x²+y²=a². z is unrestricted; this is not a filled finite cylinder. At a=0 only the axis remains.

Earlier distance rules therefore identify figures that need not be polyhedra.

Branch · Beyond the courseOptional · a brief look ahead only

Outlook: a four-dimensional cube

A square uses two coordinates between 0 and 1; a cube uses three. With four, take 0≤xᵢ≤1, i=1,2,3,4. At vertices each coordinate is 0 or 1, giving 2⁴=16 vertices. This is the four-dimensional cube, or tesseract. Its three-dimensional projection is not a sixth regular three-dimensional solid; it loses geometric information, just as a plane drawing of a cube does. Full four-dimensional geometry is outside this lesson.

When comparing frames, name faces, plane and dihedral angles and the class of solids. For a locus state the property and prove both directions: a drawing gives an example, a condition defines the entire figure.

Sources and limits

Merzliak and coauthors, advanced grade10 geometry, section17, PDF176–178: polyhedral angles; Nelin, grade10 geometry, §17, PDF160: spatial loci. Euclid, Elements XIII.18, Remark and Cambridge NRICH: five regular solids and angle restrictions. John Baez, UC Riverside, August26,2019: sixteen 4D-cube vertices; our coordinate count is original. Symmetry checks and coordinate examples are our own; no physical extra dimensions are claimed.