A certain polyhedron is constructed such that each vertex is the intersection of five triangles. How many vertices are there?

Was that easy? Try these.

...such that each vertex is the intersection of...
1. three squares and a triangle
2. three triangles and a square
3. four triangles and a square
4. a triangle, square, pentagon, and square in that order
5. a decagon, hexagon, and square

Notice any patterns?

The regular icosahedron is a Platonic solid that is constructed with each vertex being the intersection of five triangles.  The icosahedron has 12 vertices, 20 faces and 30 edges.

The (small) rhombicuboctahedron is an Archemedian solid with three squares and one triangle that meet at each vertex.  The rhombicuboctahedron has 24 vertices, 26 faces and 48 edges.

The square antiprism is a semiregular polyhedra with one square and three triangles at each vertex.  The square antiprism has 8 vertices, 10 faces and 16 edges.

Both enantiomorphs of the snub cuboctahedron, also known as snub cube, are Archemedian solids with one square and four triangles at each vertex.  The snub cube has 24 vertices, 38 faces and 60 edges.

The (small) rhombicosidodecahedron is an Archemedian solid with a triangle, square, pentagon, square at each vertex.  The polyhedron has 60 vertices, 62 faces and 120 edges.

The truncated icosidodecahedron, also known as the great rhombicosidodecahedron, is an Archemedian solid with a decagon, hexagon and square at each vertex.  This polyhedron has 120 vertices, 62 faces, and 180 edges.

I do not, yet, recognize any pattern that is sought....

Edited on July 17, 2006, 6:28 pm

Posted by Dej Mar on 2006-07-17 12:24:24