### Random Post

### Recent Posts

### Recent Comments

Jim Randell on Enigma 423: Four teams, more… | |

Jim Randell on Puzzle 61: A division sum | |

geoffrounce on Enigma 1457: Anglo-Italian… | |

Brian Gladman on Enigma 1088: That’s torn… | |

Hugh Casement on Enigma 1088: That’s torn… |

### Archives

### Categories

- article (11)
- enigma (1,122)
- misc (2)
- project euler (2)
- puzzle (31)
- site news (43)
- tantalizer (31)
- teaser (3)

### Site Stats

- 168,110 hits

By using the Euler Characteristic of simple polyhedra we can work out possible values for the number of vertices (V), faces (F) and edges (E) of the polyhedron. This Python program looks for polyhedra with up to 100 faces/vertices/edges. It runs in 34ms.

The code only finds one candidate tuple: V=9, F=9, E=16.

Looking at Wikipedia [ http://en.wikipedia.org/wiki/List_of_small_polyhedra_by_vertex_count ] for polyhedra with 9 vertices, only the elongated square pyramid and the octagonal pyramid have V=9, F=9, E=16. And only the elongated square pyramid has any quadrilateral faces, and they can all be made to be square giving 5 square faces out of the 9 faces, so more than half the faces are square, and the remaining 4 faces are triangular.

Solution:There are 9 faces. 4 of the faces are triangles.You can find larger (V, F, E) tuples that satisfy the conditions, by increasing the number of squares considered. The next one is V=2601, F=2025, E=4624, and on rec.puzzles Dean Hickerson showed how to construct a 2025-hedron with 2601 vertices that satisfies the conditions of the puzzle. [ https://groups.google.com/d/msg/rec.puzzles/RaoavExEw18/CwDAyNGzcUUJ ].

So, technically the puzzle is flawed, as there is not a unique solution. But it can easily be salvaged by putting a reasonable limit on the maximum number of faces allowed in the polyhedron.