Basic Terms for Polyhedra
Number of vertices V, number of edges E, number of faces F
Euler's Polyhedron Theorem
V − E + F = 2（holds for convex polyhedra）
Example ① (Verification with a cube)
Cube: V=8、E=12、F=6
V−E+F = 8−12+6 = 2 ✓
Cube: V=8、E=12、F=6
V−E+F = 8−12+6 = 2 ✓
Regular Polyhedra (Platonic Solids)
| Name | Shape of each face | Number of faces F | Vertices V | Edges E |
|---|---|---|---|---|
| Regular tetrahedron | Equilateral triangle | 4 | 4 | 6 |
| Regular hexahedron (cube) | Square | 6 | 8 | 12 |
| Regular octahedron | Equilateral triangle | 8 | 6 | 12 |
| Regular dodecahedron | Regular pentagon | 12 | 20 | 30 |
| Regular icosahedron | Equilateral triangle | 20 | 12 | 30 |
Example ② (Application of Euler's theorem)
Find V, E, and F for a polyhedron whose faces are all equilateral triangles and at whose every vertex five faces meet.
Each face has 3 edges → 3F = 2E (each edge is shared by 2 faces) → E = 3F/2
Five faces meet at each vertex → 5V = 2E (each edge has 2 vertices) → V = 2E/5
Euler: V−E+F=2 → (2E/5)−E+(2E/3)=2 → E(2/5−1+2/3)=2 → E(6/15−15/15+10/15)=2 → E×1/15=2 → E=30。V=12、F=20
Answer:Regular icosahedron
Find V, E, and F for a polyhedron whose faces are all equilateral triangles and at whose every vertex five faces meet.
Each face has 3 edges → 3F = 2E (each edge is shared by 2 faces) → E = 3F/2
Five faces meet at each vertex → 5V = 2E (each edge has 2 vertices) → V = 2E/5
Euler: V−E+F=2 → (2E/5)−E+(2E/3)=2 → E(2/5−1+2/3)=2 → E(6/15−15/15+10/15)=2 → E×1/15=2 → E=30。V=12、F=20
Answer:Regular icosahedron
Key points
- V−E+F=2（Euler's formula for convex polyhedra）
- There are only 5 regular polyhedra（proof: from an inequality involving the sum of face angles）
- The regular tetrahedron and regular octahedron are duals（vertices and faces are interchanged）
- The regular hexahedron and regular dodecahedron are also duals
Practice Problems
- State V, E, and F for a regular octahedron and verify Euler's theorem.
- Find the number of edges of a polyhedron whose faces are all pentagons and at whose every vertex three faces meet.
- Find V and E for a polyhedron made of 6 hexagonal faces and 2 pentagonal faces (a pencil-shaped solid).
Answers and Explanations
- Regular octahedron: V=6、E=12、F=8。6−12+8=2 ✓
- 3F=2E (2 faces per edge) and 3V=2E (3 edges per vertex). Euler: (2E/3)−E+(2E/3)=2→E(4/3−1)=2→E×1/3=2→E=30（regular dodecahedron）
- Hexagonal prism type: V=12（6 vertices each on the top and bottom）, E=18（6 top + 6 bottom + 6 lateral）, F=8（1 top + 1 bottom + 6 hexagonal faces）. Verification:12−18+8=2 ✓