1. Can a polyhedron have for its faces (a) 3 triangles?

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To determine if a polyhedron can have 3 triangular faces, we can analyze the properties of polyhedra and the conditions required for their formation.

1. **Definition of a Polyhedron**: A polyhedron is a three-dimensional solid with flat polygonal faces, straight edges, and vertices.

2. **Face Count and Vertex Configuration**: For a polyhedron to exist, the arrangement of its faces must meet certain criteria, particularly concerning the vertices where the faces meet.

3. **Triangles as Faces**: Each triangular face has 3 edges. If we have 3 triangular faces, we can calculate the total number of edges and vertices.

- **Edges**: Each triangle contributes 3 edges, but each edge is shared between two triangles. Therefore, the total number of unique edges $$ E $$ is:
     $$
     E = \frac{3 \times 3}{2} = \frac{9}{2} = 4.5
     $$
     This is not possible since the number of edges must be a whole number.

4. **Vertices**: If we consider the vertices, each triangle has 3 vertices. However, with only 3 triangles, it is impossible to arrange them in three-dimensional space without overlapping or leaving gaps.

5. **Conclusion**: Since the calculations show that we cannot have a whole number of edges and the arrangement of 3 triangular faces does not allow for a valid polyhedron structure, we conclude that:

**A polyhedron cannot have 3 triangular faces.**
A polyhedron cannot have 3 triangular faces because it's impossible to arrange them without overlapping or leaving gaps, and the calculations show that the number of edges wouldn't be a whole number.

Can a polyhedron have for its faces
(a) 3 triangles?

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