A certain polyhedral crystal has 15 faces, 5 which are quadrilaterals and 10 of which are triangles. How many vertices does it have?
I decided to just draw the shape out by just drawing 5 squares and 10 triangles, but when I counted the vertices I always got different answers. I got 15, 16, 17, and 18. I am very confused. How do I do this question?
Envision a planar pentagon. Stretch the planar pentagon to make a solid with pentagonal faces. Each face of the pentagonal solid has 5 edges and 5 vertices. Connect each vertex to a point off of each of the pentagonal faces creating 5 triangles on each side of the pentagonal solid.
The result is a polyhedron having 15 faces, 5 of which are quadrilaterals and 10 of which are triangles. There are 15 edges and 12 vertices. However, this does not satisfy Euler's famous equation v + f - e = 2 (vertices + faces - edges) or 12 + 15 - 15 does not equal 2. Perhaps this is not a legal polyhedron.
Some food for thought.
A convex polyhedron is defined as a solid with flat faces and straight edges so configured as to have every edge joining two vertices and being common to two faces.
There are many convex polyhedra, only five of which are considered regular polyhedra. Regular polyhedra satisfy three criteria. All the faces are congruent. All faces have the same number of edges equal to, or greater than, three. Each vertex joins the same number of edges. The five regular polyhedra are the tetrahedron, cube, octahedron, dodecahedron and the icosahedron.
Euler's famous equation, v - e + f = 2, applies to all convex polyhedra where v equals the number of vertices, e equals the number of edges and f equals the number of faces.
Given the number of edges of a regular polyhedron, is it possible to determine the numbers of faces and vertices knowing that v - e + f = 2. It is possible given that the given number of edges does, in fact, represent a real convex polyhedron in the first place. Given just any number, there will not always be a polyhedron with that number of edges.
The smallest number of edges possible is 6 for the 4 sided tetrahedron consisting of 4 equilateral triangles joined along their edges to form a 3 sided pyramid. Not knowing the form of the polyhedra that contains the 6 edges, it is possible to derive the specific polyhedron having 6 edges. Given that v + f - e + 2 = 8, the only possible pairs of v and f are 3-5, 4-4, 5-3. If each face has m edges, then mf = 2e = 12. Similarly, if each vertex joins n edges, nv = 2e = 12. "m" must be 3 or greater leaving us with possible m's of 3, 4, or 6 and faces of 4, 3, or 2. Having already shown that "f" must be 5, 4, or 3, we are left with 4 or 3 as the only possible values for "f". Clearly 3 faces cannot form a closed convex polyhedron, leaving us with m = 3 and 4 faces. Knowing that we have 4 faces, "v" is also 4 and our polyhedron with 6 edges has 4 faces and 4 vertices, the tetrahedron.
Following this same path for other polyhedra is not as easy. What if the number of edges e = 12? We therefore know that v + f = 14 giving us possible combinations of v and f of 3-11, 4-10, 5-9, 6-8, 7-7, 8-6, 9-5, 10-4 and 11-3.
At this point, let me introduce two other useful facts about convex polyhedra. The number of vertices is at least 2 more than half the number of faces or v = f/2 + 2. Similarly, the number of faces is at least 2 more than half the number of vertices or f = v/2 + 2. Applying these two rules to our list of possible v's and f's, we end up with only 6-8 and 8-6 as viable candidates. Now we must take note of another piece of information that we can deduce very easily from what we have already explored. We already derived the only regular convex polyhedron with 3 edged (triangular) faces. The polyhedron we seek with 12 edges must therefore have faces containing at least 4 edges. Lets assume that each face has 4 edges and see where it takes us. If each face does, indeed, have 4 edges, we can the write, as we did above, 4f = 2e = 24 making f = 6 and v = 8. What have we here? By golly, the cube, a convex polyhedron having 6 congruent faces, 8 vertices and 12 edges.
