Saturday
November 29, 2014

Homework Help: math

Posted by tyler on Tuesday, April 3, 2007 at 11:07pm.

I need to know of a shape that has 4 sides and 3 vertices.

There is no planar figure with 4 sides and 3 vertices. 3 vertices and 3 sides define the minimal planar shape of the triangle. The addition of a 4th side would require 2 additional sides and 1 vertex to create a pyramidal shape having 6 sides and 4 vertices.

An examination of three dimensional polyhedra will also lead to no solution.

Finding regular polyhedra given the number of edges

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.

Before moving on, I wonder whether you might have observed, or discovered, something that I failed to make obvious during the derivation of our cubic polyhedron? I deliberately led you down the path of concluding that the 4 edged face was the next regular convex polyhedron possible with 12 edges. Lets take a step backward for a minute and observe that the two "v" and "f" possibilities that remained after applying the two vertex and face rules were 6-8 and 8-6. By assuming that our next regular polyhedron had faces with 4 edges, we literally ignored the outside possibility that there might be another regular convex polyhedron with 3 edged faces. If I wrote 3f = 2e = 24, we would end up with f = 8 and v = 6. Is there a convex regular polyhedron with such characteristics? Yes, of course, or I wouldn't have taken you down this path in the first place. What if you created two 4 sided pyramids, each face being the same equilateral triangle, and joined them together at the square base. We end up with the octahedron having 8 faces, 6 vertices and 12 edges.

Now that we have mastered the 6 and 12 edged polyhedra, what if the number of edges e = 30?
With v + f = 32 we have candidates ranging from 4-28 to 28-2 which reduce to 12-20, 14-18, 16-16, 18-14 and 20-12 after applying the "v" and "f" rules. Before leaping ahead again, might there be another 3 edged face polyhedron? If we write 3f = 60 we end up with f = 20 and v = 12, one of our candidate polyhedrons. While not so obvious now, in reality, these values apply to the icosahedron, the regular convex polyhedron having 20 equilateral triangle faces, 12 vertices and 30 edges.

Little thought need be given to the possibility of another 4 edged face polyhedron as there is no way that additional square faces can be included within or around the cube to create another square faced polyhedron.

This leads us to move on to the 5 edged face, the pentagonal face. If we now write 5f = 60, we derive f = 12 and v = 20. Does this represent any polyhedron? Yes, the dodecahedron having 12 pentagonal faces, 20 vertices and 30 edges.

So far, we have deliberately explored the regular convex polyhedra as they were fairly easy to define. What about the irregular or prismatic polyhedra? What if we were given an e = 10? We then have v + f = 12 giving us possible "v" and "f" values of 4-8, 6-6, and 8-4 (2-10 or 10-2 are not viable as "v" must be 3 or greater as does the number of edges per face. Applying the "v" and "f" rules, the 4-8 and 8-4 drop out of the picture and we are left with the viable 6-6 candidate. If we try for triangular faces, 3f = 20 which has no solution. We know there is no other square faced solution beyond the cube. 5f = 20 leads to f = 4 and we already know that the only solution with 4 faces is the tetrahedron.

Within this general process, it is now somewhat difficult to state with any certainty what, if any, irregular polyhedron or prism satisfies these results. In other words, what, if any, polyhedron or prism has 6 faces, 6 vertices and 10 edges? It is here that I maintain that you must apply your visual acuity and 3 dimensional perception in discerning whether one exists. As it turns out, there is one, the pentagonal pyramid. This is a pyramid with a pentagonal shaped base having 6 faces, 6 vertices and 10 edges.

What about e = 9. We now have v + f = 11 giving us possible "v" and "f" values of 4-7, 5-6, 6-5 and 7-4. After applying the "v" and "f" rules, we are left with 5-6 and 6-5 as candidates. Well, 3f = 18 leads to f = 6 and v = 5 which happens to be the irregular convex pyramid of two equilateral faced pyramids joined at their bases and having 6 faces, 5 vertices and 9 edges. While not derivable using our expressions, it turns out that a truncated prism with triangular cross section has 5 faces, 6 vertices and 9 edges.

It is clear that some 3 dimensional visual acuity together with some mathematical knowledge are necessary to derive viable polyhedra given a supposedly valid number of edges.


In my meandering through the earlier material, I stumbled upon another way of deriving valid polyhedra given the number of edges. It is a bit more cumbersome but worthy of your review in case it might more easily apply to other polyhedra, which I have not pursued as yet.

Let us define our unknown polyhedron as having "f" faces, each having "m" edges, and "v" vertices, each joining "n' edges. It is clear that mf = 2e and nv = 2e. Substituting f = 2e/m and v = 2e/n into v - e + f = 2, we obtain 1/m + 1/n = 1/2 + 1/e = (e + 2)/2e.

Relationships of the form 1/x + 1/y = 1/z can be solved in the following manner.

If a value is selected for x, we then have 1/y = 1/z - 1/x = (x - z)/zx resulting in y = zx/(x - z) the solution then
becoming (x, y) = (x, zx/(x - z). We must therefore find the positive integers that make (x - z) a positive divisor of zx, where z is a given positive integer. This requires that (x - z) be = or > 1, i.e., x = or > (z + 1).
From y = zx/(x - z) we can derive y = [(x - z)z + z^2]/(x - z) = z + z^2/(x - z).
For y to be an integer, (x - z) must be a divisor of the constant z^2, noting that (x - z) must be positive.
If a positive integer k divides n^2, then setting (x - z) = k, we get the solution (x, y) = [(z + k), (z + z^2/k)].
Thus, the number of solutions is merely the number of positive divisors k of z^2.
If the prime decomposition of z is z = p1^a1(p2^a2) - - - - pk^ak, then z^2 = p1^2a1(p2^2a2) - - - - pk^2ak.
The number of solutions is then N = (2a1 + 1)(2a2 + 1) - - - - (2ak + 1).

Example: 1/x + 1/y = 1/10.
10 = 2^1 x 5^1
100 = 2^2 x 5^2
N = (2 + 1)(2 + 1) = 9
k = 1, 2, 4, 5, 10, 20, 25, 50, 100
For k = 1 x = 11 and y = 110
1/11 + 1/110 = 11/1210 + 110/1210 = 121/1210 = 1/10.
For k = 5 x = 15 and y = 30
1/15 + 1/30 = 15/450 + 30/450 = 45/450 = 1/10.
And so it goes.

How can we apply this to our problem?
Assume e = 6.
Then, 1/m + 1/n = (6 + 2)/12 = 8/12.
Dividing through by 8 yields 1/8m + 1/8n = 1/12.
12 = 2^2(3^1)
12^2 = 144 = 2^4(3^2)
N = (4 + 1)(2 + 1) = 15
k = 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144.
For k =.........8m.........8n........m.......n
........1..........13........156
........2..........14.........84
........3..........15.........60
........4..........16.........48.........2........6 (m must be 3 or greater, therefore, not viable)
........6..........18.........36
........8..........20.........30
........9..........21.........28
.......12.........24.........24.........3........3
.......16.........28.........21
.......18.........30.........20
.......24.........36.........18
.......36.........48.........16.........6........2 (m must be 3 or greater, therefore, not viable)
.......48.........60.........15
.......72.........84.........14
.....144........156........13
.
For m and n equaling 3, we have 3f = 2(6) = 12 making f = 4 and 3v = 12 making v = 4, the characteristics of a tetrahedron with 4 triangular faces, 4 vertices and 6 edges, whence 4 - 6 + 4 = 2.

What if e = 12?
Then, 1/m + 1/n = (12 + 2)/24 = 7/12.
Dividing through by 7 yields 1/7m + 1/7n = 1/12.
12 = 2^2(3^1)
12^2 = 144 = 2^4(3^2)
N = (4 + 1)(2 + 1) = 15
k = 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144.
For k =.........7m........7n.........m.......n
........1..........13........156
........2..........14.........84
........3..........15.........60
........4..........16.........48
........6..........18.........36
........8..........20.........30
........9..........21.........28..........3.......4
.......12.........24.........24
.......16.........28.........21..........4.......3
.......18.........30.........20
.......24.........36.........18
.......36.........48.........16
.......48.........60.........15
.......72.........84.........14
.....144........156........13
.
For m = 3 and n = 4, we have 3f = 2(12) = 24 making f = 8 and 4v = 24 making v = 6, the characteristics of an octahedron with 8 triangular faces, 6 vertices and 12 edges, whence 8 - 12 + 6 = 2.
For m = 4 and n = 3, we have 4f = 24 making f = 6 and 3v = 24 making v = 8, the characteristics of a cube with 6 faces, 8 vertices and 12 edges, whence 8 - 12 + 6 = 2.

What if e = 30?
Then, 1/m + 1/n = (30 + 2)/60 = 8/15.
Dividing through by 8 yields 1/8m + 1/8n = 1/15.
15 = 3^1(5^1)
15^2 = 225 = 3^2(5^2)
N = (2 + 1)(2 + 1) = 9
k = 1, 3, 5, 9, 15, 25, 45, 75, 225.
For k =.........8m........8n.........m.......n
........1...........16.......240
........3...........18........90
........5...........20........60
........9...........24........40..........3........5
.......15..........29........29
.......25..........40........24..........5........3
.......45..........60........20
.......75..........90........18
......225........240.......16
.
For m = 3 and n = 5, we have 3f = 2(30) = 60 making f = 20 and 5v = 60 making v = 12, the characteristics of an icosahedron with 20 triangular faces, 12 vertices and 30 edges, whence 12 - 30 + 20 = 2.
For m = 5 and n = 3, we have 5f = 60 making f = 12 and 3v = 60 making v = 20, the characteristics of a dodecahedron with 12 pentagonal faces, 20 vertices and 30 edges, whence 20 - 30 + 12 = 2.

As I said, not necessarily quicker.

This is not intended to be a complete treatment of the topic but it has been interesting and educatuional. I do not know if it has answered your original question but I'm sure it has shed some light on the matter. If you find any errors or false turns in the material I would appreciate your letting me know. If I have time in the future I hope to pursue it further to determine whether specific polyhedra and/or prisms can be more easily defined given only the edges, faces or vertices.

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