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Vertices in a Prism


Date: 05/12/99 at 14:09:04
From: Jason Bergner
Subject: Number of vertices in a prism

I teach sixth grade math at Warrensburg Middle School in Warrensburg, 
Missouri.

My students in my advanced math class have been trying to figure out 
a formula for finding the number of vertices in a PRISM. We were 
discussing the formula for finding the number of vertices in a polygon 
([(p-3)p]/2, where p = the number of sides in the polygon) when the 
question naturally arose. We've been tinkering with a formula as 
follows:

([(p-3)p]x2)+(px2)), where p = no. of vertices in the base of the 
prism

Can you help?

Thanks, 
Jason


Date: 05/12/99 at 15:51:03
From: Doctor Jaffee
Subject: Re: Number of vertices in a prism

Hi Jason,

The number of vertices of any polygon is the same as the number of 
sides of the polygon. For example, if you pick any six points around 
a circle: A,B,C,D,E,F in that order, then connect them, you will have 
a six-sided polygon with the six points being the vertices. Actually, 
I think you made a typographical error when you submitted this problem 
and typed in "vertices" when you meant "diagonals."  I make mistakes 
like that all the time.

The diagonals of the 6-sided polygon in the paragraph above are
AC,AD,AE,BD,BE,BF,CE,CF,DF. There are 9 of them. According to your 
formula (p-3)p/2 there should be (6-3)6/2 = 9, so your formula is 
right on target.

Now, if you have a prism and you define a diagonal of a prism as a 
segment that connects two vertices, but does not connect two adjacent 
vertices, nor is it contained in one of the faces of the prism, then 
you can generate the following chart where the first row is the 
number of sides of the base and the second row is the number of 
diagonals for that prism.

   p   3   4   5   6   7   8         ... p

   d   0   4   10  18  28  40            (p-3)p

My reasoning is this.  From each vertex of the base polygon, a 
diagonal has to end up at a vertex of the parallel polygon, but it 
can't go to 3 of those vertices: the one directly above it, nor each 
vertex to the left or right of the vertex above the chosen vertex.  
However, there are an equal number of vertices, p-3, from each of the 
p vertices, so the total is (p-3)p.

I hope this explanation makes sense. Write back if you need any 
clarification. This sounds like a great project for 6th graders.  

- Doctor Jaffee, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Permutations and Combinations

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