Jessica Waggener-
Counting Problems

Count the diagonals in a 50-gon (= a polygon with 50 sides).

I began with the idea that each point was attached to n-3 diagonals, i.e. there was a diagonal to every point except itself and its immediate neighbors. Using the octagon as an example, I counted the NEW diagonals originating from each point.

For both odd and even n-gons, I soon discovered the following pattern that helped with the summation of terms. Consider the following sequence of addends. The number beside each point represents how many NEW diagonals originate from that point.

 
Point
New diagonals in 7-gon
New diagonals in 8-gon
A
4
5
B
4
5
C
3
4
D
2
3
E
1
2
F
0
1
G
0
0
H
 
0
Sum of diagonals
= 4 + 4 + 4 + 2 = 14
= 4*3 +2
= (n-3)3 + (n-3)/2
= (n-3)(3 +1/2)
= (n-3)(3.5)
= (n-3)(n/2)
= 5 + 5 + 5 + 5 = 20
= 5 * 4
= (n-3)(n/2)
Although the generalization for the 7-gon is based on the evaluation of 3.5 as n/2, this phenomena appears in several other odd-gons. I'm not exactly sure how to justify without using specific numbers at this point.

The formula can be explained in terms of the n-gons very easily, however. The n-3 factor represents the fact that from each point there is a diagonal to every other point EXCEPT itself and its immediate neighbors, i.e. n-3. The n/2 factor represents the fact that every diagonal joins two points. So, if we were directly counting the number of diagonals radiating from each point (n-3 each), we would have to divide by two to account for the duplication.

To understand the recursive formula, consider the following sequences:
 

Point
New diagonals in 5-gon
New diagonals in 6-gon
New diagonals in 7-gon
New diagonals in 8-gon
A
2
3
4
5
B
2
3
4
5
C
1
2
3
4
D
0
1
2
3
E
0
0
1
2
F
0
0
1
G
0
0
H
0
Sum:
= 5
= 9
= 5+4
= D5 + n-2
= 14
= 9+5
= D6 + n-2
= 20
= 14+6
= D7 + n-2
Look carefully at what happens with each successive n-gon. At the top of each sequence a number is repeated. With the next n-gon, the first term is increased by 1 and a new term, n-3 is added. Together, 1 + n-3 = n-2. This is the total increase each time, i.e. Dn = Dn-1 + (n-2).

One other interesting perspective offered by the class was to view this as a version of the handshake problem. Consider that every line adjoining any two points is like a handshake. To count the diagonals, we would count all "handshakes" but subtract those that represented sides of the polygon. Since an n-gon is defined by having n sides, Dn = Hn - n.

SUMMARY OF SOLUTIONS:

Numerical solution

D50 = 1175

Recursive formula

Dn = Dn-1 + n-2

Closed formula

Dn = n(n-3)/2


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