Handshakes and Polygon DiagonalsDate: 09/12/2001 at 20:42:42 From: Brendan Subject: If a polygon has # sides, how many diagonals will it have? Dear Dr. Math, I can't figure this one out! If a polygon has 42 sides, how many diagonals does it have? Date: 09/13/2001 at 05:40:04 From: Doctor Pete Subject: Re: If a polygon has # sides, how many diagonals will it have? Hi, Let's look at a hexagon, with 6 sides. Pick a vertex. How many diagonals come from that vertex? (Three.) How many vertices in a hexagon? (Six.) So there should be 3*6 = 18 diagonals, but in fact if you count them there are only half as many. Why? Well, since a single diagonal joins exactly two vertices, for each vertex we counted the same diagonal twice. So look at a 42-gon. How many diagonals come from a single vertex? How many vertices are there in a 42-gon? How many diagonals are in a 42-gon? Let's look at something similar to this. If you have 100 people at a party, and everyone shakes hands with everyone else, how many handshakes take place? Well, pick a person. He shakes hands with 99 others. In fact, each person shakes hands with 99 others, so there should be 100*99 = 9900 handshakes; but again, it takes two people to shake hands, so this is double the correct amount, which is 4950 handshakes. By now the pattern should be clear. If there are n people at a party, the number of handshakes is n(n-1)/2. Now, there is a different way to count these handshakes. The first person shakes hands with 99 others. The second person shakes hands with 98 others, not including the first person, whom we have already counted. The third person shakes hands with 97 others, not including the first two. And so on, until the 99th person, who shakes hands with the 100th person. So there are 99 + 98 + 97 + ... + 2 + 1 handshakes. But we have seen that this is equal to 100*99/2, and this suggests a formula. In general, if there are n people at a party, then counting the number of handshakes in two different ways gives the formula 1 + 2 + ... + (n-2) + (n-1) = n(n-1)/2. Finally, we can prove this summation in yet another way. Write S = 1 + 2 + ... + (n-2) + (n-1) S = (n-1) + (n-2) + ... + 2 + 1 ---------------------------------------- 2S = n + n + ... + n + n, and since there are n-1 "n"'s, then we have 2S = n(n-1), or S = n(n-1)/2. However, the handshake analogy isn't exact. At a party, each person gets to shake hands with everyone but himself. But in a polygon, each vertex doesn't get to form a diagonal with himself... or with the two vertices nearest him. (He's already connected to them by edges.) So for a polygon, we have to subtract the number of edges from the total for the handshake problem: S = n(n-1)/2 - n = [n(n-1) - 2n]/2 = [n(n - 1 - 2)]/2 = n(n-3)/2 - Doctor Pete, The Math Forum http://mathforum.org/dr.math/ |
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