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by yeroco on January 13th, 2009
This is fairly easy to determine. You know there are eight vertices. If you connect all eight vertices to all of the other seven vertices with a line, that would be all of the sides plus the diagonals, right?
Well, you might start by thinking that each of the 8 vertices can be connected to each of the other seven. So that would be 8 times 7. However, if you start with one vertex and draw seven lines, when you move to the next vertex, you've already counted one of them. So on the second vertex, there are only 6 lines to draw... on the third vertex there are only 5 lines to draw, and so on, where at the second to the last vertex, there is only a single line to draw, and the last vertex, there are none.
So it's 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0, which is 28. But that counts the sides too. So subtract off the sides (8) and you have 20 diagonals.
This is an arithmetic series, so you can simplify the rule a bit instead of doing that adding by hand. If you add all the numbers from 1 to n you get n(n+1)/2 as a result. For this, we are adding all of the numbers up to the number of vertexes (= sides) - 1. If n is the number of sides, it would be (n-1)((n-1)+1)/2 = n(n-1)/2
The number of diagonals is then n(n-1)/2 - n, which reduces down to (n^2 - 3n)/2
Let's see if that works:
(8^2 - 3*8)/2 = (64 - 24)/2 = 20
Check.
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