n! --------- - n 2! (n-2)! n (n-1) = ------- - n 2 = n^2-n - 2n ---------- 2 = n^2 - 3n -------- 2 = n(n-3) ------ 2So there are 13(13-3)/2 = 65 diagonals for the 13-gon. This formula n(n-3)/2 I'd keep it in my treasure chest where I put other not-so-obvious formulas like volumne/surface area of a sphere, volume of a pyramid, etc. Waita minute, although my initial observation is not going anywhere but somewhat of a good start. For each point on the polygon, do not count the 2 neighbors including the point itself, so n-3 diagonals for each point. Go around each point of the polygon, so it is n(n-3), but by that time you would have counted twice the number of diagonals, so divide by 2. There you are: n(n-3)/2.
Saturday, January 27, 2024
How many diagonals does a polygon have?
I came across another problem I never learned how to solve... how many diagonals does a 13-gon have?
Geez, I nerver learned the formula of how many diagonals in a polygon, let's see if I can derive on the fly.. Start with the triangle. nope, no diagonals. The square: 2. The pentagon: 5. The hexagon: 9. It is kinda hard to draw and count bigger ones.
So I have a f(3)=0, f(4)=2, f(5)=5, f(6)=9.. and yikes I can't easily obsere a pattern and derive this formula on the fly.
But I kinda observe, something to do with n-2... because can't connect a point to its neighbors and call it diagonal... and sadly I have to give up and look the formula up. Yes I admit.
And it takes I think a bit genius to reach this observation:
The nunber of diagonals is simply number of segments connecting every pair of points, EXCEPT the edges of the polygon.
So it is the Combination n choose 2, minus n:
nC2 - n
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