During my idle doodling, I draw polygons and I start with a triangle and I keep adding a side until I can’t draw them anymore.
Then I draw all the diagonals inside the polygon. (Hey, it’s doodling, it’s not supposed to be productive.)
Is there a formula that tells you how many diagonals inside a polygon (assuming that there are no obtuse angles)?
Yes.
Finishing my question … is there a formula that tells you how many diagonals can be drawn inside of one polygon (starting with equilateral ones)?
Yes.
Yeah, each diagonal is formed by two corners of the polygon. So, if the polygon has n sides (n corners) the number of diagonals = the number of different ways to choose two of the n corners minus the n SIDES, which is [n(n-1)/2] - n.
Thank you. But only one of you.
I figure I had a 50/50 shot at a correct (if incomplete) answer. 
In otherwards it approaches a number but never really reaches it. Calculus. Is this correct?
I’m only your wildest fear, from the corners of your darkest thoughts.
Now let’s calculate PI using triangles to fill in a circle.
No, it doesn’t have anything to do with calculus; it doesn’t approach anything, it’s just the number of diagonals. For example, a hexagon would have (6*5/2)-6 = 9 diagonals.
“In otherwards it approaches a number but never really reaches it.”
Gee, that sounds like my social life.
Well thanks Cabbage. I didn’t want to spend anytime thinking on that one and you saved me the trouble.
tbea925
Would that be 01000101.
Can you believe that using that for a user name got me booted from a virtual reality world in just a few seconds? They must have put alot of extra coding into the guardian bot.
I’m only your wildest fear, from the corners of your darkest thoughts.
Do you get the same answer with the formula:
n(n-3)/2
where “n” is the number of sides or corners. (n-3) would equal the number of diagonals posible from a given corner. ( n - itself - 2 adjacent corners)
Does that work out for all polygons?
Yeah, that would work, too.