# Tag Archives: polygon

# Euclidean Construction of a Four-Part Compound Eye

## Hexacontagon Molecule

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## Tessellation in Four Colors #2

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## Tessellation in Four Colors #1

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## Mandala in Four Colors

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## A Ring of 26 Rings of 26 Triskaidecagons

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## A Polygon with “n” Sides — How Many Diagonals Does It Have?

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Please do not thing of the figure above as a decagon, but as a polygon with *n* sides — an *n*-gon, in other words. How many diagonals does it have? Well, first, there are* n* vertices for diagonals to come from, and three vertices they cannot go to — themselves, and the two immediately on either side of them, since the segments to those vertices are sides, not diagonals. That’s* n* vertices firing diagonals at *n*-3 other vertices, or (*n*)(*n-*3). However, that counts each diagonal exactly twice (once from each side), so the actual number of diagonals is half that: ** d=(n)(n-3)/2**.

Now we can look at the polygon above, and use it to check this formula, by “remembering” that it is a decagon. With* n* = 10, *d* = (*n*)(*n*-3)/2 = (10)(7)/2 = 35.

Are there really 35 diagonals in the decagon above? Well, I made those of the same length into color-groups, to make them easier to count. There are five green ones, ten yellow ones, ten red ones, and ten pink ones, which does indeed total 35.

Suppose you know a polygon has, say, 104 diagonals. Can this formula be used to find the number of sides? Yes! Substituting 104 for *d* leads to this: 104 = (*n*)(*n*-3)/2, which then becomes 208 = (*n*)(*n*-3) = *n²* – 3*n*. To set this up for the quadratic formula, I’m rearranging it to* n² *– 3*n* – 208 = 0. The quadratic formula then states that* n* = (3 ± sqrt(9 – (4)(1)(-208)))/2 = (3 ± sqrt(9 + 832))/2 = (3 ± sqrt(841))/2 = (3 ± 29)/2 = (32 or -26)/2 = 16 or -13, and only one of these answers, 16, can be the number of sides of a polygon. Voila!

## Elongated Dodecagon with Sides and Diagonals Extended As Lines

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## Tetrakaiicosagon with Diagonals

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How many diagonals does a polygon with 24 sides have?

First, consider that there are 24 vertices for diagonals to come from, and they each have 21 places to go, since they can’t go to themselves, or to the adjacent vertices. (24)(21) therefore equals twice the number of diagonals, since I just counted each one twice (once per endpoint). There are therefore (24)(21)/2 = 252 diagonals.