## New sections on nspheres, sum of a polygon's internal angles

An enlightening discussion about pi and tau.

### New sections on nspheres, sum of a polygon's internal angles

Mike, I recently added two new sections to my tau website on topics that were addressed in the Pi Manifesto. When you have a moment, would you take a look and tell me what you think? (Anyone else still here, feel free to chime in too.) Thanks.

"The sum of the internal angles of a polygon - A rebuttal"
"A different pair of formulas for every dimension. Their common link?"
PI is the SEMICIRCLE constant, not the circle constant . . . http://sites.google.com/site/taubeforeitwascool
josephlindenberg
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### Re: New sections on nspheres, sum of a polygon's internal an

Very nice discussion on "uninternal angles". I'm not sure if that's the best name for them but can't think of anything better -- too bad external angles is already taken! However, since we're changing the circle constant we might as well change the definition of 'external angle' as well!

The next section is nicely written too. The diagram (Nspheres.png) was confusing at first until I realized the red things were arrows (which it does say after the diagram). Let me think about it for a bit and see if I can come up with a clever rebuttal, hmmm....
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SpikedMath

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### Re: New sections on nspheres, sum of a polygon's internal an

How about "outside angle"? Or, if we want to be really clever, "outernal angle"?

xander

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### Re: New sections on nspheres, sum of a polygon's internal an

Since it has done without one this long, it would appear mathematics doesn't really need a name for the "uninternal angles". I just needed something to call them for purposes of this discussion, and I intentionally chose an awkward name. I considered using "outside angles", but I didn't want people getting too comfortable with the name and then absent-mindedly swapping in synonyms like "external". You know how our brains can work -- outside, external, same thing. That's also partly the reason I color-coded the names.
PI is the SEMICIRCLE constant, not the circle constant . . . http://sites.google.com/site/taubeforeitwascool
josephlindenberg
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### Re: New sections on nspheres, sum of a polygon's internal an

It's nice, though. And fits my thoughts on the subject. The formula for the angles of a polygon is not real obvious or simple to my mind, but going by the uninternal angles makes it very simple: the sum is 720° or τ, no matter how many angles are involved.
bmonk
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### Re: New sections on nspheres, sum of a polygon's internal an

Thanks for pointing out the problem with me not explaining the diagram, Mike. I've rewritten that n-spheres section, and I added a better introduction of the diagram.
PI is the SEMICIRCLE constant, not the circle constant . . . http://sites.google.com/site/taubeforeitwascool
josephlindenberg
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### Re: New sections on nspheres, sum of a polygon's internal an

bmonk wrote:It's nice, though. And fits my thoughts on the subject. The formula for the angles of a polygon is not real obvious or simple to my mind, but going by the uninternal angles makes it very simple: the sum is 720° or τ, no matter how many angles are involved.

You forgot already that $$\tau$$ = 360 degrees, because $$\pi$$ is wrong.
rdococ
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### Re: New sections on nspheres, sum of a polygon's internal an

rdococ wrote:
bmonk wrote:It's nice, though. And fits my thoughts on the subject. The formula for the angles of a polygon is not real obvious or simple to my mind, but going by the uninternal angles makes it very simple: the sum is 720° or τ, no matter how many angles are involved.

You forgot already that $$\tau$$ = 360 degrees, because $$\pi$$ is wrong.

True. But the point is that the sum of any polygon's external angles is 2$$\tau$$. If you want the sum of the internal angles, then you need to calculate a bit more.
bmonk
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### Re: New sections on nspheres, sum of a polygon's internal an

bmonk wrote:
rdococ wrote:
bmonk wrote:It's nice, though. And fits my thoughts on the subject. The formula for the angles of a polygon is not real obvious or simple to my mind, but going by the uninternal angles makes it very simple: the sum is 720° or τ, no matter how many angles are involved.

You forgot already that $$\tau$$ = 360 degrees, because $$\pi$$ is wrong.

True. But the point is that the sum of any polygon's external angles is 2$$\tau$$. If you want the sum of the internal angles, then you need to calculate a bit more.

The sum of any polygon's external angles is just tau, not 2τ, and just 360°, not 720°. Are you thinking of something else?

In addition to external, internal, and "uninternal" angles, there are also "unexternal" angles, which always sum to (n-1)τ. I don't know when they would be important, but I just thought of them randomly.
The formula for the measure of one external angle of a regular n-gon is τ/n. The measure of an interior angle is equally simple with either, (1/2-1/n)τ or (1-2/n)π. The formula for the measure of each "uninternal" angle would be (1/2+1/n)τ or (1+2/n)π, and the formula for each "unexternal" angle would be (1-1/n)τ.

I also found it interesting that for polygons that cross over themselves the external angle sum is always a multiple of tau. I also think the fact that the internal angle is only found by taking a straight angle and subtracting the exterior angle, as well as the fact that when going around the polygon's perimeter, you pivot at the same angle as the exterior angle, not the interior angle, shows that exterior angles are just as important as interior angles. This also brings up another point; the interior angle formula comes from taking n half turns/straight angles, nπ, and subtracting the exterior angles, yielding nπ-τ. It seems like both constants are equally relevant to this formula.

I also have a comment on the Pi Manifesto's argument when it mentions that the sum of internal angles of a triangle is pi. Aren't squares more important than triangles, being the shape we use to measure area, having opposite sides parallel, and being the only polygon whose interior and exterior angles are equal? The sum of both the interior and exterior angles of a square is tau. Isn't that a stronger argument for tau than the sum of the interior angles of a triangle being pi (but the sum of the exterior angles still being tau) is for pi?
τ>π
1=0
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