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It seems only fair that I now post my own no-paper-required solution… - delta_november

Feb. 6th, 2013

10:35 pm

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It seems only fair that I now post my own no-paper-required solution to the tetrahedral angle question.


Consider unit vectors normal (and pointing out) from the four sides. Call them A, B, C and D. It is clear from inspection that:
A+B+C+D = 0

thus:
A+B+C = -D

Let's sit the tetrahedron on its base such that:
D = [0 0 -1]

It is clear from inspection that the vertical components of vectors A, B and C are equal. Since the three vertical components sum to unity, each one must be 1/3.

So:
A = [x y 1/3] where x and y are at this point unknown, and irrelevant to this problem.

Since these are all unit vectors, their dot products are equal to the cosine of the angles between them. The angle between side A and the vertical (-D) is:
A dot -D = cos(theta)

thus:
[x y 1/3] dot [0 0 1] = cos(theta)

thus:
cos(theta) = 1/3

thus:
theta = acos(1/3) which is a hair more than 70 degrees

Comments:

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From:kiwano
Date:February 7th, 2013 05:57 pm (UTC)
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That's a nice slick solution. I like it. I probably would've wound up using paper if my recreation time last night didn't get eaten by making backup childcare plans for tomorrow if the snowstorm buggers up my flight home.
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From:delta_november
Date:February 8th, 2013 12:32 am (UTC)
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Safe travels!

The interesting thing about this proof is that while it does not extend to other platonic solids, I believe it does work for all dimensions.

An equilateral triangle has angle acos (1/2).
A tetrahedron has angle acos (1/3).
I believe that for higher dimension N, the angle should be acos (1/N).

While it breaks my brain to visualize, the vectors don't care. There are still more of them adding to make unity, so the fraction decreases. In four dimensions we're talking about a vector normal to a solid rather than a surface, but dot products still obey their usual rules.
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From:kiwano
Date:February 10th, 2013 09:59 pm (UTC)
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Thanks; I managed to rebook to an earlier flight that got me home before the storm.

The higher-dimension generalization you point out looks correct. I'd probably refer to the vectors as being normal to a hypersurface or hyperplane rather than a solid, but that's probably because I'm still amused by that quirk of nomenclature where a hypersomething is like a something, but with more dimensions than you'd expect..
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