==============================================================================
[Note: The quoted post included below is proving algebraically the
3-dimensional analogoue of the Pythagorean theorem: that on a tetrahedron
with three faces contained in mutually perpendicular planes, the sum of the
squares of the areas of the triangles in those planes equals the square
of the area of the remaining traingular face.]
==============================================================================
From: ceaton@net1plus.com
Subject: Re: 3D Pythagoras
Date: Fri, 21 Nov 1997 19:33:19 -0600
Newsgroups: sci.math
In article <1997Nov20.021849.7718@lafn.org>,
ba137@lafn.org (Brian Hutchings) wrote:
>
>
> In a previous article, ba137@lafn.org (Brian Hutchings) says:
>
> it occurred to me that this might have been a troll
> -- although I'm not accusing you!!!! --
> for your putative solution to FNT
>
> >http://ourworld.compuserve.com/homepages/grok/ --
>
> myself, I had not checked out the forementioned book
> in which a proof was listed, but I *had* perused it, so that
> it's possible that I'd read the italicized theorem; but
There is a geometrical prooof that in four dimensions, the sum of the
squares of the volumes of four right tetrahedrons on the axes of the four
dimensional cartesian cordinate system is equal to the square of the
volume of the tetrahedron formed by their diagonal faces. The
n-dimensional Pythagoras has been proven by C. H. Edwards, Jr., Advanced
Calculus, Academic Press, 1973, pp326-330. I can snail-mail the 4-d
geometric proof if interested.
> I recall the near-mystical experience of "visualizing" the problem,
> with no thought of the reading.
> anyway, here is what I wan't to know of your proof:
> how did you get from over the place where your text is broken?...
> I know, what it might involve, but since I'm not that good
> at algebra, you could just show us!
>
> >Square of the area of the largest face
> >
> >= areaPQR**2
> >
> >= ((s(s-a)(s-b)(s-c))**(1/2))**2 [s=(1/2)(p+q+r)]
> >
> >= s(s-a)(s-b)(s-c)
> >[s=(1/2)((a**2+b**2)**(1/2)+(b**2+c**2)**(1/2)+(c**2+a**2)**(1/2))]
>
> >= (ab/2)**2+(bc/2)**2+(ca/2)**2
> >
> >= areaOPQ**2+areaOQR**2+areaORP**2
> >
> >= Sum of the squares of the areas of the 3 smaller faces of a right
> >tetrahedron
> >
> >-- very good!... (note: a,b,c are the legs of the *tetrahedron*,
> >not the sides of any trigon PQR, obviously, but neatly !-)
>
> note, this is referred to as a "trirectangular tetrah.", in below,
> or you can call it a hexahedron-corner (cube-corner).
>
> >>I don't recall it, but it's in here, somewhere;
> >>it's also in Altshiller-Court's _Modern Pure Solid Geometry_, although
> >>I'd like to see a (3D) "tiling" proof a la the ancients.
>
> as for the other "proof", please, there are some possible ambiguities:
> what does p2**i mean -- same as p(2^i), or (p2)^i, and so on. also,
> what is the cryptical "m-> 999999. ...", or what ever, supposed to mean
> -- 10's complimentation?... p-adicism?
>
> --
> quothing The Raven: nev'mo'better!
>
> (Brian Hutchings, Living Space Programs, Santa Monica College)