Thursday, March 22, 2012

More Pythagoras than Pythagoras

So, a nifty theorem: that in a tetrahedron of which the three angles at one vertex are right angles, the square of the opposite face is the sum of the squares on the three adjacent faces. By homogeneity, it follows that the square of any plane area in 3-space is the sum of the squared areas of its three orthogonal shadows on any three orthogonal planes. Here, by homogeneity I mean that orthogonal shadows of parallel plane figures are proportional to eachother as those plane figures are.

A sketch of the hypothesis, as stated:

The "opposite" face is left implicit; the adjacent faces coloured blue, red, and grey.

There are, of course, several approaches to this theorem.

One might calculate relations among the sides and areas, and proceed algebraically to conclude that the two quantities described have equivalent expressions (e.g., in terms of the three adjecent edge lengths).

Alternatively, one might take as prior what we have stated as corollary, and argue (by suitable means) that the sum-of-squared-shadows is independent of which three orthogonal planes are chosen for catching the shadows; then, since the equation obviously holds when our plane figure lies in one of the three planes --- for then it is its own shadow on its own plane, and its other two shadows have no area, and the full proposition follows.

I rather like circles, lately, so I'll do something between the two: a special plane shape (indeed a circular disc) has three orthogonal shadows on three orthogonal planes, and their squares sum to the square on the circle; one returns to general plane figures by homogeneity again.

The niftyness of this arrangement begins with the observation that the orthogonal shadow of a circular disc is an ellipse; and moreover that the major axis of this ellipse equals the diameter of the circle. Indeed, the shadow doesn't depend on where the disc lies, only on its relative attitude. Supposing the disc's center were on the shadow plane, then the shadow and disc meet in a line, precisely this major axis. Since the area of an ellipse is universally proportional to the product of its principal axes, this reduces the problem to arguing that the squares on the minor axes of the three elliptical shadows sum to the square of the circle's diameter.

To get there, we must be sneaky: consider a line segment perpendicular to our circular disc, and equal to a diameter. With some thought, one can see (this means "prove", but it's not too hard) that the minor axis and the shadow of this perpendicular lie on the same line; but they are shadows of equal and perpendicular line segments, so their squares must sum to the square on the diameter --- once in each ellipse.
At the same time, the three orthogonal shadows of the perpendicular segment squared sum to twice the square on the diameter (this is a fun exercise) adding up six squared line shaddows in three ellipses should give three times the squared diameter (since each ellipse gives one), so the three squared minor axes sum to a single squared diameter, as promised.


Belfry Bat said...

(restored consistency, fixed spelling)

It's handy to give a suggestive name to the sort of tetrahedron described in the first paragraph: call it an axis tetrahedron, because three adjacent edges lie on coordinate axes. It also has a natural generalization to any dimension: an axis triangle (aka 2-simplex) is just a right triangle; an axis 4-simplex as four axis-3-simplices meeting at one vertex, and a fifth 3-simplex opposite that vertex, etc.

We then have the Intruiguing Side-Track: while one can obain any acute triangle as the opposite face of an axis tetrahedron, there are equations relating the edges of the opposite 3-simplex of an axis 4-simplex; specifically, in any tetrahedron, there are three pairs of disjoint edges, $(p,P),(q,Q),(r,R)$, say. In the opposite tetrahedron of an axis 4-simplex, the sum of the squares of these disjoint edges must be the same in all three pairs:
$ p^2 + P ^2 = q^2 + Q ^2 = r^2 + R^2 $.

Post a Comment