Sunday, March 18, 2012

A little more, about quadrilaterals

Before getting quite lost in why the isoperimetric problem should have exactly one solution (the answer, in brief, is relative convexity, but that's a mouthful, before we know how to spell it), a little bit about Steiner's proof, adapted slightly.

Now, last time it was pointed out that rotating a figure preserves both its area and its perimeter; and this in fact remains true of separate pieces of the figure; so, for instance, pertaining to the area within the black curve below:
one may rotate the mauve or blueish pieces, preserving their area as well as the lengths of both black and red curve or line segments. This is clearly a nifty thing to consider, as it means also that if there's a quadrilateral with the same sides as the red quadrilateral, but with greater area, then aranging the mauve and green bits around that gives a new shape with the same black perimeter, and greater area (the mauve and bluey bits will be the same, and the transparent bit in the middle can be made greater).

It just so happens that the best arrangement of vertices for the red quadrilateral is (surprise!) as a cyclic quadrilateral, aka "chord" quadrilateral, which is to say, such that the vertices are all on a circle. Perhaps the other surprise is that this is always doable. There is a bit of annoyance about proving the optimality of a cyclic arrangement, in that it works out to be more natural to maximize the square of the area --- in a calculation attributed to one Bretschneider, this is a sum of two terms, one of which is determined entirely by the four side lengths, the other proportional to their product and otherwise depending only the sum of opposite vertex angles... you can see how this is getting messy, yes? The calculus is quite straight-forward, but the geometry gets to be rather icky. I think there ought to be a 4-dimensional scissors congruence proof of this fact, considering how similar it is to Heron's formula. But we mustn't jump to conclusions! (Also, such things are difficult to read).

Anyways. Suppose, then, that the curve we had wasn't a circle. Then there must be some four points of our curve that weren't cyclic. (Three points determine a circle, a fourth is either on that circle or off it). And this in turn means there was then another curve with the same perimeter, and containing greater area.

Steiner's original proof relied on a simpler case of this quadrilateral argument, specifically that of parallelograms with fixed sides. It's easier to see that the greatest such area is a rectangle; but reducing to this case uses slightly more surgical trickery beforehand. I'm undecided, just now, which approach is tidier.


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