Dear Mr. Abrams,
You win!
Sincerely,
a fan
Wednesday, March 28, 2012
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.
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.
Monday, March 19, 2012
Since we're re-writing actual real positable Law like nobody's business...
Dear Sir Isaac,
I just had a bit of brie fall, if you will believe it, to the flooring of my computer's tower case (er... I keep it open because, being an eleven-year-old obsolescence it overheats otherwise... Joseph Fourier and William Thomson, you're next on my list!)
And so I'm writing to protest, sir, this so-called "law" of so-called "universal" gravitation. Obviously we don't want cheese in our computers (any more than we can avoid it) so obviously, it's time we extended to dairy curd the same exceptions that Helium and small birds seem to have been enjoying for centuries!
Yours, most gravely,
a physical progressive
I just had a bit of brie fall, if you will believe it, to the flooring of my computer's tower case (er... I keep it open because, being an eleven-year-old obsolescence it overheats otherwise... Joseph Fourier and William Thomson, you're next on my list!)
And so I'm writing to protest, sir, this so-called "law" of so-called "universal" gravitation. Obviously we don't want cheese in our computers (any more than we can avoid it) so obviously, it's time we extended to dairy curd the same exceptions that Helium and small birds seem to have been enjoying for centuries!
Yours, most gravely,
a physical progressive
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.
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.
Tuesday, March 13, 2012
Dear Orville,
Oatmeal.
I didn't like it much growing up. I still don't like it now. But I'm beginning to appreciate that it has, at least, an inoffensive nutritiousness. Then as now, it's more about the fruity things that can be included to break up its blandness. Also, I see why those very practical Scots formed a notable habit of salting it --- it's perfectly sensible!
Anyway, hope that dried maize thing works out for you,
a cook-to-eat-er
Oatmeal.
I didn't like it much growing up. I still don't like it now. But I'm beginning to appreciate that it has, at least, an inoffensive nutritiousness. Then as now, it's more about the fruity things that can be included to break up its blandness. Also, I see why those very practical Scots formed a notable habit of salting it --- it's perfectly sensible!
Anyway, hope that dried maize thing works out for you,
a cook-to-eat-er
Monday, March 5, 2012
A Little Bit about Isoperimetric Inequalities
So, in Crelle XVIII, Jakob Steiner published the earliest known solution of the ordinary isoperimetric problem --- that is, among measurable plane figures with fixed perimeter, find which has the greatest area. I'm sure something like this result had been known considerably earlier; Euler and later Lagrange had formulated and solved rather general extremum problems for curves, and the isoperimetric example isn't difficult to solve. Variational problems were posed by one of the Bernoulis --- most notably the brachistochrone, of which several solutions were published simultaneously (Oh! For the good old days!). My source for Steiner's priority also claims he's the first person to have formulated the problem (or at least to have published such a formulation). So I'm inclined to suspect that Newton and Leibniz and them all thought the thing so painfully obvious that they never mentioned it.
The isoperimetric problem is considerably simpler than the brachistochrone. In fact, there are a number of similar problems which have strikingly similar solutions; for instance, among rectangles having fixed perimeter, that with the greatest area is a square, which is practically a proposition in Euclid, and quickly generalizes to the elsewhere-famous Cauchy-Shwartz-Буняковский inequality. If we said "triangles" instead of "rectangles", we'd get an equliateral triangle as the solution, and if you want to keep guessing at the best answers, you'll see this post is actually a love-song to symmetry.
The isoperimetric problem itself has rotational symmetry. Since rotating a plane figure changes neither its perimeter nor its area, it should be clear that if we had any extremal isoperimetic figure, then we could get more extremal figures by rotating the first one. That is, we could get more if there were more --- if our extremal isoperimetric figure didn't have the same symmetry as the isoperimetric problem itself. But there aren't more, there's only one. And there's a good reason there's only one, but we'll get to that another time.
For now, pleasant dreams!
The isoperimetric problem is considerably simpler than the brachistochrone. In fact, there are a number of similar problems which have strikingly similar solutions; for instance, among rectangles having fixed perimeter, that with the greatest area is a square, which is practically a proposition in Euclid, and quickly generalizes to the elsewhere-famous Cauchy-Shwartz-Буняковский inequality. If we said "triangles" instead of "rectangles", we'd get an equliateral triangle as the solution, and if you want to keep guessing at the best answers, you'll see this post is actually a love-song to symmetry.
The isoperimetric problem itself has rotational symmetry. Since rotating a plane figure changes neither its perimeter nor its area, it should be clear that if we had any extremal isoperimetic figure, then we could get more extremal figures by rotating the first one. That is, we could get more if there were more --- if our extremal isoperimetric figure didn't have the same symmetry as the isoperimetric problem itself. But there aren't more, there's only one. And there's a good reason there's only one, but we'll get to that another time.
For now, pleasant dreams!
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