hmm... john huerta asked for my opinion on "lockhart's lament"... i told him that i'd try to read it...

i have a couple of reactions... not sure how many of them i'll get around to mentioning here...

at one point lockhart sets up as a bad example a presumably somewhat traditional proof that an angle "inscribed in a semi-circle" is a right angle, and following that as a contrasting good example a proof of the same fact, more or less found by one of lockhart's seventh grade students... so of course before i read the good proof i decide to think the problem over myself, in part to see whether i could guess what the alleged good proof would be...

as it turns out i guessed correctly. in fact i had a bit of an aha moment while thinking about it, and correctly suspected that i was on the track of lockhart's preferred proof. roughly the "good" proof is as follows: complete the semi-circle to a full circle by reflection through the center of the circle; now you've got a parallelogram inscribed in a circle, and it's intuitively obvious (in a way that can pretty easily be fleshed out into a proof) that such a parallelogram is a rectangle.

one of my excuses for mentioning thia now is that for various reasons i'm interested in trying to reconstruct the genesis of my aha moment...

this is probably a digression from the main issues that john huerta was interested in my opinion on... i hope that i get back to that, but no promises...

i have somewhat of a history with this particular problem, and i should describe some of that history if i get around to it... my aha moment of a few minutes ago is i'm pretty sure a new addition to that history...

let me also mention that some time ago i read another article somewhere which lockhart's lament reminds me of... i'm wondering now whether that was lockhart too...

in reconstructing my thoughts it helps that i have here the actual literal back of an envelope on which i sketched a few ideas...

i think that first i sketched my thoughts about the "bad" proof... pretty vigorously disagreeing that it's bad, rather that lockhart just presented it unfairly... which he subsequently pretty much admits is what he did, though i hadn't read that admission at that point... anyway more about this later, if i get around to it...

fishing around for other ideas, i sketched a few other pictures...

in one picture, instead of drawing a line segment from the apex of the angle to the center of the circle (as in the "bad" proof), i tried "dropping an altitude" from that apex to the boundary diameter of the semi-circle. i had some vague ideas about this... vaguely to do with connecting the original angle-inscribed-in-semicircle picture to the traditional "angle at the center of the unit circle" picture of trigonometry... which probably requires introducing another circle of twice the radius ...

in another picture i drew inside the triangle-in-semicircle the half-scale similar triangle gotten by connecting the midpoints of the sides... not sure whether there was any interesting idea there...

i returned to the idea of making the triangle-in-semicircle into a traditional "trigonometric" right-triangle inside the unit circle (again of twice the radius of the original circle)... and when i started fleshing out this picture, the parallelogram began to automatically emerge somehow, which led to the aha moment... but it's still somewhat mysterious to me as to how this happened... whether it was actually a sensible working out of my vague unit-circle idea, or just a random visual coincidence that i happened to draw that parallelogram...

not sure how worth it it would be to pursue this further...

about the "bad" proof... roughly, my idea of the good presentation of it is that you draw a line segment from the apex of the angle to the center of the circle, dividing the original triangle into two manifestly isosceles triangles; then you write down the obvious system of equations where the variables are all the angles of all three of the triangles; then you solve that system of equations (somewhat gingerly because of the clockish "cyclical" aspect of the angular variables; perhaps treating them very "geometrically" if you want to disguise the noticeably "algebraic" nature of the proof). nobody could care too much about the details of this system of equations, which presumably feeds lockhart's rhetoric about how ugly the proof is. but to me that's the nature of this problem: you identify the tool that's used to solve it, namely the particular system of angular equations, and then actually solving the system is just a foreordained anticlimax. i guess that i'm thinking of it as an example of "reducing a problem to an already solved problem", namely the solution of systems of equations involving the addition of angles.

(one good thing about this "bad" proof is how it generalizes so straightforwardly to the case of an angle inscribed in some fraction of a circle other than one half; does the "good" proof generalize in any interesting way to that case??)

from this viewpoint you can think of the parallelogram proof as really just supplementing the two isosceles triangles with two more, obtained as their mirror reflections (through the center), giving a larger system of angular equations but with the solution more intuitively transparent...

lockhart seems to like proofs that turn on suddenly (as in "aha!") seeing some (pretty much literally) overlooked part of the picture; for example the complementary semi-circle in the case of the "good" proof, or the center of the circle in the case of the good presentation of the "bad" proof...


for some reason i feel like mentioning that this morning i was trying to recall einstein's semi-famous proof of the pythagorean theorem, and was annoyed that i couldn't remember it even though i could remember little bits of it... i actually gave up and looked it up... roughly it's like this:

first, notice that although it's often stated as a theorem about squares, you could equivalently use almost anything else in place of squares, say triangles, or elephants:

"the area of the elephant on the hypotenuse is equal to the sum of the areas of the elephants on the other two sides".

(surface area if your elephants are 3d.)

since it's indifferent what we use, let's use right triangles similar to the original right triangle, with the one erected on the hypotenuse being the original right triangle itself (thus pointing "inward" rather than "outward"). then the theorem is obvious.

(it would probably help a lot if i'd ever gotten around to working out some easy way to post pictures here...)

more explicitly: when the figure erected on the hypotenuse is taken to be the original triangle itself, then not merely do the areas of the smaller figures erected on the other two sides add up to its area, but moreover those smaller figures themselves add up to it itself (with only negligible lower-dimensional boundary overlap).

in some sense this is the ultimate "dissection proof" because no steps need to be taken to enact the dissection; the figure on the hypotenuse comes automatically pre-dissected into the two smaller figures.