Adrakhonic Theorem

There is an Anathem wikia (inevitable I suppose and quite useful for my purpose) but unfortunately it doesn't have a picture of the proof of Pythagoras's theorem that the aliens put on the outside of their spaceship. So here it is:

i-bf476d399bfc0666d843522e15339ba0-and-1.PNG

[Note: the colours are mine; and I have reconstructed the picture from memory of the book and working out what it is trying to prove; hopefully I've got it right.I mean, I know I've got the proof-picture right; I'm less sure that this is exactly whats in the book. The line that looks orthogonal inside the larger square, is.]

The proof isn't critical to the story (at least not up to p 500, which is where I've got to) and isn't explained (ditto). The main character puzzles over it for a bit, then gives up. I haven't seen it before, so thought I would try to puzzle it out myself. It turns out to be Euclids basic proof, so I think the book is being unreasonable by having the character fail to get it; he should recognise it instantly, because I think that (unlike me) they would have had a proper classical geometrical education.

How does it work? Not in the way I first though. I thought that double-counting all the triangle areas would do it, but only 2 pairs of triangles are similar. Lets colour them in:

i-d1a4ccd4f642cb2942deb632ad6f29b1-and-2.PNG i-0b5ef59acab55177273dcecaa6f49f44-and-3.PNG

The two red triangels are similar because they have sides of the same length (being erected on squares) and have a common angle. So you can rotate them on top of each other. Ditto the blue.

Next step is to draw some extra lines, which the book doesn't do, because that would give the game away.

i-683768ebb278dcb3a60e6edf6d58ab53-and-4.PNG

Then you realise that the red triangles from the previous pic are each of equal area to a right angled triangle (half-base-times-height, and so you can slide the apex along a line parallel to the base) as shown here, one of which is clearly half the area of the square, and the other of which is clearly half one of the rectangles that comprises the larger square. Ditto for the blue. So half of the large square is equal to the sum of half the sum of the smaller squares, WWWWW.

Wiki, of course, has far more of this. One interesting point is how this fails to work in non-euclidean geometry.

[Note that seed doesn't have a maths channel, and it clearly isn't tech or science, so I'll call it chatter]

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In school I was one of three people doing a special level add on to A level when this came up. I came up with the solution that had most construction lines - ie long and complicated. Simon's proof was simpler and I think it probably was this. Andrew's proof was by far the simplest he drew a right angled triangle and a single construction line. Probably the only benefit of my approach was that you could be sure I hadn't copied it from anywhere - but wikipedia didn't exist then. :)

>" but only 2 pairs of triangles are similar"
This is incorrect, there are lots - I gave up counting when I got to 8 pairs and a threesome. Why don't you say identical when you mean identical? ;)

>"Next step is to draw some extra lines, which the book doesn't do, because that would give the game away."

Yes but I would draw four lines - two of yours and two different lines ;)

This proof uses the formula that area of a triangle is equal to half base times height. How far are you allowed to go before saying something is a diagramatic proof? Why would aliens use something complex like this rather than a right angled triangle and a single construction line?

[I don't know the proof you mean -W]

>[I don't know the proof you mean -W]

It leads the proofs section on wikipedia so you have already linked to it. :) Due to this I didn't set it as a challenge when I wrote that post. But if you don't know the proof I mean, your challenge should you... :)

[I see it now. It doesn't look like pure geometry to me -W]

>"I think the book is being unreasonable by having the character fail to get it"

Yes the point I was trying to get to was that it is within reach of a very good A level candidate to discover the proof without the help of seeing the construction lines. Perhaps that is pushing it a bit as Simon did go on to study maths at Oxford.

Euclids construction lines make it messy. Wouldn't aliens be more likely to have a clearer diagram? I am sure there are lots of possibilities like:

Maybe one of these will get through.

wwwdotboincforum.info/boincuser/Crandles/pythagdiagramproof.PNG

While I was still studying math, I'd seen three separate proofs, counting this one, and haven't checked Wiki yet. Of the three I know, Euclid's is arguably the most complex, relying on area and congruence theorems and numerous construction lines. The first proof I saw (back in middle school) constructed the hypoteneuse square and copied the triangle three times, producing a square with area (a+b)^2 = c^2+4*(1/2 ab), which simplifies to Pythagoras.

[Yeeees... that may be cheating, though, since its not geometrical entirely -W]

The simplest proof I've seen simply adds an altitude through the right-angle, which provides two similar triangles (the smallest interior triangle and the original triangle share two angles and a side) and a proof emerges almost immediately.

[That one also doesn't look to be pure geometry to me. You have to multiply through -W]

Checking Wiki now, I see it's the first proof listed there (proof by similar triangles), with the earlier one I mentioned showing up as the algebraic proof.

Having not read Anathem, I don't know how these compare to their mentality; why would they have chosen Euclid's instead of, say, the simplest similarity argument? (That said, I do agree -- anyone reasonably familiar with geometric proofs should be able to work their way through Euclid's without much difficulty.)

[Only the first diagram I showed was given, no working -W]

[Yeeees... that may be cheating, though, since its not geometrical entirely -W]

I did say I saw it back in middle school, in which analytic proofs were the only type covered, on the rare occasions they showed up. Lubos' version is the purely geometric equivalent of the same proof, which I agree is much simpler. (Imagine that, I agree with Lubos for once.)

[That one also doesn't look to be pure geometry to me. You have to multiply through -W]

I've seen a purely geometric version of this one as well, but I'd have to find my reference to express it.

Dear William, I actually created and posted the picture a few months ago, but later than the date indicates - just for the people who search via search engines.

My blog has technology on it so that it crashes the browsers of anti-technological luddites as efficiently as possible. Sometimes it may crash the browsers of anti-technological conservatives, too, but that's their fault, not mine ;-).

I agree with Brian that at least my picture qualifies as "pure geometry". This term means that one doesn't need any analysis. Saying that the square with side "c" has "c squared" is not analysis: it is, by definition, a geometric player in the Pythagorean theorem.

On the other hand, I don't even need you to know that the rectangle "a times b" has area "a times b" because that would be too much of difficult analysis for you, too. The picture just moves some four triangles from place to another place, to rearrange the empty brown area "square with side a" plus "square with side b" into the area "square with side c", so the areas on both pictures are equal, which is nothing less and nothing more than what the theorem states.