Three pretty geometric theorems, proved by complex numbers

Bottema'a theorem, van Aubel's theorem and Napoleon's theorem, are all brilliant. If you have never seen them before, you will be amazed! I prove then all by working in the complex plan, as van Aubel did in 1878.

[lugiagaurdien]

Absolutely gorgeous video. We need more of this on YouTube.

[peterhall]

A ripsnorter. I knew Fourier did bureaucratic jobs for Napoleon but wasn't aware that Laplace had taught him. Imagine the dinner scene over a pretentious Bordeaux: Laplace: Emperor, may we move on from world domination for a moment to this cheeky little theorem? Napoleon: How is your Russian?

[tunguyenhoc]

Beautiful. Brilliant. So enjoyable lesson. Thanks.

[gregorycampbell]

One of my favorite things about complex numbers is the surprising degree to which they:
-simplify complexity in 'Real' situations (not to mention geometrical ones, amongst many others);
-can be applied to, seemingly, ANY type of analysis, usually (probably always) in meaningful, useful ways;
-can inform and support one's intuition, despite being rooted in the wholly unintuitive concept of negative even roots.

We should have called them Ironic Numbers ... except then they wouldn't be as ironic ... which, uh, is actually kind of ironic? Hmm... 🤔

[wickedpawn]

I believe the last theorem is where Symmetrical Components in Electrical Engineering come from. Representing a 3-phase unbalanced system by the superposition of three balanced systems.

[renesperb]

Very nice geometric problems . It is also good to see, especially for people who didn't like complex numbers, how much you can achieve
by a clever application of them.

[twanvanderschoot]

Nice demonstration. But in case of Napoleon's Theorem, it is sufficient to prove that the lengths of the side of the triangle are equal, you don't need to introduce rotation over 120º.

[GregShyBoy]

For the last part of the third theorem, there's a way to avoid "knowing" the cubic root of 1. But it'll require calculating a module of a complex number. Calculations may seem longer but (arguably) less knowledge is required.

Overall — awesome job! This lesson totally answers the scholar's question "Who even needs these imaginary numbers?"

[gamespotlive]

Brilliant man, this is a great video.

[alikaperdue]

I used complex numbers miners to find all the 3rd points of similar triangles with two fixed points. It was a Riemann sphere.

[abc]

Thank you! It is pretty simple. I thought it would have been harder to understand.

[tissuepaper]

My opinion of this video can be summed up by the following: complex numbers fucking rock! Repurposing algebraic machinery to perform geometric calculations, the elegance of it is astounding.

[pauselab]

I used to prove random common Euclidean geometry problems with complex coordinates. Works much better than real coordinates most of the time though not always better than using theorems

[_faraz]

Totally love it. Can you recommend any book for geometrical theorems like these?

[DesmondFitzpatrick-vvdp]

amazing these prrofs are not better known (at least to me:) )

[agytjax]

Is this the same billionaire Jim Simons who founded the Medallion fund ?

[quantumgaming]

Question, just how from these 3 hard geometrical construction we could prove them using symmetric complex relations, can we do it just as easy the other way around?
If we have a collection of symmetrical complex relations (with some constraints on them, possibly. I don't believe any symmetrical relations would work) can we always find a geometrical construction associated with them?

[tinafeyalien]

I don't get the 1/sqrt(3) bit. Half the height of an equilateral triangle of side length 2(y-x) is [sqrt(3)/2]*(y-x)

[kailashvardhan]

Are you the man who solved the Market. If yes please teach me sir. It's needed.

[landsgevaer]

Some comment should mention the Petr-Douglas-Neumann theorem in this context. Allow me to.

I knew about the theorem but had forgotten its name. Thanks for triggering me to look it up again. Wonderful little tidbit that!