An Unusual Way to Prove Napoleon's Theorem

[@tselmegrim101]

Vectors are so powerful in geometric problems. Thank you sir for highlighting this in a intriguing way.

[@Colonies_Dev]

to me it seems like in code..it's easier to break it into more of a formulaic type of way.. and then solve variables. otherwise u end up with this weirdness where vectors are in worldspace pointing a rlly long way. not sure if that makes it easy to for example find intersections or not between vectors as lines e,g,. if u wanted to easily find intersect between two vectors as lines you'd probably turn it into an equation describing a line also that travels along the vector.. then solve variables as i mentioned

[@konradhanninenpedersen8705]

I actually laughed out loud a bit, when I got it. Really a nice proof. Thanks for another awesome video.

[@allseeingeyeoftheenemy]

Besides 3b1b, this is probably the best Math channel I've ever found. HOLY SHIT do I love your videos!

[@tamirerez2547]

Very interesting problem with amazing graphic animation!! GREAT SOLUTION in clear explanation!! How didn't I know about this channel before?? BIG LIKE 👍♥️

[@CedarAce1000]

Amazing video! No clue why it's not getting as much attention as the others...

[@MathyJaphy]

Thank you! I just posted it a few hours ago. I’m hoping it will get a bit more attention with time.

[@Acuzzio]

@@MathyJaphy It is truly amazing content and will be noticed, do not worry. It popped up in my feed immediately. Keep up the good work, man!!!

[@renesperb]

A very nice presentation of this proof !

[@joydebroy8]

Wow. Amazing proof. I would like to ask you for some animated presentation on Apollonius Circle theorem and its proof. Thank you so much for your efforts to build better understanding of the subject.

[@MathyJaphy]

Thanks for the kind words, and for the suggestion. I will look into it.

[@mike_the_tutor1166]

Another beautiful video. Thank you for sharing. If you ever figure out how to prove the Petr-Douglass-Neumann theorem in this way, I would love to see it. If you know of any other interesting geometric proofs using vectors, I would love to know about them. I just read an article from MIT that proves the sum of a triangle's medians is the zero vector, which I think is a nice starting point for those getting into geometric proofs using vectors.

[@MathyJaphy]

Thanks! Proving that the medians sum to zero would make a nice KZbin Short video. Seems pretty straight-forward to prove; a good starting exercise as you said. I found what I think is the article you mention. Even that simple answer seems more complicated than it needs to be. Is it this one? ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-a-vectors-determinants-and-planes/session-1-vectors/MIT18_02SC_we_4_comb.pdf

[@mike_the_tutor1166]

@@MathyJaphy That's the one! If you have a better explanation I'm all ears!

[@MathyJaphy]

Define vectors A, B and C that coincide with the triangle's sides. The median vectors can be expressed as (A+B/2), (B+C/2) and (C+A/2). Adding them all together and rearranging gives (A+B+C) + (A+B+C)/2. And (A+B+C) is zero because triangle. It's the same as their answer. Just a bit clearer, no?

[@mike_the_tutor1166]

@@MathyJaphy Yes, that's wonderful! Frankly, I should have thought of it myself, but didn't. Thank you for the explanation! I'm adding it to my notes right now!

[@yourpal1685]

Where's your barbershop quartet ending? I missed that

[@MathyJaphy]

Glad you like the barbershop. I don't want to overuse it, though. It felt right with Ptolemy after the wordless proof, but here it didn't fit with the P-D-N animation outro.

[@d.e.p.-j.7106]

Very nice!

[@rivkahlevi6117]

Like the PDN animations!

[@MathyJaphy]

Thanks! I’m still hoping to do a whole PDN video in the near future.

[@rivkahlevi6117]

@@MathyJaphy I'd like to see that!

[@ranjanachaudhary2110]

Soooooo coollllll

[@じーちゃんねる-v4n]

I proved it on the complex plane. 1+ω+ω^2=0 triangle ABC A(0) B(1) C(z) equilateral triangle BCX ACY ABZ ∴X(1-ω(z-1)) Y(-ω^2z) Z(-ω) center of BCX ACY ABZ are L M N respectively ∴L(((1-ω^2)+(1-ω)z)/3) M((1-ω^2)z/3) N((1-ω)/3) ∴NM=(1-ω)((1+ω)z-1)/3 NL=(1-ω)(z+ω)/3 ∴NM/NL=((1+ω)z-1)/(z+ω)=(1+ω)(z+ω)/(z+ω)=1+ω=-ω^2=cos60°+(sin60°)i ∴L, M, N are equilateral

[@Yavuzhans]

First

[@MathyJaphy]

Congrats!

[@Yavuzhans]

@@MathyJaphy Thank yu