a quasi-Pythagorean identity

Playing with triangles: a quasi-Pythagorean identity. I highlight a beautiful identity coming from geometry, which has to do with equilateral triangles and complex numbers. This has been inspired by a Tweet by Steven Strogatz from Cornell University. For this, we use Euler's formula and rotations, and some very simple algebra. This is a must see for anyone who likes math and education and hard geometry problems with elegant solutions. It is reminiscent of the Pythagorean theorem a^2 + b^2 + c^2 and the binomial formula
anti-Pythagorean theorem: • the anti Pythagorean t...
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[@pwmiles56]

Slightly less elegantly one can demonstrate rotational and translational invariances. Translate by arbitrary d and we have the identity
(a+d)^2 + (b+d)^2 + (c+d)^2 - (a+d)(b+d) - (b+d)(c+d)- (c+d)(a+d)
= a^2 + b^2 + c^2 - ab - bc - ca
= 0
For rotational and scaling invariance multiply each point by arbitrary w. We have the identity
(wa)^2 + (wb)^2 + (wc)^2 - (wa)(wb) - (wb)(wc) - (wc)(wa)
= w^2(a^2 + b^2 + c^2 - ab - bc - ca)
= 0
So we can take the case a = 1, b=1/2 + i sqrt(3)/2, c=1/2 - i sqrt(3)/2
a^2 = a
b^2 = c
c^2 = b
ab = b
bc = a
ca = c
and the result follows

[@synaestheziac]

I think part of the reason some people are getting confused is that hearing the word “Pythagorean” and seeing the squares of a, b, and c might make it seem like those are the side lengths

[@JR13751]

It works for triangle made with 3 roots of unity.
It works for any rotation/dilation about origin. Or multiplying by a constant.
It works for any translation. Or adding a constant to each point.

[@Drbob369]

Compile all these math procedures and processes into a book or website

[@blackpenredpen]

😮

[@General12th]

Hi Dr. Peyam!
Pythagoras's mind would have exploded if he could have seen complex numbers and how they help relate geometry to algebra.

[@misterdubity3073]

Neat! I always laugh at the "marker drop"

[@FromTheMountain]

I think it would have been helpful to state when each of the two cases at the start applies. If I understood correctly, the first case happens when the points a, b, c occur in clockwise order in the triangle, whereas the second case happens when they occur in counterclockwise order. Either way, the transformation you need to perform to get from y to x is the same as the transformation needed to go from z to y, which is why x/y = y/z. I wonder if there is also a purely algebraic approach, starting from simply |a - b| = |a - c| = |b - c| and then working your way to the identity.

[@romanbohac6722]

Very nice ;-)

[@minimath5882]

Pythagoras would be proud!

[@tw5718]

Weird. If it's an equilateral triangle, all angles are 60 deg. a^2 + b^2 + c^2 = 3a^2 = 3b^2 = 3c^2 = .5a^2 -17.5b^2 + 2ab + pi*ac +(18-pi)bc

[@drpeyam]

But a b c are not the length of the sides, just the coordinates of the points

[@tw5718]

Ahh. I rewatched and I originally misunderstood the statement at the beginning about the vertices and using angles. I thought you were saying a b c were the angles. My brain went duh. Thanks for the reply and videos.

[@godfreypigott]

*WRONG.* You introduced a, b, c, x, y, z as complex numbers, and then treated them as real.

[@itssurge7946]

I don’t get it

[@drpeyam]

?

[@TheNetkrot]

@@drpeyam I don't get it either, I cant understand the vector addition you propose in the beginning, it seem odd or incorrect which I doubt coming from you. Thanks for the video anyway.

[@drpeyam]

It’s correct, here a b c are coordinates of the vertices

[@itssurge7946]

What in the world is a complex number

[@drpeyam]

😂

[@sonnypring2151]

Am I the only person who knows 4pi/3 is larger than pi the angle of a straight line. So there is some mistake

[@drpeyam]

No mistake here