[2/2] … lies in the identity:
(1+cos(2θ))/2 = (cos θ)^2
The squared distance is created by ensuring that an angle in the blue parallelogram is *double* that in the red parallelogram!
The originals by @ChocoLinkage are:
https://twitter.com/ChocoLinkage/status/1212285259598794752 …https://twitter.com/ChocoLinkage/status/1222858247566061569 …
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[1/2] If anyone’s still bamboozled, the angle-doubling arises from pairs of non-convex quadrilaterals with pairs of equal sides. ABCD has side lengths 1/4,1/2,1/4,1/2. AFGB has side lengths 1/8,1/4,1/8,1/4. Because ∠ABG = ∠ABC, these quads are similar, and ∠BAF = ∠DAB.pic.twitter.com/eXF1SQaDfX
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[2/2] So ∠DAF=2∠DAB, and because AD=FJ, AF=DJ: ∠HDJ=2∠DAB Put ∠DAB=θ, ∠HDJ=2θ. The red & blue rhombi have sides 1/2, 1/4, so: AK=cos θ DH=cos(2θ)/2 Since AD=1/2, the identity (1+cos(2θ))/2 = (cos θ)^2 ⇒ AD+DH = (AK)^2. Other links impose some symmetries we assumed.
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I like the way the yellow and grey links force the red rhombus to sit symmetrically, with K lying on the horizontal line through AD, and the brown and grey links do the same for the blue rhombus and point H.
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Kraj razgovora
Novi razgovor -
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this is sorcery
Hvala. Twitter će to iskoristiti za poboljšanje vaše vremenske crte. PoništiPoništi
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@paulportesi@nntaleb@normonics@AhsanDeliri@yaneerbaryam Ahh so satisfying to watch!#MathPorn#EngineeringPornHvala. Twitter će to iskoristiti za poboljšanje vaše vremenske crte. PoništiPoništi
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