[1/2] Now that the ingenious @ChocoLinkage has posted linkages that compute both squares and cubes, I thought it would be fun to reverse-engineer the squaring one.
If you want to work through this in detail yourself, note that all lengths are multiples of 1/16.
The magic here…pic.twitter.com/K9xVNUuWSC
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Greg Egan je proslijedio/a tweet korisnika/ce上木 敬士郎/Keishiro Ueki
[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 …Greg Egan je dodan/na,
6 proslijeđenih tweetova 35 korisnika označava da im se sviđaPrikaži ovu nit -
[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.
Čini se da učitavanje traje već neko vrijeme.
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