[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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[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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