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Replying to @JDHamkins
Now the obvious exercise is: use a computer algebra system and elimination theory to derive an (implicit) algebraic equation for this curve.
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Replying to @gro_tsen @JDHamkins
It should consist of setting up 2 indeterminates for the coords of each point, one equation for each link fixing the distance between extremities, and eliminate everything except the one point being plotted. But I find too many variables or not enough equations, so I'm confused.
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Replying to @gro_tsen @JDHamkins
I suspect the equation you're missing is the one that constrains these 3 points to be collinear.pic.twitter.com/t8Gowm5bv6
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Er, a better way to put it is that the joint in the center is determined by the the two joints on the ends.
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... and I forgot to point out that there are actually 2 of these gadgets.
pic.twitter.com/L844x6BppE
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Replying to @blockspins @JDHamkins
Right! So now the count is correct. Le me see what Sage makes of it…
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Assuming the two fixed points have coordinates (0,0) and (0,−2), the teardrop's equation is: x⁴ + 2·x²·y² + y⁴ + 10·x² − 6·y² + 8·y − 3 = 0. The sage commands used to perform the computation are here:https://gist.github.com/Gro-Tsen/35c2e3ed4ac8a3b04229f489f5b195df …
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And here is the teardrop's shape computed from this equation:pic.twitter.com/8OU5XadIjT
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That's a much neater equation than I expected! I wonder if there's a nice parametric form using elliptic functions à la https://arxiv.org/abs/1501.07157 (see e.g. Lemma 3.11)
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Replying to @blockspins @JDHamkins
Well, one can compute the curve's genus. I'm a bit too tired to attempt this right now. Another thing would be to understand why the equation given by elimination theory had another component (a circle with center (0,−1) and radius 2): this is probably obvious, but IDC.
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Replying to @gro_tsen @JDHamkins
The other component arises if you "flip" the smallest rhombus, collapsing the blue point to one of the other joints.pic.twitter.com/Jaew3aF7Vo
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