Testwiki:Reference desk/Archives/Mathematics/2023 July 24

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July 24

Square unfolding function.

Let A,B,C,D & E be points and AB,BC,CD & DE be line segments. A is fixed at (0,0) and B at (1,0). Angle ABC, BCD and CDE are always identical in measure. As the angles increase from theta = pi/2 to pi, E moves from (0,0) to (4,0). Is there a good function *either* for y=f(x) or y=f(theta)?Naraht (talk) 17:46, 24 July 2023 (UTC)

Are the four line segments all the same length (1)? What is x and what is y? Are they the coordinates of E? CodeTalker (talk) 20:32, 24 July 2023 (UTC)

It's not clear what theta is meant to be, but it's not hard to find the curves traced out by C, D and E in parametric form, assuming the segments are all the same length or at least have known lengths. If θ is the angle measured externally, so θ = 0 represents a straight angle, then C is the point (1+cosθ, sinθ) and the curve traced out is a circle. Next, D is the point (1+cosθ+cos2θ, sinθ+sin2θ) and the curve is a Limaçon trisectrix. If CD is not the same length as BC then the result is a general Limaçon. Either way it's a quartic curve and you can find Cartesian and Polar equations in the respective articles. Finally E is the point (1+cosθ+cos2θ+cos3θ, sinθ+sin2θ+sin3θ) and the curve traced out is a sextic known as Freeth's nephroid. We don't have an article on it because notability is dubious, but you can find more information, including Cartesian and polar equations, at mathcurve.com. --RDBury (talk) 22:36, 24 July 2023 (UTC)

Let $\phi =\pi -\vartheta$ be the supplement of the interior angle $\vartheta ,$ so $\phi =0$ means that the points lie on a straight line. We identify the point $(x,y)$ with the complex value $x+iy.$ Assuming all segments have length 1, and putting $z=\exp (i\phi ),$ we have $E=1+z+z^2+z^3.$ Now $\exp (i\phi )=\cos \phi +i\sin \phi .$ Using the shorthand notation $z=c+is,$ we find:

$E=2c(c+1)(2c-1)+i(2sc(2c+1)).$

When $\phi =\pi ,$ $c=0$ and $s=1,$ so $E=0.$ When $\phi =0,$ $c=1$ and $s=0,$ so $E=4.$ Halfway, at $\phi =\pi /2,$ $c=s=\sqrt{1/2},$ so $E=1+i(1+\sqrt{2}).$  --Lambiam 22:44, 24 July 2023 (UTC)

OP here. AB, BC, CD and DE are all of length 1. Angles are *interior* , when at Pi/2, A and E are at the same point. When at Pi, the line is straight.Naraht (talk) 00:32, 25 July 2023 (UTC)