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

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

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 $\varphi =\pi -\vartheta$ be the supplement of the interior angle $\vartheta ,$ so $\varphi =0$ 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 $z=\exp(i\varphi ),$ we have $E=1+z+z^{2}+z^{3}.$ Now $\exp(i\varphi )=\cos \varphi +i\sin \varphi .$ Using the shorthand notation $z=c+is,$ we find:

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

When $\varphi =\pi ,$ $c=0$ and $s=1,$ so $E=0.$ When $\varphi =0,$ $c=1$ and $s=0,$ so $E=4.$ Halfway, at $\varphi =\pi /2,$ $c=s={\sqrt {1/2}},$ 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)