The idea of the rooty-helix is very simple. One starts e.g. with the length of 1, adds a right angle with the length of 1 and the hypotenuse equals sqrt{2}. If one continues with sqrt{2} repeating the procedure (adding a right angle with the length of 1) the hypotenuse is sqrt{3}. And so on and so on. At some point due to the iterations some triangles are overpainted. Because of this, one repaints the overpainted triangles
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Edit and compile if you like:
% Rooty helix % Author: Felix Lindemann \documentclass{minimal} \usepackage{tikz} \usetikzlibrary{calc} \begin{document} \pagestyle{empty} \pgfdeclarelayer{background} \pgfdeclarelayer{foreground} \pgfsetlayers{background,main,foreground} \xdefinecolor{darkgreen}{RGB}{175, 193, 36} \newcounter{cntShader} \newcounter{cntRoot} \setcounter{cntShader}{20} \def\couleur{darkgreen} \begin{tikzpicture} \foreach \y in {86,38,15}{ \setcounter{cntShader}{1} \coordinate (a) at (0,0); \coordinate (b) at (0:1); \foreach \x in {1,...,\y}{% \coordinate (c) at ($ (b)!1cm!270:(a) $); \begin{pgfonlayer}{background} \draw[fill=\couleur!\thecntShader] (a)--(b)--(c)--cycle; \end{pgfonlayer} \setcounter{cntRoot}{\x} \addtocounter{cntRoot}{1} \node[fill=white,draw,circle,inner sep=1pt] at (c) {$\sqrt{\thecntRoot}$}; \coordinate (b) at (c); \pgfmathsetcounter{cntShader}{\thecntShader+4} \setcounter{cntShader}{\thecntShader} } } \node[fill=white,draw,circle,inner sep=1pt] at (0:1) {$\sqrt{1}$}; \end{tikzpicture} \end{document}
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