# swirling hexagon [closed]

Can someone show me how to use tikz to produce a "swirling hexagon" using grey-scale to differentiate triangles at distinct arms of the spiral pattern?
A figure for this pattern can be found googling "swirling hexagon".

• You even want us to google it. That's new. – percusse Jun 20 '16 at 17:14
• -1 from me, this really is a new low. At the very least, add an image of the thing you want us to create from scratch – cmhughes Jun 20 '16 at 17:56
• I find this picture: is this what you want? A well deserved downvote, sorry. – egreg Jun 20 '16 at 17:57
• Anyway, J.M.Stern, Welcome to Tex.SE. You will find that if you post questions here which shows some effort, either into doing a bit a research, or telling us what you have already tried, you will get serious answers. Post and MWE, and a picture which shows what you want us to do. Make it easier for us to help you. – Runar Jun 20 '16 at 18:08
• I sincerely hope that upon closing, this post and the answers it spawned do not go away completely :) That extremely dry humor by percusse, as the first comment, and the slapstick that occurred in the answer that was not accepted, really should be preserved somehow as an example of TeX.SE humor. – A Feldman Jun 22 '16 at 9:31

Is this what you wanted? Please at least include an image of what you want to achieve next time. It's understandable that not everyone knows how to produce a starting code, but make our job easier by providing as much help as you can.

## Code

\documentclass[margin=10pt]{standalone}
\usepackage{tikz}

\usetikzlibrary{calc, shapes}

\tikzset{
box/.style={
regular polygon,
regular polygon sides=6,
minimum size=1cm,
inner sep=0mm,
outer sep=0mm,
rotate=90,
draw
}
}

\begin{document}
\begin{tikzpicture}

\foreach \x [
evaluate=\x as \grad using int(100-(\x*15)),
evaluate=\x as \deg using int(\x*60)
] in {1,...,6}{
\coordinate (n\x) at ({90+60*\x}:5mm);
\begin{scope}[rotate=\deg]
\fill[black!\grad] (0,0) --++ (150:5mm) --++ (210:5mm) --++ (270:1cm) --++ (330:1.5cm) --++ (390:1.5cm) --++ (90:5mm) --++ (210:1.5cm) --++ (150:1cm) --++ (90:5mm) -- cycle;
\end{scope}
\draw (0,0) -- ({90+60*\x}:5mm);
}
\node[box] at (0,0) {};
\end{tikzpicture}
\end{document}

• not swirly enough. the OP wants it swirly, you know... :D ;) (joke aside. I like it.) – naphaneal Jun 20 '16 at 18:23
• @naphaneal :D This is the only "swirling hexagon" that made sense to me. – Alenanno Jun 20 '16 at 18:23
• +1 for even helping the last of us (with the least amount of information given)! – Joel Duscha Jun 20 '16 at 18:29

This question really is too vague. At least attach an image of what you want. Googling the image returns many very different results, and different depending on who is searching. I found the following picture:

So I created this:

\documentclass{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\draw [fill=green] (0,0) -- (2,0) -- (1,1.5) -- cycle;
\draw [fill=green] (3,0) rectangle (4,3);
\draw [fill=green] (5,0) rectangle (6,5);
\draw [fill=green] (7,0) rectangle (8,5);
\end{tikzpicture}
\end{document}


• That's a triangle, not a hexagon. :P – Alenanno Jun 20 '16 at 17:42
• well, google gave me this, and a girl in a sweater, among other things. Should i draw the girl? I'm confused. Better lie down. – Runar Jun 20 '16 at 17:46
• Disappointed. Insufficient swirling. Pls more swirling. – Peter LeFanu Lumsdaine Jun 20 '16 at 18:26
• @PeterLeFanuLumsdaine Some would even settle for "more cowbell." Did I mention, you can google it. – Steven B. Segletes Jun 20 '16 at 18:57
• This is sincerely hilarious. +1 for making me smile! If I could accept it as the answer I would. – A Feldman Jun 21 '16 at 0:24