Drawing outward normal vectors to a regular polygon without specifying the coordinates

Question

I would like to replicate the following figure:

And I tried using rotate to circumvent using a specific coordinate together with the predefined anchor for regular polygon, and apparently it doesn't work.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{shapes}
\begin{document}
\begin{tikzpicture}
\node[regular polygon, regular polygon sides=3, draw, minimum size=4cm]
(m) at (0,0) {};

\fill [black] (m.corner 1) circle (3pt);
\draw [black] (m.corner 1) circle (6pt);

\fill [black] (m.corner 2) circle (3pt);
\draw [black] (m.corner 2) circle (6pt);

\fill [black] (m.corner 3) circle (3pt);
\draw [black] (m.corner 3) circle (6pt);

\draw [black, ->, rotate=-90] (m.side 1) -- (m.corner 1);

\end{tikzpicture}
\end{document}

It appears to me that if I named the starting point and ending point for draw using the anchors, then the rotate wouldn't work?

Answer 1

You can use the calc library with the partway syntax, which accepts an optional rotation argument.

The expression

($(A)!0.5!90:(B)$)

specifies the point that lies half the distance between (A) and (B) away from (A) in the direction 90 degrees rotated from the direction between (A) and (B).

In your case, you could draw the normal to side 1 using

\draw [-latex, thick] (m.side 1) -- ($(m.side 1)!0.5!90:(m.corner 1)$); 

\documentclass{article}
\usepackage[margin=1cm]{geometry}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric, calc}

\begin{document}
\begin{tikzpicture}
\node[regular polygon, regular polygon sides=3, draw, minimum size=4cm]
(m) at (0,0) {};

\fill [black] (m.corner 1) circle (3pt);
\draw [black] (m.corner 1) circle (6pt);

\fill [black] (m.corner 2) circle (3pt);
\draw [black] (m.corner 2) circle (6pt);

\fill [black] (m.corner 3) circle (3pt);
\draw [black] (m.corner 3) circle (6pt);

\draw [-latex, thick] (m.side 1) -- ($(m.side 1)!0.5!90:(m.corner 1)$);
\draw [-latex, thick] (m.side 2) -- ($(m.side 2)!0.5!90:(m.corner 2)$);
\draw [-latex, thick] (m.side 3) -- ($(m.side 3)!0.5!90:(m.corner 3)$);

\end{tikzpicture}

\end{document}

Answer 2

I think these arrow tips are not predefined in TikZ. But can be defined from scratch if important.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric}
\begin{document}
\begin{tikzpicture}
\node[regular polygon, regular polygon sides=3, draw, minimum size=4cm,thick] (m) at (0,0) {};

\foreach \x[count=\xi from 0] in {1,2,3}{
\fill [black,thick] (m.corner \x) circle (3pt);
\draw [black,thick] (m.corner \x) circle (6pt);
\draw [black,thick,->] (m.side \x) -- (\xi*120+150:2cm);
};

\end{tikzpicture}
\end{document}

By adding the calc library, one can also use the geometric center with similar syntax given in Jake's answer for rotation-awareness (also works for more sides too).

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric, calc}
\begin{document}
\begin{tikzpicture}[rotate=45,transform shape]
\node[regular polygon, regular polygon sides=3, draw, minimum size=4cm,thick] (m) at (0,0) {};
\foreach \x in {1,2,3}{
\fill [black,thick] (m.corner \x) circle (3pt);
\draw [black,thick] (m.corner \x) circle (6pt);
\draw [black,thick,->] (m.side \x) -- ($(m.side \x)!-1!(m.center)$);
};
\end{tikzpicture}
\end{document}

Answer 3

Solution with tikz-euclide, can of course be improved =)

\documentclass{standalone}
\usepackage{tkz-euclide}
\usetkzobj{all}

\begin{document}

\begin{tikzpicture}[rotate=10]
\def\radius{10} % Defines the size (sidelength of the triangle)
\def\T{0.20}     % Defines the thickness of the figure
\pgfmathsetmacro{\E}{\radius/16} % Macro for drawing the filled circles
\pgfmathsetmacro{\r}{\radius/8}   % Macro for drawing the circles.

\tkzDefPoint(0,0){A}
\tkzDefPoint(\radius,0){B}
\tkzDefPointBy[rotation= center A angle 60](B)
\tkzGetPoint{C}
\tkzCentroid(A,B,C)    \tkzGetPoint{O}
\tkzDefMidPoint(A,B)  \tkzGetPoint{M1}
\tkzDefMidPoint(A,C)  \tkzGetPoint{M2}
\tkzDefMidPoint(B,C)  \tkzGetPoint{M3}
\tkzInterLC(A,O)(O,B) \tkzGetSecondPoint{T1}
\tkzInterLC(B,O)(O,B) \tkzGetSecondPoint{T2}
\tkzInterLC(C,O)(O,C) \tkzGetSecondPoint{T3}

\tkzDrawPolygon[line width=\T cm](A,B,C) 
\tkzDrawCircle[R,line width=\T cm](A,\r cm) \tkzDrawCircle[R, fill=black](A,\E cm)
\tkzDrawCircle[R,line width=\T cm](B,\r cm) \tkzDrawCircle[R, fill=black](B,\E cm)
\tkzDrawCircle[R,line width=\T cm](C,\r cm) \tkzDrawCircle[R, fill=black](C,\E cm)
\tkzDrawSegments[-stealth,line width = \T cm](M3,T1 M2,T2 M1,T3)
\end{tikzpicture}

\end{document}

Answer 4

I don't understand the request : without-specifying-the-coord You need to use coordinates and in the first answers ($(A)!0.5!90:(B)$) or ($(m.side \x)!-1!(m.center)$) uses coordinates. A simple possibility is with some coordinates , without libraries:

\documentclass{standalone}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture} [-latex,ultra thick]
 \draw[fill] (-30:4) circle (3pt) -- (90:4) circle (3pt) -- (210:4) circle (3pt) -- (-30:4);
 \draw (-30:4) circle (6pt)  (90:4) circle (6pt)  (210:4) circle (6pt) ; 
 \draw[->] (30:2)  -- (30:4);
 \draw[->] (150:2) -- (150:4); 
 \draw[->] (270:2) -- (270:4);  
\end{tikzpicture}
\end{document}