六边形内接三角形

Question

我刚刚开始使用 TIKZ，我需要帮助尝试绘制一个内接于等边三角形的六边形。

Answer 1

一种解决方案（但还有许多其他解决方案）tkz-euclide。它只用于绘制您想要的内容，但如果您想使用一些顶点，那么最好使用\tkzDefPointBy[或使用获取坐标\tkzDefTriangle[equilateral]

\documentclass{standalone} 
\usepackage{tkz-euclide}
\begin{document}       
\begin{tikzpicture}
\pgfmathsetmacro{\c}{6} 
\tkzDefPoints{0/0/A,\c/0/B}
\tkzDefTriangle[equilateral](A,B)\tkzGetPoint{C}
\tkzDefPointBy[homothety=center A ratio 1./3](B) \tkzGetPoint{a}
\tkzDefPointsBy[homothety=center A ratio 2./3](B,C){b,c} 
\foreach \x / \y in {A/B,A/a,C/c,b/B}
{\tkzDrawTriangle[equilateral](\x,\y)}
\end{tikzpicture}    
\end{document}

现在有一个有趣的解决方案

\documentclass{standalone} 
\usepackage{tkz-euclide}
\begin{document} 

 \begin{tikzpicture}
 \pgfmathsetmacro{\c}{6} 
 \tkzDefPoints{0/0/A,\c/0/B}
 \tkzDefTriangle[equilateral](A,B)\tkzGetPoint{C}
 \tkzDefSpcTriangle[medial](A,B,C){a,b,c}
 \tkzDefMidPoint(A,a)\tkzGetPoint{ma}
 \tkzDefMidPoint(B,b)\tkzGetPoint{mb}
 \tkzDefMidPoint(C,c)\tkzGetPoint{mc}
 \tkzInterLL(C,ma)(A,B) \tkzGetPoint{c1}
 \tkzInterLL(C,mb)(A,B) \tkzGetPoint{c2}
 \tkzInterLL(A,mb)(B,C) \tkzGetPoint{a1}
 \tkzInterLL(A,mc)(B,C) \tkzGetPoint{a2}
 \tkzInterLL(B,ma)(A,C) \tkzGetPoint{b1}
 \tkzInterLL(B,mc)(A,C) \tkzGetPoint{b2}
 \tkzDrawPolygon(A,B,C)
 \tkzDrawSegments[dashed](A,a B,b C,c C,c1 C,c2 A,a1 A,a2 B,b1 B,b2)
\tkzDrawPolygon[red,thick](a1,a2,b2,b1,c1,c2)
 \end{tikzpicture}

\end{document}

第三种解决方案

\documentclass{standalone} 
\usepackage{tkz-euclide} % soon with "elements"
\begin{document} 

\begin{tikzpicture}
 \pgfmathsetmacro{\c}{6} 
 \tkzDefPoints{0/0/A,\c/0/B}
 \tkzDefTriangle[equilateral](A,B)\tkzGetPoint{C}
 \tkzDefTriangleCenter[centroid](A,B,C) \tkzGetPoint{I}
 \tkzDefPointBy[homothety=center A ratio 1./3](B) \tkzGetPoint{c1}

 \tkzInterLC(B,C)(I,c1) \tkzGetPoints{a1}{a2}
 \tkzInterLC(A,C)(I,c1) \tkzGetPoints{b1}{b2}
 \tkzInterLC(A,B)(I,c1) \tkzGetPoints{c1}{c2}
 \tkzDrawPolygon(A,B,C)
 \tkzDrawCircle[thin,orange](I,c1)
 \tkzDrawPolygon[red,thick](a2,a1,b2,b1,c2,c1)
\end{tikzpicture}
\end{document}

Answer 2

你只需要shapes.geometric图书馆。圆周率日快乐！

\documentclass[tikz, margin=3mm]{standalone}
\usetikzlibrary{shapes.geometric}
\begin{document}
\begin{tikzpicture}[reg/.style={draw,transform shape,
    regular polygon,regular polygon sides=#1}]
 \path[scale=pi]  (0,0) node[reg=3]{} node[reg=6]{}
 (3/4,0) node[reg=3]{} node[reg=6,rotate=30,scale={cos(30)}]{};
\end{tikzpicture}
\end{document}

Answer 3

只是为了好玩：

\documentclass[tikz,border=2mm]{standalone}
\begin{document}
\begin{tikzpicture}
\foreach\i in {0,120,240}
  \draw[rotate=\i] (-60:1) --++ (-2,0) --++ (60:1) --++ (-60:1);
\end{tikzpicture}
\end{document}

Answer 4

纯粹为了娱乐价值——元帖子，使用luamplib，因此用进行编译lualatex。

\documentclass[border=5mm]{standalone}
\usepackage{luamplib}
\begin{document}
\begin{mplibcode}
beginfig(1);
path t;     t := up; t := for i=0, 1, 2: t rotated 120i -- endfor cycle;
t := t shifted 2 up; t := for i=0, 1, 2: t rotated 120i -- endfor cycle;
draw t scaled 42;
endfig;
\end{mplibcode}
\end{document}