TikZ — 对数螺旋中的黄金三角形

Question

我正在尝试重新绘制此图像（来源) 在 TikZ 中。

下面是我的具有任意底边长度的三角形的代码（只需直接构造）：

\begin{tikzpicture}[x=1mm, y=1mm]

    % Base length
    \def\l{40}      

    \draw[line width=0.1mm] (0, 0) -- (\l, 0) -- (0.5*\l, {0.5*\l*tan(72)}) -- cycle;

    % y = tan(36)x              
    \draw[line width=0.1mm, domain=0:{0.25*(1+sqrt(5))*\l}] plot(\x, {tan(36)*\x});

    % y = tan(144)x - tan(144)*\l
    \draw[line width=0.1mm, domain={0.5*\l}:\l] plot(\x, {tan(144)*\x - tan(144)*\l});

    % y = tan(252)x + (tan(36)*0.25*(1+sqrt(5))*\l - tan(252)*(0.25*(1+sqrt(5))*\l)
    \pgfmathsetmacro{\p}{((tan(36)*0.25*(1+sqrt(5))*\l - tan(252)*(0.25*(1+sqrt(5))*\l) + tan(144)*\l)/(tan(144) - tan(252))}
    \pgfmathsetmacro{\q}{0.25*(1+sqrt(5))*\l}               
    \draw[line width=0.1mm, domain={\p}:{\q}] plot(\x, {tan(252)*\x + tan(36)*0.25*(1+sqrt(5))*\l -  tan(252)*0.25*(1+sqrt(5))*\l});

    % y = tan(36)*0.5*\l
    \pgfmathsetmacro{\p}{(tan(36)*0.5*\l + tan(252)*0.25*(1+sqrt(5))*\l - tan(36)*0.25*(1+sqrt(5))*\l)/(tan(252))}
    \draw[line width=0.1mm, domain={0.5*\l}:{\p}] plot(\x, {tan(36)*0.5*\l});

    % y = tan(108)x + ((tan(144)*(tan(36)*0.25*(1+sqrt(5))*\l - tan(252)*0.25*(1+sqrt(5))*\l + tan(144)*\l)/(tan(144) - tan(252)) - tan(144)*\l) - tan(108)*(tan(36)*0.25*(1+sqrt(5))*\l - tan(252)*0.25*(1+sqrt(5))*\l + tan(144)*\l)/(tan(144) - tan(252)))       
    \pgfmathsetmacro{\m}{tan(108)}
    \pgfmathsetmacro{\b}{((tan(144)*(tan(36)*0.25*(1+sqrt(5))*\l - tan(252)*0.25*(1+sqrt(5))*\l + tan(144)*\l)/(tan(144) - tan(252)) - tan(144)*\l) - tan(108)*(tan(36)*0.25*(1+sqrt(5))*\l - tan(252)*0.25*(1+sqrt(5))*\l + tan(144)*\l)/(tan(144) - tan(252)))}   
    \pgfmathsetmacro{\p}{(tan(36)*0.5*\l - \b)/\m}      
    \pgfmathsetmacro{\q}{(-\b - tan(144)*\l)/(\m - tan(144))}       
    \draw[line width=0.1mm, domain={\p}:{\q}] plot(\x, {\m*\x + \b});
\end{tikzpicture}

“有点”混乱，但确实有效。因此，问题在于将具有正确参数的螺旋放在正确的位置。

我计算了原始图像中两条红线的交点（螺旋的极点）的坐标，基长为 40，并移动了旋转的螺旋：

    \pgfmathsetmacro{\cx}{(-sqrt(0.5*(5+sqrt(5)))*20 - (1/3)*sqrt(5+2*sqrt(5))*40)/(((-1/3)*sqrt(5+2*sqrt(5)) - sqrt(0.5*(5+sqrt(5)))))}
    \pgfmathsetmacro{\cy}{(-1/3)*sqrt(5+2*sqrt(5))*((-sqrt(0.5*(5+sqrt(5)))*20 - (1/3)*sqrt(5+2*sqrt(5))*40)/(((-1/3)*sqrt(5+2*sqrt(5)) - sqrt(0.5*(5+sqrt(5)))))) + (1/3)*sqrt(5+2*sqrt(5))*40}
    \pgfmathsetmacro{\a}{(5/(3*pi))*ln((1+sqrt(5))/2)}
    \begin{scope}[rotate around={-270:(\cx, \cy)}]
        \draw[line cap=round, line width=0.2mm, domain=-3*pi:4*pi, variable=\t, samples=1000]
            plot ({\cx + 2.04666*exp(\a*\t)*cos(deg(\t))}, 
                                            {\cy - 2.04666*exp(\a*\t)*sin(deg(\t))});
    \end{scope}

输出看起来不错，但估计参数 2.04666 让我很烦恼。有什么方法可以计算它吗？有没有更好的方法将正确的螺旋放在正确的位置？

Answer 1

我把螺旋画反了，我想这是错误的做法。

\documentclass[tikz,border=5]{standalone}
\usetikzlibrary{calc,intersections,math}
\begin{document}
\begin{tikzpicture}[line join=round, line cap=round, x=1pt, y=1pt]
\tikzmath{%
  coordinate \a, \b, \c, \o;
  int \i;
  \g = (1 + sqrt(5)) / 2;
  \a = (0,0);
  \b = (100,0);
  for \i in {1,...,6}{
    if (\i > 1) then {
      \a = (\b);
      \b = ($(\b)!1-1/\g!(\c)$);
    };
    \n = veclen(\bx-\ax, \by-\ay) * \g;
    \c = ($(\b)!\g!-72:(\a)$);
    {
      \fill [draw=black!50, fill opacity=1/10] 
        (\a) coordinate (a-\i) --
        (\b) coordinate (b-\i) --
        (\c) coordinate (c-\i) -- cycle;
    };
  };
}

\draw [red, dashed, name path=A] ($(a-1)!0.5!(c-1)$) -- (b-1);
\draw [red, dashed, name path=B] ($(a-2)!0.5!(c-2)$) -- (b-2);
\path [name intersections={of=A and B, by=O}];
\tikzmath{%
  \c1 = (c-1);
  \c2 = (c-2);
  \o = (O);
  \r1 = veclen(\cx1-\ox, \cy1-\oy);
  \r2 = veclen(\cx2-\ox, \cy2-\oy);
  \th1 = Mod(atan2(\cy1-\oy, \cx1-\ox), 360);
  \th2 = Mod(atan2(\cy2-\oy, \cx2-\ox), 360);
  \ph = \th2 > \th1 ? \th2 - \th1 : \th1 - \th2;
  \k = ln(1/\g)/\ph;
  \n = 20 * \ph;
}
\draw [blue!50!cyan, shift=(O)] 
  plot [smooth, domain=0:\n, samples=200, variable=\t]
    (\th1+\t:{\r1*exp(\k*\t)});
\end{tikzpicture}
\end{document}

Answer 2

一个代码高尔夫解决方案;)

\documentclass[tikz,border=7mm]{standalone}
\def~{(36:1)--(0,0)--(1,0)edge[bend left=54,blue](0,0)
      [shift={(1,0)},rotate=108,scale=0.61803398875]}
\begin{document}
  \tikz[scale=4]\draw~~~~~~;
\end{document}

编辑：arc代替的“更好”的近似值edge是：

\documentclass[tikz,border=7mm]{standalone}
\def~{(0,0)edge[black](36:1)edge[black](1,0)arc(-144:-36:0.61803398875)       
      [shift={(1,0)},rotate=108,scale=0.61803398875]}
\begin{document}
  \tikz[blue,scale=3]\draw~~~~~~;
\end{document}

在这种情况下，“精确”螺旋可以用以下方式绘制：

\draw[red,shift={(0.675186453011,0.333224435406)}] 
    plot[domain=-pi:3*pi,samples=300] (\x r:{0.379556295255/exp(0.255290802108*\x)});

如果我们将其叠加，我们就可以看到近似误差。