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帕普斯链

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帕普斯链

我画了一个圆圈帕普斯链在 arbelos 上。如何绘制其他圆圈？

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\documentclass[10pt]{article}
\usepackage[a4paper,margin=2cm]{geometry}
\usepackage[x11names]{xcolor}
\usepackage{fouriernc}
\usepackage{tikz}
\usetikzlibrary{calc,intersections}

\begin{document}

\centering

\begin{tikzpicture}[scale=1.5,line join=round,font=\small]
\coordinate[label=below:$A$] (A) at (0,0);
\coordinate[label=below:$B$] (B) at (4,0);
\coordinate[label=below:$C$] (C) at (6,0);
\draw[thick,lightgray](0,0)--(6,0);
\draw[Burlywood4,thick] 
(6,0) arc [radius=3, start angle=0, delta angle=180] -- 
(0,0) arc [radius=2, start angle=180, delta angle=-180]--(4,0) arc [radius=1, start angle=-180, delta angle=-180]--cycle;

\path[name path=b1] 
(6,0) arc [radius=3, start angle=0, delta angle=180];
\path[name path=b2] (0,0) arc [radius=2, start angle=180, delta angle=-180];
\path[name path=b3] (4,0) arc [radius=1, start angle=-180, delta angle=-180];

\draw[olive!50,name path=b4] (4,0) arc (225:60:{sqrt(2)});
\draw[violet!50,name path=b5] (4,4) arc (45:-40:{2*sqrt(2)});
\path [name intersections={of = b4 and b5, by={D}}];
\path [name intersections={of = b2 and b4, by={T,E}}];
\path [name intersections={of = b3 and b5, by={F}}];
\coordinate (K) at ($(E)!.5!(F)$);
\coordinate (L) at ($(D)!.5!(F)$);
\coordinate (M) at ($(K)!1cm!90:(F)$);
\coordinate (N) at ($(L)!1cm!90:(D)$);
\path[name path=g1] (K)--(M);
\path[name path=g2] (L)--(N);
\path [name intersections={of = g1 and g2, by={O}}];
\draw[thick,Red4]
  let 
  \p1=( $(E)-(O) $ )
  in
  (O) circle ({veclen(\x1,\y1)});
\foreach \p in {D,E,F,O}
\draw[fill,DodgerBlue4] (\p) circle(1pt);
%\foreach \p in {D,E,F}
%\node[above] at (\p) {$\p$};
\end{tikzpicture}
\end{document}

答案1

使用链接中提供的坐标：

\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\newcommand\RadiiBig{4}
\newcommand\RadiiSmall{3}
\begin{document}

\begin{tikzpicture}
  \coordinate[label=below:A](A) at (0,0);
  \coordinate[label=below:B](B) at (\RadiiBig*2,0);
  \coordinate[label=below:C](C) at (\RadiiSmall*2,0);
  \pgfmathsetmacro{\r}{\RadiiSmall/\RadiiBig}
  \draw[gray!50]($(A)+(-1,0)$)--($(B)+(1,0)$);
  \draw(\RadiiBig,0) circle (\RadiiBig);
  \draw(\RadiiSmall,0) circle (\RadiiSmall);
  % 
  \draw ($(C)+(\RadiiBig-\RadiiSmall,0)$) circle (\RadiiBig-\RadiiSmall);
  \foreach \n in {1,...,15}{%
    \pgfmathsetmacro{\denom}{\n*\n*(1-\r)*(1-\r)+\r}
    \pgfmathsetmacro{\x}{2*\RadiiBig*\r*(1+\r)/(2*\denom)}
    \pgfmathsetmacro{\y}{2*\RadiiBig*\n*\r*(1-\r)/\denom}
    \pgfmathsetmacro{\Radn}{2*\RadiiBig*\r*(1-\r)/(2*\denom)}
    \draw(\x,\y) circle (\Radn);
    \draw(\x,-\y) circle (\Radn);
  }
\end{tikzpicture}
\end{document}

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编辑

按照 Thruston 的想法，将圆反转应用于 Tikz。使用点 P 通过 AP*AP'=r^2 反转为 P' 的定义，其中 r 是镜像圆，我得到了以下结果。我只做了一个反转点的函数，因此圆的镜像中心必须从中得出。

\begin{tikzpicture}[
  declare function={CircInv(\APp,\r)=\r*\r/\APp;}
  ]
  \coordinate[label=below:A](A) at (0,0);
  \coordinate[label=below:B](B) at (\RadiiBig*2,0);
  \coordinate[label=below:C](C) at (\RadiiSmall*2,0);
  \pgfmathsetmacro{\Rp}{\RadiiBig-\RadiiSmall}%% R'
  \pgfmathsetmacro{\r}{2*\RadiiBig}%% r: mirror circle
  \pgfmathsetmacro{\R}{(CircInv(2*\RadiiSmall,\r)-\r)/2}%% R: Mirror image of first circle
  \coordinate[label=below:$P'$](Pp) at ($(C)+(\Rp,0)$);%% Center first circle
  \coordinate[label=below:$P$](P) at ($(B)+(\R,0)$);%% Center of mirror circle of first circle
  %%
  \draw[gray!50]($(A)+(-1,0)$)--($(B)+(3,0)$);
  \draw(\RadiiBig,0) circle (\RadiiBig);
  \draw(\RadiiSmall,0) circle (\RadiiSmall);
  \draw[red](A) (-20:\r) arc (-20:50:\r);\draw[red](A) -- +(-15:\r) node[pos=0.5,anchor=south]{$r$};
  \draw[blue](Pp) circle (\Rp);\draw[blue] (Pp) -- +(45:\Rp) node[pos=0.5,anchor=south east]{$R'$};
  \draw[gray](P) circle (\R);\draw[gray] (P) -- +(45:\R) node[pos=0.5,anchor=south east]{$R$};
  %%
  \foreach \n in {1,2}{%% First loop showing the column circles
    \pgfmathsetmacro{\alpha}{atan(2*\n*\R/(\r+\R))}
    \pgfmathsetmacro{\X}{2*\n*\R/sin(\alpha)}%% Length from (A) to center of circle
    \pgfmathsetmacro{\Rpn}{(CircInv(\X-\R,\r)-CircInv(\X+\R,\r))/2}%% Size of mirror circle
    \pgfmathsetmacro{\Y}{CircInv(\X+\R,\r)+\Rpn}%% Length (A to center mirror circle
    \draw[gray](A) -- (\alpha: \X);
    \draw[gray](P |- {(0,2*\n*\R)}) circle (\R);
    \draw[blue](A) -- (\alpha: \Y) circle (\Rpn);
  }
  \foreach \n in {3,4,...,30}{%% Same as previous loop but without gray stuff
    \pgfmathsetmacro{\alpha}{atan(2*\n*\R/(\r+\R))}
    \pgfmathsetmacro{\X}{2*\n*\R/sin(\alpha)}
    \pgfmathsetmacro{\Rpn}{(CircInv(\X-\R,\r)-CircInv(\X+\R,\r))/2}
    \pgfmathsetmacro{\Y}{CircInv(\X+\R,\r)+\Rpn}
    \draw[blue](A) (\alpha: \Y) circle (\Rpn);    
  }
\end{tikzpicture}

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答案2

你可以用圆翻转也一样。

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我不确定如何使用 TikZ 进行反转，但这里有一个元帖子版本，包含在 中luamplib。使用 进行编译lualatex。

\RequirePackage{luatex85}
\documentclass[border=5mm]{standalone}
\usepackage{luamplib}
\begin{document}
\mplibtextextlabel{enable}
\begin{mplibcode}
% invert path or pair P in circle C
vardef invert(expr P, C) = 
  save I, r; pair I; numeric r;
  I = center C;
  r = abs(point 0 of C shifted -I); 
  if pair P: if abs(P-I) > 0: unitvector(P-I) scaled (r/abs(P-I)*r) shifted fi I
  elseif path P:
     save T; numeric T;
     T = length P;
     for t=0 upto T-1: invert(point t of P, C) .. endfor if cycle P: cycle else: invert(point T of P, C) fi
  fi
enddef;  
beginfig(1);

    pair A,B,C;
    numeric r; 

    A = origin;
    C = (10cm,0);

    r = 3/4;
    B = r[A,C];

    path c[];
    c1 = fullcircle scaled 2 abs(A-C); % large circle for the inversions
    c2 = fullcircle scaled abs(A-C) shifted 1/2[A,C];
    c3 = fullcircle scaled abs(A-B) shifted 1/2[A,B];
    c4 = fullcircle scaled abs(B-C) shifted 1/2[B,C];
    c5 = invert(c4,c1);

    numeric d; d = abs(point 0 of c5-point 4 of c5);

    for i=1 upto 42:
        draw invert(c5 shifted (0,i*d), c1);
    endfor

    draw subpath(0,4) of c2 withcolor 2/3 blue;
    draw subpath(0,4) of c3 withcolor 2/3 blue;
    draw subpath(0,4) of c4 withcolor 2/3 blue;

    draw A--C;
    dotlabel.bot("$A$", A);
    dotlabel.bot("$B$", B);
    dotlabel.bot("$C$", C);

endfig;
\end{mplibcode}
\end{document}

如果我多画出几个构造部分，这个方法会更容易理解一些：灰色圆圈是c5我的代码；粉色圆弧是的一部分，它是列中的 s 被反转的c1圆圈。c5

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答案3

这是一个简单的解决方案，使用tkz-euclide v5最新版本的tkz-euclide： 看第47.25节手册。

\documentclass[tikz,border=3mm]{standalone}
\usepackage[svgnames]{xcolor}
\usepackage{tkz-euclide}% v5

\pgfmathsetmacro{\xB}{6}
\pgfmathsetmacro{\xC}{9}
\pgfmathsetmacro{\xD}{(\xC*\xC)/\xB}
\pgfmathsetmacro{\xJ}{(\xC+\xD)/2}
\pgfmathsetmacro{\r}{\xD-\xJ}
\pgfmathsetmacro{\nc}{16}
\begin{document}

\begin{tikzpicture}
\tkzDefPoints{0/0/A,\xB/0/B,\xC/0/C,\xD/0/D} 
\tkzDefCircle[diameter](A,C)\tkzGetPoint{x} 
\tkzDrawCircle[fill=teal!30](x,C) 
\tkzDefCircle[diameter](A,B)\tkzGetPoint{y} 
\tkzDrawCircle[fill=teal!30](y,B)
  \foreach \i in {-\nc,...,0,...,\nc}
  {\tkzDefPoint(\xJ,2*\r*\i){J}
   \tkzDefPoint(\xJ,2*\r*\i-\r){H}
   \tkzDefCircleBy[inversion = center A through C](J,H)
   \tkzDrawCircle[fill=teal](tkzFirstPointResult,tkzSecondPointResult)}
\end{tikzpicture}

\end{document}

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