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如何自动绘制位于两个倾斜镜子之间的一点的图像？

tag-icon • tag-icon tikz-pgf pstricks metapost asymptote postscript
如何自动绘制位于两个倾斜镜子之间的一点的图像？

考虑位于两个倾斜镜子之间的一点，X如下所示。

\documentclass[pstricks,border=20pt,12pt]{standalone}
\usepackage{pst-eucl}

\def\Angle{37}% extreme case in which 360/\Angle-1 is not an integer.
\def\R{5}

\begin{document}
\begin{pspicture}[showgrid](-\R,-\R)(\R,\R)
    \pstGeonode[PointSymbol=none,PosAngle={45,-90},PointNameSep=24pt](\R;\Angle){B}(0,0){O}(\R;0){A}
    \psframe*[linecolor=lightgray](O|0,-12pt)(A|0,0)
    \rput{\Angle}(0,0){\psframe*[linecolor=lightgray](O)(B|0,12pt)}
    \psline(B)(O)(A)
    \pstMarkAngle[MarkAngleRadius=1.6,LabelSep=.8]{A}{O}{B}{$\theta$}
    \pstGeonode(3,1){X}
\end{pspicture}
\end{document}

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X我们如何才能自动绘制任意角度的点的图像\Angle？欢迎使用 PSTricks（首选）、Aysmptote、Metapost、TikZ 甚至原始 PostScript 的解决方案。

编辑

哦，我的鬼，现有的答案没有回答，因为你可能误解了我的问题，但我的问题实际上非常非常清楚。

考虑以下的草图\Angle=60。

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注： 的证明360/angle-1仍在进行中这里。

答案1

\documentclass[pstricks,border=20pt,12pt]{standalone}
\usepackage{pst-eucl,pst-plot}
\def\Angle{37}
\def\R{5}

\begin{document}
\begin{pspicture}[showgrid](-\R,-\R)(\R,\R)
    \pstGeonode[PointSymbol=none,PosAngle={45,-90},PointNameSep=24pt](\R;\Angle){B}(0,0){O}(\R;0){A}
    \pcline[linecolor=lightgray,linewidth=12pt,offset=-6pt,strokeopacity=0.6](O)(A)
    \pcline[linecolor=lightgray,linewidth=12pt,offset=6pt,strokeopacity=0.6](O)(B)
    \psline(B)(O)(A)
    \pstMarkAngle[MarkAngleRadius=1.6,LabelSep=.8]{A}{O}{B}{$\theta$}
    \pstGeonode(!3 \Angle\space 2 div PtoC){X}
    \pstOrtSym{O}{A}{X}[I_1] \psline[linestyle=dotted](X)(I_1)
    \newif\ifA \Afalse
    \pstFPDiv\No{360}{\Angle}
    \multido{\iA=1+1,\iB=2+1}{\numexpr\No-1}{%
      \ifA\pstOrtSym{O}{A}{I_\iA}[I_\iB]\else\pstOrtSym{O}{B}{I_\iA}[I_\iB]\fi
      \psline[linestyle=dotted](I_\iA)(I_\iB)
      \ifA\Afalse\else\Atrue\fi
    }%
    \pscircle[linestyle=dotted](O){3}
\end{pspicture}

\end{document}

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并且两个方向：

\documentclass[pstricks,border=20pt,12pt]{standalone}
\usepackage{pst-eucl,pst-plot}
\def\Angle{37}
\def\R{5}

\begin{document}
\begin{pspicture}[showgrid](-\R,-\R)(\R,\R)
    \pstGeonode[PointSymbol=none,PosAngle={45,-90},PointNameSep=24pt](\R;\Angle){B}(0,0){O}(\R;0){A}
    \pcline[linecolor=lightgray,linewidth=12pt,offset=-6pt,strokeopacity=0.6](O)(A)
    \pcline[linecolor=lightgray,linewidth=12pt,offset=6pt,strokeopacity=0.6](O)(B)
    \psline(B)(O)(A)
    \pstMarkAngle[MarkAngleRadius=1.6,LabelSep=.8]{A}{O}{B}{$\theta$}
    \pstGeonode(!3 \Angle\space 2 div PtoC){X}
    \color{blue}\psset{linecolor=blue}%
    \pstOrtSym{O}{A}{X}[I_1] \psline[linestyle=dotted](X)(I_1)
    \newif\ifA \Afalse
    \pstFPDiv\No{360}{\Angle}
    \multido{\iA=1+1,\iB=2+1}{\numexpr\No-1}{%
      \ifA\pstOrtSym{O}{A}{I_\iA}[I_\iB]\else\pstOrtSym{O}{B}{I_\iA}[I_\iB]\fi
      \psline[linestyle=dotted](I_\iA)(I_\iB)
      \ifA\Afalse\else\Atrue\fi
    }%
    \color{red}\psset{linecolor=red}%
    \pstOrtSym{O}{B}{X}[I_1] \psline[linestyle=dotted](X)(I_1)
    \newif\ifA \Atrue
    \pstFPDiv\No{360}{\Angle}
    \multido{\iA=1+1,\iB=2+1}{\numexpr\No-1}{%
      \ifA\pstOrtSym{O}{A}{I_\iA}[I_\iB]\else\pstOrtSym{O}{B}{I_\iA}[I_\iB]\fi
      \psline[linestyle=dotted](I_\iA)(I_\iB)
      \ifA\Afalse\else\Atrue\fi
    }%
    \pscircle[linestyle=dotted](O){3}
\end{pspicture}

\end{document}

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答案2

另一次尝试使用 MetaPost，经过了大幅修改。如果 k-1=floor(360/theta)，它会检查第“k”幅图像是否等于原始点 X，如果是，则在该点旁边打印“成功！”。否则，它会打印“失败！”。可以进行任意次数的尝试，只需修改“theta”参数即可。因此，很容易检查，如果 360/theta 是整数，则图像数量确实是 360/theta - 1。但我仍然不确定这是否是想要的……

input latexmp; 
setupLaTeXMP(options = "12pt",  
    mode = rerun, 
    textextlabel = enable, 
    multicolor=enable,
    packages="SIunits");

u:=1.5cm; % for scaling
% Angle
numeric theta; theta = 30; 
% Supposed number of images
hypothesis := 360/theta-1;
number_of_images := floor(hypothesis);
% mirrors
numeric eps; eps = 0.4*u; % mirrors thickness
pair A, B, X; 
path mirror[]; 
A := (5u, 0); 
mirror1 = origin -- A -- A-(0, eps) -- (0, -eps) -- cycle; 
mirror2 = mirror1 reflectedabout(origin, A) rotated theta;
B:= point 1 of mirror2; 

beginfig(0);
% Grid and labels
drawoptions(dashed withdots);
for i = -5 upto 5:
    draw (-5, i)*u -- (5, i)*u;
    draw (i, -5)*u -- (i, 5)*u;
    label.bot("$" & decimal(i) & "$", (i*u, -5*u));
    label.lft("$" & decimal(i) & "$", (-5u, i*u));
endfor;
% mirrors drawing
drawoptions();
draw mirror1 ; fill mirror1 withcolor 0.8white ; draw origin -- A;
draw mirror2 ; fill mirror2 withcolor 0.8white; draw origin--B;
label.rt("$A$", A); label.rt("$B$", B);
% The angle drawing and its label
draw anglebetween(origin--A, origin--B, "$\theta =" & decimal(theta) & "\degree$");
% X
pair X[]; X0 = (3, 1)*u; freedotlabel("$X$", X0, origin);
draw X0 withpen pencircle scaled 3bp;
% Placement of floor(360/theta-1) images of X 
drawoptions(dashed evenly);
k:=1; 
forever:
    exitunless k <= number_of_images;
    X[k] = X[k-1] reflectedabout(origin,
        if (k = round(k/2)*2): A else: B fi);
    draw X[k-1]--X[k];
    draw X[k] withpen pencircle scaled 3bp;
    freedotlabel("$X_{" & decimal(k) & "}$", X[k], origin);
    k := k+1;
endfor;
% Success or no success?
X[k]= X[k-1] reflectedabout(origin, if (k = round(k/2)*2): A else: B fi);
if (abs(X[k]-X0)< 1e-12):
    draw X[k-1] -- X0;
    freedotlabel("$X$\ \textcolor{red}{Success!}", X0, origin)
else:
    draw X[k-1] -- X[k];
    freedotlabel("$X$", X0, origin);
    freedotlabel("\textcolor{red}{Failed!}", X[k], origin)
fi; 
% The circle
draw fullcircle scaled (2*abs(X1)) dashed withdots;
endfig;
end.

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答案3

我尝试使用 MetaPost，但它是值得的：它有点匆忙完成，而且我不确定我是否正确理解了你想要什么。

编辑：正如赫伯特所做的那样，我在图像所在的位置添加了圆圈。

使用MetaFun格式和浮点算法进行处理：

mpost --mem=metafun --numbersystem=double angles.mp

input latexmp; 
setupLaTeXMP(options = "12pt",  mode = rerun, textextlabel = enable);
u:=1cm; % for scaling
numeric my_angle; 
my_angle = 37;
numeric eps; eps = 0.4*u; % mirror thickness
pair A, B, X; 
path mirror[]; 
A := (5u, 0); 
mirror1 = origin -- A -- A-(0, eps) -- (0, -eps) -- cycle; 
mirror2 = mirror1 reflectedabout(origin, A) rotated my_angle;
B:= point 1 of mirror2; 

beginfig(0);
% Grid and labels
drawoptions(dashed withdots);
for i = -5 upto 5:
    draw (-5, i)*u -- (5, i)*u;
    draw (i, -5)*u -- (i, 5)*u;
    label.bot("$" & decimal(i) & "$", (i*u, -5*u));
    label.lft("$" & decimal(i) & "$", (-5u, i*u));
endfor;
% mirrors drawing
drawoptions();
draw mirror1 ; fill mirror1 withcolor 0.8white ; draw origin -- A;
draw mirror2 ; fill mirror2 withcolor 0.8white; draw origin--B;
% The angle drawing and its label
draw anglebetween(origin--A, origin--B, textext("$\theta$"));
% The images
pair X[]; X0 = (3, 1)*u; label.rt("$X$", X0);
draw X0 withpen pencircle scaled 3bp;
for k= 0 upto 360/my_angle:
    X[k+1] = X[k] reflectedabout(origin,
        if (k = round(k/2)*2): A else: B fi);
    draw X[k]--X[k+1] dashed evenly;
    draw X[k+1] withpen pencircle scaled 3bp;
    label.rt("$X_{" & decimal(k+1) & "}$", X[k+1]);
endfor;
% Circle (as Herbert did)
draw fullcircle xyscaled (abs(X0)*2) withpen pencircle scaled 0.7bp dashed withdots; 
endfig;
end.

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