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如何绘制与球体相切的三角形？

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如何绘制与球体相切的三角形？

我正在尝试绘制问题 21 的图形 https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_21

一个球体，中心为 $O$，半径为 6。一个三角形，边长分别为 $15$、$15$ 和 $24$，位于空间中，其每条边都与球体相切。我试过

\documentclass[border=2mm,12pt,tikz]{standalone}
\usepackage{tikz-3dplot}
\begin{document}
    \tdplotsetmaincoords{60}{140}
    \begin{tikzpicture}[tdplot_main_coords,scale=1/2] 
    \path 
    (0,0,0)  coordinate  (O);

    \begin{scope}
    \fill[ball color=gray!10,tdplot_screen_coords] (O)  circle[radius=6]; 
    \end{scope}
    \end{tikzpicture}
\end{document}

我怎样才能画三角形？

答案1

这是使用 pgf 方法从输入计算点的版本。输出远不如minthien_2016 的回答很好，我（在这个答案中）没有区分三角形和圆形的可见部分和隐藏部分。步骤如下：

  1. 计算r内切圆半径；
  2. h计算包含三角形的平面的高度；
  3. 计算内切圆与三角形的切点到三角形角的距离以及三角形的角度；
  4. 添加角。一条边可以与 x 轴平行。因此两个角分别位于 和 处z=h，y=-r并且x距离彼此相切。然后可以根据角度重建第三个角。

结果如下：

\documentclass[border=2mm,12pt,tikz]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{fpu}
\def\pgfmathsetmacroFPU#1#2{\begingroup% 
\pgfkeys{/pgf/fpu,/pgf/fpu/output format=fixed}% 
\pgfmathsetmacro{#1}{#2}% 
\pgfmathsmuggle#1\endgroup}% 
\begin{document}
\begin{tikzpicture}[scale=1/2,declare
    function={a=15;b=15;c=24;R=6;}] 
 \path (0,0,0)  coordinate  (O);
 % compute radius of incircle
 \pgfmathsetmacroFPU{\saux}{(a+b+c)/2}
 \pgfmathsetmacroFPU{\inradius}{sqrt(\saux*(\saux-a)*(\saux-b)*(\saux-c))/\saux}
 % compute height of circle
 \pgfmathsetmacro{\haux}{sqrt(R*R-\inradius*\inradius)}
 % compute angles of triangle
 \pgfmathsetmacro{\alphaaux}{acos((b*b+c*c-a*a)/(2*b*c))}
 \pgfmathsetmacro{\betaaux}{acos((a*a+c*c-b*b)/(2*a*c))}
 \pgfmathsetmacro{\gammaaux}{acos((a*a+b*b-c*c)/(2*a*b))}
 % compute distances from the corners of the triangle to the points where
 % the triangle touches the circle
 \pgfmathsetmacro{\touchA}{(b+c-a)/2}
 \pgfmathsetmacro{\touchB}{(a+c-b)/2}
 \pgfmathsetmacro{\touchC}{(a+b-c)/2}
 \tdplotsetmaincoords{60}{140}
 \begin{scope}[tdplot_main_coords]
  \fill[ball color=gray!10,tdplot_screen_coords] (O)  circle[radius=6]; 
  \path (-\touchA,-\inradius,\haux) coordinate[label=right:$A$](A)
   (\touchB,-\inradius,\haux) coordinate[label=left:$B$](B)
   ({-\touchA+cos(\alphaaux)*b},{-\inradius+sin(\alphaaux)*b},\haux) 
     coordinate[label=right:$C$](C)
   (0,-\inradius,\haux) coordinate (TAB)
   ({-\touchA+cos(\alphaaux)*\touchA},{-\inradius+sin(\alphaaux)*\touchA},\haux)
    coordinate (TAC)
   ({\touchB+cos(180-\betaaux)*\touchB},{-\inradius+sin(180-\betaaux)*\touchB},\haux) 
    coordinate (TBC)
    (0,0,\haux) coordinate (M);
  \draw (A) -- (B) -- (C) -- cycle;
  \begin{scope}[canvas is xy plane at z=\haux]
   \draw[blue,thick] (M) circle[radius=\inradius];
  \end{scope}
  \foreach \X in {A,B,C,M,TAB,TAC,TBC}
  {\fill (\X) circle[radius=3pt];}
 \end{scope}
\end{tikzpicture}
\end{document}

在此处输入图片描述

答案2

经过一些计算，我得到了这个代码

\documentclass[border=2mm,12pt,tikz]{standalone}
\usepackage{tikz-3dplot-circleofsphere}
\usepackage{fouriernc}
\begin{document}
\tdplotsetmaincoords{60}{140}
\begin{tikzpicture}[tdplot_main_coords,scale=1/2,line join = round, line cap = round,declare function={R=6;r=4;h=2*sqrt(5); Angle=acos(r/R);}] 
\path 
 (-3, -4,h) coordinate (A)
(12, -4,h) coordinate (B)
(-36/5, 52/5,h)  coordinate (C)
(0,0,h)  coordinate  (I)
(0,0,0)  coordinate  (O)
(0,-4,h)  coordinate  (H)
(-96/25, -28/25, h)  coordinate  (K)
;
\begin{scope}
  \fill[ball color=gray!10,tdplot_screen_coords] (O)  circle (R); 
   \end{scope}
   \begin{scope}[shift={(O)}]
  \tdplotCsDrawLatCircle[blue, thick]{R}{{Angle}}
  \end{scope}
   \draw[thick] (A) -- (B) -- (C) -- cycle;
  \draw[dashed] (O) -- (I) -- (H) -- cycle (I) -- (K) -- (O);
  \foreach \p in {A,B,C,I,O,H,K}
  \draw[fill=black] (\p) circle (2.5pt);
  \foreach \p/\g in {A/90,B/-90,C/-90,I/90,O/-90,H/90,K/90}
  \path (\p)+(\g:6mm) node{$\p$};
  \end{tikzpicture}
\end{document}

在此处输入图片描述

答案3

在此处输入图片描述

渐近线版本，去除了球体的顶部，适用于一般三角形：

//
// file tri-tan-sphere.asy
// run "asy -f png -render=4 tri-tan-sphere.asy"
// to get tri-tan-sphere.png
//
import graph3; size(200,0);
currentprojection=orthographic(camera=(-20,-25,27),up=(0.2,0.9,1));
triple f(pair t){return (cos(t.y)*cos(t.x),cos(t.y)*sin(t.x),sin(t.y));}
real R=6, a=15, b=20, c=24,
     rho=(a+b+c)/2,
     r=1/4*sqrt(4*a^2*b^2-(a^2+b^2-c^2)^2)/rho,
     alpha=2*aTan(r/(rho-a)),
     h=sqrt(R^2-r^2);
triple O,A,B,C,I,At,Bt,Ct;
A=O; B=(c,0,0); C=(b*Cos(alpha),b*Sin(alpha),0);
I=(a*A+b*B+c*C)/2/rho;
transform3 t=shift(Z*h-I); 
A=t*A; B=t*B; C=t*C; I=t*I;
guide3 gt=A--B--C--cycle;
At=B+(rho-b)/a*(C-B); Bt=C+(rho-c)/b*(A-C); Ct=A+(rho-a)/c*(B-A);
surface s=surface(gt);
draw(s,lightyellow+opacity(0.5));
draw(circle(I,r)^^gt,deepblue);
triple[] P={O,A,B,C,I,At,Bt,Ct,};
string[] L={"O","A","B","C","I","A_t","B_t","C_t",};
triple[] T={Z+Y,A-I,B-I,C-I,Z+Y,I-A,I-B,I-C,};
for(int i=0;i<P.length;++i){dot(P[i]); label("$"+L[i]+"$",P[i],2unit(T[i]));}
surface sc=scale3(R)*surface(f,(0,-pi/2),(2pi,acos(r/R)),Spline);
draw(sc,lightgray,meshpen=nullpen,render(merge=true));

或者，命令使用交互式 3D 矢量 WebGL 图形asy -f html tri-tan-sphere.asy 生成。tri-tan-sphere.html

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