How to draw a sphere and a rectangular solid? - TeX

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Anonymous 1123

You can try this code
```
\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools
\usepackage{tikz-3dplot-circleofsphere}
\begin{document}
	\pgfdeclarelayer{background} 
	\pgfdeclarelayer{foreground}
	\pgfsetlayers{background,main,foreground}  
	\tikzset{reverseclip/.style={overlay,insert path={(-12383.99999pt,-12383.99999pt) rectangle (12383.99999pt,12383.99999pt)}}}
	\begin{tikzpicture}[3d/install view={phi=70,theta=80},scale=1,line cap=butt,
		line join=round,visible/.style={draw,solid},
		hidden/.style={draw,very thin,cheating dash},declare function={r = 3; h = -r/sqrt(2);}] 
		\path
		(0,0,0) coordinate (O)
		(-r,0,0) coordinate (A)
		(0,r,0)  coordinate (B)
		(r,0,0) coordinate (C)
		(0,-r,0) coordinate (D)
		(-r,0,h) coordinate (A')
		(0,r,h)  coordinate (B')
		(r,0,h) coordinate (C')
		(0,-r,h) coordinate (D')
		;
		\path[3d coordinate ={(O')=0.5*(A') + 0.5*(C')}];
		\path[3d/circumcircle center={A={(A)},B={(B)},C={(C)}}] coordinate (T); 
		\pgfmathsetmacro{\myR}{sqrt(TD("(A)-(B)o(A)-(B)"))/2} ;
		\draw[hidden]  (O') circle[radius=\myR];
		\path  pic{3d/circle on sphere={R=r,C={(O)}, P={(T)}}};
		\draw[3d/screen coords,hidden] (O) circle[radius=r];
		\begin{scope}
			\clip (D) -- (D') -- (C')  -- (C) --cycle [reverseclip];
			\draw[3d/screen coords] (O) circle[radius=r];
		\end{scope}
	
	\begin{scope}[tdplot_screen_coords]
	\clip  (O) circle[radius=r][reverseclip];
	\draw[visible] (B) -- (B');
\end{scope}

\begin{scope}[tdplot_screen_coords]
\clip  (O) circle[radius=r];
	\draw[hidden] (B) -- (B');
\end{scope}
	
	\foreach \p in {A,B,C,D,A',B',C',D',O,O'}
		{\draw[fill=black] (\p) circle (1.2 pt);}
		\foreach \p/\g in {A/90,B/120,C/60,D/60,A'/150,B'/-90,C'/-90,D'/-90,O/60,O'/210}
		{\path (\p)+(\g:3mm) node{$\p$};}
		\draw[hidden] 
		(A) -- (B) -- (C) -- (D) -- cycle
		(D') -- (A') -- (B') (A) -- (A') (A) -- (C) (B) -- (D) (A') -- (C') (B') -- (D');
		\draw[visible]  (C) -- (C') (D) -- (D') (D') -- (C') -- (B');
	\end{tikzpicture}
\end{document} 
```
![ScreenHunter 1105.png](/image?hash=84fd4f0ee43990e8a27726aea5ff18304ce09d856708e4313ef4096cb9402d3d)

``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools \usepackage{tikz-3dplot-circleofsphere} \begin{document} \pgfdeclarelayer{background} \pgfdeclarelayer{foreground} \pgfsetlayers{background,main,foreground} \tikzset{reverseclip/.style={overlay,insert path={(-12383.99999pt,-12383.99999pt) rectangle (12383.99999pt,12383.99999pt)}}} \begin{tikzpicture}[3d/install view={phi=85,theta=80},scale=1,line cap=butt, line join=round,visible/.style={draw,solid}, hidden/.style={draw,very thin,cheating dash},declare function={r = 3; h = -r/sqrt(2);}] \path (0,0,0) coordinate (O) (-r,0,0) coordinate (A) (0,r,0) coordinate (B) (r,0,0) coordinate (C) (0,-r,0) coordinate (D) (-r,0,h) coordinate (A') (0,r,h) coordinate (B') (r,0,h) coordinate (C') (0,-r,h) coordinate (D') ; \path[3d coordinate ={(O')=0.5*(A') + 0.5*(C')}]; \path[3d/circumcircle center={A={(A)},B={(B)},C={(C)}}] coordinate (T); \pgfmathsetmacro{\myR}{sqrt(TD("(A)-(B)o(A)-(B)"))/2} ; \draw[hidden] (O') circle[radius=\myR]; \path pic{3d/circle on sphere={R=r,C={(O)}, P={(T)},fore layer=foreground,back layer=background}}; \draw[3d/screen coords,dashed] (O) circle[radius=r]; \begin{scope} \clip (D) -- (D') -- (B') -- (B) -- (C) -- (D) --cycle [reverseclip]; \draw[3d/screen coords] (O) circle[radius=r]; \end{scope} \foreach \p in {A,B,C,D,A',B',C',D',O,O'} {\draw[fill=black] (\p) circle (1.2 pt);} \foreach \p/\g in {A/90,B/0,C/60,D/180,A'/150,B'/-90,C'/-90,D'/-90,O/60,O'/210} {\path (\p)+(\g:3mm) node{$\p$};} \draw[hidden] (A) -- (B) -- (C) -- (D) -- cycle (D') -- (A') -- (B') (A) -- (A') (A) -- (C) (B) -- (D) (A') -- (C') (B') -- (D'); \draw[visible] (B) -- (B') (C) -- (C') (D) -- (D') (D') -- (C') -- (B'); \end{tikzpicture} \end{document} ```

Please give me an advice about this code. Is it true? ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools \usepackage{tikz-3dplot-circleofsphere} \begin{document} \pgfdeclarelayer{background} \pgfdeclarelayer{foreground} \pgfsetlayers{background,main,foreground} \tikzset{reverseclip/.style={insert path={(current bounding box.south west)rectangle (current bounding box.north east)} }} \begin{tikzpicture}[3d/install view={phi=85,theta=80},scale=1,line cap=butt, line join=round,visible/.style={draw,solid}, hidden/.style={draw,very thin,cheating dash},declare function={r = 3; h = -r/sqrt(2);}] \path (0,0,0) coordinate (O) (-r,0,0) coordinate (A) (0,r,0) coordinate (B) (r,0,0) coordinate (C) (0,-r,0) coordinate (D) (-r,0,h) coordinate (A') (0,r,h) coordinate (B') (r,0,h) coordinate (C') (0,-r,h) coordinate (D') ; \path[3d coordinate ={(O')=0.5*(A') + 0.5*(C')}]; \path[3d/circumcircle center={A={(A)},B={(B)},C={(C)}}] coordinate (T); \pgfmathsetmacro{\myR}{sqrt(TD("(A)-(B)o(A)-(B)"))/2} ; \draw[hidden] (O') circle[radius=\myR]; \path pic{3d/circle on sphere={R=r,C={(O)}, P={(T)},fore layer=foreground,back layer=background}}; \draw[3d/screen coords,dashed] (O) circle[radius=r]; \begin{scope} \clip (D) -- (D') -- (B') -- (B) -- (C) -- (D) --cycle [reverseclip]; \draw[3d/screen coords] (O) circle[radius=r]; \end{scope} \foreach \p in {A,B,C,D,A',B',C',D',O,O'} {\draw[fill=black] (\p) circle (1.2 pt);} \foreach \p/\g in {A/90,B/0,C/60,D/180,A'/150,B'/-90,C'/-90,D'/-90,O/60,O'/210} {\path (\p)+(\g:3mm) node{$\p$};} \draw[hidden] (A) -- (B) -- (C) -- (D) -- cycle (D') -- (A') -- (B') (A) -- (A') (A) -- (C) (B) -- (D) (A') -- (C') (B') -- (D'); \draw[visible] (B) -- (B') (C) -- (C') (D) -- (D') (D') -- (C') -- (B'); \end{tikzpicture} \end{document} ``` ```

Is this better? ``` \documentclass[tikz,border=3mm]{standalone} \usepackage{tikz-3dplot-circleofsphere} \begin{document} \tdplotsetmaincoords{80}{110} \begin{tikzpicture}[tdplot_main_coords,ultra thin, declare function={r=3;r2=r/sqrt(2);}, visible/.style={solid,line width=0.5pt},hidden/.style={densely dashed,thin}, bullet/.style={circle,draw,fill,inner sep=0.3pt,label={[scale=0.5]#1}}] \draw[ball color=gray,opacity=.5,tdplot_screen_coords] (0,0,0) circle (r); \draw[hidden](-r,0,0)node[bullet={above:{$A$}}] (A){} --(0,r,0)node[bullet={above:{$B$}}] (B){} --(r,0,0)node[bullet={above:{$C$}}] (C){} --(0,-r,0)node[bullet={above:{$D$}}] (D){} --(-r,0,0)--(r,0,0) (0,r,0)--(0,-r,0) (0,r,-r/sqrt 2)--(-r,0,-r2)node[bullet={below:{$A'$}}] (A'){} --(0,-r,-r2)node[bullet={below:{$D'$}}] (D'){} (-r,0,-r2)--(r,0,-r2) (0,r,-r2)--(0,-r,-r2) (-r,0,0)--(-r,0,-r2) (0,-r,0)--(0,-r,-.6*r2) ; \draw[visible] (0,r,-r2)node[bullet={below:{$B'$}}] (B'){} --(r,0,-r2)circle(.6pt)node[scale=.5,below]{$C'$}--(0,-r,-r2) (r,0,0)--(r,0,-r2) (0,r,0)--(0,r,-r2) (0,-r,-r2)--(0,-r,-.6*r2); \tdplotCsDrawLatCircle[tdplotCsFront/.style={visible},tdplotCsBack/.style={hidden}]{r}{0} \tdplotCsDrawLatCircle[tdplotCsFront/.style={hidden},tdplotCsBack/.style={hidden}]{r}{-45} \end{tikzpicture} \end{document} ```

The lower circle in your screen shot has a larger radius than in the OP's. But I agree that in the OP's version it should be all dashed. I agree with the second version but the boundary of the sphere is not drawn in the OP's version.

1 Answer