How to draw a sphere is tangent with every edge of a tetrahedron without using Maple? - TeX

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Top Answer

user 3.14159

Here is a first version. The tetrahedron is constructed similarly to [the post you link to](https://topanswers.xyz/tex?q=1349) but now more constructions have been added to the [`3dtools` library](https://github.com/marmotghost/tikz-3dtools) so this is easier. Then we can construct the incircles of each face, and in the construction of these the touching points are automatically determined. We can then construct the sphere by taking four touching points that do not lie in a plane (which requires a bit trial and error). The intersection of the sphere with the faces can then be drawn with the `3d/circle on sphere` key.

```
\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools
\begin{document}
\pgfdeclarelayer{background} 
\pgfdeclarelayer{foreground}
\pgfsetlayers{background,main,foreground} 
\begin{tikzpicture}[line cap=round,line join=round,
	dot/.style={circle,inner sep=1pt,fill},
	declare function={lAB=5;lAC=7;lBC=6;lAD=5;lBD=4;lCD=6;}]
	% ^^^ the lengths of the tetrahedron	
 \begin{scope}[3d/install view={phi=70,psi=0,theta=70}]
  % construct a triangle of three vertices in the xy plane 
  % using the cosine law	
  \pgfmathsetmacro{\mytheta}{acos(-1*(lBC*lBC-lAC*lAC-lAB*lAB)/(2*lAB*lAC))}	
  \path (0,0,0) coordinate (A)
   (lAB,0,0) coordinate (B)
   ({lAC*cos(\mytheta)},{lAC*sin(\mytheta)},0) coordinate (C);
%  cross checks   
%   \pgfmathsetmacro{\dAB}{tddistance("(A)","(B)")}%
%   \pgfmathsetmacro{\dAC}{tddistance("(A)","(C)")}%
%   \pgfmathsetmacro{\dBC}{tddistance("(B)","(C)")}%
%   \typeout{\dAB,\dBC,\dAC}
  \path[overlay,3d/aux keys/i1=D,3d/aux keys/i2=D',
  	3d/intersection of three spheres={rA=lAD,rB=lBD,rC=lCD}]; 
  \path[pic actions/.append style={draw=none}]
   foreach \X/\Y/\Z in {A/B/C,A/B/D,A/C/D,B/C/D}
   {pic{3d incircle={A={(\X)},B={(\Y)},C={(\Z)},
   center name=i\X\Y\Z,projection prefix=in\X\Y\Z}}};
  \path[3d/circumsphere center={A={(inABCb)},B={(inABDc)},C={(inACDb)},D={(inBCDa)}}]
  coordinate (I) (I) node[dot,label=above:{$I$}]{} ;	
  \pgfmathsetmacro{\myR}{tddistance("(I)","(inABCb)")}%	radius of sphere
  \draw[3d/screen coords] (I) circle[radius=\myR];
  \path[overlay,3d coordinate={(O')=0.25*(A)+0.25*(B)+0.25*(C)+0.05*(D)}];
  \tikzset{3d/polyhedron/.cd,O={(O')},
    fore layer=foreground,back layer=background,
	fore/.append style={fill opacity=0.4,thick}, back/.append style={fill opacity=0.4},
	draw face with corners={{(A)},{(B)},{(C)}},
	draw face with corners={{(A)},{(B)},{(D)}},
 	draw face with corners={{(A)},{(C)},{(D)}},
 	draw face with corners={{(B)},{(C)},{(D)}}}  
  \foreach \X/\Y/\Z in {A/B/C,A/B/D,A/C/D,B/C/D}
  {\path pic{3d/circle on sphere={C={(I)},R=\myR,P={(i\X\Y\Z)}}};}	
 \end{scope}
\end{tikzpicture}
\end{document}
```
![Screen Shot 2021-01-29 at 10.40.27 PM.png](/image?hash=5a23e60e0484629d7f08ed01d05ae800f8505dad705b96fe198b646adaad0e13)

A version that has no colored faces and distinguishes between hidden and visibe stretches:
```
\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{calc,decorations.pathreplacing,3dtools} % https://github.com/marmotghost/tikz-3dtools
\begin{document}
\pgfdeclarelayer{background} 
\pgfdeclarelayer{foreground}
\pgfsetlayers{background,main,foreground} 
\begin{tikzpicture}[line cap=round,line join=round,
	dot/.style={circle,inner sep=1pt,fill},
	declare function={lAB=5;lAC=7;lBC=6;lAD=5;lBD=4;lCD=6;}]
	% ^^^ the lengths of the tetrahedron	
 \begin{scope}[3d/install view={phi=70,psi=0,theta=70}]
  % construct a triangle of three vertices in the xy plane 
  % using the cosine law	
  \pgfmathsetmacro{\mytheta}{acos(-1*(lBC*lBC-lAC*lAC-lAB*lAB)/(2*lAB*lAC))}	
  \path (0,0,0) coordinate (A)
   (lAB,0,0) coordinate (B)
   ({lAC*cos(\mytheta)},{lAC*sin(\mytheta)},0) coordinate (C);
%  cross checks   
%   \pgfmathsetmacro{\dAB}{tddistance("(A)","(B)")}%
%   \pgfmathsetmacro{\dAC}{tddistance("(A)","(C)")}%
%   \pgfmathsetmacro{\dBC}{tddistance("(B)","(C)")}%
%   \typeout{\dAB,\dBC,\dAC}
  \path[overlay,3d/aux keys/i1=D,3d/aux keys/i2=D',
  	3d/intersection of three spheres={rA=lAD,rB=lBD,rC=lCD}]; 
  \path[pic actions/.append style={draw=none}]
   foreach \X/\Y/\Z in {A/B/C,A/B/D,A/C/D,B/C/D}
   {pic{3d incircle={A={(\X)},B={(\Y)},C={(\Z)},
   center name=i\X\Y\Z,projection prefix=in\X\Y\Z}}};
  \path[3d/circumsphere center={A={(inABCb)},B={(inABDc)},C={(inACDb)},D={(inBCDa)}}]
  coordinate (I) (I) node[dot,label=above:{$I$}]{} ;	
  \pgfmathsetmacro{\myR}{tddistance("(I)","(inABCb)")}%	radius of sphere
  \begin{scope}
   \clip[overlay,convex hull of={A,B,C,D},generous outside path'];
   \draw[3d/screen coords,3d/visible] (I) circle[radius=\myR];
  \end{scope}
  \begin{scope}
   \clip[overlay,convex hull of={A,B,C,D}];
   \draw[3d/screen coords,3d/hidden] (I) circle[radius=\myR];
  \end{scope}
  \path[overlay,3d coordinate={(O')=0.25*(A)+0.25*(B)+0.25*(C)+0.05*(D)}];
  \tikzset{3d/polyhedron/.cd,O={(O')},
    fore layer=foreground,back layer=background,
	fore/.append style={fill opacity=0,thick}, back/.append style={fill opacity=0},
	draw face with corners={{(A)},{(B)},{(C)}},
	draw face with corners={{(A)},{(B)},{(D)}},
 	draw face with corners={{(A)},{(C)},{(D)}},
 	draw face with corners={{(B)},{(C)},{(D)}}}  
  \foreach \X/\Y/\Z in {A/B/C,A/B/D,A/C/D,B/C/D}
  {\pgfmathsetmacro{\itest}{TD("(nscreenx,nscreeny,nscreenz)o(i\X\Y\Z)-(I)")>0?1:0}
   \ifnum\itest=1
    \path[%3d/visible/.append style={store path=\X\Y\Z-vis},
		3d/hidden/.append style={postaction={store path=\X\Y\Z-hid}}] 
		pic{3d/circle on sphere={C={(I)},R=\myR,P={(i\X\Y\Z)}}};
	\pgfmathtruncatemacro{\np}{tikztdindexoflastcoordinate}	
	\ifnum\np>0
	 \draw[3d/screen coords,3d/visible] 
	 	[stored path/first coordinate of=\X\Y\Z-hid] coordinate (tmpA)
	 	[stored path/last coordinate of=\X\Y\Z-hid] coordinate (tmpB)
		let \p1=($(tmpA)-(I)$),\p2=($(tmpB)-(I)$),
			\n1={atan2(\y1,\x1)},\n2={atan2(\y2,\x2)},
			\n3={(abs(\n1-\n2)>180?\n2+360:\n2)}
		in (tmpA) arc[start angle=\n1,end angle=\n3,radius=\myR];
	\fi
	\tikzset{stored path/reset coordinate index}
   \else
    \path[3d/visible/.style={3d/hidden}] pic{3d/circle on sphere={C={(I)},R=\myR,P={(i\X\Y\Z)}}};
   \fi}	
 \end{scope}
\end{tikzpicture}
\end{document}
```
![Screen Shot 2021-01-30 at 8.07.43 AM.png](/image?hash=3cfa40fe1d7c03d9dd92b261a11fc9db4e4fa01ed450f58d1fa4574f26c3b57e)

Answer #2

Anonymous 1123

This code borrow some code from marmot. 
```
\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools
\pgfdeclarelayer{background} 
\pgfdeclarelayer{foreground}
\pgfsetlayers{background,main,foreground}  
\tikzset{reverseclip/.style={overlay,insert path={(-12383.99999pt,-12383.99999pt) rectangle (12383.99999pt,12383.99999pt)}}}
\begin{document}
 \begin{tikzpicture}[3d/install view={phi=70,theta=70},scale=1,line cap=butt,
	line join=round,visible/.style={draw,solid},
	hidden/.style={draw,very thin,cheating dash},
	declare function={lAB=5;lAC=7;lBC=6;lAD=5;lBD=4;lCD=6;}]
	% ^^^ the lengths of the tetrahedron	
	\begin{scope}[3d/install view={phi=70,psi=0,theta=70}]
		% construct a triangle of three vertices in the xy plane 
		% using the cosine law	
		\pgfmathsetmacro{\mytheta}{acos(-1*(lBC*lBC-lAC*lAC-lAB*lAB)/(2*lAB*lAC))}	
		\path (0,0,0) coordinate (A)
		(lAB,0,0) coordinate (B)
		({lAC*cos(\mytheta)},{lAC*sin(\mytheta)},0) coordinate (C);
		%  cross checks   
		%   \pgfmathsetmacro{\dAB}{tddistance("(A)","(B)")}%
		%   \pgfmathsetmacro{\dAC}{tddistance("(A)","(C)")}%
		%   \pgfmathsetmacro{\dBC}{tddistance("(B)","(C)")}%
		%   \typeout{\dAB,\dBC,\dAC}
		\path[overlay,3d/aux keys/i1=D,3d/aux keys/i2=D',
		3d/intersection of three spheres={rA=lAD,rB=lBD,rC=lCD}]; 
		\path pic[red,hidden]{3d incircle={% 
				A={(A)},B={(B)},C={(C)}, center name = X}};
		
		\path pic[orange,visible]{3d incircle={% 
				A={(A)},B={(B)},C={(D)}, center name = Y}};
		\path pic[blue,hidden]{3d incircle={% 
				A={(A)},B={(D)},C={(C)}}};	
		\path pic[orange,visible]{3d incircle={% 
				A={(C)},B={(B)},C={(D)}}};
		\path [overlay, 
		3d coordinate={(myn1)=(A)-(B)x(A)-(C)},
		3d coordinate={(myn2)=(D)-(B)x(D)-(A)}];
		\path[3d/line with direction={(myn1) through (X) named d1},
		3d/line with direction={(myn2) through (Y) named d2}];	
		\path[3d/intersection of={d1 with d2}] coordinate (Z);
		
		\path[3d/line through={(A) and (B) named lAB}];
		\path[3d/project={(Z) on lAB}] coordinate (W);	
		\pgfmathsetmacro{\myR}{sqrt(TD("(Z)-(W)o(Z)-(W)"))}
		%\shade[ball color=white,3d/screen coords] (Z) circle[radius=\myR];
	\draw[3d/screen coords] (Z) circle[radius=\myR];
\begin{scope}
			\clip[3d/screen coords]  (Z) circle[radius=\myR][reverseclip];
			\draw[visible] (D) -- (C);
		\end{scope}
		
		\begin{scope}
			\clip[3d/screen coords]  (Z) circle[radius=\myR];
			\draw[hidden] (D) -- (C);
		\end{scope}
		
		\draw[hidden] 
		(A) -- (C)	(X) -- (Z) -- (Y);
		\draw [visible] (D) -- (A) -- (B) -- (C)  (D) -- (B);

		\foreach \p in {A,B,C,D,Z,X,Y}
		{\draw[fill=black] (\p) circle (1.2 pt);}
		\foreach \p/\g in {A/-90,B/-90,C/-90,D/90,Z/90,X/-90,Y/90}
		{\path (\p)+(\g:3mm) node{$\p$};}
		 
	\end{scope}
\end{tikzpicture}

\end{document}  
```
![ScreenHunter 1124.png](/image?hash=14993c858b657cc0d60a87ef776b0de91134c6fcea302e0a49460efd877965a0)

You need to newest update of `3dtools` to be able to compile this. ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{3dtools}%https://github.com/marmotghost/tikz-3dtools \begin{document} \begin{tikzpicture} \begin{scope}[3d/install view={phi=110,psi=0,theta=60}] \pgfmathsetmacro{\myr}{sqrt(1+tan(30)*tan(30))} \pgfmathsetmacro{\myh}{2*sin(acos(\myr/2))} \path (-1,{-tan(30)},1) coordinate (C1) (1,{-tan(30)},1) coordinate (C2) (0,{tan(60)-tan(30)},1) coordinate (C3) (0,0,{1+\myh}) coordinate (C4); \let\mylistd\empty \tikzset{add screen depth/.code={% \pgfmathsetmacro{\mycoor}{TD("(#1)")}% \pgfmathsetmacro{\mysd}{screendepth(\mycoor)}% \ifx\mylistd\empty \edef\mylistd{\mysd}% \else \edef\mylistd{\mylistd,\mysd}% \fi}} \tikzset{add screen depth/.list={C1,C2,C3,C4}} \pgfkeys{/my lists/.cd, my initial array/.is array={\mylistd}, my values/.initial=\pgfkeysvalueof{/my lists/my initial array/content},% my values/.sort numeric list={\temp}{\templ},% sort yields sorted list and index my sorted array/.is array/.expanded={\temp}, my index machinery/.is array/.expanded={\templ}}% \tikzset{ball 1/.style={ball color=red},ball 2/.style={ball color=blue}, ball 3/.style={ball color=green!70!black},ball 4/.style={ball color=orange}} \foreach \X in \templ { \shade[3d/screen coords,ball \X] (C\X) circle[radius=1];} \end{scope} \end{tikzpicture} \end{document} ```

Well, this depends on what you want to draw. Maybe something like this? ![Screen Shot 2021-03-09 at 10.29.02 AM.png](/image?hash=b3153bcf9810671eb4217b9f7d4dc7f6d438db50cb762db0dc5aefad3f279d32) Notice that, as usual, if the view is to be free, the code is longer than what fits into comments.

As you probably wanted to imply, the answer is not entirely accurate. One possible way to parametrize the curves is using `cosh` and `sinh` (since `cosh(x)^2-sinh(x)^2=1`). Note that the `perspective` library uses slightly different conventions for the view angles, so if you want to use `tikz-3dplot` or `3dtools`, you need to replace `20` by `70` to get the same view. ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{calc,perspective} \begin{document} \begin{tikzpicture}[scale=3,variable=\t,fill opacity=0.5, >=stealth,semithick] \path (1,0) coordinate (ex) (0,1) coordinate (ey); \begin{scope}[3d view={110}{20}] \path (0,1,0) coordinate (ey'); \path let \p1=(ey'),\n1={atan2(\y1,\x1)} in [left color=teal!50,right color=teal,middle color=white, shading angle=\n1] plot[domain=-1:1] ({-cosh(\t)},1,{sinh(\t)}) -- plot[domain=1:-1] ({-cosh(\t)},-1,{sinh(\t)}) -- cycle; \path[x={(ex)},y={(ey)},ball color=orange] (0,0,0) circle[radius=1]; \path let \p1=(ey'),\n1={atan2(\y1,\x1)} in [left color=teal!50,right color=teal,middle color=white, shading angle=\n1] plot[domain=-1:1] ({cosh(\t)},1,{sinh(\t)}) -- plot[domain=1:-1] ({cosh(\t)},-1,{sinh(\t)}) -- cycle; \draw[->] (1,0,0) -- (1.5,0,0) node[pos=1.2]{$x$}; \draw[->] (0,1,0) -- (0,1.5,0) node[pos=1.2]{$y$}; \draw[->] (0,0,1) -- (0,0,1.5) node[pos=1.2]{$z$}; \end{scope} \end{tikzpicture} \end{document} ``` ![Screen Shot 2021-03-07 at 1.47.15 PM.png](/image?hash=d125f675e232063cac373f8eda9d616d6755c07a7e6b50df01f34cc91cf54db9)

I do not know what the correct result is. Maybe this one? ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{3dtools} % https://github.com/marmotghost/tikz-3dtools \begin{document} \begin{tikzpicture}[scale=1,line cap=butt, line join=round,declare function={a=5;R=1/2*a;h=a; angB =-30;r=a/2;}] \begin{scope}[3d/install view={phi=60,psi=0,theta=70}] \path (0,0,0) coordinate (A) (0,a,0) coordinate (C) ({ R*cos(angB)}, {a/2 + R*sin(angB)},0) coordinate (B) (0,0,h) coordinate (S); \path[3d/line through={(S) and (B) named lSB}]; \path[3d/line through={(S) and (C) named lSC}]; \path[3d/project={(A) on lSB}] coordinate (H); \path[3d/project={(A) on lSC}] coordinate (K); \path[3d/circumsphere center={A={(A)},B={(B)},C={(C)},D={(H)}}] coordinate (O); \pgfmathsetmacro{\myR}{sqrt(TD("(O)-(A)o(O)-(A)"))} ; % \path pic{3d/circle on sphere={R=\myR,C={(O)}, P={(O)}}}; \path[3d/circumcircle center={A={(H)},B={(B)},C={(C)}}] coordinate (J); \path pic{3d/circle on sphere={R=\myR,C={(O)}, P={(J)}}}; % \path[3d/circumcircle center={A={(A)},B={(H)},C={(K)}}] coordinate (X); \begin{scope} \clip[overlay,even odd clip,convex hull of={A,H,K,S},generous outside path']; \path pic{3d/circle on sphere={R=\myR,C={(O)}, P={(X)}}}; \end{scope} \begin{scope} \clip[overlay,convex hull of={A,H,K,S}]; \path[3d/visible/.style={3d/hidden}] pic{3d/circle on sphere={R=\myR,C={(O)}, P={(X)}}}; \end{scope} % \path[3d/screen coords] (O) circle[radius=\myR]; \begin{scope} \clip[overlay,even odd clip,convex hull of={S,A,B,C},generous outside path']; \draw[3d/screen coords,3d/visible] (O) circle[radius=\myR]; \end{scope} \begin{scope} \clip[overlay,convex hull of={S,A,B,C}]; \draw[3d/screen coords,3d/hidden] (O) circle[radius=\myR]; \end{scope} \foreach \p in {A,B,C,O,S,H,K} {\draw[fill=black] (\p) circle (1.2 pt);} \foreach \p/\g in {A/-90,B/-90,C/-90,O/90,H/90,K/90,S/90} {\path (\p)+(\g:3mm) node{$\p$};} % \draw[3d/visible] (S) --(A) (S) -- (H) (S)--(K) ; \draw[3d/hidden] (A) -- (B) (A) -- (C) (B) -- (H) (C) -- (K) (B) -- (C) (A) -- (H) (A) -- (K) (H) -- (K); \end{scope} \end{tikzpicture} \end{document} ```

As far as I can see the circle around `O`, i.e. the boundary of the sphere, is clipped correctly. You can also use ``` \begin{scope} \clip[overlay,even odd clip,convex hull of={S,A,B,C},generous outside path']; \draw[3d/screen coords,3d/visible] (O) circle[radius=\myR]; \end{scope} ``` but this yields the same result. The problem seems to be that other circles are not clipped correctly. You could clip them in the same way as above.

@marmot, re: [your answer](#a1868), I used ``` \begin{scope} \clip[overlay,convex hull of={S,A,B,C},generous outside path']; \draw[3d/screen coords,3d/visible] (O) circle[radius=\myR]; \end{scope} \begin{scope} \clip[overlay,convex hull of={S,A,B,C}]; \draw[3d/screen coords,3d/hidden] (O) circle[radius=\myR]; \end{scope} ``` in full code ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{3dtools} \usetikzlibrary{calc,decorations.pathreplacing,3dtools} % \pgfdeclarelayer{background} \pgfdeclarelayer{foreground} \pgfsetlayers{background,main,foreground} \tikzset{reverseclip/.style={overlay,insert path={(-12383.99999pt,-12383.99999pt) rectangle (12383.99999pt,12383.99999pt)}}} \begin{document} \begin{tikzpicture}[scale=1,line cap=butt, line join=round,declare function={a=5;R=1/2*a;h=a; angB =-30;r=a/2;}] \begin{scope}[3d/install view={phi=60,psi=0,theta=70}] \path (0,0,0) coordinate (A) (0,a,0) coordinate (C) ({ R*cos(angB)}, {a/2 + R*sin(angB)},0) coordinate (B) (0,0,h) coordinate (S); \path[3d/line through={(S) and (B) named lSB}]; \path[3d/line through={(S) and (C) named lSC}]; \path[3d/project={(A) on lSB}] coordinate (H); \path[3d/project={(A) on lSC}] coordinate (K); \path[3d/circumsphere center={A={(A)},B={(B)},C={(C)},D={(H)}}] coordinate (O); \pgfmathsetmacro{\myR}{sqrt(TD("(O)-(A)o(O)-(A)"))} ; \path[3d/circumcircle center={A={(A)},B={(B)},C={(C)}}]; \path pic{3d/circle on sphere={R=\myR,C={(O)}, P={(O)}}}; \path[3d/circumcircle center={A={(H)},B={(B)},C={(C)}}] coordinate (J); \path pic{3d/circle on sphere={R=\myR,C={(O)}, P={(J)}}}; \path[3d/circumcircle center={A={(A)},B={(H)},C={(K)}}] coordinate (X); \path[overlay] pic{3d/circle on sphere={R=\myR,C={(O)}, P={(X)}}}; \begin{scope} \clip[overlay,convex hull of={S,A,B,C},generous outside path']; \draw[3d/screen coords,3d/visible] (O) circle[radius=\myR]; \end{scope} \begin{scope} \clip[overlay,convex hull of={S,A,B,C}]; \draw[3d/screen coords,3d/hidden] (O) circle[radius=\myR]; \end{scope} \foreach \p in {A,B,C,O,S,H,K} {\draw[fill=black] (\p) circle (1.2 pt);} \foreach \p/\g in {A/-90,B/-90,C/-90,O/90,H/90,K/90,S/90} {\path (\p)+(\g:3mm) node{$\p$};} \draw[3d/visible] (S) --(A) (S) -- (H) (S)--(K) ; \draw[3d/hidden] (A) -- (B) (A) -- (C) (B) -- (H) (C) -- (K) (B) -- (C) (A) -- (H) (A) -- (K) (H) -- (K); \end{scope} \end{tikzpicture} \end{document} ```

This code get correct result ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools \begin{document} \pgfdeclarelayer{background} \pgfdeclarelayer{foreground} \pgfsetlayers{background,main,foreground} \begin{tikzpicture}[3d/install view={phi=70,theta=70},scale=1,line cap=butt, line join=round,visible/.style={draw,solid}, hidden/.style={draw,very thin,cheating dash},declare function={R=3;a=R;}] \path (a,a,a) coordinate (A) (a,-a,-a) coordinate (B) (-a,a,-a) coordinate (C) (-a,-a,a) coordinate (D) (0,0,0) coordinate (O) ; \path pic[red,hidden]{3d incircle={% A={(A)},B={(B)},C={(C)}, center name = X}}; \path pic[orange,visible]{3d incircle={% A={(C)},B={(B)},C={(D)}, center name = Y}}; \path pic[orange,visible]{3d incircle={% A={(A)},B={(D)},C={(C)}}}; \path pic[blue,hidden]{3d incircle={% A={(A)},B={(B)},C={(D)}}}; \path [overlay, 3d coordinate={(myn1)=(A)-(B)x(A)-(C)}, 3d coordinate={(myn2)=(D)-(B)x(D)-(C)}]; \path[3d/line with direction={(myn1) through (X) named d1}, 3d/line with direction={(myn2) through (Y) named d2}]; \path[3d/intersection of={d1 with d2}] coordinate (Z); \path[3d/line through={(A) and (B) named lAB}]; \path[3d/project={(Z) on lAB}] coordinate (W); \pgfmathsetmacro{\myR}{sqrt(TD("(Z)-(W)o(Z)-(W)"))} %\shade[ball color=white,3d/screen coords] (Z) circle[radius=\myR]; \draw[3d/screen coords] (Z) circle[radius=\myR]; \foreach \p in {A,B,C,D,Z,X,Y} {\draw[fill=black] (\p) circle (1.2 pt);} \foreach \p/\g in {A/-90,B/-90,C/-90,D/90,Z/0,X/0,Y/0} {\path (\p)+(\g:3mm) node{$\p$};} \draw[hidden] (A) -- (B) (X) -- (Z) -- (Y); \draw [visible] (D) -- (A) -- (C) -- (B) -- cycle (D) -- (C); \end{tikzpicture} \end{document} ```

To determine center of sphere. 1) First, you find the line passing throught incircle of the triangle `ABC` and pepenicular to the plane `(ABC)`, named the line is `l1`; 2) second, you find the line passing throught incircle of the triangle `DAC` and pepenicular to the plane `(DAC)`, named the line is `l2`; 3) Center of sphere is intersection of two the lines `l1` and `l2`.

In this code, you try rename `I` by `O`. The point `I` incorrect, `O` is correct. ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools \begin{document} \pgfdeclarelayer{background} \pgfdeclarelayer{foreground} \pgfsetlayers{background,main,foreground} \begin{tikzpicture}[3d/install view={phi=70,theta=70},scale=1,line cap=butt, line join=round,visible/.style={draw,solid}, hidden/.style={draw,very thin,cheating dash},declare function={a=3;R=a;}] \path (a,a,a) coordinate (A) (a,-a,-a) coordinate (B) (-a,a,-a) coordinate (C) (-a,-a,a) coordinate (D) (0,0,0) coordinate (I) ; \shade[ball color=white,3d/screen coords] (I) circle[radius=R]; \path pic[red,hidden]{3d incircle={% A={(A)},B={(B)},C={(C)}}}; \path pic[blue,hidden]{3d incircle={% A={(A)},B={(B)},C={(D)}}}; \path pic[orange,visible]{3d incircle={% A={(A)},B={(D)},C={(C)}}}; \path pic[orange,visible]{3d incircle={% A={(D)},B={(B)},C={(C)}}}; \foreach \p in {A,B,C,D,I} {\draw[fill=black] (\p) circle (1.2 pt);} \foreach \p/\g in {A/-90,B/-90,C/-90,D/90,I/90} {\path (\p)+(\g:3mm) node{$\p$};} \draw[hidden] (A) -- (B) ; \draw [visible] (D) -- (A) -- (C) -- (B) -- cycle (D) -- (C); \end{tikzpicture} \end{document} ```

``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools \begin{document} \pgfdeclarelayer{background} \pgfdeclarelayer{foreground} \pgfsetlayers{background,main,foreground} \begin{tikzpicture}[ dot/.style={circle,inner sep=1pt,fill}, declare function={l1=5;l2=6;l3=7;l4=5;l5=5;l6=6;}] % ^^^ the lengths of the tetrahedron \begin{scope}[3d/install view={phi=70,psi=0,theta=70}] % construct a triangle of three vertices in the xy plane % using the cosine law \pgfmathsetmacro{\mytheta}{acos((l3*l3-l1*l1-l2*l2)/(2*l1*l2))} \path (0,0,0) coordinate (A) (l1,0,0) coordinate (B) ({l2*cos(\mytheta)},{l2*sin(\mytheta)},0) coordinate (C); \path[overlay,3d/aux keys/i1=D,3d/aux keys/i2=D',3d/intersection of three spheres= {rA=l4,rB=l5,rC=l6}]; \begin{scope}[pic actions/.append style={dashed}] \path pic{3d incircle={A={(A)},B={(B)},C={(C)},center name=iABC,projection prefix=inABC}}; \path pic{3d incircle={A={(A)},B={(C)},C={(D)},center name=iACD,projection prefix=inACD}}; \path pic{3d incircle={A={(A)},B={(B)},C={(D)},center name=iABD,projection prefix=inABD}}; \path pic{3d incircle={A={(B)},B={(C)},C={(D)},center name=iBCD,projection prefix=inBCD}}; \end{scope} \path[3d/circumsphere center={A={(inABCa)},B={(inABDc)},C={(inACDa)},D={(inBCDa)}}] coordinate (I) (I) node[circle,inner sep=1.2pt,fill,label=above:{$I$}]{} ; \path[overlay,3d coordinate={(O')=0.25*(A)+0.25*(B)+0.25*(C)+0.05*(D)}]; \tikzset{3d/polyhedron/.cd,O={(O')}, fore layer=foreground,back layer=background, fore/.append style={fill opacity=0.4,thick}, back/.append style={fill opacity=0.4}, draw face with corners={{(A)},{(B)},{(C)}}, draw face with corners={{(A)},{(B)},{(D)}}, draw face with corners={{(A)},{(C)},{(D)}}, draw face with corners={{(B)},{(C)},{(D)}}} \end{scope} \end{tikzpicture} \end{document} ``` ![Screen Shot 2021-01-29 at 7.50.25 PM.png](/image?hash=08fb3513113af77448c749325f6754a30ffa3fd10621491a380361c8b3a94e7a)

@Anonymous, re: [your question](#question), I played a little bit and have a similar problem to what you seem to encounter, i.e. `I` is off. To get going, I guessed the sixth length from the prescription that "the sums of the lengths of pairs of opposite edges are equal;" and got `l6=6;`, which is a bit different from what you have for the distance between `C` and `D`. And `I` is even worse. I will post the code in a separate comment but am wondering if you are 100% sure that the statements in the linked paper are fully accurate.

2 Answers