Draw three circles tangent to the orther two, and tangent to each side of a triangle. - TeX

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user 3.14159

There are many possibilities. Here we let Ti*k*Z find the corners. We know that the inner triangle has edge length `2*r`, and the outer ones are the intersections of the tangents to two circles each.
```
\documentclass[tikz,border=2mm]{standalone}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[declare function={r=1;},
	c/.style={circle,draw,inner sep=0pt,minimum size=r*2cm}]
	\draw foreach \X in {1,2,3}
	 {++ (-120+120*\X:2) node[c](c\X){}} 
	 (c1.center)-- (c2.center) -- (c3.center) -- cycle
	 (intersection of c1.30--c2.30 and c2.150--c3.150) --
	 (intersection of c2.150--c3.150 and c3.-90--c1.-90) --
	 (intersection of c1.30--c2.30 and c3.-90--c1.-90)
	 -- cycle;
\end{tikzpicture}
\end{document}
```
![Screen Shot 2021-03-02 at 7.48.50 PM.png](/image?hash=f6d3749c0db6019476b581bb5c514ccc274b268e5feab4d4bd578eb6758dff12)

One can also just compute the relevant coordinates.
```
\documentclass[tikz,border=2mm]{standalone}
\begin{document}
\begin{tikzpicture}[declare function={r=1;},
	pics/tricircl/.style={code={%
	 \path foreach \YY in {1,...,#1}
	  {foreach \XX in {1,...,\YY}
	  {({2*r*(\XX-\YY/2-1/2)},{-2*\YY*r*sin(60)}) node[c](c-\XX-\YY){}}}; 
	 \draw (c-1-1.center) -- (c-#1-#1.center) -- (c-1-#1.center) -- cycle
 	  ({2*r*(1/2-#1/2)-r/tan(30)},{-2*r*#1*sin(60)-r}) --
	  ({2*r*(#1/2-1/2)+r/tan(30)},{-2*r*#1*sin(60)-r}) --
	  (0,{-2*r*sin(60)+r/sin(30)}) --cycle;
	}},pics/tricircl/.default=2,
	c/.style={circle,draw,inner sep=0pt,minimum size=r*2cm}]
	\path pic{tricircl=2} (8,0) pic{tricircl=3}
	 (4,-8) pic{tricircl=4};
\end{tikzpicture}
\end{document}
```
![Screen Shot 2021-03-02 at 9.01.27 PM.png](/image?hash=01415797456176f986f9311f2f34b8858bc89d6fc5b8a5163a5928d0968bfc67)

We have to multiply the outer radius by `cos(360/(2*n)`, where `n` is the number of corners. ``` \documentclass[tikz]{standalone} \usetikzlibrary{shapes.geometric,calc} \begin{document} \begin{tikzpicture}[c/.style={circle,fill,inner sep=1pt}] \node[regular polygon,regular polygon sides=6,draw,minimum height=5cm] (a) at (0,0) {}; \draw[red] let \p1=($(a.corner 1)-(a.center)$), \n1={veclen(\x1,\y1)} in circle[radius=\n1]; \foreach \x[count=\xi] in {A,B,...,F}{ \node (a-\xi) at ([shift={({90+(\xi-1)*360/6}:3mm)}]a.corner \xi) {\x}; } \draw[blue] let \p1=($(a.corner 1)-(a.center)$), \n1={veclen(\x1,\y1)*cos(360/6/2)} in circle[radius=\n1]; \path foreach \p/\g in {a/-90} {(\p)node[c]{}+(\g:3mm) node{$\p$}}; \end{tikzpicture} \end{document} ``` ![Screen Shot 2021-03-30 at 7.44.52 PM.png](/image?hash=232682be52a15f95fc855dd489cbe696a826df42136cf28e101ef1dbd821147c)

@marmot, re: [your answer](#a1909), How to draw incircle of this polygon? ``` \documentclass[tikz]{standalone} \usetikzlibrary{shapes.geometric,calc} \begin{document} \begin{tikzpicture}[c/.style={circle,fill,inner sep=1pt}] \node[regular polygon,regular polygon sides=6,draw,minimum height=5cm] (a) at (0,0) {}; \draw[red] let \p1=($(a.corner 1)-(a.center)$), \n1={veclen(\x1,\y1)} in circle (\n1); \foreach \x[count=\xi] in {A,B,...,F}{ \node (a-\xi) at ([shift={({90+(\xi-1)*360/6}:3mm)}]a.corner \xi) {\x}; } \path foreach \p/\g in {a/-90} {(\p)node[c]{}+(\g:3mm) node{$\p$}}; \end{tikzpicture} \end{document} ```

This is not so easy, see https://en.wikipedia.org/wiki/Circle_packing_in_a_circle. That is, the general case is not known. For this case you could do ``` \documentclass[tikz,border=3mm]{standalone} \begin{document} \begin{tikzpicture}[declare function={r=1;}, c/.style={circle,draw,inner sep=0pt,minimum size=r*2cm}] \draw[rotate=45] foreach \X in {-1,1} {foreach \Y in {-1,1} {(\X,\Y) node[c]{}}} (0,0) circle[radius={r+sqrt(2)*r}]; \end{tikzpicture} \end{document} ```

@marmot, re: [your answer](#a1909), How can I replace the triangle by circle. Like this ``` \documentclass[border=2mm]{standalone} \usepackage{tikz} \begin{document} \begin{tikzpicture} \draw (0,0) circle[radius=1 + sqrt(2)]; \draw ({sqrt(2)},0) circle[radius=1]; \draw (0,{sqrt(2)}) circle[radius=1]; \draw ({-sqrt(2)},0) circle[radius=1]; \draw (0,{-sqrt(2)}) circle[radius=1]; \end{tikzpicture} \end{document} ``` ![ScreenHunter 22.png](/image?hash=8985b23680deda21084472028ae1918a1f45a442e332c68236d2145dcb8f5d14) Ingenenal, `n` circles in a large circle.

``` \documentclass[tikz,border=2mm]{standalone} \usetikzlibrary{calc} \begin{document} \begin{tikzpicture}[declare function={r=1;}, pics/tricircl/.style={code={% \path foreach \YY in {1,...,#1} {foreach \XX in {1,...,\YY} {({2*r*(\XX-\YY/2-1/2)},{-2*\YY*r*sin(60)}) node[c](c-\XX-\YY){}}}; \draw (c-1-1.center) -- (c-#1-#1.center) -- (c-1-#1.center) -- cycle (intersection of c-#1-#1.30--c-1-1.30 and c-1-1.150--c-1-#1.150) -- (intersection of c-1-1.150--c-1-#1.150 and c-1-#1.-90--c-#1-#1.-90) -- (intersection of c-1-#1.-90--c-#1-#1.-90 and c-#1-#1.30--c-1-1.30) -- cycle; }},pics/tricircl/.default=2, c/.style={circle,draw,inner sep=0pt,minimum size=r*2cm}] \path pic{tricircl=2} (8,0) pic{tricircl=3} (4,-8) pic{tricircl=4}; \end{tikzpicture} \end{document} ```

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