Inscribing multiple incircles to a triangle in tikz

What is the easiest way to have TikZ or tikz-euclide generate a picture with successive incircles to a triangle without needing to calculate the points for each smaller triangle?

I have the following code that has two incircles to a triangle. What if I want to continue or automate this drawing for different triangles?

\begin{tikzpicture}

\draw ({-sqrt(3)},-1) -- ({(1-sqrt(3))/2}, {(1+sqrt(3))/2}) -- ({2+sqrt(3)},-1) -- ({-sqrt(3)},-1) ;

\tkzDefPoint(3.7,-1){A}
\tkzDefPoint(1.16452,0.482362){B}
\tkzDefPoint(0.767327,-1){C}
\tkzDrawSegments(A,B B,C C,A)

\tkzDefCircle[in](A,B,C)\tkzGetPoint{I}\tkzGetLength{rIN}
\tkzDrawCircle[R](I,\rIN pt)

\end{tikzpicture}

Thanks!

I interpreted your question so that the new triangles are similar to the old one. Your example was different in that point.

With a little work, you can create an image that has that recursive done at every side and also in the circle itself.

Code

\documentclass[tikz]{standalone}
\usetikzlibrary{calc,through}
\makeatletter
\pgfmathdeclarefunction{tikzveclen}{2}{%
\begingroup
%    \pgfpointdiff{\tikz@scan@one@point\pgfutil@firstofone(#1)}
%                 {\tikz@scan@one@point\pgfutil@firstofone(#2)}%
\pgfpointdiff{\pgfpointanchor{#1}{center}}{\pgfpointanchor{#2}{center}}%
\edef\pgfmath@temp{{\pgf@sys@tonumber\pgf@x}{\pgf@sys@tonumber\pgf@y}}%
\expandafter\pgfmathveclen@\pgfmath@temp
\pgfmath@smuggleone\pgfmathresult
\endgroup}
\makeatother
\begin{document}
\begin{tikzpicture}[scale=4]
\path (0,0) coordinate (A) + (0:3) coordinate (B) +(60:1) coordinate (C);
\draw (A) -- (B) -- (C) -- cycle;
\foreach \cnt in {1,...,10}{
\pgfmathsetmacro\triA{tikzveclen("B","C")}
\pgfmathsetmacro\triB{tikzveclen("C","A")}
\pgfmathsetmacro\triC{tikzveclen("A","B")}
%\pgfmathsetmacro\triS{.5*\triA+.5*\triB+.5*\triC}
\path (barycentric cs:A=\triA,B=\triB,C=\triC) coordinate (M)
node [draw, circle through=($(A)!(M)!(C)$)] (M) {};
\draw ($(C)-(A)$) coordinate (vecB)
(M.60-90) coordinate (@) % or: ([shift=(60-90:\triRadius)] M) coordinate (@)
(intersection of @--[shift=(vecB)]@ and B--C) coordinate (C) --
(intersection of @--[shift=(vecB)]@ and B--A) coordinate (A);
}
\end{tikzpicture}
\end{document}

Output • This is fantastic! Thank you! What if I wanted to have all the triangles similar except for the largest one? Is there an easy way to do that? Nov 27 '13 at 21:05
• @DavidCaliri Well, then just do the first, non-regular step separately and enter the pseudo-recursive algorithm/the foreach loop after that. Nov 28 '13 at 1:00
• That worked great! Thank you! I also worked it to get the circles going down each corner of the triangle, but I'm not sure if I did this in the most efficient way. I apologize I am new to this site and not sure how I should post the new code. In a new question, or an answer to this question? Dec 1 '13 at 22:39

With math library (cvs-version) it's easy:

\documentclass[tikz,border=2mm]{standalone}
\usetikzlibrary{math}
\begin{document}
\begin{tikzpicture}

\foreach \i [remember=\r (initially 3)] in {1,2,...,5}{
\draw (0,0) circle (\r);
\draw (90:\r) -- (-30:\r)--(210:\r)--cycle;
\tikzmath{
\r=\r*sin(30);
}
}
\end{tikzpicture}
\end{document} EDIT

And this is a valid solution for TiKZ 2.10. It uses through library.

\documentclass[border=2mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{through}
\begin{document}
\begin{tikzpicture}
\draw (0,0) circle (3cm);
\coordinate (a) at (90:3);
\coordinate (b) at (-30:3);
\coordinate (c) at (210:3);
\foreach \in in {1,2,...,5}
{
\node[circle through=(a),draw] {};
\draw (a)--(b)--(c)--cycle;
\coordinate(aux) at (a);
\path (a)--(b) coordinate[pos=.5] (a);
\path (b)--(c) coordinate[pos=.5] (b);
\path (c)--(aux) coordinate[pos=.5] (c);
}
\end{tikzpicture}

\end{document} 