ecco il risultato finale
`
\documentclass{article}
\usepackage{etex}
\usepackage{tikz,pgf}
\usetikzlibrary{calc,through,intersections}
\begin{document}
\begin{tikzpicture}
\coordinate (a) at (0,0);
\coordinate (b) at (6,0);
\coordinate (c) at (9,6);
\node(A)[label=$A$] at (a) {$\bullet$};
\node(B)[label=50:$B$] at (b) {$\bullet$};
\node(C)[label=$C$] at (c) {$\bullet$};
\draw[thick] (a)–(b)–(c)–cycle;
\node(c1) at (a)[name path=c1,circle through=(b)] {};
\path[name path=a–c] (a) — (c);
\path[name intersections={of=c1 and a–c}];
\coordinate (ac) at (intersection-1);
\node(c2) at (ac)[name path=c2,circle through=(b)] {};
\node(c3) at (b)[name path=c3,circle through=(ac)] {};
\path[name intersections={of=c2 and c3}];
\coordinate (bta) at (intersection-2);
\path[name path=b–c] (b) — (c);
\path[name path=a–bta] (a) — (bta);
\path[name intersections={of=a–bta and b–c}];
\coordinate (bisa) at (intersection-1);
\draw(a)–(bisa);
\node(c4) at (b)[name path=c4,circle through=(a)] {};
\path[name path=b–c] (b) — (c);
\path[name intersections={of=c4 and b–c}];
\coordinate (bc) at (intersection-1);
\node(c5) at (bc)[name path=c5,circle through=(a)] {};
\node(c6) at (a)[name path=c6,circle through=(bc)] {};
\path[name intersections={of=c5 and c6}];
\coordinate (btb) at (intersection-1);
\path[name path=a–c] (a) — (c);
\path[name path=b–btb] (b) — (btb);
\path[name intersections={of=b–btb and a–c}];
\coordinate (bisb) at (intersection-1);
\draw(b)–(bisb);
\node(c7) at (c)[name path=c7,circle through=(b)] {};
\path[name path=a–c] (a) — (c);
\path[name intersections={of=c7 and a–c}];
\coordinate (ca) at (intersection-1);
\node(c8) at (ca)[name path=c8,circle through=(b)] {};
\node(c9) at (b)[name path=c9,circle through=(ca)] {};
\path[name intersections={of=c8 and c9}];
\coordinate (btc) at (intersection-2);
\path[name path=a–b] (a) — (b);
\path[name path=c–btc] (c) — (btc);
\path[name intersections={of=c–btc and a–b}];
\coordinate (bisc) at (intersection-1);
\draw(c)–(bisc);
% %
\coordinate(inc) at(intersection 1 of c–btc and a–bta);
\coordinate(piede) at($(a)!(inc)!(c)$);
\node(A)[label=$I$] at (inc) {$\bullet$};
\node(c10) at (inc)[name path=c10,draw,circle through=(piede)] {};
\end{tikzpicture}
\end{document}
`
trovato l’incentro !
questo è il codice con l’altro metodo
`
\documentclass{minimal}
\usepackage{tikz}
\usetikzlibrary{shapes,calc,arrows,trees,positioning,through,intersections,fadings}
\begin{document}
\begin{tikzpicture}
\coordinate (a) at (0,0);
\coordinate (b) at (6,0);
\coordinate (c) at (9,6);
\node(A)[label=$A$] at (a) {$\bullet$};
\node(B)[label=50:$B$] at (b) {$\bullet$};
\node(C)[label=$C$] at (c) {$\bullet$};
\draw[thick] (a)–(b)–(c)–cycle;
\node(c1) at (a)[circle through=(b)] {};
\coordinate(ac) at(intersection 1 of c1 and a–c);
\node(c2) at (ac)[circle through=(b)] {};
\node(c3) at (b)[circle through=(ac)] {};
\coordinate(bta) at(intersection 1 of c2 and c3 );
% \node(xx)[label=$C$] at (bta) {$\bullet$};
\coordinate(bisa) at(intersection 1 of a–bta and b–c);
\draw (a)–(bisa);
\node(c4) at (b)[circle through=(a)] {};
\coordinate(bc) at(intersection 1 of c4 and b–c);
\node(c5) at (bc)[circle through=(a)] {};
\node(c6) at (a)[circle through=(bc)] {};
\coordinate(btb) at(intersection 1 of c5 and c6 );
% \node(xx)[label=$C$] at (btb) {$\bullet$};
\coordinate(bisb) at(intersection 1 of b–btb and a–c);
\draw (b)–(bisb);
\node(c7) at (c)[circle through=(b)] {};
\coordinate(ca) at(intersection 1 of c7 and c–a);
%\node(xx)[label=$bbbb$] at (ca) {$\bullet$};
\node(c8) at (ca)[circle through=(b)] {};
\node(c9) at (b)[circle through=(ca)] {};
\coordinate(btc) at(intersection 1 of c8 and c9 );
%\node(xx)[label=$C$] at (btc) {$\bullet$};
\coordinate(bisc) at(intersection 1 of c–btc and a–b);
\draw (c)–(bisc);
\coordinate(inc) at(intersection 1 of c–btc and a–bta);
\coordinate(piede) at($(a)!(inc)!(c)$);
\node(A)[label=$I$] at (inc) {$\bullet$};
\node(c10) at (inc)[draw,circle through=(piede)] {};
\end{tikzpicture}
\end{document}
`
la resa non è uguale tuttavia
il primo da
[attachment=187]prova6.pdf[/attachment]
il secondo da
[attachment=188]prova4.pdf[/attachment]
il primo è migliore del secondo o no?
grazie claudio