\documentclass[pstricks,border=15pt]{standalone}
\usepackage{pstricks-add,pst-eucl}%
\begin{document}
\begin{pspicture}[showgrid](0,-3)(8,4)
\pnodes(3,3){A}(1,-1){B}(7,-1){C}
\pstMiddleAB[PosAngle=135]{A}{B}{M}
\pstMiddleAB{A}{C}{N}
\pstMiddleAB{M}{N}{I}
\def\figA{\pspolygon(A)(M)(N)}
\figA%
%\psrotate(I){-50}{\pspolygon(3,3)(2,1)(5,1)}%
\psrotate(I){-50}{\figA}
\end{pspicture}
\end{document}
Com \psrotate(I){-50}{\pspolygon(3,3)(2,1)(5,1)}
, eu consigo
Com \psrotate(I){-50}{\figA}
, eu não ganho nada
Assuma isso A=(3.2783,3.5876), B=(1.07368,-1.235), C=(7.777,-1.3336)
.
Como obter a saída sem calcular M e N?
Responder1
\psrotate
não funciona com nós (nós são pontos fixos no sistema atual).
\documentclass[pstricks,border=15pt]{standalone}
\usepackage{pstricks-add,pst-eucl}%
\begin{document}
\begin{pspicture}[showgrid](0,-3)(8,4)
\pnodes(3,3){A}(1,-1){B}(7,-1){C}
\pstMiddleAB[PosAngle=135]{A}{B}{M}
\pstMiddleAB{A}{C}{N}
\pstMiddleAB{M}{N}{I}
\def\figA{\pspolygon(A)(M)(N)}
\figA
\pstRotation[RotAngle=-50]{I}{A,M,N}[a,m,n]
\pspolygon[linecolor=red](a)(m)(n)
\end{pspicture}
\end{document}