3

The following MWE produces an output shown below.

\documentclass[pstricks,12pt]{standalone}
\usepackage{pst-eucl}
\begin{document}
\begin{pspicture}(-1,-2)(8,6)
\psset{linejoin=2,dotsize=2pt,labelsep=3pt,PointNameSep=9pt}
\pstGeonode[PointSymbol={*,*,none},PointName={B,C,none},PosAngle={-110,-60}](0,0){B}(7,0){C}(-1,0){Z}(5,0){Y}
\pstInterCC[PosAngleA=90]{B}{Y}{C}{Z}{A}{M2}
\pspolygon(B)(A)(C)
\pstTriangleIC{A}{B}{C} \uput[-10](IC_O){$I$} \psdot(IC_O)
\pstProjection{A}{B}{IC_O}[P]
\pstProjection{A}{C}{IC_O}[M]
\pstProjection{C}{B}{IC_O}[K]

\psset{PointSymbol=none,PointName=none}
\pstSymO{IC_O}{A,B}[P1,P2]
\pstLineAB{P1}{P2}

\pstInterLC{A}{IC_O}{IC_O}{P}{J1}{J2}
\uput[90](J1){$N$} \psdot(J1)
\pstTranslation{P}{M}{J1}[J3]
\pstInterLL{J1}{J3}{A}{B}{Q1}
\pstInterLL{J1}{J3}{A}{C}{Q2}
\pstTriangleIC{Q1}{Q2}{A} \uput[45](IC_O){$J$} \psdot(IC_O)

\pstInterLL{P1}{P2}{A}{C}{W1}
\pstInterLL{P1}{P2}{B}{C}{W2}
\pstTriangleIC{W1}{W2}{C} \uput[-10](IC_O){$L$} \psdot(IC_O)
\uput[0](4,3){$(l)$}
\end{pspicture}
\end{document}

enter image description here

Three points I, J, L, each is defined as IC_O by \pstTriangleIC. It means IC_O for I gets overridden for J which in turn gets overridden for L. So the last invocation makes IC_O refers to L. Can anyone help me to draw triangle IJL by using IC_O?

  • If I am not wrong the A point is not defined – Hafid Boukhoulda Jan 31 at 10:33
  • @HafidBoukhoulda Oh, I load it normally with a up-to-date pst-eucl... – Trong Vuong Jan 31 at 11:12
  • What I mean is that unlike the points B and C the point A coordinates are not defined – Hafid Boukhoulda Jan 31 at 11:13
  • In the example you provided the coordinates of B and C are (0,0) and (7,0) right. What about the point A ? – Hafid Boukhoulda Jan 31 at 11:20
  • @HafidBoukhoulda "A" is one of the intersection points of two circle with two centers B,C and radius 5,8 cm. – Trong Vuong Jan 31 at 11:37
3

Cache\pnode(IC_O){I} for I before IC_O gets overridden for J.

\documentclass[pstricks,12pt]{standalone}
\usepackage{pst-eucl}
\begin{document}
\begin{pspicture}(-1,-2)(8,6)
\psset{linejoin=2,dotsize=2pt,labelsep=3pt,PointNameSep=9pt}
\pstGeonode[PointSymbol={*,*,none},PointName={B,C,none},PosAngle={-110,-60}](0,0){B}(7,0){C}(-1,0){Z}(5,0){Y}
\pstInterCC[PosAngleA=90]{B}{Y}{C}{Z}{A}{M2}
\pspolygon(B)(A)(C)
\pstTriangleIC{A}{B}{C} \uput[-10](IC_O){$I$} \psdot(IC_O)
\pstProjection{A}{B}{IC_O}[P]
\pstProjection{A}{C}{IC_O}[M]
\pstProjection{C}{B}{IC_O}[K]
\pnode(IC_O){I}
\psset{PointSymbol=none,PointName=none}
\pstSymO{IC_O}{A,B}[P1,P2]
\pstLineAB{P1}{P2}
%
\pstInterLC{A}{IC_O}{IC_O}{P}{J1}{J2}
\uput[90](J1){$N$} \psdot(J1)
\pstTranslation{P}{M}{J1}[J3]
\pstInterLL{J1}{J3}{A}{B}{Q1}
\pstInterLL{J1}{J3}{A}{C}{Q2}
\pstTriangleIC{Q1}{Q2}{A} \uput[45](IC_O){$J$} \psdot(IC_O)
\pnode(IC_O){J}
%
\pstInterLL{P1}{P2}{A}{C}{W1}
\pstInterLL{P1}{P2}{B}{C}{W2}
\pstTriangleIC{W1}{W2}{C} \uput[-10](IC_O){$L$} \psdot(IC_O)
\uput[0](4,3){$(l)$}
\pnode(IC_O){L}
\pspolygon(I)(J)(L)
\end{pspicture}
\end{document}

enter image description here

  • Unbelievable, the solution is very simple..... – Trong Vuong Jan 31 at 12:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.