6

\documentclass[12pt,border=15pt,pstricks]{standalone} \usepackage{pst-eucl} \begin{document} \begin{pspicture}[showgrid,PointSymbol=none](-1,-3)(11,5) \pstGeonode[PointName=none](2,4){A}(0,0){B}(6,0){C} \pstCircleABC[PosAngle=180]{A}{B}{C}{O} \pstGeonode[PosAngle=60](10,2){M} \pstMiddleAB[PointName=none]{O}{M}{M'} \pstInterCC[DiameterB=\pstDistAB{O}{M},...


6

Edit (based on @chishimotoji's comment) Only PointSymbolA=none is enough to delete the black bullet. But, that is strange to me since the bullet is at point C in coordinates (0,8) (note the change of units). Pass the options PointSymbolA=none (and for other 2 points) to the \pstTriangle command. \documentclass[border=15pt,pstricks,12pt]{standalone} \...


5

Not sure if this is the easiest. But it works. \documentclass[pstricks,border=1cm]{standalone} \usepackage{pst-eucl} \begin{document} \begin{pspicture}(8,-6) \pstTriangle(0,-6){B}(8,-6){A}(2,0){C} \pstMarkAngle{A}{B}{C}{} \pstGeonode[PosAngle=180]([nodesep=4]{B}C){E} \pstInterLC[PointSymbol=none,PointName=none]{C}{A}{C}{E}{G}{F} \...


5

Here is a TikZ solution: \documentclass[margin=2mm]{standalone} \usepackage{tikz} \usetikzlibrary{calc} \begin{document} \begin{tikzpicture} \draw[green] (0,0) -- (0,2) coordinate (b); \draw[red] (0,0) -- (4,3) coordinate (a); \draw[blue] let \p1=(a), \p2=(b) in (0,0) -- (\y2/\y1*\x1,0); \end{tikzpicture} \end{document}


5

Step 1 \documentclass[pstricks,12pt,border=15pt]{standalone} \usepackage{pst-eucl} \begin{document} \begin{pspicture}[showgrid](-1,-3)(7,5) \pstTriangle(2,4){A}(0,0){B}(6,0){C} \pstCircleABC[PosAngle=60]{A}{B}{C}{O} \pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666 \end{pspicture} \end{document} Note: HomCoef cannot accept RPN 2 3 div so I have to ...


5

\documentclass{standalone} \usepackage{pst-eucl,pst-text,pstricks-add,pst-grad} \usepackage{mathptmx} \DeclareFixedFont{\RM}{T1}{ptm}{b}{n}{1.7cm}% \begin{document} \begin{pspicture}[showgrid=false](-5.2,-5.2)(10,5.2) \psframe*[fillstyle=solid,linecolor=black!40,opacity=.9](-5.2,-5.2)(10,5.2) \pscircle*[dimen=inner](0,0){5} {% \psset{fillstyle=solid,...


4

\documentclass[12pt,pstricks]{standalone} \usepackage{pst-eucl} \begin{document} \begin{pspicture}[showgrid](-4,-2)(4,6) \pstGeonode[PosAngle={180,0,-90,90}](-3,0){A}(3,0){C}(-1,-1){B}(0,5){S} \pstTranslation[PosAngle=45]{B}{A}{C}[D] \pstCircleABC[linestyle=none]{S}{A}{C}{I} \pstInterLL[PosAngle=-60]{A}{C}{B}{D}{H} \psline(S)(A)(B)(S)(C)(B) \bgroup \...


4

The simpler the code the more challenging it becomes. \documentclass[pstricks,12pt]{standalone} \usepackage{pst-eucl} \begin{document} \begin{pspicture}[PointSymbolB=none,PointNameB=](7,5) \pstTriangle(1,1){B}(2.5,4){C}(6,1){A} \pstTranslation[PointName=none,PointSymbol=none]{B}{A}{C}[C'] \pstInterLC[PosAngle=45]{A}{C}{C}{B}{N}{x} \pstInterLC[PosAngle=-90]{...


4

I think making double arcs for angle marks has not been implemented yet. You can mimic it by making two angle marks in which one is without label as follows. You can rotate the angle label (about its center of gravity) with \rput because LabelMarkOffset is not for rotating angle label (about its center of gravity). \documentclass[border=15pt,pstricks,12pt]{...


3

Not a reliable solution because the labels are sometimes inside the polygon. \documentclass[pstricks,border=15pt]{standalone} \usepackage{pst-eucl} \pstVerb { /rv {rand 36000 mod 100 div} def % random number from 0.00 to 359.99 } \begin{document} \psLoop{30}{% \begin{pspicture}[showgrid=false](-2,-2)(2,2) \pscircle[linestyle=dashed,linecolor=...


3

I got the solution even though it is not elegant enough. \documentclass[border=0pt,pstricks]{standalone} \usepackage{pst-eucl} \psset{PointName=none,PointSymbol=none} \begin{document} \begin{pspicture}[showgrid=false](6,6) \pstGeonode[CurveType=polyline](1,1){A}(5,5){B}(4,1){C} \pstMarkAngle[MarkAngleRadius=1.6,LabelSep=1]{A}{B}{C}{\rput{(B)}(0,0){\...


3

I've frequently had this problem. There seems to be something in the way that key PointName is implemented in pst-eucl that's getting in the way. The way \pstGeonode is setting the node names is through an artifact of how \psnode works---which is expecting something expandable in \edef to create string for the internal name for the node. When you're ...


3

Your code without an intermediate node doesn't work, because \pstRotation internally uses \rput. When a node is shifted with \rput, the coordinates saved with saveNodeCoors cannot be used together with other coordinates of nodes defined outside of the \rput (see How do we explain the behavior of \rput, \psGetNodeCenter and saveNodeCoors? for a more ...


3

An alternative tikz version: \documentclass[tikz,border=12pt]{standalone} \usetikzlibrary{calc,bending,arrows.meta,quotes,angles} \usepackage{textcomp} \begin{document} \begin{tikzpicture}[dot/.style={fill,circle,inner sep=1.2pt},>= Stealth]\footnotesize \pgfmathsetmacro\BC{sqrt(1.5*1.5+3*3)} \pgfmathsetmacro\AC{sqrt(3.5*3.5+3*3)} \draw (1,2)coordinate[...


3

\documentclass[pstricks,12pt]{standalone} \usepackage{pst-eucl} \def\b{3 } \begin{document} \begin{pspicture}(-.6,-1)(4.85,4) \pstGeonode[PosAngle={-135,135,-45,45}](0,0){A}(!0 \b){B}(!\b 2 sqrt mul 0){D}(D|B){C} \pstMiddleAB{B}{C}{M} \pstInterLL[PointName=none]{A}{C}{M}{D}{E} \pspolygon[fillstyle=solid,fillcolor=green](E)([nodesep=6pt]{C}E)([offset=6pt,...


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