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-rw-r--r--Master/texmf-dist/doc/generic/pst-eucl/Changes20
-rw-r--r--Master/texmf-dist/doc/generic/pst-eucl/pst-eucl-doc.pdfbin1189759 -> 1341324 bytes
-rw-r--r--Master/texmf-dist/doc/generic/pst-eucl/pst-eucl-doc.tex829
3 files changed, 710 insertions, 139 deletions
diff --git a/Master/texmf-dist/doc/generic/pst-eucl/Changes b/Master/texmf-dist/doc/generic/pst-eucl/Changes
index f05be18cb40..585f6936d30 100644
--- a/Master/texmf-dist/doc/generic/pst-eucl/Changes
+++ b/Master/texmf-dist/doc/generic/pst-eucl/Changes
@@ -4,13 +4,25 @@ pst-eucl.pro --------
pst-eucl.tex --------
+1.67 2019/10/28 - add macros to add/subtract/divide the length of two segment, \pstDistAdd, \pstDistSub, and \pstDistDiv etc.
+ - add macros to reduce or enlarge the distance, \pstDistMul, \pstDistCoef.
+ - add macro to get the distance from point C to line AB, \pstDistABC.
+ - add macro to get the definite proportion node on segment, \pstProportionNode.
+ - add macro to get the fourth harmonic point, \pstFourthHarmonicNode.
+ - add macro to put the label for the segment, \pstLabelAB.
+ - add macro to locate a point on segment with specified length, \pstLocateAB.
+ - add macro to extend a segment to a new point with specified length, \pstExtendAB.
+ - add macro to get the inversion mapping of a point to the inversion center, \pstInversion.
+ - add macro to get the Geometric Mean of two segment, \pstGeometricMean.
+ - add macro to get the Harmonic Mean of two segment, \pstHarmonicMean.
+ - add macro to get the radical axis of two circles, \pstCircleRadicalAxis.
1.66 2019/10/20 - add macros to operate the node coordinates, \pstAbscissa, \pstOrdinate, \pstMoveNode etc.
- add optional parameters angleA and angleB for \pstCircleOA and \pstCircleAB.
- add optional parameters to output the inner circle center and outer circle center for \pstTriangleIC and \pstTriangleOC.
- add macros to draw the tangent line and tangent node of circle.
- add macros to draw the external and internal common tangent lines of two circles.
- add macros to draw conics (ellipse, parabola and hyperbola) and their geometrical elements, such as focus, directrix and intersections.
-1.65 2019/08/19 - new type for angle
+1.65 2019/08/19 - new type for angle
1.64 2019/01/31 - fix for PointName and pstInterCC
1.63 2019/01/27 - fix for PointSymbol=none for pstTriangle
1.62 2019/01/13 - added fillstyle for angles
@@ -23,10 +35,10 @@ pst-eucl.tex --------
1.58 2018/08/07 - allow PointSymbol?=none
1.57 2017/11/28 - fix bug with StandardSyml->StandardSymL
1.56 2017/04/18 - \psGetAngleABC:
- - added dec -1 le { /dec 15 def } if
+ - added dec -1 le { /dec 15 def } if
- added \pst@usecolor\pslinecolor in line 1616
-1.55 2016/10/11 - fix for \pstRightAngle
-1.54 2016/09/01 - added MarkArrow, MarkArroww,MarkArrowww
+1.55 2016/10/11 - fix for \pstRightAngle
+1.54 2016/09/01 - added MarkArrow, MarkArroww,MarkArrowww
1.53 2016/05/03 - revert changes of CodeFig(A|B)
1.52 2015/10/19 - added more optional arguments (ts)
1.51 2014/05/17 - added two new functions for angles and distances
diff --git a/Master/texmf-dist/doc/generic/pst-eucl/pst-eucl-doc.pdf b/Master/texmf-dist/doc/generic/pst-eucl/pst-eucl-doc.pdf
index 7f37a4e7ae7..1766af0c738 100644
--- a/Master/texmf-dist/doc/generic/pst-eucl/pst-eucl-doc.pdf
+++ b/Master/texmf-dist/doc/generic/pst-eucl/pst-eucl-doc.pdf
Binary files differ
diff --git a/Master/texmf-dist/doc/generic/pst-eucl/pst-eucl-doc.tex b/Master/texmf-dist/doc/generic/pst-eucl/pst-eucl-doc.tex
index 25aefc74877..dc97768b52d 100644
--- a/Master/texmf-dist/doc/generic/pst-eucl/pst-eucl-doc.tex
+++ b/Master/texmf-dist/doc/generic/pst-eucl/pst-eucl-doc.tex
@@ -216,7 +216,7 @@ sometimes it results the numbers more than 9 fraction digits,
which are not supported good by \PST\space with '! number too big' issue.} to generate the numerical values,
or the expandable command \Lcs{fpeval}\footnote{Provided by package \texttt{xfp},
it can truncate the fraction part digits using the \texttt{trunc} function perfectly,
-e.g. \texttt{\textbackslash{}fpeval\{trunc(18/7,3)\}}.} to get a purely numerical result,
+e.g. \texttt{\textbackslash{}fpeval\{trunc(18/7,3)\}}.} to get a purely numerical result.
The macro \Lcs{pstMoveNode} use them to move node $A$ by abscissa increment $dx$
and ordinate increment $dy$ to get the target node $B$.
@@ -316,26 +316,60 @@ controled by the parameter \Lkeyword{MarkAngle} (default is 45). Their width and
depends of the width and color of the line when the drawing is done, as shown is the
next example.
-
-
\begin{LTXexample}[width=5cm,pos=l]
\begin{pspicture}[showgrid=true](-2,-2)(2,2)
- \rput{18}{%
- \pstGeonode[PosAngle={0,90,180,-90}](2,0){A}(2;72){B}
- (2;144){C}(2;216){D}(2;288){E}}
- \pstSegmentMark[SegmentSymbol=none]{A}{B}
- \pstSegmentMark[linecolor=green]{B}{C}
- \psset{linewidth=2\pslinewidth}
- \pstSegmentMark[linewidth=2\pslinewidth]{C}{D}
- \pstSegmentMark[MarkAngle=90]{D}{E}
- \pstSegmentMark{E}{A}
+\rput{18}{%
+ \pstGeonode[PosAngle={0,90,180,-90}](2,0){A}(2;72){B}
+ (2;144){C}(2;216){D}(2;288){E}}
+\pstSegmentMark[SegmentSymbol=none]{A}{B}
+\pstSegmentMark[linecolor=green]{B}{C}
+\psset{linewidth=2\pslinewidth}
+\pstSegmentMark[linewidth=2\pslinewidth]{C}{D}
+\pstSegmentMark[MarkAngle=90]{D}{E}
+\pstSegmentMark{E}{A}
\end{pspicture}
\end{LTXexample}
-
The length and the separation of multiple hases can be set by \Lkeyword{MarkHashLength} and \Lkeyword{MarkHashSep}.
+\subsection{Segment labels}
+
+According to the manual of \PST, you can use the macros \Lcs{naput}, \Lcs{ncput} and \Lcs{nbput}
+to put the label \textit{above}, \textit{cover}, \textit{below} the segment. The macro \Lcs{pstLabelAB}
+just use them to draw a ruler bar and put the label on the ruler bar.
+
+\begin{BDef}
+\Lcs{pstLabelAB}\OptArgs\Largb{A}\Largb{B}\Largb{label}
+\end{BDef}
+
+You can use the parameters of \Lcs{ncline} to control the ruler bar,
+such as \Lkeyword{linestyle}, \Lkeyword{linecolor}, \Lkeyword{linewidth},
+\Lkeyword{arrows}, \Lkeyword{nodesep} etc; and use the parameters of \Lcs{ncput}
+to control the label position, such as \Lkeyword{nrot}, \Lkeyword{npos} etc;
+there is another parameter \Lkeyword{offset} to control the separation between
+the rule bar and the segment.
+
+It does not display the ruler bar as default, and you need to setup \Lkeyword{linestyle}
+to display it.
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-1,-1)(4,4)
+\psset{dotscale=0.5}\psset{PointSymbol=*}\footnotesize
+\pstGeonode[PosAngle=-90](0.5,1.5){A}
+\pstGeonode[PosAngle=-90](2.5,1.5){B}\pstLineAB{A}{B}
+\pstLabelAB{A}{B}{$\sqrt{a^2+b^2}$}
+\pstGeonode[PosAngle=0](0,0.5){C}
+\pstGeonode[PosAngle=0](0,3.5){D}\pstLineAB{C}{D}
+\pstLabelAB[linestyle=dashed]{C}{D}{$\sqrt{a^2+b^2}$}
+\pstGeonode[PosAngle=190](-1,-1){E}
+\pstGeonode[PosAngle=10](3,0){F}\pstLineAB{E}{F}
+\pstLabelAB[linestyle=dashed,arrows=|-|,offset=10pt,linecolor=blue!50]{E}{F}{$\sqrt{a^2+b^2}$}
+\pstLabelAB[linestyle=dashed,arrows=|<->|,offset=10pt,nrot=:D]{F}{E}{$\sqrt{a^2+b^2}$}
+\pstGeonode[PosAngle=100](0,4){G}
+\pstGeonode[PosAngle=-50](4,2){H}\pstLineAB{G}{H}
+\pstLabelAB[linestyle=solid,linecolor=red!50,arrows=|-|,offset=15pt,nrot=:U,npos=0.7]{G}{H}{\textcolor{red!50}{$\dfrac{a}{b}$}}
+\end{pspicture}
+\end{LTXexample}
\subsection{Triangles}
@@ -536,6 +570,8 @@ parameters is equal to 0.
\end{pspicture}
\end{LTXexample}
+\vspace{10pt}
+
The macro \Lcs{pstLine} draws a new line with two nodes, or two coordinates
or one node and one coordinate. This macro is similar with \Lcs{pstLineAB},
but more compatible.
@@ -547,6 +583,8 @@ but more compatible.
\Lcs{pstLine}\OptArgs\cAny\cAny
\end{BDef}
+\vspace{10pt}
+
The macros \Lcs{pstLineAA} and \Lcs{pstLineAS} draw a new line with one node,
the slope \texttt{angle} between the line and the horizontal axis, or the
slope \texttt{gradient} of the line, and create a new node $B$ on the line.
@@ -572,6 +610,8 @@ Here are some examples:
\end{pspicture}
\end{LTXexample}
+\vspace{10pt}
+
The macro \Lcs{pstLineAbsNode} creates a new node $C$ whose abscissa
is the given value $x_1$ on the line $AB$. The macro \Lcs{pstLineOrdNode} creates a new node $C$ whose ordinate is the given value $y_1$ on the line $AB$.
You can input $x_1$ or $y_1$ as any number(e.g, 2.0),
@@ -596,65 +636,460 @@ For example,
\end{pspicture}
\end{LTXexample}
+\vspace{10pt}
+
+The macro \Lcs{pstProportionNode} creates the nodes $C$ and $C'$ on segment $AB$ which are satisfied $|AC|:|BC|=\lambda,\;(\lambda>0)$.
+The node $C$ is inside the segment $AB$ and the node $C'$ is outside the segment $AB$, we have
+\begin{equation*}
+\left\{
+\begin{array}{l}
+x_{C}=\dfrac{x_{A}+\lambda{}x_{B}}{1+\lambda}\\
+y_{C}=\dfrac{y_{A}+\lambda{}y_{B}}{1+\lambda}
+\end{array}
+\right.
+\quad\text{and}\quad
+\left\{
+\begin{array}{l}
+x_{C'}=\dfrac{x_{A}-\lambda{}x_{B}}{1-\lambda}\\
+y_{C'}=\dfrac{y_{A}-\lambda{}y_{B}}{1-\lambda}
+\end{array}
+\right.
+\end{equation*}
+
+\begin{BDef}
+\Lcs{pstProportionNode}\OptArgs\Largb{A}\Largb{B}\Largb{$\lambda$}\Largb{C}\Largb{C'}
+\end{BDef}
+
+You can use \Lcs{pstDistDiv} to get the ratio of two segments to $\lambda$,
+we will introduce \Lcs{pstDistDiv} later.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-1,-1)(4,4)
+\pstGeonode[PosAngle=-40,PointSymbol=|](0.5,1.5){A}
+\pstGeonode[PosAngle=-40,PointSymbol=|](3.0,3.0){B}
+\pstGeonode[PosAngle=90,linecolor=purple!60,CurveType=polyline](3,0){X}(4,0){Y}
+\pstGeonode[PosAngle=90,linecolor=brown!60,CurveType=polyline](1.5,-1){X'}(4,-1){Y'}
+\pstLineAB[linecolor=red,nodesep=-2.5]{A}{B}
+\psset{PosAngle=-40,PointSymbol=*,dotscale=1.5}
+\pstProportionNode[linecolor=yellow]{A}{B}{3.0}{C}{C'}
+\pstProportionNode[linecolor=blue]{A}{B}{1.0}{D}{D'}
+\pstProportionNode[linecolor=green]{A}{B}{0.2}{E}{E'}
+\pstProportionNode[linecolor=brown]{A}{B}{\pstDistDiv{X}{Y}{X'}{Y'}}{F}{F'}
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
+The four collinear points $A,B,C,D$ are called \texttt{Harmonic Conjugation Points} if their cross ratio is $-1$,
+that is
+$$(AB,CD)=\dfrac{AC}{BC}:\dfrac{AD}{BD}=-1$$
+If given three collinear points $A,B,C$, how can we get the fourth harmonic point?
+The following macro \Lcs{pstFourthHarmonicNode} is used to get the fourth harmonic point.
+It create a new node $X$ on the same line, but when $A,B,C$ are not collinear, we put it at origin.
+
+\begin{BDef}
+\Lcs{pstFourthHarmonicNode}\OptArgs\Largb{A}\Largb{B}\Largb{C}\Largb{X}
+\end{BDef}
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-2,-2)(3,3)\footnotesize
+\psset{unit=0.6cm}\psset{dotscale=0.5}\psset{PointSymbol=*}
+\pstGeonode[PosAngle=-90](-2,-2){A}(5,-2){I}(-1,-2){J}
+\pstLineAA[linestyle=none,PointName=none,PointSymbol=none]{A}{38}{A'}
+\pstLineAbsNode[PosAngle=-80]{A}{A'}{0}{B}
+\pstLineAbsNode[PosAngle=20]{A}{A'}{2.5}{C}
+\pstFourthHarmonicNode[PosAngle=180,PointNameSep=0.2]{A}{B}{C}{X}
+% check if A,N,M are also collinear.
+\pstInterLL[PosAngle=90]{J}{X}{I}{C}{P}
+\pstLineAB[linestyle=dashed]{J}{P}
+\pstLineAB[linestyle=dashed]{I}{P}
+\pstInterLL[PosAngle=20]{J}{B}{I}{P}{M}
+\pstInterLL[PosAngle=140]{I}{B}{J}{P}{N}
+\pstLineAB[linestyle=dashed]{J}{M}
+\pstLineAB[linestyle=dashed]{I}{N}
+\pstLineAB[linestyle=dashed]{A}{M}
+\pstLineAB[linestyle=dashed]{A}{I}
+\pstLineAB{A}{C}
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
+If you want to draw a node like \textsf{'Given $EF$, please find node $C$ on $AB$ such that $AC=EF$'},
+you can use the macro \Lcs{pstLocateAB} to do this, it can seek the node $C$ from $A$ to $B$ with the
+specified length, which can be got from \Lcs{pstDist}, \Lcs{pstDistConst}, \Lcs{pstDistAdd}, \Lcs{pstDistSub},
+etc.
+
+\begin{BDef}
+\Lcs{pstLocateAB}\OptArgs\Largb{A}\Largb{B}\Largb{$L$}\Largb{C}
+\end{BDef}
+
+Note that seek from $B$ will get the node $C$ in the reverse order, for example,
+
+\begin{LTXexample}[width=5cm,pos=l]
+\begin{pspicture}[showgrid=true](-3,-3)(3,3)\footnotesize
+\psset{unit=0.5cm}\psset{dotscale=0.5}\psset{PointSymbol=*}
+\pstGeonode[PosAngle=90,CurveType=polyline](-2,0){A}(-1,0){B}
+\pstGeonode[PosAngle=90,CurveType=polyline](-2,1){A'}(0,1){B'}
+\pstLocateAB[PosAngle=90]{A}{B}{\pstDist{A'}{B'}}{C}
+\pstLocateAB[PosAngle=90]{B}{A}{\pstDist{A'}{B'}}{C'}
+\psset{linestyle=dashed,SegmentSymbol=MarkHashh,MarkAngle=90}
+\pstSegmentMark{A'}{B'}\pstSegmentMark{B}{C'}\pstSegmentMark{A}{C}
+\pstGeonode[PosAngle=90,CurveType=polyline](-3,-2){D}(3,-4){E}
+\pstGeonode[PosAngle=90,CurveType=polyline](-4,2){D'}(-3,4){E'}
+\pstLocateAB[PosAngle=90]{D}{E}{\pstDist{D'}{E'}}{F}
+\pstLocateAB[PosAngle=90]{E}{D}{\pstDist{D'}{E'}}{F'}
+\psset{linestyle=dashed,SegmentSymbol=MarkHashhh,MarkAngle=90}
+\pstSegmentMark{D'}{E'}\pstSegmentMark{E}{F'}\pstSegmentMark{D}{F}
+\pstGeonode[PosAngle=0,CurveType=polyline](2,0){I}(2,1){J}
+\pstGeonode[PosAngle=0,CurveType=polyline](3,2){I'}(3,4){J'}
+\pstLocateAB[PosAngle=0]{I}{J}{\pstDist{I'}{J'}}{K}
+\pstLocateAB[PosAngle=0]{J}{I}{\pstDist{I'}{J'}}{K'}
+\psset{linestyle=dashed,SegmentSymbol=MarkHashhh,MarkAngle=45}
+\pstSegmentMark{I'}{J'}\pstSegmentMark{J}{K'}\pstSegmentMark{I}{K}
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
+If you want to draw a node like \textsf{'Given $EF$, please extend $AB$ to $C$ such that $BC=EF$'},
+you can use the macro \Lcs{pstExtendAB} to do this, it can extend $AB$ from $B$ to one node with the
+specified length, which can be got from \Lcs{pstDist}, \Lcs{pstDistConst}, \Lcs{pstDistAdd}, \Lcs{pstDistSub},
+etc.
+
+\begin{BDef}
+\Lcs{pstExtendAB}\OptArgs\Largb{A}\Largb{B}\Largb{$L$}\Largb{C}
+\end{BDef}
+
+Note that extend $BA$ to $C$ will get the node $C$ in the reverse order, for example,
+
+\begin{LTXexample}[width=5cm,pos=l]
+\begin{pspicture}[showgrid=true](-3,-3)(3,3)\footnotesize
+\psset{unit=0.5cm}\psset{dotscale=0.5}\psset{PointSymbol=*}
+\pstGeonode[PosAngle=90,CurveType=polyline](-2,0){A}(-1,0){B}
+\pstGeonode[PosAngle=90,CurveType=polyline](-2,1){A'}(0,1){B'}
+\pstExtendAB[PosAngle=90]{A}{B}{\pstDist{A'}{B'}}{C}
+\pstExtendAB[PosAngle=90]{B}{A}{\pstDist{A'}{B'}}{C'}
+\psset{linestyle=dashed,SegmentSymbol=MarkHashh,MarkAngle=90}
+\pstSegmentMark{A'}{B'}\pstSegmentMark{B}{C}\pstSegmentMark{A}{C'}
+\pstGeonode[PosAngle=90,CurveType=polyline](-2,-2){D}(0,-3){E}
+\pstGeonode[PosAngle=90,CurveType=polyline](-4,2){D'}(-3,4){E'}
+\pstExtendAB[PosAngle=90]{D}{E}{\pstDist{D'}{E'}}{F}
+\pstExtendAB[PosAngle=90]{E}{D}{\pstDist{D'}{E'}}{F'}
+\psset{linestyle=dashed,SegmentSymbol=MarkHashhh,MarkAngle=90}
+\pstSegmentMark{D'}{E'}\pstSegmentMark{E}{F}\pstSegmentMark{D}{F'}
+\pstGeonode[PosAngle=0,CurveType=polyline](2,0){I}(2,1){J}
+\pstGeonode[PosAngle=0,CurveType=polyline](3,2){I'}(3,4){J'}
+\pstExtendAB[PosAngle=0]{I}{J}{\pstDist{I'}{J'}}{K}
+\pstExtendAB[PosAngle=0]{J}{I}{\pstDist{I'}{J'}}{K'}
+\psset{linestyle=dashed,SegmentSymbol=MarkHashhh,MarkAngle=45}
+\pstSegmentMark{I'}{J'}\pstSegmentMark{J}{K}\pstSegmentMark{I}{K'}
+\end{pspicture}
+\end{LTXexample}
+
+You can find the node $C$ on segment $AB$ satisfied $|AC|$:$|AB|$=\Lkeyword{DistCoef}
+using \Lcs{pstTranslation}, but it can't do the same thing like \Lcs{pstLocateAB} and \Lcs{pstExtendAB}
+when the given segment $EF$ is not parallel with $AB$, it will be introduced in the later sections.
+
+\vspace{10pt}
+
+If you want to find the inversion point $C'$ of $C$ to the inversion center $O$ with inversion raduis $R$,
+that is, the point $C'$ is satisfied the inversion transform equation
+$$|OC|\times|OC'|=R^2$$
+you can use the macro \Lcs{pstInversion} to do this work.
+In fact, we use the macro \Lcs{pstLocateAB} to implement this macro
+by passing the value $\dfrac{R^2}{|OC|}$ to parameter length.
+
+\begin{BDef}
+\Lcs{pstInversion}\OptArgs\Largb{O}\Largb{A}\Largb{C}\Largb{C'}
+\end{BDef}
+
+It is possible to omit the parameter $A$ and then to specify the inversion radius or
+the inversion diameter using the parameters \Lkeyword{Radius} and \Lkeyword{Diameter},
+which will be introduced in the next section.
+
+It is clear that the inversion mapping of a line is a circle, and the inversion mapping
+of a point on the inversion circle is itself.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-1,-2)(4,3)
+\psset{dotscale=0.5}\psset{PointSymbol=*}\footnotesize
+\def\ra{1.5}
+\pstGeonode[PosAngle=180](1,1){O}
+\pstCircleOA[linecolor=red!50,Radius=\pstDistVal{\ra}]{O}{}
+\pstCircleRotNode[PosAngle=180,RotAngle=180,Radius=\pstDistVal{\ra}]{O}{}{A}
+\pstInversion[PosAngle=0,Radius=\pstDistVal{\ra}]{O}{}{A}{A'}
+\pstGeonode[PosAngle=0](3,3){C}
+\pstInversion[PosAngle=100,Radius=\pstDistVal{\ra}]{O}{}{C}{C'}
+\pstLineAB{O}{C}\pstLineAB{O}{C'}
+\pstGeonode[PosAngle=0](3,1.5){D}
+\pstInversion[PosAngle=90,Radius=\pstDistVal{\ra}]{O}{}{D}{D'}
+\pstLineAB{O}{D}\pstLineAB{O}{D'}
+\pstGeonode[PosAngle=0](3,0){E}
+\pstInversion[PosAngle=-90,Radius=\pstDistVal{\ra}]{O}{}{E}{E'}
+\pstLineAB{O}{E}\pstLineAB{O}{E'}
+\pstGeonode[PosAngle=0](3,-2){F}
+\pstInversion[PosAngle=-120]{O}{A}{F}{F'}
+\pstLineAB{O}{F}\pstLineAB{O}{F'}
+\pstLineAB[linecolor=black!50]{C}{F}
+\pstCircleABC[linestyle=dashed,linecolor=blue!40,PosAngle=0]{C'}{D'}{E'}{O'}
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
+If you want to find the node $C$ from $A$ to $B$, such that $AC$ is the geometric mean of two
+given segments $DE$ of $FG$, that is,
+$$|AC|^2=|DE|\times|FG|$$
+you can use the macro \Lcs{pstGeometricMean} to do this work.
+It also can be used to draw a circle when given two points on the circle,
+and a line tangents to the circle.
+
+\begin{BDef}
+\Lcs{pstGeometricMean}\OptArgs\Largb{A}\Largb{B}\Largb{$L_1$}\Largb{$L_2$}\Largb{C}
+\end{BDef}
+
+In fact, we use the macro \Lcs{pstLocateAB} to implement this macro
+by passing the value $\sqrt{L_1\times{}L_2}$ to parameter length.
+The length $L_1$ and $L_2$ can be got from \Lcs{pstDist}, \Lcs{pstDistConst},
+\Lcs{pstDistAdd}, \Lcs{pstDistSub}, etc.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](0,-3)(4,3)
+\psset{dotscale=0.5}\psset{PointSymbol=*}\footnotesize
+\pstGeonode[PosAngle=90](0,1){A}
+\pstGeonode[PosAngle=0](3.2,2){C}(3.2,1){D}(3.2,-2){E}
+\pstGeometricMean[PosAngle=90]{C}{A}{\pstDistAB{C}{D}}{\pstDistAB{D}{E}}{B}
+\pstCircleABC[linecolor=gray!60]{B}{D}{E}{O}
+\pstLineAB[linecolor=red!40]{C}{D}
+\pstLineAB[linecolor=blue!40]{D}{E}
+\psset{linestyle=dashed}
+\pstLineAB[linecolor=purple!80]{C}{A}
+\pstLineAB{D}{B}\pstLineAB{E}{B}
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
+If you want to find the node $C$ from $A$ to $B$, such that $AC$ is the harmonic mean of two
+given segments $DE$ of $FG$, that is,
+$$\dfrac{1}{|AC|}=\dfrac{1}{2}(\dfrac{1}{|DE|}+\dfrac{1}{|FG|})$$
+you can use the macro \Lcs{pstHarmonicMean} to do this work.
+
+\begin{BDef}
+\Lcs{pstHarmonicMean}\OptArgs\Largb{A}\Largb{B}\Largb{$L_1$}\Largb{$L_2$}\Largb{C}
+\end{BDef}
+
+In fact, we use the macro \Lcs{pstLocateAB} to implement this macro
+by passing the value $\dfrac{2L_1L_2}{L_1+L_2}$ to parameter length.
+The length $L_1$ and $L_2$ can be got from \Lcs{pstDist}, \Lcs{pstDistConst},
+\Lcs{pstDistAdd}, \Lcs{pstDistSub}, etc.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](0,-3)(4,3)
+\psset{dotscale=0.5}\psset{PointSymbol=*}\footnotesize
+\pstGeonode[PosAngle=90](1,2){A}
+\pstGeonode[PosAngle=-90](0,-2){C}(2.5,-2){D}(4,-2){E}
+\pstHarmonicMean[PosAngle=60]{D}{A}{\pstDistAB{C}{D}}{\pstDistAB{D}{E}}{B}
+\pstLineAB[linecolor=red!40]{C}{D}
+\pstLineAB[linecolor=blue!40]{D}{E}
+\pstLineAB[linecolor=purple!80]{A}{D}
+\pstLineAB[linestyle=dashed]{C}{B}
+\pstLineAB[linestyle=dashed]{E}{B}
+\pstLineAB{C}{A}\pstLineAB{E}{A}
+\end{pspicture}
+\end{LTXexample}
+
+\subsection{Distance}
+Like as coordinates, the distance works at the PostScript level,
+that is, it should be used where the code is interpreted by PostScript engine,
+but not \TeX\ engine. There were three macros to operate the distance before v1.66:
+
+\begin{BDef}
+\Lcs{pstDistAB}\Largb{A}\Largb{B}\\
+\Lcs{pstDistVal}\Largb{l}\\
+\Lcs{pstDistCalc}\Largb{expr}
+\end{BDef}
+
+The first specifies a distance between two points. The second macro can be used to
+specify an explicit numerical value $l$, which is in \texttt{User coordinate}.
+The third one uses the \Lcs{pscalculate} to calculate
+the result of the input expression, which is in \texttt{User coordinate} too.
+The parameter \Lkeyword{DistCoef} can be used to specify
+a coefficient to reduce or enlarge the result distance.
+This parameter will come into effect if it is specified before these macros.
+
+After v1.66, We provide three macros which disable the effect of parameter \Lkeyword{DistCoef}
+one to one as following:
+
+\begin{BDef}
+\Lcs{pstDist}\Largb{A}\Largb{B}\\
+\Lcs{pstDistConst}\Largb{l}\\
+\Lcs{pstDistExpr}\Largb{expr}
+\end{BDef}
+
+We provide the macro \Lcs{pstDistCoef} to reduce or enlarge a given distance explicitly,
+for example: \verb|\pstDistCoef{\pstDist{A}{B}}|, or use macro \Lcs{pstDistMul} to multiply
+the input coefficient.
+
+\vspace{10pt}\noindent{}{\large{\textbf{Note}}}:
+The series of macros \verb|\pstDist*| get the length result in the \texttt{Screen coordinate},
+so you need to convert the length to the \texttt{User coordinate} by macro \Lcs{pstUserDist},
+when use them where need the user coordinate numbers, e.g,
+
+\begin{lstlisting}
+\pnode(! 1 \pstUserDist{\pstDistAdd{A}{B}{C}{D}}){A}
+\pstMoveNode(0,\pstUserDist{\pstDistAdd{A}{B}{C}{D}}){A}{E}
+\end{lstlisting}
+
+You can convert the distance in \texttt{User coordinate} to \texttt{Screen coordinate} by
+macro \Lcs{pstScreenDist}, it is just another name of \Lcs{pstDistConst}. As we said before,
+macros \Lcs{pstAbscissa} and \Lcs{pstOrdinate} give the coordinate of one node in
+\texttt{User coordinate}, so if you want to draw a circle using them, you should type:\\[8pt]
+\verb|\pstCircleOA[Radius=\pstDistConst{\pstAbscissa{A}}]{A}{}|
+
+\vspace{10pt}
+
+It is possible to use the raw PostScript command to make more complex arithmetic operations.
+In order to hide the lower level Postscipt language, we add more macros for distance
+addition and subtraction, such as \Lcs{pstDistAdd}[Val/Coef] and \Lcs{pstDistSub}[Val/Coef], etc.
+These macros can be used to calculate the Radius or Diameter to define a circle.
+
+The macros \Lcs{pstDistAdd} and \Lcs{pstDistSub} are used to get the addition and subtraction
+of the given segments $AB$ and $CD$. The macro \Lcs{pstDistDiv} is used to
+get the length ratio of the given segments $AB$ and $CD$, you can pass the ratio to
+macro \Lcs{pstProportionNode}, or setup the ratio to parameter \Lkeyword{DistCoef} in
+macro \Lcs{pstTranslation}, or pass the ratio to any \verb|\pstDist|* macros which need
+a $\lambda$ parameter.
+
+\begin{BDef}
+\Lcs{pstDistMul}\Largb{A}\Largb{B}\Largb{$\lambda$}\\
+\Lcs{pstDistAdd}\Largb{A}\Largb{B}\Largb{C}\Largb{D}\\
+\Lcs{pstDistAddVal}\Largb{A}\Largb{B}\Largb{$\lambda$}\Largb{$L$}\\
+\Lcs{pstDistAddCoef}\Largb{A}\Largb{B}\Largb{$\lambda_1$}\Largb{C}\Largb{D}\Largb{$\lambda_2$}\\
+\Lcs{pstDistSub}\Largb{A}\Largb{B}\Largb{C}\Largb{D}\\
+\Lcs{pstDistSubVal}\Largb{A}\Largb{B}\Largb{$\lambda$}\Largb{$L$}\\
+\Lcs{pstDistSubCoef}\Largb{A}\Largb{B}\Largb{$\lambda_1$}\Largb{C}\Largb{D}\Largb{$\lambda_2$}\\
+\Lcs{pstDistDiv}\Largb{A}\Largb{B}\Largb{C}\Largb{D}
+\end{BDef}
+
+In these macros, the length $L$ is a numerical value in the \texttt{Screen Coordinate},
+so it is possible to pass the result of any macros like \verb|\pstDist| to it.
+$\lambda$ is a numerical value to multiply, and most important is that the parameter
+\Lkeyword{DistCoef} doesn't take effect any more.
+It is better to describe in formula:
+\\
+- macro \Lcs{pstDistAB} get the screen length of $\text{DistCoef}*|AB|$\\
+- macro \Lcs{pstDistVal} get the screen length of $\text{DistCoef}*l$\\
+- macro \Lcs{pstDistCalc} get the screen length of $\text{DistCoef}*\text{expr}$\\
+- macro \Lcs{pstDistCoef} get the screen length of $\text{DistCoef}*\text{<arg>}$\\
+- macro \Lcs{pstDist} get the screen length of $|AB|$\\
+- macro \Lcs{pstDistConst} get the screen length of $l$\\
+- macro \Lcs{pstDistExpr} get the screen length of $\text{expr}$\\
+- macro \Lcs{pstDistMul} get the screen length of $\lambda{}|AB|$\\
+- macro \Lcs{pstDistAdd} get the screen length of $|AB|+|CD|$\\
+- macro \Lcs{pstDistAddVal} get the screen length of $\lambda{}|AB|+L$\\
+- macro \Lcs{pstDistAddCoef} get the screen length of $\lambda_1{}|AB|+\lambda_2{}|CD|$\\
+- macro \Lcs{pstDistSub} get the screen length of $abs(|AB|-|CD|)$\\
+- macro \Lcs{pstDistSubVal} get the screen length of $abs(\lambda{}|AB|-L)$\\
+- macro \Lcs{pstDistSubCoef} get the screen length of $abs(\lambda_1{}|AB|-\lambda_2{}|CD|)$\\
+- macro \Lcs{pstDistDiv} get the the ratio of length $|AB|:|CD|$
+
+For example, the following one draw a circle with radius length $2|AB|+3|CD|+4|EF|$,
+it shows how to operate more than two distances.
+\begin{lstlisting}
+\pstCircleOA[Radius=\pstDistAddVal{A}{B}{2.0}{\pstDistAddCoef{C}{D}{3.0}{E}{F}{4.0}}]{A}{}
+\end{lstlisting}
+
+Another example is for \Lcs{pstDistMul}, the old code like as
+\begin{lstlisting}
+\pstCircleOA[DistCoef=1 3 div,Radius=\pstDistAB{A}{B}]{O}{}
+\pstCircleOA[DistCoef=1 3 div,Radius=\pstDistAB{A}{B}]{A}{B}{O}{}{I}{J}
+\pstInterCC[DistCoef=1 3 div,RadiusA=\pstDistAB{A}{B},DistCoef=none,RadiusA=\pstDistAB{C}{D}]{O1}{}{O2}{}{I}{J}
+\end{lstlisting}
+could be simplified to
+\begin{lstlisting}
+\pstCircleOA[Radius=\pstDistMul{A}{B}{1 3 div}]{O}{}
+\pstInterLC[Radius=\pstDistMul{A}{B}{1 3 div}]{A}{B}{O}{}{I}{J}
+\pstInterCC[RadiusA=\pstDistMul{A}{B}{1 3 div},RadiusA=\pstDistAB{C}{D}]{O1}{}{O2}{}{I}{J}
+\end{lstlisting}
+
+\vspace{10pt}\noindent{}{\Large{\textbf{Important}}}!
+We recommend that you should use the distance macros which disable the parameter \Lkeyword{DistCoef}
+instead of \verb|\pstDistAB|, \verb|\pstDistVal| or \verb|\pstDistCalc|,
+when you need to pass their result into \Lcs{pstDistAddVal} or \Lcs{pstDistSubVal},
+as it will give you the error result sometimes.
+For example, the following code \\[8pt]
+\verb|\pstDistAddVal{A}{B}{2.0}{\pstDistAB{C}{D}}|\\[8pt]
+is expected to get the length of $2|AB|+|CD|$. If current \Lkeyword{DistCoef} is $\lambda$,
+then it will give the error result as $2|AB|+\lambda|CD|$.
+The right way is \\[8pt]
+\verb|\pstDistAddVal{A}{B}{2.0}{\pstDist{C}{D}}|
+
+\vspace{10pt}
+At last, we provide a macro named \Lcs{pstDistABC} to get the distance from $C$ to line $AB$.
+
+\begin{BDef}
+\Lcs{pstDistABC}\Largb{A}\Largb{B}\Largb{C}
+\end{BDef}
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-1,-1)(3,3)
+\psset{dotscale=0.5}\psset{PointSymbol=*}\footnotesize
+\pstGeonode[PosAngle=-90](0,0){O}
+\pstGeonode[PosAngle=-90](2,0){A}
+\pstGeonode[PosAngle=90](1,1.5){B}
+\pstCircleOA[linecolor=red,Radius=\pstDistABC{O}{A}{B}]{B}{}
+\pstCircleOA[linecolor=blue,Radius=\pstDistABC{B}{O}{A}]{A}{}
+\pstCircleOA[linecolor=green,Radius=\pstDistABC{A}{B}{O}]{O}{}
+\pstLineAB[linecolor=red]{O}{A}
+\pstLineAB[linecolor=blue]{B}{O}
+\pstLineAB[linecolor=green]{A}{B}
+\end{pspicture}
+\end{LTXexample}
+
\subsection{Circles}
A circle can be defined either with its center and a point of its
-circumference, or with two diameterly opposed points. There is two
-commands :
+circumference, or with two diameterly opposed points. There are two
+commands:
\begin{BDef}
\Lcs{pstCircleOA}\OptArgs\Largb{O}\Largb{A}\OptArg{angleA}\OptArg{angleB}\\
-\Lcs{pstCircleAB}\OptArgs\Largb{O}\Largb{A}\OptArg{angleA}\OptArg{angleB}
+\Lcs{pstCircleAB}\OptArgs\Largb{A}\Largb{B}\OptArg{angleA}\OptArg{angleB}
\end{BDef}
-\Lcs{pstCircleOA} draws the circle of center $O$ crossing $A$ from \Lkeyword{angleA} to \Lkeyword{angleB}, going counter clockwise.
-Possible options are \Lkeyword{Radius} and \Lkeyword{Diameter}.
+\Lcs{pstCircleOA} draws the circle of center $O$ crossing $A$ from \Lkeyword{angleA} to \Lkeyword{angleB},
+going counter clockwise. Possible options are \Lkeyword{Radius} and \Lkeyword{Diameter}.
\Lcs{pstCircleAB} draws the circle of diameter $AB$ with the same options.
-
For the first macro, it is possible to omit the second point and then
to specify a radius or a diameter using the parameters \Lkeyword{Radius}
-and \Lkeyword{Diameter}. The values of these parameters must be specified
-with one of the two following macros :
-
-\begin{BDef}
-\Lcs{pstDistAB}\OptArgs\Largb{A}\Largb{B}\\
-\Lcs{pstDistVal}\OptArgs\Largb{x}
-\end{BDef}
+\footnote{The package \texttt{pst-fractal} also defines an optional key
+named \texttt{Radius}, if you need to use this package with \texttt{pst-eucl},
+you need to setup the key \texttt{Radius} as following:
+\texttt{\textbackslash{}psset[pst-eucl]\{Radius=\textbackslash{}pstDistVal\{3\}\}}.}
+and \Lkeyword{Diameter}. The values of these parameters can be specified
+with one of the \verb|\pstDist|* series macros.
-%\Lcs{pstDistAB} Specifies distance $AB$ for the parameters
-% \Lkeyword{Radius}, \Lkeyword{Diameter} and \Lkeyword{DistCoef}.
-%
-%\Lcs{pstDistVal} Specifies a numerical value for the parameters
-% \Lkeyword{Radius}, \Lkeyword{Diameter}, and \Lkeyword{DistCoef}.
-
-The first specifies a distance between two points. The parameter
-\Lkeyword{DistCoef} can be used to specify a coefficient to reduce or
-enlarge this distance. To be taken into account this last parameter
-must be specified before the distance. The second macro can be used to
-specify an explicit numeric value.
-%
We will see later how to draw the circle crossing three points.
-%
- With this package, it becomes possible to draw:
- \begin{compactitem}
- \item {\color{red} the circle of center $A$ crossing $B$;}
- \item {\color{green} the circle of center $A$ whose radius is $AC$;}
- \item {\color{blue} the circle of center $A$ whose radius is $BC$;}
- \item {\color{Sepia} the circle of center $B$ whose radius is $AC$;}
- \item {\color{Aquamarine} the circle of center $B$ of diameter $AC$;}
- \item {\color{RoyalBlue} the circle whose diameter is $BC$.}
- \end{compactitem}
-
-\clearpage
+With this package, it becomes possible to draw:
+\begin{compactitem}
+\item {\color{red} the circle of center $A$ crossing $B$;}
+\item {\color{green} the circle of center $A$ whose radius is $AC$;}
+\item {\color{blue} the circle of center $A$ whose radius is $BC$;}
+\item {\color{Sepia} the circle of center $B$ whose radius is $AC$;}
+\item {\color{Aquamarine} the circle of center $B$ of diameter $AC$;}
+\item {\color{RoyalBlue} the circle whose diameter is $BC$.}
+\end{compactitem}
-\begin{LTXexample}[width=\linewidth,pos=t]
-\begin{pspicture}[showgrid](-4,-3.3)(4,3)
-\psset{linewidth=2\pslinewidth}
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid](-3,-3)(3,3)\footnotesize
+\psset{unit=0.65cm}\psset{dotscale=0.5}\psset{PointSymbol=*}
\pstGeonode[PosAngle={0,-135,90},PointSymbol={*,*,square}](1,0){A}(-2,-1){B}(0,1){C}
\pstCircleOA[linecolor=red]{A}{B}
-\pstCircleOA[linecolor=green, DistCoef=2 3 div, Radius=\pstDistAB{A}{C}]{A}{}
+\pstCircleOA[linecolor=green, Radius=\pstDistMul{A}{C}{2 3 div}]{A}{}
\pstCircleOA[linecolor=blue, Radius=\pstDistAB{B}{C}]{A}{}[45][270]
\pstCircleOA[linecolor=Sepia, Radius=\pstDistAB{A}{C}]{B}{}
\pstCircleOA[linecolor=Aquamarine, Diameter=\pstDistAB{A}{C}]{B}{}[80][320]
@@ -662,6 +1097,35 @@ We will see later how to draw the circle crossing three points.
\end{pspicture}
\end{LTXexample}
+\vspace{10pt}
+
+The following example show how to use the more complex distance macros,
+and the parameter to fill the circle.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-3,-3)(3,3)\footnotesize
+\psset{unit=0.65cm}\psset{dotscale=0.5}\psset{PointSymbol=*}
+\pstGeonode[PosAngle=90,CurveType=polyline](0,0){A}(1,0){B}
+\pstGeonode[PosAngle=90,CurveType=polyline](0,1){A'}(2,1){B'}
+\pstCircleOA[linecolor=gray,Radius=\pstDistAdd{A}{B}{A'}{B'}]{A}{}
+\pstCircleOA[linecolor=red,Radius=\pstDistAddVal{A}{B}{1.0}{\pstDistConst{0.5}}]{A}{}
+\pstCircleOA[linecolor=blue,Radius=\pstDistAddCoef{A}{B}{0.5}{A'}{B'}{1.5}]{A}{}
+\pstCircleOA[linecolor=green,Radius=\pstDistSub{A}{B}{A'}{B'}]{B'}{}
+\pstCircleOA[linecolor=brown,Radius=\pstDistSubCoef{A}{B}{1.8}{A'}{B'}{0.5}]{A}{}
+\pnode(-1.5,-2){D}
+\pstCircleOA[linecolor=pink,fillstyle=solid,fillcolor=pink!40,Radius=\pstDistMul{A}{B}{0.8}]{D}{}
+\psdot(D)\uput{0.2}[-45](D){$D$}
+\pstCircleOA[linecolor=purple,Radius=\pstDistConst{\pstAbscissa{D}} abs]{D}{}
+\end{pspicture}
+\end{LTXexample}
+
+The last row set the absolute value of the abscissa of node $D$ to \Lkeyword{Radius},
+and then draw a circle at center $D$. Note that it does not work before v1.67,
+as the \Lcs{pstCircleOA} and \Lcs{pstCircleAB} were implemented with a \Lcs{rput} command,
+which will set the center $D$'s coordinate to origin, it causes that the Radius was set to zero
+and none circle will be draw out, so we remove the \Lcs{rput} code in v1.67,
+and everything works well now.
+
\subsection{Circle arcs}
\begin{BDef}
@@ -669,7 +1133,6 @@ We will see later how to draw the circle crossing three points.
\Lcs{pstArcnOAB}\OptArgs\Largb{O}\Largb{A}\Largb{B}
\end{BDef}
-
These two macros draw circle arcs, $O$ is the center, the radius
defined by $OA$, the beginning angle given by $A$ and the final angle
by $B$. Finally, the first macro draws the arc in the direct way,
@@ -688,6 +1151,97 @@ two points are at the same distance of $O$.
\end{pspicture}
\end{LTXexample}
+\subsection{Circle nodes}
+
+Do you want to draw a point on the circle? A point can be positioned on a circle
+using its rotation angle by macro \Lcs{pstCircleNode} or \Lcs{pstCircleRotNode}.
+The first \Lcs{pstCircleNode} requires an explicit parameter angle $\theta$ to calculate the point;
+but the second \Lcs{pstCircleRotNode} requires an implicit parameter \Lkeyword{RotAngle} to calculate the point,
+If you not set \Lkeyword{RotAngle}, the default value is $60^\circ$.
+
+The circle is defined by center $O$ and point $A$ on the circle or \Lkeyword{Radius} or \Lkeyword{Diameter} in parameter.
+
+\begin{BDef}
+\Lcs{pstCircleNode}\OptArgs\Largb{O}\Largb{A}\Largb{$\theta$}\Largb{X}\\
+\Lcs{pstCircleRotNode}\OptArgs\Largb{O}\Largb{A}\Largb{X}
+\end{BDef}
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-1,-1)(4,4)
+\psset{dotscale=0.5}\psset{PointSymbol=*}\footnotesize
+\psset{Radius=\pstDistVal{2.0}}
+\pstGeonode[PosAngle=0](1.5,1.5){O}
+\pstCircleOA[linecolor=red]{O}{}
+\pstCircleRotNode[PosAngle=0,RotAngle=0]{O}{}{A}
+\pstCircleRotNode[PosAngle=60]{O}{}{B} % default 60 degree
+\pstCircleRotNode[PosAngle=90,RotAngle=90]{O}{}{C}
+\pstCircleRotNode[PosAngle=150,RotAngle=\pscalculate{3*360/7}]{O}{}{D}
+\pstCircleRotNode[PosAngle=180,RotAngle=180]{O}{}{E}
+\pstCircleRotNode[PosAngle=230,RotAngle=230]{O}{}{F}
+\pstCircleRotNode[PosAngle=270,RotAngle=270]{O}{}{G}
+\pstCircleNode[PosAngle=-45]{O}{}{-45}{H}
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
+A point can be positioned on a circle using its absolute abscissa or ordinate too.
+You can input $x_1$ or $y_1$ as any number(e.g, 2.0), or use \Lcs{pscalculate} or \Lcs{fpeval} to generate the value,
+or use \Lcs{pstAbscissa} and \Lcs{pstOrdinate} to get the abscissa and ordinate of any other node.
+
+\begin{BDef}
+\Lcs{pstCircleAbsNode}\OptArgs\Largb{O}\Largb{A}\Largb{$x_1$}\Largb{C}\Largb{D}\\
+\Lcs{pstCircleOrdNode}\OptArgs\Largb{O}\Largb{A}\Largb{$y_1$}\Largb{C}\Largb{D}
+\end{BDef}
+
+for example,
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-1,-1)(4,4)
+\pstGeonode[PosAngle=60](1.5,1.5){O}
+\pstGeonode[PosAngle=-30](2.5,0){A}
+\pstCircleOA[linecolor=red]{O}{A}
+\pstCircleAbsNode[PosAngleA=-60,PosAngleB=60,PointSymbol=*]{O}{A}{1.0}{C}{D}
+\pstCircleOrdNode[PosAngleA=150,PosAngleB=30,PointSymbol=*]{O}{A}{1.0}{E}{F}
+\pstLineAB[linestyle=dashed,linecolor=gray!40,nodesep=-0.5]{C}{D}
+\pstLineAB[linestyle=dashed,linecolor=gray!40,nodesep=-0.5]{E}{F}
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
+A point can be positioned on a circle using its curved abscissa, that is,
+the arc length from a given node.
+
+\begin{BDef}
+\Lcs{pstCurvAbsNode}\OptArgs\Largb{O}\Largb{A}\Largb{B}\Largb{Abs}
+\end{BDef}
+
+\begin{sloppypar}
+Possible optional arguments are \Lkeyword{PointSymbol}, \Lkeyword{PosAngle},
+ \Lkeyword{PointName}, \Lkeyword{PointNameSep}, \Lkeyword{PtNameMath}, and \Lkeyword{CurvAbsNeg}.
+%
+The point \Argsans{$B$} is positioned on the circle of center
+\Argsans{$O$} crossing \Argsans{$A$}, with the curved abscissa
+\Argsans{Abs}. The origin is \Argsans{$A$} and the direction is
+anti-clockwise by default. The parameter \Lkeyword{CurvAbsNeg}
+\DefaultVal{false} can change this behavior.
+\end{sloppypar}
+
+If the parameter \Lkeyword{PosAngle} is not specified, the point label is put
+automatically in oirder to be alined with the circle center and the point.
+
+\begin{LTXexample}[width=5cm,pos=l]
+\begin{pspicture}[showgrid](-2.5,-2.5)(2.5,2.5)
+\pstGeonode{O}(2,0){A}
+\pstCircleOA{O}{A}
+\pstCurvAbsNode{O}{A}{M_1}{\pstDistVal{5}}
+\pstCurvAbsNode[CurvAbsNeg=true]%
+ {O}{A}{M_2}{\pstDistAB{A}{M_1}}
+\end{pspicture}
+\end{LTXexample}
+
+
+
\subsection{Circle tangent}
The macro \Lcs{pstCircleTangentLine} is used to draw a tangent line $AT$ from a point $A$ on the circle,
@@ -756,85 +1310,84 @@ You also can use \Lkeyword{DiameterA} and \Lkeyword{DiameterB} to define the two
\end{pspicture}
\end{LTXexample}
-\subsection{Curved abscissa}
-
-A point can be positioned on a circle using its curved abscissa.
+\subsection{Circle radical axis}
+If you want to draw the \texttt{Radical Axis} of two given circles, read the following sentenses.
+For given $\odot{O_1}$ with radius $r_1$ and $\odot{O_2}$ with radius $r_2$, and the center
+$O_1(x_1,y_1)$, $O_2(x_2,y_2)$, then any point $P(x,y)$ on the \texttt{Radical Axis} is satisfied:
+$$(x-x_1)^2+(y-y_1)^2-r_1^2=(x-x_2)^2+(y-y_2)^2-r_2^2$$
+It can be simplified to a equation of a line:
+$$2(x_2-x_1)x+2(y_2-y_1)y=(x_2^2+y_2^2-r_2^2)-(x_1^2+y_1^2-r_1^2)$$
+It is clear that the circles with same center have no radical axis,
+and the radical axis is perpendicular to the line of centers.
+We provide the macro \Lcs{pstCircleRadicalAxis} to draw the \texttt{Radical Axis} of two given circles.
+It can handler every position relations of circles such as separation, intersection and inclusion.
\begin{BDef}
-\Lcs{pstCurvAbsNode}\OptArgs\Largb{O}\Largb{A}\Largb{B}\Largb{Abs}
+\Lcs{pstCircleRadicalAxis}\OptArgs\Largb{$O_1$}\Largb{A}\Largb{$O_2$}\Largb{B}\Largb{C}\Largb{D}
\end{BDef}
-\begin{sloppypar}
-Possible optional arguments are \Lkeyword{PointSymbol}, \Lkeyword{PosAngle},
- \Lkeyword{PointName}, \Lkeyword{PointNameSep}, \Lkeyword{PtNameMath}, and \Lkeyword{CurvAbsNeg}.
-%
-The point \Argsans{$B$} is positioned on the circle of center
-\Argsans{$O$} crossing \Argsans{$A$}, with the curved abscissa
-\Argsans{Abs}. The origin is \Argsans{$A$} and the direction is
-anti-clockwise by default. The parameter \Lkeyword{CurvAbsNeg}
-\DefaultVal{false} can change this behavior.
-\end{sloppypar}
+Both parameter $A$ and $B$ can be omitted and then to specify the each radius or
+diameter using the parameters \Lkeyword{RadiusA}, \Lkeyword{DiameterA}, and \Lkeyword{RadiusB}, \Lkeyword{DiameterB}.
+This macro create two new nodes $C$ and $D$ on the radical axis, you can find them in following examples.
-If the parameter \Lkeyword{PosAngle} is not specified, the point label is put
-automatically in oirder to be alined with the circle center and the point.
+When they are intersected, we can see the radical axis is the intersected chord line.
-\begin{LTXexample}[width=5cm,pos=l]
-\begin{pspicture}[showgrid](-2.5,-2.5)(2.5,2.5)
-\pstGeonode{O}(2,0){A}
-\pstCircleOA{O}{A}
-\pstCurvAbsNode{O}{A}{M_1}{\pstDistVal{5}}
-\pstCurvAbsNode[CurvAbsNeg=true]%
- {O}{A}{M_2}{\pstDistAB{A}{M_1}}
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-1,-1)(4,4)
+\psset{dotscale=0.5}\psset{PointSymbol=*}\footnotesize
+\def\ra{1.2}\def\rb{2.0}
+\pstGeonode[PosAngle=0](0,1){O1}(1.5,1.5){O2}
+\pstCircleOA[linecolor=red!50,Radius=\pstDistVal{\ra}]{O1}{}
+\pstCircleOA[linecolor=blue!50,Radius=\pstDistVal{\rb}]{O2}{}
+\pstCircleRadicalAxis[PosAngle={0,0},RadiusA=\pstDistVal{\ra},RadiusB=\pstDistVal{\rb},nodesep=-1,linecolor=brown]{O1}{}{O2}{}{A}{B}
\end{pspicture}
\end{LTXexample}
+When they are tangent, we can see the radical axis is the common tangent line.
-A point can be positioned on a circle using its absolute abscissa or ordinate too.
-You can input $x_1$ or $y_1$ as any number(e.g, 2.0), or use \Lcs{pscalculate} or \Lcs{fpeval} to generate the value,
-or use \Lcs{pstAbscissa} and \Lcs{pstOrdinate} to get the abscissa and ordinate of any other node.
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-1,-1)(3,4)
+\psset{dotscale=0.5}\psset{PointSymbol=*}\footnotesize
+\def\ra{1.2}\def\rb{2.0}
+\pstGeonode[PosAngle=-90,PointName={O_1,O_2}](1,1){O1}(1,1.8){O2}
+\pstCircleOA[linecolor=red!50,Radius=\pstDistVal{\ra}]{O1}{}
+\pstCircleOA[linecolor=blue!50,Radius=\pstDistVal{\rb}]{O2}{}
+\pstCircleRadicalAxis[nodesep=-2,PosAngle={-90,-90},RadiusA=\pstDistVal{\ra},RadiusB=\pstDistVal{\rb}]{O1}{}{O2}{}{A}{B}
+\pstLineAB[linecolor=red,nodesep=-3]{A}{B}
+\end{pspicture}
+\end{LTXexample}
-\begin{BDef}
-\Lcs{pstCircleAbsNode}\OptArgs\Largb{O}\Largb{A}\Largb{$x_1$}\Largb{C}\Largb{C}\\
-\Lcs{pstCircleOrdNode}\OptArgs\Largb{O}\Largb{A}\Largb{$y_1$}\Largb{C}\Largb{C}
-\end{BDef}
+When one of them contains the other, the radical axis is out of the circles.
-for example,
\begin{LTXexample}[width=6cm,pos=l]
-\begin{pspicture}[showgrid=true](-1,-1)(4,4)
-\pstGeonode[PosAngle=60](1.5,1.5){O}
-\pstGeonode[PosAngle=-30](2.5,0){A}
-\pstCircleOA[linecolor=red]{O}{A}
-\pstCircleAbsNode[PosAngleA=-60,PosAngleB=60,PointSymbol=*]{O}{A}{1.0}{C}{D}
-\pstCircleOrdNode[PosAngleA=150,PosAngleB=30,PointSymbol=*]{O}{A}{1.0}{E}{F}
-\pstLineAB[linestyle=dashed,linecolor=gray!40,nodesep=-0.5]{C}{D}
-\pstLineAB[linestyle=dashed,linecolor=gray!40,nodesep=-0.5]{E}{F}
+\begin{pspicture}[showgrid=true](-1,-2)(4,3)
+\psset{dotscale=0.5}\psset{PointSymbol=*}\footnotesize
+\def\ra{1.2}\def\rb{2.0}
+\pstGeonode[PosAngle=0](1.2,1){O1}(1.5,1.5){O2}
+\pstCircleOA[linecolor=red!50,Radius=\pstDistVal{\ra}]{O1}{}
+\pstCircleOA[linecolor=blue!50,Radius=\pstDistVal{\rb}]{O2}{}
+\pstCircleRadicalAxis[PosAngle={-90,90},RadiusA=\pstDistVal{\ra},RadiusB=\pstDistVal{\rb},nodesepA=-1,nodesepB=-3,linecolor=brown]{O1}{}{O2}{}{A}{B}
\end{pspicture}
\end{LTXexample}
-A point can be positioned on a circle using its rotation angle by macro \Lcs{pstCircleRotNode}.
-The rotation angle should be passed by the \Lkeyword{RotAngle} in the \texttt{Options}.
-The circle is defined by center $O$ and point $A$ on the circle or \Lkeyword{Radius} in parameter.
-If you not set \Lkeyword{RotAngle}, the default value is $60^\circ$.
-
-\begin{BDef}
-\Lcs{pstCircleRotNode}\OptArgs\Largb{O}\Largb{A}\Largb{X}
-\end{BDef}
+When they are separated, the radical axis is between of the circles.
\begin{LTXexample}[width=6cm,pos=l]
-\begin{pspicture}[showgrid=true](-1,-1)(4,4)
+\begin{pspicture}[showgrid=true](-1,-2)(4,3)
\psset{dotscale=0.5}\psset{PointSymbol=*}\footnotesize
-\psset{Radius=\pstDistVal{2.0}}
-\pstGeonode[PosAngle=0](1.5,1.5){O}
-\pstCircleOA[linecolor=red]{O}{}
-\pstCircleRotNode[PosAngle=0,RotAngle=0]{O}{}{A}
-\pstCircleRotNode[PosAngle=60]{O}{}{B} % default 60 degree
-\pstCircleRotNode[PosAngle=90,RotAngle=90]{O}{}{C}
-\pstCircleRotNode[PosAngle=150,RotAngle=\pscalculate{3*360/7}]{O}{}{D}
-\pstCircleRotNode[PosAngle=180,RotAngle=180]{O}{}{E}
-\pstCircleRotNode[PosAngle=230,RotAngle=230]{O}{}{F}
-\pstCircleRotNode[PosAngle=270,RotAngle=270]{O}{}{G}
-\pstCircleRotNode[PosAngle=-45,RotAngle=-45]{O}{}{H}
+\def\ra{1.2}\def\rb{2.0}
+\pstGeonode[PosAngle=0](-1,0){O1}(2.5,1.5){O2}
+\pstCircleOA[linecolor=red!50,Radius=\pstDistVal{\ra}]{O1}{}
+\pstCircleOA[linecolor=blue!50,Radius=\pstDistVal{\rb}]{O2}{}
+\pstCircleRadicalAxis[PosAngle={-90,90},RadiusA=\pstDistVal{\ra},RadiusB=\pstDistVal{\rb},nodesep=-1,linecolor=brown]{O1}{}{O2}{}{A}{B}
+\psset{linestyle=dashed,linecolor=gray!40}
+\pstCircleTangentNode[Radius=\pstDistVal{\ra},PosAngle={90,200}]{O1}{}{A}{P}{Q}
+\pstCircleTangentNode[Radius=\pstDistVal{\rb},PosAngle={10,100}]{O2}{}{A}{X}{Y}
+\pstCircleOA{A}{P}
+\pstCircleTangentNode[Radius=\pstDistVal{\ra},PosAngle={210,200}]{O1}{}{B}{I}{J}
+\pstCircleTangentNode[Radius=\pstDistVal{\rb},PosAngle={10,-10}]{O2}{}{B}{R}{S}
+\pstCircleOA{B}{I}
\end{pspicture}
\end{LTXexample}
@@ -873,6 +1426,8 @@ used to modify the increment from a point to the next one
\end{pspicture}
\end{LTXexample}
+\clearpage
+
\section{Conics}
\subsection{Standard Ellipse}
The Standard Ellipse $E$ with coordinate translation is defined by center $O(x_0,y_0)$,
@@ -902,7 +1457,7 @@ the macro will draw the whole ellipse.
\begin{LTXexample}[width=6cm,pos=l]
\begin{pspicture}[showgrid=true](0,0)(4,4)
\psset{dotscale=0.5}\psset{PointSymbol=*}\footnotesize
-\def\ra{2.4}\def\rb{0.8}\def\rot{56}
+\def\ra{2.4}\def\rb{0.8}
\pstGeonode[PosAngle=-90,PointNameSep=0.2](2,2){O}
%\psellipse[linecolor=red!60](O)(\ra,\rb)
\pstEllipse[linecolor=red!60](O)(\ra,\rb)[0][120]
@@ -1226,8 +1781,8 @@ Using macro \Lcs{pstGeneralEllipseFocusNode} to find the two focus nodes, and ma
\Lcs{pstGeneralEllipseDirectrixLine} to get the two directrix lines.
\begin{BDef}
-\Lcs{pstGeneralEllipseFocusNode}\OptArgs\Largr{O}\Largr{$a,\,b$}\OptArg{$\theta$}\Largb{$t$}\Largb{A}\\
-\Lcs{pstGeneralEllipseDirectrixLine}\OptArgs\Largr{O}\Largr{$a,\,b$}\OptArg{$\theta$}\Largb{A}
+\Lcs{pstGeneralEllipseFocusNode}\OptArgs\Largr{O}\Largr{$a,\,b$}\OptArg{$\theta$}\Largb{$t$}\Largb{$F_1$}\Largb{$F_2$}\\
+\Lcs{pstGeneralEllipseDirectrixLine}\OptArgs\Largr{O}\Largr{$a,\,b$}\OptArg{$\theta$}\Largb{$L_x$}\Largb{$L_y$}\Largb{$R_x$}\Largb{$R_y$}
\end{BDef}
for example,
@@ -1303,8 +1858,6 @@ please refer to Theorem \ref{EllipseTangentPointTheorem}.
\end{pspicture}
\end{LTXexample}
-\clearpage
-
\subsection{Standard Parabola}
The Standard Parabola $P$ with coordinate translation is defined by vertex $O(x_0,y_0)$,
the half of the focus chord axis $abs(p)$.
@@ -1434,7 +1987,7 @@ If you don't know the focus $F$, or the directrix line, we will find them automa
\pstParabolaFocusNode[linecolor=red!40](O){\p}{F}
\pstLine[linecolor=gray!40,nodesepA=-0.8,nodesepB=-0.8]{0,2}{4,1}
\pstParabolaLineInter[linecolor=gray!40,PosAngle={120,210}](O){\p}{0,2}{4,1}{P}{Q}
-% if you know focus F, but don't known directrix line
+% if you know focus F, but don't know directrix line
\pstParabolaPolarNode[linecolor=purple!40,PosAngle=-90](O){\p}(F){P}{Q}{T}
\end{pspicture}
\end{LTXexample}
@@ -2204,7 +2757,7 @@ The macro \Lcs{pstHyperbolaTangentNode} is used to find the tangent point $A$ an
We use the following theorem to find the tangent points $A$ and $B$ of $T$:
\begin{theorem}\label{HyperbolaTangentPointTheorem}
Let $T$ is a point out of the hyperbola, for any two chords $TPQ$ and $TRS$ of the hyperbola, suppose $PR$ and $QS$ intersect at $X$,
-$RQ$ and $PS$ intersect at $Y$, then the intersection point $A$ and $B$ of $XY$ and the hyperbola are the tangent points from $T$.
+$RQ$ and $PS$ intersect at $Y$, then the intersection points $A$ and $B$ of $XY$ and the hyperbola are the tangent points from $T$.
\end{theorem}
\begin{LTXexample}[width=6cm,pos=l]
@@ -2356,7 +2909,7 @@ The macro \Lcs{pstIHyperbolaTangentNode} is used to find the tangent point $A$ a
\Lcs{pstIHyperbolaTangentNode}\OptArgs\Largr{O}\Largr{$a,\,b$}\Largb{$T$}\Largb{$A$}\Largb{$B$}
\end{BDef}
-We also use the following theorem \ref{HyperbolaTangentPointTheorem} to find the tangent points $A$ and $B$ of $T$.
+We also use the theorem \ref{HyperbolaTangentPointTheorem} to find the tangent points $A$ and $B$ of $T$.
\begin{LTXexample}[width=6cm,pos=l]
\begin{pspicture}[showgrid=true](-2,-1)(4,3)
@@ -2539,7 +3092,7 @@ The macro \Lcs{pstGeneralHyperbolaTangentNode} is used to find the tangent point
\Lcs{pstGeneralHyperbolaTangentNode}\OptArgs\Largr{O}\Largr{$a,\,b$}\OptArg{$\theta$}\Largb{$T$}\Largb{$A$}\Largb{$B$}
\end{BDef}
-We also use the following theorem \ref{HyperbolaTangentPointTheorem} to find the tangent points $A$ and $B$ of $T$.
+We also use the theorem \ref{HyperbolaTangentPointTheorem} to find the tangent points $A$ and $B$ of $T$.
\begin{LTXexample}[width=6cm,pos=l]
\begin{pspicture}[showgrid=true](-2,-1)(4,3)
@@ -2725,7 +3278,7 @@ The macro \Lcs{pstGeneralIHyperbolaTangentNode} is used to find the tangent poin
\Lcs{pstGeneralIHyperbolaTangentNode}\OptArgs\Largr{O}\Largr{$a,\,b$}\OptArg{$\theta$}\Largb{$T$}\Largb{$A$}\Largb{$B$}
\end{BDef}
-We also use the following theorem \ref{HyperbolaTangentPointTheorem} to find the tangent points $A$ and $B$ of $T$.
+We also use the theorem \ref{HyperbolaTangentPointTheorem} to find the tangent points $A$ and $B$ of $T$.
\begin{LTXexample}[width=6cm,pos=l]
\begin{pspicture}[showgrid=true](-2,-1)(4,3)
@@ -3204,8 +3757,8 @@ the circle of centre $O$ and with radius $OC$.
The circle is specified with its center and either a point of its
circumference or with a radius specified with parameter \Lkeyword{radius}
or its diameter specified with parameter \Lkeyword{Diameter}. These two
-parameters can be modify by coefficient \Lkeyword{DistCoef}.
-
+parameters can be specified by macros \Lcs{pstDist},\Lcs{pstDistMul},\Lcs{pstDistAdd},
+\Lcs{pstDistSub} etc.
The position of the wo points is such that the vectors $\vec{AB}$ abd
$\vec{M_1M_2}$ are in the same direction. Thus, if the points
@@ -3222,9 +3775,9 @@ at the center of the circle.
\pstCircleOA[linecolor=red]{O}{A}
\pstInterLC[PosAngle=-80]{C}{B}{O}{A}{D}{E}
\pstInterLC[PosAngleB=60, Radius=\pstDistAB{O}{D}]{I}{C}{O}{}{F}{G}
-\pstInterLC[PosAngleB=180,DistCoef=1.3,Diameter=\pstDistAB{O}{D}]
+\pstInterLC[PosAngleB=180,Diameter=\pstDistMul{O}{D}{1.3}]
{I}{B}{O}{}{H}{J}
-\pstCircleOA[linecolor=red,DistCoef=1.3,Diameter=\pstDistAB{O}{D}]{O}{}
+\pstCircleOA[linecolor=red,Diameter=\pstDistMul{O}{D}{1.3}]{O}{}
\psset{nodesep=-1}
\pstLineAB[linecolor=green]{E}{C}
\pstLineAB[linecolor=cyan]{I}{C}
@@ -3269,21 +3822,27 @@ specification using radius and diameter. For such specifications it
exists the parameters \Lkeyword{RadiusA}, \Lkeyword{RadiusB},
\Lkeyword{DiameterA} and \Lkeyword{DiameterB}.
+The macro \Lcs{pstInterCC} will not display the intersections as default, if you want to display
+the label or symbol of the intersections, you must setup the parameters \Lkeyword{PosAngleA}
+and \Lkeyword{PosAngleB} to change the default behavior.
+
\begin{LTXexample}
\begin{pspicture}[showgrid](-3,-4)(7,3)
\pstGeonode[PointName={\Omega,O}](3,-1){Omega}(1,-1){O}
\pstGeonode[PointSymbol=square, PosAngle={-90,90}](0,3){A}(2,2){B}
\psset{PointSymbol=o}
-\pstCircleOA[linecolor=red, DistCoef=1 3 10 div add, Radius=\pstDistAB{A}{B}]{O}{}
+\pstCircleOA[linecolor=red, Radius=\pstDistMul{A}{B}{1 3 10 div add}]{O}{}
\pstCircleOA[linecolor=Orange, Diameter=\pstDistAB{A}{B}]{O}{}
\pstCircleOA[linecolor=Violet, Radius=\pstDistAB{A}{B}]{Omega}{}
\pstCircleOA[linecolor=Purple, Diameter=\pstDistAB{A}{B}]{Omega}{}
-\pstInterCC[DistCoef=1 3 10 div add, RadiusA=\pstDistAB{A}{B},
- DistCoef=none, RadiusB=\pstDistAB{A}{B}]{O}{}{Omega}{}{D}{E}
-\pstInterCC[DiameterA=\pstDistAB{A}{B}, RadiusB=\pstDistAB{A}{B}]{O}{}{Omega}{}{F}{G}
-\pstInterCC[DistCoef=1 3 10 div add, RadiusA=\pstDistAB{A}{B},
- DistCoef=none, DiameterB=\pstDistAB{A}{B}]{O}{}{Omega}{}{H}{I}
-\pstInterCC[DiameterA=\pstDistAB{A}{B}, DiameterB=\pstDistAB{A}{B}]{O}{}{Omega}{}{J}{K}
+\pstInterCC[RadiusA=\pstDistMul{A}{B}{1 3 10 div add},
+ RadiusB=\pstDistAB{A}{B}]{O}{}{Omega}{}{D}{E}
+\pstInterCC[DiameterA=\pstDistAB{A}{B}, RadiusB=\pstDistAB{A}{B},
+ PosAngleA=90,PosAngleB=-90]{O}{}{Omega}{}{F}{G}
+\pstInterCC[RadiusA=\pstDistMul{A}{B}{1 3 10 div add}, DiameterB=\pstDistAB{A}{B},
+ PosAngleA=90,PosAngleB=-90]{O}{}{Omega}{}{H}{I}
+\pstInterCC[DiameterA=\pstDistAB{A}{B}, DiameterB=\pstDistAB{A}{B},
+ PosAngleA=90,PosAngleB=-90]{O}{}{Omega}{}{J}{K}
\end{pspicture}
\end{LTXexample}
generated by cgit v1.2.3 (git 2.39.1) at 2024-12-29 06:05:45 +0000