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authorKarl Berry <karl@freefriends.org>2022-07-15 21:46:25 +0000
committerKarl Berry <karl@freefriends.org>2022-07-15 21:46:25 +0000
commita37835eecfb8c04bd99ca2420cc72418dec4198b (patch)
treea1c9effe63509d07db09207df1153ef0ed5ba255 /Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-pointby.tex
parent0c65eda87a3a65cdb4bcf42a2912bd9721816bd7 (diff)
tkz-euclide (15jul22)
git-svn-id: svn://tug.org/texlive/trunk@63907 c570f23f-e606-0410-a88d-b1316a301751
Diffstat (limited to 'Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-pointby.tex')
-rw-r--r--Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-pointby.tex89
1 files changed, 36 insertions, 53 deletions
diff --git a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-pointby.tex b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-pointby.tex
index a2f5584670a..67e357c625f 100644
--- a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-pointby.tex
+++ b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-pointby.tex
@@ -1,4 +1,4 @@
-\section{Definition of points by transformation : \tkzcname{tkzDefPointBy} }
+\section{Definition of points by transformation}
These transformations are:
\begin{itemize}
@@ -11,11 +11,13 @@ These transformations are:
\item inversion with respect to a circle.
\end{itemize}
+\subsection{\tkzcname{tkzDefPointBy}}
The choice of transformations is made through the options. There are two macros, one for the transformation of a single point \tkzcname{tkzDefPointBy} and the other for the transformation of a list of points \tkzcname{tkzDefPointsBy}. By default the image of $A$ is $A'$. For example, we'll write:
\begin{tkzltxexample}[]
\tkzDefPointBy[translation= from A to A'](B)
\end{tkzltxexample}
The result is in \tkzname{tkzPointResult}
+
\medskip
\begin{NewMacroBox}{tkzDefPointBy}{\oarg{local options}\parg{pt}}%
The argument is a simple existing point and its image is stored in \tkzname{tkzPointResult}. If you want to keep this point then the macro \tkzcname{tkzGetPoint\{M\}} allows you to assign the name \tkzname{M} to the point.
@@ -44,11 +46,11 @@ options & & examples \\
\bottomrule
\end{tabular}
-The image is only defined and not drawn.
+\medskip
+\emph{The image is only defined and not drawn.}
\end{NewMacroBox}
-\subsection{Examples of transformations}
-\subsubsection{translation}
+\subsubsection{\tkzname{translation}}
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[>=latex]
@@ -61,7 +63,7 @@ The image is only defined and not drawn.
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{reflection (orthogonal symmetry)}
+\subsubsection{\tkzname{reflection} (orthogonal symmetry)}
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=.75]
@@ -217,7 +219,6 @@ Directly
\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[scale=.5]
\tkzDefPoints{4/0/A,6/0/P,0/0/O}
- \tkzDefCircle(O,A)
\tkzDefPointBy[inversion = center O through A](P)
\tkzGetPoint{P'}
\tkzDrawSegments(O,P)
@@ -230,10 +231,9 @@ Directly
\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[scale=.5]
\tkzDefPoints{4/0/A,6/0/P,0/0/O}
- \tkzDefCircle(O,A)
\tkzDefLine[orthogonal=through P](O,P)
\tkzGetPoint{L}
- \tkzDefTangent[from = P](O,A) \tkzGetPoints{R}{Q}
+ \tkzDefLine[tangent from = P](O,A) \tkzGetPoints{R}{Q}
\tkzDefPointBy[projection=onto O--A](Q) \tkzGetPoint{P'}
\tkzDrawSegments(O,P O,A)
\tkzDrawSegments[new](O,P O,Q P,Q Q,P')
@@ -252,7 +252,7 @@ Directly
\end{tkzexample}
-\subsubsection{Inversion of lines}
+\subsubsection{\tkzname{Inversion of lines} ex 1}
\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[scale=.5]
\tkzDefPoints{0/0/O,3/0/I,4/3/P,6/-3/Q}
@@ -262,7 +262,8 @@ Directly
\tkzGetPoint{A'}
\tkzDefPointBy[inversion = center O through I](P)
\tkzGetPoint{P'}
-\tkzDrawCircle[new,diameter](O,A')
+\tkzDefCircle[diameter](O,A')\tkzGetPoint{o}
+\tkzDrawCircle[new](o,A')
\tkzDrawLines[add=.25 and .25,red](P,Q)
\tkzDrawLines[add=.25 and .25](O,A)
\tkzDrawSegments(O,P)
@@ -270,6 +271,7 @@ Directly
\end{tikzpicture}
\end{tkzexample}
+\subsubsection{\tkzname{inversion of lines} ex 2}
\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[scale=.5]
\tkzDefPoints{0/0/O,3/0/I,3/2/P,3/-2/Q}
@@ -279,7 +281,8 @@ Directly
\tkzGetPoint{A'}
\tkzDefPointBy[inversion = center O through I](P)
\tkzGetPoint{P'}
-\tkzDrawCircle[new,diameter](O,A')
+\tkzDefCircle[diameter](O,A')\tkzGetPoint{o}
+\tkzDrawCircle[new](o,A')
\tkzDrawLines[add=.25 and .25,red](P,Q)
\tkzDrawLines[add=.25 and .25](O,A)
\tkzDrawSegments(O,P)
@@ -287,6 +290,7 @@ Directly
\end{tikzpicture}
\end{tkzexample}
+\subsubsection{\tkzname{inversion of lines} ex 3}
\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[scale=.5]
\tkzDefPoints{0/0/O,3/0/I,2/1/P,2/-2/Q}
@@ -296,7 +300,8 @@ Directly
\tkzGetPoint{A'}
\tkzDefPointBy[inversion = center O through I](P)
\tkzGetPoint{P'}
-\tkzDrawCircle[new,diameter](O,A')
+\tkzDefCircle[diameter](O,A')
+\tkzDrawCircle[new](I,A')
\tkzDrawLines[add=.25 and .75,red](P,Q)
\tkzDrawLines[add=.25 and .25](O,A')
\tkzDrawSegments(O,P')
@@ -304,31 +309,31 @@ Directly
\end{tikzpicture}
\end{tkzexample}
-
-\subsubsection{Inversion of circle}
-\begin{tkzexample}[latex=6cm,small]
-\begin{tikzpicture}[scale=.5]
+\subsubsection{\tkzname{inversion} of circle and \tkzname{homothety} }
+\begin{tkzexample}[latex=7cm,small]
+\begin{tikzpicture}[scale=.75]
\tkzDefPoints{0/0/O,3/2/A,2/1/P}
-\tkzDefTangent[from = O](A,P) \tkzGetPoints{T}{X}
-\tkzDefPointsBy[homothety=center O ratio 1.25](A,P,T){}
-\tkzInterCC(A,P)(A',P') \tkzGetPoints{C}{D}
+\tkzDefLine[tangent from = O](A,P) \tkzGetPoints{T}{X}
+\tkzDefPointsBy[homothety = center O%
+ ratio 1.25](A,P,T){}
+\tkzInterCC(A,P)(A',P') \tkzGetPoints{C}{D}
\tkzCalcLength(A,P)
\tkzGetLength{rAP}
-\tkzDefPointOnCircle[R= angle 190 center A radius \rAP]
+\tkzDefPointOnCircle[R= center A angle 190 radius \rAP]
\tkzGetPoint{M}
\tkzDefPointBy[inversion = center O through C](M)
\tkzGetPoint{M'}
-\tkzDrawCircles(A,P A',P')
+\tkzDrawCircles[new](A,P A',P')
\tkzDrawCircle(O,C)
\tkzDrawLines[add=0 and .5](O,T' O,A' O,M' O,P')
\tkzDrawPoints(A,A',P,P',O,T,T',M,M')
-\tkzLabelPoints(O,T,T')
-\tkzLabelPoints[above left](M,M')
+\tkzLabelPoints(O,T,T',M,M')
\tkzLabelPoints[below](P,P')
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Inversion of Triangle with respect to the Incircle}
+
+\subsubsection{\tkzname{inversion} of Triangle with respect to the Incircle}
\begin{tkzexample}[latex=6cm,small]
\begin{tikzpicture}[scale=1]
\tkzDefPoints{0/0/A,5/1/B,3/6/C}
@@ -348,15 +353,15 @@ Directly
\tkzDrawPolygon(A,B,C)
\tkzDrawCircle(O,b)\tkzDrawPoints(A,B,C,O)
\tkzDrawCircles[dashed,gray](Ja,y Jb,x Jc,z)
-\tkzDrawArc[line width=1pt,orange](Jb,x)(z)
-\tkzDrawArc[line width=1pt,orange](Jc,z)(y)
-\tkzDrawArc[line width=1pt,orange](Ja,y)(x)
+\tkzDrawArc[line width=1pt,orange,delta=0](Jb,x)(z)
+\tkzDrawArc[line width=1pt,orange,delta=0](Jc,z)(y)
+\tkzDrawArc[line width=1pt,orange,delta=0](Ja,y)(x)
\tkzLabelPoint[below](A){$A$}\tkzLabelPoint[above](C){$C$}
\tkzLabelPoint[right](B){$B$}\tkzLabelPoint[below](O){$O$}
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Inversion: orthogonal circle with inversion circle}
+\subsubsection{\tkzname{inversion}: orthogonal circle with inversion circle}
The inversion circle itself, circles orthogonal to it, and lines through the inversion center are invariant under inversion. If the circle meets the reference circle, these invariant points of intersection are also on the inverse circle. See I and J in the next figure.
\begin{tkzexample}[latex=5cm,small]
@@ -364,7 +369,7 @@ The inversion circle itself, circles orthogonal to it, and lines through the inv
\tkzDefPoint(0,0){O}\tkzDefPoint(1,0){A}
\tkzDefPoint(-1.5,-1.5){z1}
\tkzDefPoint(1.5,-1.25){z2}
-\tkzDefCircleBy[orthogonal through=z1 and z2](O,A)
+\tkzDefCircle[orthogonal through=z1 and z2](O,A)
\tkzGetPoint{c}
\tkzDrawCircle[new](c,z1)
\tkzDefPointBy[inversion = center O through A](z1)
@@ -379,28 +384,6 @@ The inversion circle itself, circles orthogonal to it, and lines through the inv
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{\tkzname{Inversion} and \tkzname{homothety} }
-\begin{tkzexample}[latex=7cm,small]
-\begin{tikzpicture}[scale=.75]
-\tkzDefPoints{0/0/O,3/2/A,2/1/P}
-\tkzDefTangent[from = O](A,P) \tkzGetPoints{T}{X}
-\tkzDefPointsBy[homothety = center O%
- ratio 1.25](A,P,T){}
-\tkzInterCC(A,P)(A',P') \tkzGetPoints{C}{D}
-\tkzCalcLength(A,P)
-\tkzGetLength{rAP}
-\tkzDefPointOnCircle[R= angle 190 center A radius \rAP]
-\tkzGetPoint{M}
-\tkzDefPointBy[inversion = center O through C](M)
-\tkzGetPoint{M'}
-\tkzDrawCircles[new](A,P A',P')
-\tkzDrawCircle(O,C)
-\tkzDrawLines[add=0 and .5](O,T' O,A' O,M' O,P')
-\tkzDrawPoints(A,A',P,P',O,T,T',M,M')
-\tkzLabelPoints(O,T,T',M,M')
-\tkzLabelPoints[below](P,P')
-\end{tikzpicture}
-\end{tkzexample}
For a more complex example see \tkzname{Pappus} \ref{pappus}
@@ -477,7 +460,7 @@ options & & examples \\
The points are only defined and not drawn.
\end{NewMacroBox}
-\subsubsection{Example of translation}
+\subsubsection{\tkzname{translation} of multiple points}
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[>=latex]
\tkzDefPoints{0/0/A,3/0/B,3/1/A',1/2/C}
@@ -493,7 +476,7 @@ The points are only defined and not drawn.
\end{tikzpicture}
\end{tkzexample}
-\subsubsection{Example of symmetry: an oval}
+\subsubsection{\tkzname{symmetry} of multiple points: an oval}
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=0.4]