diff options
Diffstat (limited to 'Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-presentation.tex')
-rw-r--r-- | Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-presentation.tex | 72 |
1 files changed, 36 insertions, 36 deletions
diff --git a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-presentation.tex b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-presentation.tex index 0073d788032..027cb7ea0dd 100644 --- a/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-presentation.tex +++ b/Master/texmf-dist/doc/latex/tkz-euclide/TKZdoc-euclide-presentation.tex @@ -51,27 +51,27 @@ Here are some comparisons between \tkzname{\TIKZ } and \tkzname{\tkznameofpack} Explanation : -The fourth tutorial of the \emph{PgfManual} is about geometric constructions. \emph{T. Tantau} proposes to get the drawing with its beautiful tool Ti\emph{k}Z. Here I propose the same construction with \emph{tkz-elements}. The color of the Ti\emph{k}Z code is orange and that of \emph{tkz-elements} is red. +The fourth tutorial of the \emph{PgfManual} is about geometric constructions. \emph{T. Tantau} proposes to get the drawing with its beautiful tool Ti\emph{k}Z. Here I propose the same construction with \emph{tkz-elements}. The color of the Ti\emph{k}Z code is green!50!black and that of \emph{tkz-elements} is red. \medskip -\hspace*{1cm}\vbox{\color{orange} |\usepackage{tikz}|\\ +\vbox{\color{green!50!black} |\usepackage{tikz}|\\ |\usetikzlibrary{calc,intersections,through,backgrounds}|} \medskip -\hspace*{1cm}\vbox{\color{red} |\usepackage{tkz-euclide}|} +\vbox{\color{red} |\usepackage{tkz-euclide}|} \medskip How to get the line AB ? To get this line, we use two fixed points.\\ \medskip -\hspace*{1cm}\vbox{\color{orange} +\vbox{\color{green!50!black} |\coordinate [label=left:$A$] (A) at (0,0);|\\ |\coordinate [label=right:$B$] (B) at (1.25,0.25);|\\ |\draw (A) -- (B);|} \medskip -\hspace*{1cm}\vbox{\color{red} +\vbox{\color{red} |\tkzDefPoint(0,0){A}|\\ |\tkzDefPoint(1.25,0.25){B}|\\ |\tkzDrawSegment(A,B)|\\ @@ -81,14 +81,14 @@ How to get the line AB ? To get this line, we use two fixed points.\\ We want to draw a circle around the points $A$ and $B$ whose radius is given by the length of the line AB. \medskip -\hspace*{1cm}\vbox{\color{orange} +\vbox{\color{green!50!black} |\draw let \p1 = ($ (B) - (A) $),|\\ |\n2 = {veclen(\x1,\y1)} in|\\ | (A) circle (\n2)|\\ | (B) circle (\n2);|} \medskip -\hspace*{1cm}\vbox{\color{red} +\vbox{\color{red} |\tkzDrawCircles(A,B B,A)| } @@ -96,27 +96,27 @@ The intersection of the circles $\mathcal{D}$ and $\mathcal{E}$ \medskip -\hspace*{1cm}\vbox{\color{orange} +\vbox{\color{green!50!black} |draw [name path=A--B] (A) -- (B);|\\ |node (D) [name path=D,draw,circle through=(B),label=left:$D$] at (A) {}; |\\ |node (E) [name path=E,draw,circle through=(A),label=right:$E$] at (B) {};|\\ -|path [name intersections={of=D and E, by={[label=above:$C$]C, [label=below:$C'$]C'}}]; |\\ +|path [name intersections={of=D and E, by={[label=above:$C$]C,[label=below:$C'$]C'}}]; |\\ |draw [name path=C--C',red] (C) -- (C');|\\ |path [name intersections={of=A--B and C--C',by=F}];|\\ |node [fill=red,inner sep=1pt,label=-45:$F$] at (F) {};|\\} \medskip -\hspace*{1cm}\vbox{\color{red} |\tkzInterCC(A,B)(B,A) \tkzGetPoints{C}{X}|\\} +\vbox{\color{red} |\tkzInterCC(A,B)(B,A) \tkzGetPoints{C}{X}|\\} How to draw points : \medskip -\hspace*{1cm}\vbox{\color{orange} |\foreach \point in {A,B,C}|\\ +\vbox{\color{green!50!black} |\foreach \point in {A,B,C}|\\ |\fill [black,opacity=.5] (\point) circle (2pt);|\\} \medskip -\hspace*{1cm}\vbox{\color{red}| \tkzDrawPoints[fill=gray,opacity=.5](A,B,C)|\\} +\vbox{\color{red}| \tkzDrawPoints[fill=gray,opacity=.5](A,B,C)|\\} \subsubsection{Complete code with \pkg{tkz-euclide}} @@ -124,12 +124,12 @@ We need to define colors |\colorlet{input}{red!80!black} |\\ |\colorlet{output}{red!70!black}|\\ -|\colorlet{triangle}{orange!40} | +|\colorlet{triangle}{green!50!black!40} | \begin{tkzexample}[vbox,small] \colorlet{input}{red!80!black} \colorlet{output}{red!70!black} - \colorlet{triangle}{orange!40} + \colorlet{triangle}{green!50!black!40} \begin{tikzpicture}[scale=1.25,thick,help lines/.style={thin,draw=black!50}] \tkzDefPoint(0,0){A} \tkzDefPoint(1.25+rand(),0.25+rand()){B} @@ -164,7 +164,7 @@ Explanation In the first part, we need to find the midpoint of the straight line $AB$. With \TIKZ\ we can use the calc library \medskip -\hspace*{1cm}\vbox{\color{orange} |\coordinate [label=left:$A$] (A) at (0,0);|\\ +\vbox{\color{green!50!black} |\coordinate [label=left:$A$] (A) at (0,0);|\\ |\coordinate [label=right:$B$] (B) at (1.25,0.25);|\\ |\draw (A) -- (B);|\\ |\node [fill=red,inner sep=1pt,label=below:$X$] (X) at ($ (A)!.5!(B) $) {};|\\} @@ -173,40 +173,40 @@ With \pkg{tkz-euclide} we have a macro \tkzcname{tkzDefMidPoint}, we get the poi \medskip -\hspace*{1cm}\vbox{\red |\tkzDefPoints{0/0/A,0.75/0.25/B,1/1.5/C}|\\ -|\tkzDefMidPoint(A,B) \tkzGetPoint{X}|}\\ +\vbox{\red |\tkzDefPoints{0/0/A,0.75/0.25/B,1/1.5/C}|\\ +|\tkzDefMidPoint(A,B) \tkzGetPoint{X}|} \medskip Then we need to construct a triangle equilateral. It's easy with \pkg{tkz-euclide} . With TikZ you need some effort because you need to use the midpoint $X$ to get the point $D$ with trigonometry calculation. \medskip -\hspace*{1cm}\vbox{\color{orange} +\vbox{\color{green!50!black} |\node [fill=red,inner sep=1pt,label=below:$X$] (X) at ($ (A)!.5!(B) $) {}; | \\ |\node [fill=red,inner sep=1pt,label=above:$D$] (D) at | \\ |($ (X) ! {sin(60)*2} ! 90:(B) $) {}; | \\ |\draw (A) -- (D) -- (B); | \\ -} \\ +} \medskip -\hspace*{1cm}\vbox{\color{red} |\tkzDefTriangle[equilateral](A,B) \tkzGetPoint{D}|}\\ +\vbox{\color{red} |\tkzDefTriangle[equilateral](A,B) \tkzGetPoint{D}|} We can draw the triangle at the end of the picture with \medskip -\hspace*{1cm}\vbox{\color{red} |\tkzDrawPolygon{A,B,C}|} +\vbox{\color{red} |\tkzDrawPolygon{A,B,C}|} \medskip We know how to draw the circle $\mathcal{H}$ around $B$ through $C$ and how to place the points $E$ and $F$ \medskip -\hspace*{1cm}\vbox{\color{orange} +\vbox{\color{green!50!black} |\node (H) [label=135:$H$,draw,circle through=(C)] at (B) {};| \\ |\draw (D) -- ($ (D) ! 3.5 ! (B) $) coordinate [label=below:$F$] (F);| \\ -|\draw (D) -- ($ (D) ! 2.5 ! (A) $) coordinate [label=below:$E$] (E);|} \\ +|\draw (D) -- ($ (D) ! 2.5 ! (A) $) coordinate [label=below:$E$] (E);|} \medskip -\hspace*{1cm}\vbox{\color{red} |\tkzDrawCircle(B,C)|\\ +\vbox{\color{red} |\tkzDrawCircle(B,C)|\\ |\tkzDrawLines[add=0 and 2](D,A D,B)|} \medskip @@ -216,32 +216,32 @@ Intersecting a Line and a Circle : here we search the intersection of the circle The infinite straight line $DB$ intercepts the circle but with \TIKZ\ we need to extend the lines $DB$ and that can be done using partway calculations. We get the point $F$ and $BF$ or $DF$ intercepts the circle \medskip -\hspace*{1cm}\vbox{\color{orange}| \node (H) [label=135:$H$,draw,circle through=(C)] at (B) {}; | \\ +\vbox{\color{green!50!black}| \node (H) [label=135:$H$,draw,circle through=(C)] at (B) {}; | \\ |\path let \p1 = ($ (B) - (C) $) in| \\ | coordinate [label=left:$G$] (G) at ($ (B) ! veclen(\x1,\y1) ! (F) $); | \\ -|\fill[red,opacity=.5] (G) circle (2pt);|} \\ +|\fill[red,opacity=.5] (G) circle (2pt);|} \medskip Like the intersection of two circles, it's easy to find the intersection of a line and a circle with \pkg{tkz-euclide}. We don't need $F$ \medskip -\hspace*{1cm}\vbox{\color{red} | \tkzInterLC(B,D)(B,C)\tkzGetFirstPoint{G}|} +\vbox{\color{red} | \tkzInterLC(B,D)(B,C)\tkzGetFirstPoint{G}|} \medskip There are no more difficulties. Here the final code with some simplications. We draw the circle $\mathcal{K}$ with center $D$ and passing through $G$. It intersects the line $AD$ at point $L$. $AL = BC$. -\hspace*{1cm}\vbox{\color{red} | \tkzDrawCircle(D,G)|} -\hspace*{1cm}\vbox{\color{red} | \tkzInterLC(D,A)(D,G)\tkzGetSecondPoint{L}|} +\vbox{\color{red} | \tkzDrawCircle(D,G)|} +\vbox{\color{red} | \tkzInterLC(D,A)(D,G)\tkzGetSecondPoint{L}|} \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=1.5] \tkzDefPoint(0,0){A} \tkzDefPoint(0.75,0.25){B} \tkzDefPoint(1,1.5){C} -\tkzDefTriangle[equilateral](A,B) \tkzGetPoint{D} -\tkzInterLC[near](D,B)(B,C) \tkzGetSecondPoint{G} -\tkzInterLC[near](A,D)(D,G) \tkzGetFirstPoint{L} +\tkzDefTriangle[equilateral](A,B)\tkzGetPoint{D} +\tkzInterLC[near](D,B)(B,C) \tkzGetSecondPoint{G} +\tkzInterLC[near](A,D)(D,G) \tkzGetFirstPoint{L} \tkzDrawCircles(B,C D,G) \tkzDrawLines[add=0 and 2](D,A D,B) \tkzDrawSegment(A,B) @@ -399,7 +399,7 @@ Let's analyze the figure \[\widehat{BCA}=90^\circ -\alpha/2 \] \item Finally \[\widehat{CBD}=\alpha=36^\circ \] - the triangle $CBD$ is a "golden" triangle. + the triangle $CBD$ is a \code{golden} triangle. \end{enumerate} \vspace*{24pt} @@ -448,7 +448,7 @@ After building the golden triangle $BCD$, we build the point $A$ by noticing tha \subsubsection{Part II: two others methods with golden and euclid triangle} -\tkzname{\tkznameofpack} knows how to define a "golden" or "euclide" triangle. We can define $BCD$ and $BCA$ like gold triangles. +\tkzname{\tkznameofpack} knows how to define a \code{golden} or \code{euclide} triangle. We can define $BCD$ and $BCA$ like gold triangles. \begin{center} @@ -529,7 +529,7 @@ $IB=a$, $AI=1$ \tkzDefMidPoint(A,B) \tkzGetPoint{M} \tkzDefPointWith[orthogonal](I,M) \tkzGetPoint{H} \tkzInterLC(I,H)(M,B) \tkzGetFirstPoint{C} - \tkzDrawSegment[style=orange](I,C) + \tkzDrawSegment[style=purple](I,C) \tkzDrawArc(M,B)(A) \tkzDrawSegment[dim={$1$,-16pt,}](A,I) \tkzDrawSegment[dim={$(a-1)/2$,-10pt,}](I,M) @@ -587,7 +587,7 @@ The following code consists of several parts: \item The third one includes the different drawings; \begin{tkzltxexample}[] - \tkzDrawSegment[style=orange](I,H) + \tkzDrawSegment[style=purple](I,H) \tkzDrawPoints(O,I,A,B,M) \tkzDrawArc(M,A)(O) \tkzDrawSegment[dim={$1$,-16pt,}](A,I) |