3

In the following example, why there is no blue triangle ?

\documentclass[a4paper]{article}
\usepackage[marginparsep=3pt, top=2cm, bottom=1.5cm, left=3cm, right=1.5cm]{geometry}

\usepackage{tikz}
\usetikzlibrary{calc,intersections}

\tikzset{small dot/.style={fill=black,circle,scale=0.3},}

\begin{document}
\begin{tikzpicture}

\coordinate (A0) at (0,0) ;
\coordinate (A1) at (1,0) ;
\coordinate (A2) at (0,1) ;

\draw[fill=red] (A0)--(A1)--(A2)--cycle ;

\draw ($ (A0)!.5!(A1) $) node (E0) {};  
\draw ($ (A1)!.5!(A2) $) node (E1) {};  
\draw ($ (A2)!.5!(A0) $) node (E2) {};  

\draw[fill=blue] (E0)--(E1)--(E2)--cycle ;

\end{tikzpicture}
\end{document}

enter image description here

It is just a simple sample of the problem I have. As I need to use let...in... to calculate between ! ! coefficient, I use this definition of nodes for the Ex nodes. If there is another way or how to correct the blue triangle drawing.

One can tcheat :

\draw ($ (A0)!.5!(A1) $) node (E0) {};  
\draw ($ (A1)!.5!(A2) $) node (E1) {};  
\draw ($ (A2)!.5!(A0) $) node (E2) {};  

\coordinate (B0) at (E0) ;
\coordinate (B1) at (E1) ;
\coordinate (B2) at (E2) ;


\draw[fill=blue] (B0)--(B1)--(B2)--cycle ;

works fine, but why i doesn't work at first time ?

Tarass
  • 16,912

3 Answers3

4

Define node's as coordinate:

\documentclass[a4paper]{article}
\usepackage[marginparsep=3pt, top=2cm, bottom=1.5cm, left=3cm, right=1.5cm]{geometry}

\usepackage{tikz}
\usetikzlibrary{calc,intersections}

%\tikzset{small dot/.style={fill=black,circle,scale=0.3},}

\begin{document}
\begin{tikzpicture}

\coordinate (A0) at (0,0) ;
\coordinate (A1) at (1,0) ;
\coordinate (A2) at (0,1) ;

\draw [fill=red] (A0)--(A1)--(A2)--cycle ;

\coordinate (E0) at ($ (A0)!.5!(A1) $);  
\coordinate (E1) at ($ (A1)!.5!(A2) $);  
\coordinate (E2) at ($ (A2)!.5!(A0) $)  ;  

\draw [fill=blue] (E0)--(E1)--(E2)--cycle ;

\end{tikzpicture}
\end{document}
  • I was my firt thaught but I need to use let...in to calculate the coordinates and none of this works :
    \coordinate let \p1=(A0) in (E0) at ($ (A0)!.5!(A1) $);
    \coordinate (E1) at \p1=(A0) in ($ (A1)!.5!(A2) $);
    . This examples are useless, but it's just for example.     – Tarass Feb 26 '14 at 13:04
  • But one can tcheat, I add an edit. – Tarass Feb 26 '14 at 13:07
  • Sorry I edited your message instead of mine, excuse me please. And thank you. – Tarass Feb 26 '14 at 13:12
  • 1
    I erroneously approved Tarass edit. @Tarass: the modification you're suggesting is working, but not in your original example. Though, even your original example could work: add the option [coordinate] to all nodes. Then, have a look to the difference between nodes and coordinates: on this site there are good references. – Claudio Fiandrino Feb 26 '14 at 13:12
  • @Claudio Fiandrino: I check no difference between my cheat and your good solution. I keep yours ;-) thank you. – Tarass Feb 26 '14 at 13:18
4

nodes are not geometric points but occupy space even if they don't have content. This is because they have inner sep and outer separations. When you artie

\draw[fill=blue] (E0)--(E1)--(E2)--cycle ;

you are not giving from which point of node the lines should be drawn as no anchors are specified in the above line of code. Please note that a coordinate is a geometric point and thus you can draw straight from it. Therefore to draw from a node, a proper anchor must be specified like (E0.center). On the other hand, instead of a node, you can define a coordinate straight away as ferahfeza did in his answer, or simply define node as a coordinate in the following manner:

node[coordinate] (E0) {};

or 

node[inner sep=0pt,outer sep=0pt] (E0) {};

Both of these work. But if you adopt the latter, you still have to give the anchors like (E0.center).

A sample code:

\documentclass[a4paper]{article}
\usepackage[marginparsep=3pt, top=2cm, bottom=1.5cm, left=3cm, right=1.5cm]{geometry}

\usepackage{tikz}
\usetikzlibrary{calc,intersections}

\tikzset{small dot/.style={fill=black,circle,scale=0.3},}

\begin{document}
\begin{tikzpicture}

\coordinate (A0) at (0,0) ;
\coordinate (A1) at (1,0) ;
\coordinate (A2) at (0,1) ;

\draw[fill=red] (A0)--(A1)--(A2)--cycle ;

\path ($ (A0)!.5!(A1) $) node[inner sep=0pt,outer sep=0pt] (E0) {};
\path ($ (A1)!.5!(A2) $) node[inner sep=0pt,outer sep=0pt] (E1) {};
\path ($ (A2)!.5!(A0) $) node[inner sep=0pt,outer sep=0pt] (E2) {};

\draw[fill=blue] (E0.north)--(E1.center)--(E2.east)--cycle ;

\end{tikzpicture}
\end{document}

enter image description here

Please note that, instead of giving center as the anchor from which to draw, I have chosen appropriate anchors (in a failed attempt) so that the blue triangle corners don't protrude out of the red triangle. Moral is, when you want coordinate, define one and not nodes.

3

A compact solution without calc based on default options.

\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[scale=5]
\draw[fill=red]   (0,0) -- coordinate (E0)(1,0) 
                        -- coordinate (E1)(0,1) 
                        -- coordinate (E2) 
                           cycle ;   
\draw[fill=blue] (E0)--(E1)--(E2)--cycle ;
\end{tikzpicture}
\end{document}

enter image description here

Alain Matthes
  • 95,075
  • Thank you Alain. The solution must but let ... in syntaxly compatible. The best, for what I have to do, was \draw[coordinate] ($ (A0)!.5!(A1) $) node (E0) {};. – Tarass Feb 26 '14 at 15:02