How to draw an equilateral triangle with TikZ?

Question

I'm trying to draw a little picture with TikZ. It should show an equilateral triangle with length 1 and circles with radius 1/2 on each corner of the triangle. Next, the intersections of the circles should be highlighted with a dot. This is what I tried so far:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections}

\begin{document}
    \begin{tikzpicture}[scale=1.5]
    \coordinate (a) at (0,0);
    \coordinate (b) at (1,0);
    \coordinate (c) at (0.5,0.866);

    \draw[dashed] (a)--(b)--(c)--(a);
    \draw[name path=circleA] (a) circle (0.5);
    \draw[name path=circleB] (b) circle (0.5);
    \draw[name path=circleC] (c) circle (0.5);

    \fill (0.5,0) circle (1pt);
    \path [name intersections={of=circleA and circleC,name=AC}];
    \fill (AC-1) circle (1pt);
    \path[name intersections={of=circleB and circleC,name=BC}];
    \fill (BC-1) circle (1pt);
    \end{tikzpicture}
\end{document}

As one can see, the intersections aren't placed exactly where they should be, because the triangle isn't perfectly equilateral. How can I do that?

Answer 1

Intersection computations rely on numerically sensitive mechanisms and it does not always give you the precision you are looking for especially if tangents are in question (they get very close and might trigger and early/late true signal). But since the problem is relatively easy geometrically you can get away with many options. Here is another one

\documentclass[tikz]{standalone}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
  \draw[dashed](0,0) coordinate (a)--+(1,0) coordinate (b)--+(60:1) coordinate (c) --cycle;
  \foreach\x[remember=\x as \lastx (initially c)] in{a,b,c}{
    \node[fill,circle,inner sep=1pt] at($(\x)!0.5!(\lastx)$)  (\x-\lastx) {};
    \draw (\x)circle (0.5);
  }
\end{tikzpicture}
\end{document}

Answer 2

Not sure this is the best way, but...

\documentclass[tikz,border=10pt]{standalone}
\usetikzlibrary{shapes.geometric,calc,through}
\begin{document}
  \begin{tikzpicture}
    \node (tri) [regular polygon, regular polygon sides=3, draw, densely dashed, minimum width=50mm] {};
    \foreach \i/\j in {1/2,2/3,3/1}
    {
      \node [draw] at (tri.corner \i) [circle through={($(tri.corner \i)!1/2!(tri.corner \j)$)}, draw] {};
      \path [fill] ($(tri.corner \i)!1/2!(tri.corner \j)$) circle (2.5pt);
    }
  \end{tikzpicture}
\end{document}

Answer 3

With tkz-euclide (made more verbose for clarity)

\documentclass[border=5mm]{standalone}
\usepackage{tkz-euclide}
\usetkzobj{all}
\begin{document}
\begin{tikzpicture}[scale=.8]
\tkzDefPoint(0,0){A}
\tkzDefPoint(5,0){B}
\tkzDefEquilateral(A,B)\tkzGetPoint{C}
\tkzDrawPolygon[color=magenta,dashed](A,B,C)
\tkzDefMidPoint(A,B) \tkzGetPoint{D}
\tkzDefMidPoint(B,C) \tkzGetPoint{E}
\tkzDefMidPoint(A,C) \tkzGetPoint{F}
\tkzDefCircle[radius](A,D)
\tkzDefCircle[radius](B,E)
\tkzDefCircle[radius](C,F)
\tkzDrawCircle(A,D)
\tkzDrawCircle(B,E)
\tkzDrawCircle(C,F)
\tkzDrawPoints[size=12,fill=green](D,E,F)
%\tkzCentroid(A,B,C)\tkzGetPoint{G}
%\tkzDrawPoint(G)
%\tkzDrawLines[add = 0 and 2/3](A,G B,G C,G)
\end{tikzpicture}
\end{document}

Answer 4

One way among others of solving this, with MetaPost for whom it may interest. The equilateral triangle is built by rotating a unit side by 60°, and the intersections are found using the intersectionpoint operator. It is interesting to notice that this operator fails with MetaPost's default fixed-point arithmetic. Only after switching to floating-point numerics (\mplibnumbersystem{double}) it is able to find the intersection points.

\documentclass[border=2bp]{standalone}
\usepackage{luamplib}
  \mplibnumbersystem{double}
\begin{document}
  \begin{mplibcode}
    numeric u; u = 5cm; path circle[];
    beginfig(1);
      z1 = origin; z2 = (u, 0); z3 = z2 rotatedaround (z1, 60);
      draw z1 -- z2 -- z3 -- cycle dashed evenly;
      for i= 1, 2, 3:
        circle[i] = fullcircle scaled u shifted z[i];
        draw circle[i];
      endfor;
      z12 = circle1 intersectionpoint circle2;
      z13 = circle1 intersectionpoint circle3;
      z23 = circle2 intersectionpoint circle3;
      for i = 12, 13, 23:
        drawdot z[i] withpen pencircle scaled 5bp;
      endfor 
    endfig;
  \end{mplibcode}
\end{document}

To be typeset with LuaLaTeX. Output: