76

First of all: don't forget that exactly 140 years ago, Albert Einstein was born; but exactly one year ago, Stephen William Hawking passed away. What a special day for science!


Question

How to draw the letter π not in the standard way (i.e. \pi)?

I mean "draw", not "type"! Today there will be no \pi, but there will be something like this

\documentclass[tikz]{standalone}
\begin{document}
\begin{tikzpicture}
\draw (0,2)--(3,2);
\draw (1,0)--(1,2);
\draw (2,0)--(2,2);
\end{tikzpicture}
\end{document}

enter image description here

(inspired by David Carlisle – I draw this in TikZ just because I don't know how to use picture ;))

Or this

\documentclass[tikz]{standalone}
\usepackage{tikzducks}
\begin{document}
\begin{tikzpicture}
\duck
\duck[xshift=1cm,yshift=2cm]
\duck[xshift=1cm,yshift=4cm]
\duck[xshift=1cm,yshift=6cm]
\duck[xshift=0cm,yshift=8cm]
\duck[xshift=-2cm,yshift=7.5cm]
\duck[xshift=2cm,yshift=8cm]
\duck[xshift=4cm,yshift=8cm]
\duck[xshift=6cm,yshift=8cm]
\duck[xshift=8cm,yshift=8cm]
\duck[xshift=5cm,yshift=6cm]
\duck[xshift=5cm,yshift=4cm]
\duck[xshift=5.5cm,yshift=2cm]
\duck[xshift=6.5cm,yshift=0cm]
\duck[xshift=8cm,yshift=1cm]
\node[font=\huge] at (4,11) {Happy $\pi$ day with \verb|tikzducks|!};
\end{tikzpicture}
\end{document}

enter image description here

They are my proudest π drawings, and as today is Pi day, I'd like to see yours!


Well, why didn't I delay the time of asking the question by 8 minutes? I asked this at 1:51:31 UTC time, and the "Pi second" of this year is at 1:59:26 today!

| improve this question | | | | |
  • @user49915 I don't think we can have the output and the code being exactly the same :)) – user156344 Mar 14 '19 at 3:04
  • JouleV are you saying @user49915 "can't have his pi and use it" look at all the other "typos" for answers :-) – user170109 Mar 14 '19 at 12:09
  • @KJO No. Just look at the C code. It is compilable -- it is like a perfect Codegolf answer for a TeX.SE question :) I don't think we can have the same thing with LaTeX: the code is compilable and it fits to a pi shape! – user156344 Mar 14 '19 at 12:11
  • @user49915 Btw if 49915 is not a special number for you, I think you should be user165772 :) user49915 is no longer here anymore. – user156344 Mar 14 '19 at 12:14
  • @user49915 well, I need an old compiler at least, I would guess. Any details on the requirements? – Baldrickk Mar 14 '19 at 14:57

25 Answers 25

97
+500

Here's one with \shapepar, with great thanks to flowframtk.

enter image description here

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{libertine}
\usepackage{shapepar}

\newcommand{\sep}{\discretionary{}{}{}}

\begin{document}

\small%
\shapepar[1.00375pt]{{137.821777}%
{0.0}b{27.0}%
\\{0.0}t{27.0}{249.0}%
\\{11.955168}t{14.0}{250.0}%
\\{23.910336}t{2.0}{249.0}%
\\{27.074219}t{0.0}{135.5}st{135.5}{111.5}%
\\{35.865504}t{57.0}{35.0}t{179.0}{35.0}%
\\{47.820672}t{56.0}{34.0}t{178.0}{34.0}%
\\{59.775841}t{55.0}{34.0}t{177.0}{34.0}%
\\{71.731009}t{53.0}{35.0}t{175.0}{35.0}%
\\{83.686177}t{52.0}{34.0}t{174.0}{34.0}%
\\{95.641345}t{51.0}{34.0}t{173.0}{34.0}%
\\{107.596513}t{49.0}{35.0}t{171.0}{35.0}%
\\{119.551681}t{48.0}{34.0}t{170.0}{34.0}%
\\{131.506849}t{47.0}{34.0}t{169.0}{34.0}%
\\{135.137695}t{46.0}{35.0}t{168.0}{35.0}%
\\{143.462017}t{45.0}{35.0}t{168.0}{34.0}%
\\{154.276367}t{44.0}{34.0}t{166.0}{35.0}%
\\{155.417186}t{44.0}{34.0}t{166.0}{35.0}%
\\{163.378906}t{43.0}{34.0}t{166.0}{34.0}%
\\{167.372354}t{43.0}{34.0}t{165.0}{35.0}t{256.0}{3.0}%
\\{171.936859}t{42.0}{34.0}t{165.0}{35.0}t{253.0}{8.0}%
\\{179.327522}t{41.0}{35.0}t{165.0}{35.0}t{248.0}{13.0}%
\\{182.050781}t{41.0}{34.0}t{165.0}{36.0}t{245.0}{15.0}%
\\{187.185547}t{40.0}{35.0}t{166.0}{36.0}t{240.0}{17.0}%
\\{191.28269}t{40.0}{34.0}t{166.0}{39.0}t{235.0}{19.0}%
\\{193.214996}t{40.0}{34.0}t{166.0}{40.0}t{232.0}{20.0}%
\\{197.688477}t{39.0}{35.0}t{167.0}{50.0}jt{217.0}{31.0}%
\\{198.388672}t{39.0}{35.0}t{167.0}{80.0}%
\\{203.237858}t{39.0}{34.0}t{169.0}{72.0}%
\\{207.880219}t{38.0}{34.0}t{171.0}{64.0}%
\\{207.958008}t{38.0}{34.0}t{171.0}{64.0}%
\\{215.193026}t{37.0}{19.0}t{175.0}{49.0}%
\\{215.660156}t{37.0}{18.0}t{176.0}{47.0}%
\\{222.195312}t{37.0}{2.0}t{186.0}{23.0}%
\\{222.195312}e{37.0}%
\\{223.362305}t{191.0}{13.0}%
\\{223.362305}e{191.0}%
}%
3\sep{}.\sep{}1\sep{}4\sep{}1\sep{}5\sep{}9\sep{}2\sep{}6\sep{}5\sep{}3\sep{}5\sep{}8\sep{}9\sep{}7\sep{}9\sep{}3\sep{}2\sep{}3\sep{}8\sep{}4\sep{}6\sep{}2\sep{}6\sep{}4\sep{}3\sep{}3\sep{}8\sep{}3\sep{}2\sep{}7\sep{}9\sep{}5\sep{}0\sep{}2\sep{}8\sep{}8\sep{}4\sep{}1\sep{}9\sep{}7\sep{}1\sep{}6\sep{}9\sep{}3\sep{}9\sep{}9\sep{}3\sep{}7\sep{}5\sep{}1\sep{}0\sep{}5\sep{}8\sep{}2\sep{}0\sep{}9\sep{}7\sep{}4\sep{}9\sep{}4\sep{}4\sep{}5\sep{}9\sep{}2\sep{}3\sep{}0\sep{}7\sep{}8\sep{}1\sep{}6\sep{}4\sep{}0\sep{}6\sep{}2\sep{}8\sep{}6\sep{}2\sep{}0\sep{}8\sep{}9\sep{}9\sep{}8\sep{}6\sep{}2\sep{}8\sep{}0\sep{}3\sep{}4\sep{}8\sep{}2\sep{}5\sep{}3\sep{}4\sep{}2\sep{}1\sep{}1\sep{}7\sep{}0\sep{}6\sep{}7\sep{}9\sep{}8\sep{}2\sep{}1\sep{}4\sep{}8\sep{}0\sep{}8\sep{}6\sep{}5\sep{}1\sep{}3\sep{}2\sep{}8\sep{}2\sep{}3\sep{}0\sep{}6\sep{}6\sep{}4\sep{}7\sep{}0\sep{}9\sep{}3\sep{}8\sep{}4\sep{}4\sep{}6\sep{}0\sep{}9\sep{}5\sep{}5\sep{}0\sep{}5\sep{}8\sep{}2\sep{}2\sep{}3\sep{}1\sep{}7\sep{}2\sep{}5\sep{}3\sep{}5\sep{}9\sep{}4\sep{}0\sep{}8\sep{}1\sep{}2\sep{}8\sep{}4\sep{}8\sep{}1\sep{}1\sep{}1\sep{}7\sep{}4\sep{}5\sep{}0\sep{}2\sep{}8\sep{}4\sep{}1\sep{}0\sep{}2\sep{}7\sep{}0\sep{}1\sep{}9\sep{}3\sep{}8\sep{}5\sep{}2\sep{}1\sep{}1\sep{}0\sep{}5\sep{}5\sep{}5\sep{}9\sep{}6\sep{}4\sep{}4\sep{}6\sep{}2\sep{}2\sep{}9\sep{}4\sep{}8\sep{}9\sep{}5\sep{}4\sep{}9\sep{}3\sep{}0\sep{}3\sep{}8\sep{}1\sep{}9\sep{}6\sep{}4\sep{}4\sep{}2\sep{}8\sep{}8\sep{}1\sep{}0\sep{}9\sep{}7\sep{}5\sep{}6\sep{}6\sep{}5\sep{}9\sep{}3\sep{}3\sep{}4\sep{}4\sep{}6\sep{}1\sep{}2\sep{}8\sep{}4\sep{}7\sep{}5\sep{}6\sep{}4\sep{}8\sep{}2\sep{}3\sep{}3\sep{}7\sep{}8\sep{}6\sep{}7\sep{}8\sep{}3\sep{}1\sep{}6\sep{}5\sep{}2\sep{}7\sep{}1\sep{}2\sep{}0\sep{}1\sep{}9\sep{}0\sep{}9\sep{}1\sep{}4\sep{}5\sep{}6\sep{}4\sep{}8\sep{}5\sep{}6\sep{}6\sep{}9\sep{}2\sep{}3\sep{}4\sep{}6\sep{}0\sep{}3\sep{}4\sep{}8\sep{}6\sep{}1\sep{}0\sep{}4\sep{}5\sep{}4\sep{}3\sep{}2\sep{}6\sep{}6\sep{}4\sep{}8\sep{}2\sep{}1\sep{}3\sep{}3\sep{}9\sep{}3\sep{}6\sep{}0\sep{}7\sep{}2\sep{}6\sep{}0\sep{}2\sep{}4\sep{}9\sep{}1\sep{}4\sep{}1\sep{}2\sep{}7\sep{}3\sep{}7\sep{}2\sep{}4\sep{}5\sep{}8\sep{}7\sep{}0\sep{}0\sep{}6\sep{}6\sep{}0\sep{}6\sep{}3\sep{}1\sep{}5\sep{}5\sep{}8\sep{}8\sep{}1\sep{}7\sep{}4\sep{}8\sep{}8\sep{}1\sep{}5\sep{}2\sep{}0\sep{}9\sep{}2\sep{}0\sep{}9\sep{}6\sep{}2\sep{}8\sep{}2\sep{}9\sep{}2\sep{}5\sep{}4\sep{}0\sep{}9\sep{}1\sep{}7\sep{}1\sep{}5\sep{}3\sep{}6\sep{}4\sep{}3\sep{}6\sep{}7\sep{}8\sep{}9\sep{}2\sep{}5\sep{}9\sep{}0\sep{}3\sep{}6\sep{}0\sep{}0\sep{}1\sep{}1\sep{}3\sep{}3\sep{}0\sep{}5\sep{}3\sep{}0\sep{}5\sep{}4\sep{}8\sep{}8\sep{}2\sep{}0\sep{}4\sep{}6\sep{}6\sep{}5\sep{}2\sep{}1\sep{}3\sep{}8\sep{}4\sep{}1\sep{}4\sep{}6\sep{}9\sep{}5\sep{}1\sep{}9\sep{}4\sep{}1\sep{}5\sep{}1\sep{}1\sep{}6\sep{}0\sep{}9\sep{}4\sep{}3\sep{}3\sep{}0\sep{}5\sep{}7\sep{}2\sep{}7\sep{}0\sep{}3\sep{}6\sep{}5\sep{}7\sep{}5\sep{}9\sep{}5\sep{}9\sep{}1\sep{}9\sep{}5\sep{}3\sep{}0\sep{}9\sep{}2\sep{}1\sep{}8\sep{}6\sep{}1\sep{}1\sep{}7\sep{}3\sep{}8\sep{}1\sep{}9\sep{}3\sep{}2\sep{}6\sep{}1\sep{}1\sep{}7\sep{}9\sep{}3\sep{}1\sep{}0\sep{}5\sep{}1\sep{}1\sep{}8\sep{}5\sep{}4\sep{}8\sep{}0\sep{}7\sep{}4\sep{}4\sep{}6\sep{}2\sep{}3\sep{}7\sep{}9\sep{}9\sep{}6\sep{}2\sep{}7\sep{}4\sep{}9\sep{}5\sep{}6\sep{}7\sep{}3\sep{}5\sep{}1\sep{}8\sep{}8\sep{}5\sep{}7\sep{}5\sep{}2\sep{}7\sep{}2\sep{}4\sep{}8\sep{}9\sep{}1\sep{}2\sep{}2\sep{}7\sep{}9\sep{}3\sep{}8\sep{}1\sep{}8\sep{}3\sep{}0\sep{}1\sep{}1\sep{}9\sep{}4\sep{}9\sep{}1\sep{}2\sep{}9\sep{}8\sep{}3\sep{}3\sep{}6\sep{}7\sep{}3\sep{}3\sep{}6\sep{}2\sep{}4\sep{}4\sep{}0\sep{}6\sep{}5\sep{}6\sep{}6\sep{}4\sep{}3\sep{}0\sep{}8\sep{}6\sep{}0\sep{}2\sep{}1\sep{}3\sep{}9\sep{}4\sep{}9\sep{}4\sep{}6\sep{}3\sep{}9\sep{}5\sep{}2\sep{}2\sep{}4\sep{}7\sep{}3\sep{}7\sep{}1\sep{}9\sep{}0\sep{}7\sep{}0\sep{}2\sep{}1\sep{}7\sep{}9\sep{}8\sep{}6\sep{}0\sep{}9\sep{}4\sep{}3\sep{}7\sep{}0\sep{}2\sep{}7\sep{}7\sep{}0\sep{}5\sep{}3\sep{}9\sep{}2\sep{}1\sep{}7\sep{}1\sep{}7\sep{}6\sep{}2\sep{}9\sep{}3\sep{}1\sep{}7\sep{}6\sep{}7\sep{}5\sep{}2\sep{}3\sep{}8\sep{}4\sep{}6\sep{}7\sep{}4\sep{}8\sep{}1\sep{}8\sep{}4\sep{}6\sep{}7\sep{}6\sep{}6\sep{}9\sep{}4\sep{}0\sep{}5\sep{}1\sep{}3\sep{}2\sep{}0\sep{}0\sep{}0\sep{}5\sep{}6\sep{}8\sep{}1\sep{}2\sep{}7\sep{}1\sep{}4\sep{}5\sep{}2\sep{}6\sep{}3\sep{}5\sep{}6\sep{}0\sep{}8\sep{}2\sep{}7\sep{}7\par
\end{document}
| improve this answer | | | | |
52

Some tessellated pi...

enter image description here

This one is done in plain Metapost, so compile with mpost.

prologues := 3;
outputtemplate := "%j%c.eps";
input colorbrewer-rgb;
beginfig(1);
    path pi; numeric t; t = 13;
    pi = (origin -- (5,0) -- (5,1) -- (4,1) -- (4,4) -- (3,4) -- (3,1) -- (2,1) -- (2,4) -- (1,4) -- (1,1) -- up -- cycle) scaled t;

    for i=1 upto 48:
        for j=1 upto 48:
            fill pi                              shifted (4t*i-2t*j, t*i+5t*j) withcolor Spectral[7][i mod 7 + 1];
            fill pi rotated 180 shifted (4t, 5t) shifted (4t*i-2t*j, t*i+5t*j) withcolor Spectral[7][(3+i) mod 7 + 1];
       endfor
    endfor
    clip currentpicture to unitsquare scaled 100t shifted (0, 32t);
endfig;
end.

You will need to load Metapost Colorbrewer for the colours...

| improve this answer | | | | |
  • 1
    Wonderful!!!! Very nice for my opinion. – Sebastiano Mar 14 '19 at 16:28
46

For Pi day, the tikzlings decided to go on holidays. Unfortunately, the snowman could not come with them, so they sent him a postcard:

enter image description here

\documentclass{standalone}
\usepackage{tikz}
\usepackage{tikzlings}

\newcommand{\palm}{%
    \foreach \x in {1.2,1.15,...,0} {
        \pgfmathsetmacro{\y}{0.2*\x*(1-\x)};
        \pgfmathsetmacro{\len}{0.3-0.11*\x};
        \pgfmathsetmacro{\angup}{-50-30*\x)};
        \pgfmathsetmacro{\angdown}{20+30*\x)};
        \fill[bottom color=green!80!black,top color=green!0!brown,shift={(\x,\y)},rotate=\angup] (0,0) -- (0.05,0) -- (0.025+0.015*rnd,\len+0.03*rnd) -- cycle;
        \fill[top color=green!80!black,bottom color=green!0!brown,shift={(\x,\y)},rotate=\angdown] (0,0) -- (0.05,0) -- (0.025+0.015*rnd,-\len+0.03*rnd) -- cycle;
    }
}
\newcommand{\palmtree}{%
    \foreach \y in {0,0.05,...,2} {
        \fill[inner color=brown!40!yellow,outer color=brown] (0.1*\y*\y,\y) ellipse({0.2-0.015*\y} and 0.1);
    }
    \foreach \angle in {-20,-10,0} {
        \begin{scope}[shift={(0.3,2)},rotate=\angle]
            \palm
        \end{scope}
        \begin{scope}[shift={(0.5,2)},rotate=-\angle,xscale=-1]
            \palm
        \end{scope}
    }
}

\begin{document}

\begin{tikzpicture}
    \pgfmathsetseed{1}
    \begin{scope}[shift={(0.7,0)}]
        \palmtree
    \end{scope}
    \begin{scope}[shift={(-0.7,0)},xscale=-1]
        \palmtree
    \end{scope}
    \bear[hat,xshift=-1.6cm,yshift=1.9cm,scale=0.25];
    \coati[tophat,xshift=-1.2cm,yshift=1.9cm,scale=0.25];
    \koala[beret,xshift=-0.8cm,yshift=1.9cm,scale=0.25];
    \marmot[strawhat,xshift=-0.4cm,yshift=1.9cm,scale=0.25];
    \moles[harlequin,xshift=-0.0cm,yshift=1.9cm,scale=0.25];
    \mouse[strawhat=blue,xshift=0.4cm,yshift=1.9cm,scale=0.25];
    \owl[beret=green!50!black,xshift=0.8cm,yshift=1.9cm,scale=0.25];
    \penguin[tophat=red,xshift=1.2cm,yshift=1.9cm,scale=0.25];
    \sloth[hat=brown!40!yellow,xshift=1.6cm,yshift=1.9cm,scale=0.25];
\end{tikzpicture}

\end{document}
| improve this answer | | | | |
44

Writing π with the digits of π - using the verbatim environment.

enter image description here

\documentclass{article}
\linespread{0.7}
\begin{document}
\begin{verbatim}
     3.141592653589793238462643383279
   5028841971693993751058209749445923
  07816406286208998628034825342117067
  9821    48086         5132     
 823      06647        09384
46        09550        58223
17        25359        4081
          2848         1117
          4502         8410
          2701         9385
         21105        55964
         46229        48954
         9303         81964
         4288         10975
        66593         34461
       284756         48233
       78678          31652        71
      2019091         456485       66
     9234603           48610454326648
    2133936            0726024914127
    3724587             00660631558
    817488               152092096
\end{verbatim}
\end{document}

Based on ascii art drawing by Jorel - https://www.flickr.com/photos/jorel314/3352784321/

| improve this answer | | | | |
  • 2
    How did you format the code? By trial/error? – Sigur Mar 14 '19 at 2:49
40

We could extract the MetaPost paths for the glyph \pi from the font and draw it using LuaTeX.

\documentclass{article}

\usepackage{fontspec}
\setmainfont{latinmodern-math.otf}

\usepackage{luacode}
\begin{luacode*}
-- We need some utilities from ConTeXt
callbacks = callbacks or {}
callbacks.supported = callbacks.supported or {}
CONTEXTLMTXMODE = CONTEXTLMTXMODE or (status.obj_ptr == nil and 2 or 1)
dofile(kpse.find_file("util-fmt.lua"))
dofile(kpse.find_file("node-ini.lua"))
dofile(kpse.find_file("font-mps.lua"))
dofile(kpse.find_file("font-shp.lua"))

-- That's a simple reimplemetation of ConTeXt's \showshape macro
function outlinepaths(character)
    local fontid      = font.current()
    local shapedata   = fonts.hashes.shapes[fontid] -- by index
    local chardata    = fonts.hashes.characters[fontid] -- by unicode
    local shapeglyphs = shapedata.glyphs or { }

    character = utf.byte(character)
    local c = chardata[character]
    if c then
        if not c.index then
            return {}
        end
        local glyph = shapeglyphs[c.index]
        if glyph and (glyph.segments or glyph.sequence) then
            local units  = shapedata.units or 1000
            local factor = 100/units
            local paths  = fonts.metapost.paths(glyph,factor)
            return paths
        end
    end
end
\end{luacode*}

\usepackage{luamplib}
\everymplib{beginfig(0);}
\everyendmplib{endfig;}

\def\mpdefineoutlines#1{\directlua{
    local char = "\luaescapestring{#1}"
    local outlines = outlinepaths("#1")
    for i, path in ipairs(outlines) do
        tex.print("fill " .. path .. ";")
    end
  }}

\begin{document} 

\begin{mplibcode}

  \mpdefineoutlines{𝜋}

\end{mplibcode}

\end{document}

Instead of using luamplib, we could also simply print the path to the log file and copy it to a MetaPost file. With some additonal formatting we get:

prologues := 3;
outputformat := "pdf";

beginfig(1)
  fill (56.70,40.70)
    .. controls (56.70,43.10) and (54.60,43.10)
    .. (52.70,43.10)
    -- (19.20,43.10)
    .. controls (17,43.10) and (13.20,43.10)
    .. (8.80,38.40)
    .. controls (5.30,34.50) and (2.70,29.90)
    .. (2.70,29.40)
    .. controls (2.70,29.40) and (2.70,28.40)
    .. (3.90,28.40)
    .. controls (4.70,28.40) and (4.90,28.80)
    .. (5.50,29.60)
    .. controls (10.40,37.30) and (16.20,37.30)
    .. (18.20,37.30)
    -- (23.90,37.30)
    .. controls (20.70,25.20) and (15.30,13.10)
    .. (11.10,4)
    .. controls (10.30,2.50) and (10.30,2.30)
    .. (10.30,1.60)
    .. controls (10.30,-0.30) and (11.90,-1.10)
    .. (13.20,-1.10)
    .. controls (16.20,-1.10) and (17,1.70)
    .. (18.20,5.40)
    .. controls (19.60,10) and (19.60,10.20)
    .. (20.90,15.20)
    -- (26.50,37.30)
    -- (37.80,37.30)
    .. controls (34.50,22.50) and (33.60,18.20)
    .. (33.60,11.50)
    .. controls (33.60,10) and (33.60,7.30)
    .. (34.40,3.90)
    .. controls (35.40,-0.50) and (36.50,-1.10)
    .. (38,-1.10)
    .. controls (40,-1.10) and (42.10,0.70)
    .. (42.10,2.70)
    .. controls (42.10,3.30) and (42.10,3.50)
    .. (41.50,4.90)
    .. controls (38.60,12.10) and (38.60,18.60)
    .. (38.60,21.40)
    .. controls (38.60,26.70) and (39.30,32.10)
    .. (40.40,37.30)
    -- (51.80,37.30)
    .. controls (53.10,37.30) and (56.70,37.30)
    .. (56.70,40.70)
    -- cycle;
endfig;
end

Or you can even use the path with TikZ.

\documentclass{article}

\usepackage{tikz}

\begin{document} 

\begin{tikzpicture}[x=1pt,y=1pt]

  \fill (56.70,40.70)
    .. controls (56.70,43.10) and (54.60,43.10)
    .. (52.70,43.10)
    -- (19.20,43.10)
    .. controls (17,43.10) and (13.20,43.10)
    .. (8.80,38.40)
    .. controls (5.30,34.50) and (2.70,29.90)
    .. (2.70,29.40)
    .. controls (2.70,29.40) and (2.70,28.40)
    .. (3.90,28.40)
    .. controls (4.70,28.40) and (4.90,28.80)
    .. (5.50,29.60)
    .. controls (10.40,37.30) and (16.20,37.30)
    .. (18.20,37.30)
    -- (23.90,37.30)
    .. controls (20.70,25.20) and (15.30,13.10)
    .. (11.10,4)
    .. controls (10.30,2.50) and (10.30,2.30)
    .. (10.30,1.60)
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    -- cycle;

\end{tikzpicture}

\end{document}

The output is rather unspectacular.

enter image description here

| improve this answer | | | | |
  • 8
    I wonder (but don’t have anywhere near the TeX-fu to attack it myself): could one combine this answer with the Liantze Lim’s shapepar-based answer, to print the digits of pi (as there) in the shape of metafont’s actual pi (as used here), instead of with the shape input by hand (as it is there)? – Peter LeFanu Lumsdaine Mar 14 '19 at 11:26
  • 4
    You can sort of do it with ConTeXt to calculate \parshape but \pi is too irregular to give good results (code, screenshot). – Henri Menke Mar 14 '19 at 20:57
34

Time for a bad joke...

\documentclass[margin=1cm]{standalone}
\usepackage{tikz}

\begin{document}

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    (308.9521,637.1660) .. (309.5391,637.1660) .. controls (310.1260,637.1660) and
    (310.7775,636.9719) .. (311.0391,636.7773) .. controls (311.6898,636.1266) and
    (308.3685,630.1331) .. (307.0000,629.3516) .. controls (305.0446,628.3755) and
    (301.5932,628.5691) .. (296.1895,630.0684) .. controls (290.1988,631.6953) and
    (290.9143,632.0231) .. (281.7305,623.0977) .. controls (275.4782,616.9762) and
    (275.1520,616.4541) .. (274.3066,612.7441) .. controls (273.1997,607.4042) and
    (273.2643,604.0163) .. (274.6328,597.1133) .. controls (275.4144,593.2726) and
    (275.9331,586.9538) .. (276.1309,578.7461) -- (276.5215,566.2402) --
    cycle(243.8203,570.0820) .. controls (244.0149,570.0820) and
    (245.6453,573.7951) .. (247.3359,578.3535) .. controls (249.0298,582.9789) and
    (251.9611,589.6208) .. (253.7188,593.2031) -- (256.9785,599.7188) --
    (256.8477,608.9023) .. controls (256.7839,614.1115) and (256.9794,619.2572) ..
    (257.4355,620.8203) .. controls (258.0225,623.1649) and (258.6725,623.7517) ..
    (262.8418,626.1602) .. controls (265.4480,627.6594) and (271.5023,631.2384) ..
    (276.3223,634.1699) -- (285.1152,639.4473) .. controls (255.9661,639.4473) and
    (227.5437,639.1752) .. (208.3731,638.7637) .. controls (208.1397,637.8716) and
    (207.5751,636.8804) .. (206.6387,635.7305) .. controls (204.8140,633.5167) and
    (204.2940,633.3228) .. (200.9062,633.1250) -- (197.1934,632.9297) --
    (197.2578,624.7891) .. controls (197.2578,617.2353) and (197.4514,616.2568) ..
    (199.3398,611.3731) .. controls (202.1406,604.2754) and (204.8119,600.4308) ..
    (209.1758,597.3047) .. controls (211.1950,595.8054) and (215.6221,592.2265) ..
    (219.0098,589.2949) .. controls (222.3975,586.3634) and (227.6701,582.1298) ..
    (230.7324,579.9160) .. controls (233.7948,577.7022) and (237.8974,574.5767) ..
    (239.8496,573.0137) .. controls (241.8050,571.3868) and (243.6257,570.0820) ..
    (243.8203,570.0820) -- cycle;
\end{tikzpicture}
\end{document}

Life of Pi

For a tiger, there's a lot of code, so it's available here. The output:

Life of Pi

| improve this answer | | | | |
  • 3
    I like the idea. Never heard of the book or the movie. I know the book with the tiger but the image shows (as far as I can see) a lion. – albert Mar 14 '19 at 17:39
  • ooh you switched the tiger to Lions of TikZ :-) that really is a fun answer I'm sorry I put your name to my poor version – user170109 Mar 14 '19 at 18:16
  • 2
    I demand a tiger… Thanks for a good laugh ;-) @albert The movie was awesome, another masterpiece by Ang Lee. – Ruixi Zhang Mar 14 '19 at 20:24
  • 1
    @albert, @ RuixiZhang, @KJO, @samcarter: updated. :) – Paulo Cereda Mar 14 '19 at 22:16
  • @PauloCereda Long live the Inkscape! – user156344 Mar 15 '19 at 2:03
30

Here is a slightly different visualization of π.

enter image description here

I re-drew this from my copy of Proofs without Words by Roger B. Nelsen. The original was published in Mathematics Magazine, 50.3, May 1977.

Here I have used Metapost using luamplib, so compile with lualatex.

\documentclass[border=5mm]{standalone}
\usepackage{luatex85}
\usepackage{luamplib}
\begin{document}
\mplibtextextlabel{enable}
\begin{mplibcode}
beginfig(0);
    path C, C', S; numeric u; u=68;
    C = fullcircle rotated 90 scaled 2u;
    C' = C rotated 180 shifted (3.14159265359u ,0); 
    S = unitsquare rotated -90 
                   scaled 1.77245385091u
                   shifted point 0 of C'; 

    z0 = (xpart point 2 of C', ypart point 0 of C');
    fill C withcolor 7/8[blue,white]; 
    fill S withcolor 7/8[blue,white]; 

    drawoptions(dashed withdots scaled 1/2);
    draw point 4 of C shifted (-u,0) -- point 3 of S shifted (1/2u,0);
    draw halfcircle rotated 180 
                    scaled abs(point 4 of C - z0)
                    shifted 1/2[point 4 of C,z0];
    drawoptions();
    forsuffixes @=C,C':
        draw @; 
        draw point 0 of @ -- center @ -- point 2 of @ dashed evenly scaled 1/2;
        drawdot point 0 of @ withpen pencircle scaled 3;
    endfor 
    draw point 2 of C' -- z0 dashed evenly scaled 1/2;
    draw S;
    drawarrow subpath(-1/2,-3/2) of C scaled 1.2 withcolor 2/3 red;

    label.top("The Rolling Circle Squares Itself — Thomas Elsner", 
                    1/2[point 0 of C, point 4 of C'] shifted 20 up);
    label.bot("$\pi$", 1/2[point 4 of C, point 0 of C']);
    label.lft("$\sqrt\pi$", 1/2[point 1 of S, point 0 of C']);
    label.rt("$1$", 1/2 point 0 of C);
endfig;
\end{mplibcode}
\end{document}
| improve this answer | | | | |
  • Beautiful! I like that you showed what Pi is rather than the symbol. (You've also inspired me to learn luamplib.) – hackerb9 Mar 15 '19 at 17:23
29

enter image description here

\documentclass[serif]{beamer}

\usepackage{pst-text,pst-eucl,pst-grad}
\usepackage[active,tightpage]{preview}
\PreviewBorder=0pt
\PreviewEnvironment{pspicture}


\DeclareFixedFont{\ps}{U}{psy}{m}{n}{12cm}% the symbol font
\DeclareFixedFont{\PS}{T1}{ptm}{m}{n}{11cm}% the times font
\DeclareFixedFont{\RM}{T1}{ptm}{b}{n}{2cm}

\def\x{3.43}

\psset
{
    PointName=none,
    PointSymbol=none,
    linestyle=none,
    fillstyle=gradient,
    gradlines=1500,
    gradangle=30,
    gradmidpoint=1,
}

\newrgbcolor{TopBegin}{0.027 0.6 0.254}
\newrgbcolor{TopEnd}{0.521 0.749 0.125}

\newrgbcolor{LeftBegin}{0 0.368 0.549}
\newrgbcolor{LeftEnd}{0 0.596 0.701}

\newrgbcolor{BottomBegin}{0.905 0.223 0.050}
\newrgbcolor{BottomEnd}{0.949 0.568 0.003}

\newrgbcolor{RightBegin}{0.513 0.117 0.380}
\newrgbcolor{RightEnd}{0.870 0.007 0.349}

\begin{document}
\begin{frame}
\begin{pspicture}[showgrid=false](-\x,\x)(\x,-\x)
    \pstGeonode(-\x,\x){TL}(-\x,-\x){BL}(\x,-\x){BR}(\x,\x){TR}
    \pstGeonode
        (-2.4,1.7){A}
        (-1.2,1.7){B}
        (-1.5,-2.2){C}
        (0.7,-2.3){D}
        (1.0,1.7){E}
    \only<6>{\psclip{\pscircle[linewidth=0,fillstyle=none]{\x}}}
        \only<1->{\pspolygon[gradbegin=TopBegin,gradend=TopEnd](TL)(A)(E)(TR)}
        \only<2->{\pspolygon[gradbegin=LeftBegin,gradend=LeftEnd](TL)(A)(B)(C)(BL)}
        \only<3->{\pspolygon[gradbegin=BottomBegin,gradend=BottomEnd](BL)(C)(B)(E)(D)(BR)}
        \only<4->{\pspolygon[gradbegin=RightBegin,gradend=RightEnd](BR)(D)(E)(TR)}
        \only<5->{\rput(-0.2,-0.45){\pscharpath[fillstyle=solid,fillcolor=white]{\ps p}}}
    \only<6>{\endpsclip}
\end{pspicture}
\end{frame}
\end{document}
| improve this answer | | | | |
24

My contribuation for pi(e)-day:

enter image description here

The thieves were already there:

enter image description here

| improve this answer | | | | |
  • 9
    Did you use LaTeX to prepare the cake? :-) – gerrit Mar 14 '19 at 8:21
  • 3
    @gerrit a bit - it was used to get a prototype for the pi and the ducks ;-) – Ulrike Fischer Mar 14 '19 at 8:41
  • 2
    You staged it, right? There is no way the Baer is faster at the cake than the marmot. – user121799 Mar 14 '19 at 21:11
  • 2
    @marmot The marmot is rather cautious and sent the Bär to check the situation. – Ulrike Fischer Mar 14 '19 at 21:58
20

One should also honor Euler a bit.

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikzlings,amsmath}
\makeatletter
\tikzset{/thing/.cd,
 pie/.code=\thing@cheesetrue\def\thing@cheese{#1}, %<-pretend you didn't see that
 pie/.default=pink!70!red}
\makeatother                            
\begin{document}
\begin{tikzpicture}[font=\sffamily]
\marmot[pie,whiskers,teeth,shadow]
\node[anchor=east,scale=5,transform shape] at (-0.6,1) {$\pi\cdot\mathsf{e}=\text{\sffamily pie}$};
\end{tikzpicture}
\end{document}

enter image description here

| improve this answer | | | | |
  • Do marmots even eat pink cheese? Hahah. All the answers including the question are fantastic. – Sebastiano Mar 14 '19 at 12:54
20

The obligatory forest solution:

\documentclass{standalone}

\usepackage[edges]{forest}
\usetikzlibrary{calc}
\forestset{
  forked edge'/.style={
    edge={rotate/.option=!parent.grow},
    edge path'={(!u.parent anchor)++(0pt,-30pt) -- (!u.parent anchor)++(0pt,30pt)  |- (.child anchor)},   
  }
}
\begin{document}
\Huge
\begin{forest}forked edges,for tree={edge={line width=4pt}}
[ [ ] [ ]]
\end{forest}
\end{document}

pi

| improve this answer | | | | |
  • 1
    Hah! I can't see the forest from the \pi's – morbusg Mar 14 '19 at 19:56
19

A few mathematical representations:

\documentclass[10pt,a4paper]{article}
\begin{document}
    \[\int_{-\infty}^{\infty}\frac{\sin x}{x}dx\]
\end{document}

enter image description here

Honory representation: Ramanujan's equation

\documentclass[10pt,a4paper]{article}
\usepackage{amsmath}
\begin{document}
    \[\sqrt{6}\cdot\prod_{p~\text{prime}}^{\infty}\frac{p}{\sqrt{p^2-1}}\]
\end{document}

enter image description here

In a computer programmer's paradigm:

\documentclass[10pt,a4paper]{article}
\usepackage{amsmath}
\begin{document}
    \[4\cdot\arctan{1}\]
\end{document}

enter image description here

  • 3
    Don't forget the computer programmer's method for obtaining it as a double precision constant: pi = 4 atan(1). – Steven B. Segletes Mar 14 '19 at 10:26
  • @StevenB.Segletes Another possibility :D – Raaja is not active on TEXSE Mar 14 '19 at 10:29
  • @Raaja Your answer, with Reda Drissi's answer, seem to be off-topic in this question :) (but don't worry, I like them all!) – user156344 Mar 14 '19 at 14:16
  • @JouleV I know, therefore, I posted this answer as a community wiki ;). IMO, these are some of the must-to-be-added answers for the sake of \pi day ;), because, their scope is beyond TeX by itself. Nevertheless, thank you for liking the answer(s). And, I still owe you a pie from TikZ which, I am baking now. – Raaja is not active on TEXSE Mar 14 '19 at 14:28
  • @Raaja You made me hungry, and your comment makes me starving :)) – user156344 Mar 14 '19 at 14:29
19

In honor of Archimedes....

enter image description here

\documentclass{article}


\usepackage[svgnames]{xcolor}
\usepackage{tikz}

\pagestyle{empty}

\begin{document}
    \noindent
\foreach \x/\y in {6/Pink,12/Yellow,24/LightGreen,48/Orange,96/Magenta}
 {
\begin{tikzpicture}
\fill[White] (0,0) circle(3.2cm);
\draw[fill=\y] (0,0) circle(3cm);
\foreach \z in {1,...,\x}
 {
  \pgfmathsetmacro\rx{3*cos(360*(\z/\x))};
  \pgfmathsetmacro\ry{3*sin(360*(\z/\x))};
  \draw (0,0)--(\rx,\ry);
  \pgfmathsetmacro\rxp{3*cos(360*((\z-1)/\x))};
  \pgfmathsetmacro\ryp{3*sin(360*((\z-1)/\x))};
  \draw (\rxp,\ryp)--(\rx,\ry);
};
\end{tikzpicture} 
}
\raisebox{3cm}{Archimedes's algorithm for computing $\pi$}
\end{document}
| improve this answer | | | | |
17
\def\Tau{\ooalign{%
  $\bigcirc$\cr
  \hskip.3em $^\circ$\cr
  \hskip.49em \vrule depth .5ex height .95ex width .4pt
}}
$$ \Tau\over 2 $$
\bye

enter image description here

| improve this answer | | | | |
  • 7
    Finally a relief from all those half-turn heretics! – Roman Odaisky Mar 14 '19 at 20:02
  • 1
    +1. We should make June 28 a thing, too. – Ruixi Zhang Mar 14 '19 at 20:28
  • ok so this is a reference to tau, but what's the robot got to do with it? – LarsH Mar 15 '19 at 1:17
  • 3
    @LarsH it‘s a Tau unit miniature from the tabletop game Warhammer 40k. – morbusg Mar 15 '19 at 11:56
15

This lua code shows how to use Monte Carlo simulation (MCS) to estimate π. Approach and Matlab solution is here.

\documentclass{article}
\usepackage{luacode}

\begin{document}
\luaexec{
tp=tex.print
N=5000000 --[[ the experiment event number ]]
r=1 --[[ the circle radius ]]
n=0 --[[ sucessful event number  ]]
for i = 1,N,1 
do 
   x=-r+2*r*math.random()
   y=-r+2*r*math.random()
   if ((x*x+y*y)<=r*r) then n = n+1  end
end
lua_pi = 4*n/N
tp("Estimated value of pi :")  tp(lua_pi)
}
\end{document}

enter image description here

| improve this answer | | | | |
  • 1
    I scrolled all the way down here to find a solution that uses the computational powers of tex to actually estimate pi, great! – HRSE Mar 15 '19 at 9:22
15

My little litle litle...........contribution...with Mathcha.

enter image description here

\documentclass[10pt]{article}
\usepackage{tikz}
\begin{document}
\tikzset{every picture/.style={line width=0.75pt}}       
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1]
\draw[line width=2mm, violet] (221,106) .. controls (261,76) and (315.5,126) .. (355.5,96) ;
\draw[line width=1.5mm, violet] (254.5,98) -- (239.5,176) ;
\draw[line width=1.5mm, violet] (318,107) -- (335.5,179) ;
\end{tikzpicture}
\end{document}

or.....this

enter image description here

\documentclass[10pt]{article}
    \usepackage{tikz}
    \begin{document}
    \begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1]
    \draw[line width=3mm, orange] (200,103) .. controls (240,73) and (267.5,123) .. (307.5,93) ;
    \draw[line width=2.3mm, orange] (226.5,160) .. controls (244.5,158) and (235.5,136) .. (235.5,94) ;
    \draw[line width=2.3mm, orange]  (284.5,160) .. controls (276.5,152) and (276.5,140.75) .. (276.13,134.75) .. controls (275.75,128.75) and (276,122) .. (274.5,103) ;
    \end{tikzpicture}
    \end{document}

.....and for my students :-) this l(circle)(r)/(2r) = π


enter image description here

\documentclass[a4paper,12pt]{article}
\usepackage{tikz,amsmath,amssymb}
\begin{document}
\tikzset{every picture/.style={line width=0.75pt}}        
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1]
\draw   (125,142.25) .. controls (125,90.2) and (167.2,48) .. (219.25,48) .. controls (271.3,48) and (313.5,90.2) .. (313.5,142.25) .. controls (313.5,194.3) and (271.3,236.5) .. (219.25,236.5) .. controls (167.2,236.5) and (125,194.3) .. (125,142.25) -- cycle ;
\draw[|-|,red]   (219.25,142.25) -- (287.54,207.96) ;
\draw (248,184) node   {$r$};
\draw (325,92) node   {$\ell(\mathcal{C})$};
\end{tikzpicture}
\end{document}
| improve this answer | | | | |
14

A bit rough, but anyway.

pi image

\documentclass{standalone}
\def\pgfsysdriver{pgfsys-\Gin@driver}
\usepackage{pgfsys}
\usepackage{pgffor}
\usepackage{pgfmath}
\usepgflibrary{fpu}
\pgfkeys{
 /pgf/fpu=true,
 /pgf/fpu/output format=fixed
}
\def\pgfpt{\dimexpr\pgfmathresult pt\relax}
\begin{document}
\Large
\hspace{5mm}%
\foreach \i in {0,...,90}{%
 \pgfmathparse{-sin(\i) * 50}%
 \raisebox{\pgfpt}{.}%
 \pgfmathparse{(cos(\i) - cos(\i - 1)) * 5 - 4}%
 \hspace{\pgfpt}%
}\\
\hspace{-2mm}%
\foreach \i in {0,...,50}{.\hspace{-3pt}}\\
\hspace{-8mm}%
\foreach \i in {0,...,120}{%
 \pgfmathparse{-sin(\i) * 50}%
 \raisebox{\pgfpt}{.}%
 \pgfmathparse{(cos(\i) - cos(\i - 1)) * 5 - 3.7}%
 \hspace{\pgfpt}%
}%
\end{document}
| improve this answer | | | | |
14

Do you know what the value of π is? Here is an answer.

\documentclass[tikz]{standalone}
\newlength{\numheight}
\settoheight{\numheight}{1}
\begin{document}
\begin{tikzpicture}[x={\numheight/2},y={\numheight/2}]
\draw (-6,-.2)|-(-5,0) (-5.6,0)--(-5.75,-1) (-5.3,0)--(-5.3,-.875) arc (180:360:.125);
\draw (-4.5,.25) to[out=60,in=-120] (-3.5,.25) (-4.5,-.25) to[out=60,in=-120] (-3.5,-.25);
\draw (-3,0) arc (-90:90:1 and 0.5) (-3,0) arc (90:-90:1 and 0.5);
\draw (-1,-1) circle (.1pt);
\draw (.5,-1)|-(0,1) (0,-1)--(1,-1);
\draw (2,1)--(1.5,0)--(2.5,0) (2.5,1)--(2.5,-1);
\draw (3.5,-1)|-(3,1) (3,-1)--(4,-1);
\draw (5.5,1)-|(4.5,0) arc (90:-90:1 and 0.5);
\draw (6.5,0)-|(7,.5) arc (0:270:0.5) (7,0) arc (0:-90:1);
\draw (7.5,.5) arc (180:-60:0.5) to[out=210,in=90] (7.5,-1)--(8.5,-1);
\draw (9,-.5)|-(9.5,0) arc (90:-180:0.5) (9,0) arc (180:90:1);
\draw (11.5,1)-|(10.5,0) arc (90:-90:1 and 0.5);
\draw (12,0) arc (-90:90:1 and 0.5) (12,0) arc (90:-90:1 and 0.5);
\draw (13.5,-.5)|-(14,0) arc (90:-180:0.5) (13.5,0) arc (180:90:1);
\end{tikzpicture}
\end{document}

enter image description here

Don't worry about the spacing. This is supposed to be in a monospaced font :))

Without code...

enter image description here

| improve this answer | | | | |
13

Happy \pi(less) day!!

enter image description here

\documentclass{report}
\begin{document}
\noindent%
\rule{30pt}{1pt}\\[-1pt]
\rule{8pt}{0pt}%
\rule{1pt}{30pt}%
\rule{12pt}{0pt}%
\rule{1pt}{30pt}
\end{document}
| improve this answer | | | | |
13

You can use different representations like :

 \documentclass[10pt,a4paper]{article}
 \usepackage{amsmath}
 \begin{document}
    \[\sqrt{6\sum\limits^{\infty}\frac{1}{n^2}}\]
 \end{document}

latex.codecogs representation

Or the gamma function :

 \documentclass[10pt,a4paper]{article}
 \usepackage{amsmath}
 \begin{document}
    \[\Gamma\left(\frac{1}{2}\right)^2\]
 \end{document}

enter image description here

You could also use Leibniz Wallis or BBP formula.

| improve this answer | | | | |
  • 10
    while this is true, I think, the answer misses the purpose of the topic... – naphaneal Mar 14 '19 at 14:11
  • 4
    @naphaneal, Yea, but it's a Pi Day post, so have an upvote. You too. – user1717828 Mar 15 '19 at 12:07
11

My small contribution with some slagroom vlaai ;-) (I really think, we should not omit this one).

\documentclass{article}
\usepackage{graphicx}
\begin{document}
How can we forget this: $\pi$.

Anyway a happy pie day:

\includegraphics[height=2in, width=2in]{a}

\end{document}

enter image description here

After baking the PIE for JouleV in tikz:

%&lualatex
% !TeX TXS-program:compile = txs:///lualatex/[--shell-escape]


\documentclass{standalone}
\usepackage{pgfplots}
\usepackage{tikz}
% % lets bake some pi with a nice recipe from: https://helloacm.com/r-programming-tutorial-how-to-compute-pi-using-monte-carlo-in-r/

\begin{document}
\begin{tikzpicture}
\begin{axis}[]

\foreach \i in {1,...,10000}{
    % Lets start baking the PI(E)
\pgfmathparse{rnd}
%VARIABLES
\pgfmathsetmacro{\x}{\pgfmathresult}

\pgfmathparse{rnd}
%VARIABLES
\pgfmathsetmacro{\y}{\pgfmathresult}

\pgfmathparse{\x*\x+\y*\y)}
%VARIABLES
\pgfmathsetmacro{\t}{\pgfmathresult}

\pgfmathparse{\t^0.5)}
%VARIABLES
\pgfmathsetmacro{\z}{\pgfmathresult}

% now the comparison
\pgfmathparse{notgreater(\z,1)} 
\ifnum\pgfmathresult=1 
\addplot[red, mark=*] coordinates {(\x,\y)};
\else
\addplot[yellow, mark=*] coordinates {(\x,\y)};
\fi

}
\end{axis}

\end{tikzpicture}
\end{document}

to get:

enter image description here

| improve this answer | | | | |
  • 2
    I don't think \includegraphics qualifies as drawing. – Henri Menke Mar 14 '19 at 9:32
  • 1
    @Raaja Hope I can eat it soon :D – user156344 Mar 14 '19 at 10:12
  • 1
    +1 for vlaai :D – Marijn Mar 14 '19 at 12:10
  • 1
    Are we nearly there yet? is it only half baked? – user170109 Mar 22 '19 at 15:03
  • 1
    @KJO That is why I specifically mentioned that you must use your oven in Lualatex mode :D – Raaja is not active on TEXSE Mar 22 '19 at 15:35
6
+100

Run both with latex->dvips->ps2pdf

\documentclass[border=15pt,pstricks]{standalone}
\usepackage{pst-plot,pst-text} 
\newcommand*\PI{%
3{,}%
1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679%
8214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196%
4428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273%
7245870066063155881748815209209628292540917153643678925903600113305305488204665213841469519415116094%
3305727036575959195309218611738193261179310511854807446237996274956735188575272489122793818301194912%
9833673362440656643086021394946395224737190702179860943702770539217176293176752384674818467669405132%
0005681271452635608277857713427577896091736371787214684409012249534301465495853710507922796892589235%
4201995611212902196086403441815981362977477130996051870721134999999837297804995105973173281609631859%
5024459455346908302642522308253344685035261931188171010003137838752886587533208381420617177669147303%
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}
\pagestyle{empty}
\begin{document}
\DeclareFixedFont{\ps}{U}{psy}{m}{n}{12cm}
\begin{pspicture}(-6,-6)(15,5.5)
  \pstextpath{\pscustom[linestyle=none]{%
  \parametricplot[plotpoints=500]{0}{3510}{t cos t 600 div mul t sin t 600 div mul}
  \psline(!3510 cos 3510 600 div mul 30 add 3510 sin 3510 600 div mul)}}{\PI}
  \rput(0,-3.2){\pscharpath[linecolor=black!20,strokeopacity=0.8,
    fillstyle=solid,fillcolor=blue!40,opacity=0.75]{\rput[b](0,0){\ps p}}}
\end{pspicture}

\end{document}

enter image description here

\documentclass[pstricks]{standalone}
\usepackage[tiling]{pst-fill}      % PSTricks package for filling/tiling
\usepackage{pst-text}              % PSTricks package for character path
\begin{document}
\DeclareFixedFont{\ptsmall}{T1}{ptm}{m}{sc}{5mm}
\DeclareFixedFont{\ps}{U}{psy}{m}{n}{8cm}
\psboxfill{\ptsmall$\pi$}
\begin{pspicture}(-3,0)(2,4.5)
    \pscharpath[fillstyle=solid,fillcolor=cyan!20,addfillstyle=boxfill,
                fillangle=30,fillsep=0.6mm]%
                {\rput[b](-0.5,0){\ps\char112}}
\end{pspicture}

\end{document}

enter image description here

| improve this answer | | | | |
5

enter image description here

\usetikzlibrary{decorations.text,calc,math}
\begin{tikzpicture}[decoration={text effects along path,
text={3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233},
text effects/.cd,
character count=\i, character total=\n,
characters={text along path, evaluate={\c=\i/\n*100;},
text=orange!\c!blue, scale=\i/\n+0.5}}]
\draw[decorate , domain=-10:1080,variable=\t,samples=100]
      plot ({(\t+300)*cos(\t)/(720)},
           {(\t+300)*sin(\t)/(720)});

\node at (0,0){\Huge$\pi$};

\end{tikzpicture}
| improve this answer | | | | |
  • Wow...it is beautiful much much...my sincere compliments. Regards. – Sebastiano Mar 17 at 11:29
4

mwe

\documentclass{article}
\usepackage{pstricks}
\begin{document}
\psset{xunit=1pt,yunit=1pt,runit=1pt}
\begin{pspicture}(800,1200)
\pscustom[linewidth=15]{\newpath\moveto(170,992)\curveto(165,928)(152,855)(107,813)}
\pscustom[linewidth=15]{\newpath\moveto(253,973)\curveto(229,913)(250,823)(282,809)}
\pscustom[linewidth=15]{\newpath\moveto(88,962)\curveto(172,1053)(223,922)(327,996)}
\end{pspicture}
\end{document}
| improve this answer | | | | |
3
\begin{tikzpicture}
\filldraw[blue] (-3.32,1.88) .. controls (-2.62,4.28) and (0.04,2.87) .. (1.48,3.41) .. controls (1.38,2.54) and (0.57,2.53) .. (0.09,2.48) .. controls (0.08,1.03) and (-0.48,-1.4) .. (0.82,-0.31) .. controls (0.78,-1.24) and (-1.71,-2.57) .. (-0.31,2.51) .. controls (-0.87,2.5) and (-0.87,2.5) .. (-1.48,2.53) .. controls (-1.49,1.31) and (-2.05,-1.77) .. (-2.93,-1.1) .. controls (-3.48,-0.36) and (-2.27,-0.74) .. (-1.88,2.55) .. controls (-2.47,2.51) and (-2.79,2.36) .. (-3.32,1.88);
\end{tikzpicture}

enter image description here

| improve this answer | | | | |

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