# Modeling randomly generated inscribed triangles

The problem I am modeling: Three points are randomly chosen on a circle. What is the probability that the triangle formed by these three points contains the center of the circle?

Conceptual understanding: Suppose we fix two of the three points, call them A and B. In order for the triangle to contain the center, the third point C must lie within the arc A'B', where A' and B' are the image of points A and B respectively under a rotation of 180 degrees.

What I want to happen: The randomly generated inscribed triangle to be filled green when it contains the center, and to fill red when it does not contain the center. I would also like to keep tally of the number of successes and failures to compute an experimental probability.

A few key things: I have access to the x and y coordinates of each point by using \pgfextractx and \pgfextracty. My method was to test whether the point C is between both the x-coordinates and y-coordinates of A and B by using \xintifboolexpr, however, this is flawed.

Minimal Working Example:

\documentclass{article}
\usepackage[left=2cm, right=2cm, top=2cm, bottom=1cm]{geometry}
\usepackage{amsmath,amsfonts,tikz,xintexpr,calc}

\newcommand\circletest{
\begin{tikzpicture}[scale=0.6]

\newdimen{\tempxa}
\newdimen{\tempya}
\newdimen{\tempxb}
\newdimen{\tempyb}
\newdimen{\tempxc}
\newdimen{\tempyc}

\draw (0,0) coordinate (O);
\pgfextractx\tempxa{\pgfpointanchor{A}{center}}
\pgfextracty\tempya{\pgfpointanchor{A}{center}}
\pgfextractx\tempxb{\pgfpointanchor{B}{center}}
\pgfextracty\tempyb{\pgfpointanchor{B}{center}}
\pgfextractx\tempxc{\pgfpointanchor{C}{center}}
\pgfextracty\tempyc{\pgfpointanchor{C}{center}}

\xintifboolexpr { (((\tempxc > -\tempxa) && (\tempxc < -\tempxb)) || ((\tempxc > -\tempxb) && (\tempxc < -\tempxa))) && (((\tempyc > -\tempya) && (\tempyc < -\tempyb)) || ((\tempyc > -\tempyb) && (\tempyc < -\tempya)))} %%I know this is grotesque
{\filldraw[color=green!80!black!100, fill=green!15] (A) -- (B) -- (C) -- cycle;} %true
{\filldraw[color=red!80!black!100, fill=red!15] (A) -- (B) -- (C) -- cycle;} %false

\draw (A) node[below]{A};
\draw (B) node[below]{B};
\draw (C) node[below]{C};
\end{tikzpicture}}

\begin{document}

\foreach \x in {0,1,...,11}{
\circletest
}

\end{document}



The issue I am having: Clearly my comparison operator \xintifboolexpr, along with my grotesque code following it is the problem. I am seeking a simpler method to tell if the point C is along the arc of the circle between (-\tempax,-\tempay) and (-\tempbx,-\tempby).

EDIT: A correct solution from Sandy G's suggestion.

\documentclass{article}
\usepackage[left=2cm, right=2cm, top=2cm, bottom=1cm]{geometry}
\usepackage{amsmath,amsfonts,tikz,xintexpr,calc}
\usepackage{xfp}

\newcommand\circletest{
\begin{tikzpicture}[scale=0.6]

\pgfmathsetmacro{\rndA}{rnd*360}
\pgfmathsetmacro{\rndB}{rnd*360}
\pgfmathsetmacro{\rndC}{rnd*360}

%defining x and y coordinates of each point

%calculating side lengths of triangle
\def\A{\fpeval{sqrt((\xb-\xc)^2 + (\yb-\yc)^2)}}
\def\B{\fpeval{sqrt((\xa-\xc)^2 + (\ya-\yc)^2)}}
\def\C{\fpeval{sqrt((\xa-\xb)^2 + (\ya-\yb)^2)}}

%calculating angles of triangle
\def\angleA{\fpeval{acosd((\B^2 + \C^2 -\A^2)/(2*\B*\C))}}
\def\angleB{\fpeval{acosd((\C^2 + \A^2 -\B^2)/(2*\C*\A))}}
\def\angleC{\fpeval{acosd((\A^2 + \B^2 -\C^2)/(2*\A*\B))}}

%defining some coordinates
\draw (0,0) coordinate (O);
\draw (\xa,\ya) coordinate (A);
\draw (\xb,\yb) coordinate (B);
\draw (\xc,\yc) coordinate (C);

%test if center is in circle
\xintifboolexpr{((\angleA < 90) && (\angleB < 90)) && (\angleC < 90)}
{\filldraw[color=green!80!black!100, fill=green!15] (A) -- (B) -- (C) -- cycle;} %true
{\filldraw[color=red!80!black!100, fill=red!15] (A) -- (B) -- (C) -- cycle;} %false

%Drawing points on top of line
\draw[fill=black] (\xa,\ya) circle(1.5pt);
\draw[fill=black] (\xb,\yb) circle(1.5pt);
\draw[fill=black] (\xc,\yc) circle(1.5pt);
\draw[fill=black] (O) circle(1.5pt);

\end{tikzpicture}}

\begin{document}

\foreach \x in {0,1,...,30}{
\circletest
}

\end{document}


• You could calculate the three angles. If any angle is obtuse (> 90°) then the center is outside the triangle. Otherwise the triangle contains the center. Mar 25, 2020 at 1:02
• Great suggestion! I brought it to fruition in my edit :) Mar 25, 2020 at 2:30
• Way cool observation. When one of the angles is 90 degrees the line connecting the two other angles runs right through the center. Apr 2, 2020 at 8:15

One can use the calc library and this prescription, which is very much like yours but perhaps a bit shorter. Using the calc library also allows us to avoid introducing new dimensions. Defining a pic has the advantage that you can use TikZ to arrange the drawings in any way you like.

\documentclass{article}
\usepackage[left=2cm, right=2cm, top=2cm, bottom=1cm]{geometry}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
What is the probability for the triangle containing the origin? Without loss of
generality we can take the angle of $A$ to be 0 (because one can rotate the
setup without changing the probability). Then the angle of $B$, $\beta$ can be
chosen to be between $0$ and $\pi$ (because one can reflect the setup at the
$x$--axis without changing the probability). Then the angle of $C$,
$\gamma$, needs to satisfy
$\pi<\gamma<\pi+\beta$
for the center to be inside the triangle, see Figure~\ref{fig:derivation}.
As $\beta$ scans the domain $[0,\pi]$, the probability for a triangle with
corners at random positions of the circle enclosing the center of the circle is
$1/4$.
\begin{figure}[ht]
\centering
\begin{tikzpicture}[dot/.style={circle,inner sep=1pt,fill},
declare function={rr=2.5;}]
\begin{scope}
\draw (0,0) circle[radius=rr] (0,0) -- (rr,0) node[dot,label=right:$A$]{};
\pgfmathsetmacro{\rndB}{rnd*90}
node[midway,anchor=180+\rndB/2,circle]{$\beta$}
(0,0) -- (\rndB:rr) node[dot,label={[anchor=\rndB+180]:$B$}]{};
\draw[dashed] (180+\rndB:rr) -- (0,0) -- (180:rr);
node[midway,anchor=\rndB/2,circle,align=right]{allowed\\ positions\\ for $C$};
\end{scope}
%
\begin{scope}[xshift=2.8*rr*1cm]
\draw (0,0) circle[radius=rr] (0,0) -- (rr,0) node[dot,label=right:$A$]{};
\pgfmathsetmacro{\rndB}{90+rnd*90}
node[midway,anchor=180+\rndB/2,circle]{$\beta$}
(0,0) -- (\rndB:rr) node[dot,label={[anchor=\rndB+180]:$B$}]{};
\draw[dashed] (180+\rndB:rr) -- (0,0) -- (180:rr);
node[midway,anchor=\rndB/2,circle,align=right]{allowed\\ positions\\ for $C$};
\end{scope}
\end{tikzpicture}
\label{fig:derivation}
\end{figure}

\begin{figure}[ht]
\centering
\begin{tikzpicture}[pics/circletest/.style={code={
\tikzset{circletest/.cd,#1}%
\def\pv##1{\pgfkeysvalueof{/tikz/circletest/##1}}%
\pgfmathsetmacro{\rndA}{rnd*360}
\pgfmathsetmacro{\rndB}{rnd*360}
\pgfmathsetmacro{\rndC}{rnd*360}
\path (\rndA:\pv{r}) coordinate[label={[anchor=\rndA+180]:$A$}] (A)
(\rndB:\pv{r}) coordinate[label={[anchor=\rndB+180]:$B$}] (B)
(\rndC:\pv{r}) coordinate[label={[anchor=\rndC+180]:$C$}] (C);
\draw let \p1=(A),\p2=(B),\p3=(C),\p0=(O),
\n1={(\x0-\x2)*(\y1-\y2)-(\x1-\x2)*(\y0-\y2)},
\n2={(\x0-\x3)*(\y2-\y3)-(\x2-\x3)*(\y0-\y3)},
\n3={(\x0-\x1)*(\y3-\y1)-(\x3-\x1)*(\y0-\y1)}
in \pgfextra{\pgfmathtruncatemacro\itest{%
((\n1 < 0) || (\n2 < 0) || (\n3 < 0)) &&
((\n1 > 0) || (\n2 > 0) || (\n3 > 0))}}
\ifnum\itest=0
[color=green!80!black!100, fill=green!15] (A) -- (B) -- (C) -- cycle
\else
[color=red!80!black!100, fill=red!15]  (A) -- (B) -- (C) -- cycle
\fi;
\fill (O) circle[radius=1pt] node[below]{$O$};
}},circletest/.cd,r/.initial=1]
\path foreach \X in {1,...,5}
{  foreach \Y in {1,...,5} {(3*\X,3*\Y) pic{circletest}}};
\end{tikzpicture}
\end{figure}

\end{document}


An alternative proposal based on intersections. Construct a ray that leaves the circle from its center. If the number of intersections with the triangle is even, the center is outside of the triangle, otherwise it is inside.

\documentclass{article}
\usepackage[left=2cm, right=2cm, top=2cm, bottom=1cm]{geometry}
\usepackage{tikz}
\usetikzlibrary{intersections}
\begin{document}
\begin{tikzpicture}[pics/circletest/.style={code={
\tikzset{circletest/.cd,#1}%
\def\pv##1{\pgfkeysvalueof{/tikz/circletest/##1}}%
\pgfmathsetmacro{\rndA}{rnd*360}
\pgfmathsetmacro{\rndB}{rnd*360}
\pgfmathsetmacro{\rndC}{rnd*360}
\path (\rndA:\pv{r}) coordinate[label={[anchor=\rndA+180]:$A$}] (A)
(\rndB:\pv{r}) coordinate[label={[anchor=\rndB+180]:$B$}] (B)
(\rndC:\pv{r}) coordinate[label={[anchor=\rndC+180]:$C$}] (C);
\path[name path=triangle] (A) -- (B) -- (C) -- cycle;
\path[name path=ray,overlay] (O) -- ({180+(\rndA+\rndB+\rndC)/3}:1.5*\pv{r});
\draw[name intersections={of=triangle and ray,total=\t}]
\ifodd\t
[color=green!80!black!100, fill=green!15] (A) -- (B) -- (C) -- cycle
\else
[color=red!80!black!100, fill=red!15]  (A) -- (B) -- (C) -- cycle
\fi;
}},circletest/.cd,r/.initial=1]
\path foreach \X in {1,...,5}
{  foreach \Y in {1,...,5} {(3*\X,3*\Y) pic{circletest}}};
\end{tikzpicture}
\end{document}


This approach is limited by the accuracy of intersections, and can fail if the triangle is to thin, i.e. essentially a line.

P.S. These distributions are consistent with the actual probability.

• You could calculate the three angles. If any angle is obtuse (> 90°) then the center is outside the triangle. Otherwise the triangle contains the center. Mar 25, 2020 at 1:04
• @SandyG Thanks! I agree. The point of setting the above code in the way it is done is that the test works for any point. That is, given a triangle ABC and a point P, the test will decide whether or not P is inside the triangle, regardless of whether it is the center of the circumcircle. But I agree that one could devise simpler tests here, such as the one you describe, or the one that is implicit in the analytic discussion at the end.
– user194703
Mar 25, 2020 at 1:09
• True. That way we can decide if a triangle chosen at random in the plane contains a random point in the plane! ;-) Just kidding. Nice solution! Mar 25, 2020 at 1:15
• This is beautiful yet daunting solution. I have much to learn! Thanks for taking the time to do this. Mar 25, 2020 at 2:31
• @LoganWeinert This is explained in section 18.3 Defining New Pic Types of pgfmanual v3.1.5. I like /.style={code={...}} better than /.pic because you can then also use /.style args={#1 and #2}{code={...}}, say. And I tried to explain why I like using pgf keys in this answer. It is basically because I always can add features without losing backward compatibility, and also because I personally can then remember the usage more easily.
– user194703
Mar 25, 2020 at 19:38

To satisfy my curiosity about the experimental probability, I did this in metapost. It seems to take about 100,000 triangles to consistently get the theoretical probability (i.e. 1/4) to 3 decimal places. If you comment the drawing commands to just print the result, then 1,000,000 runs only takes a few seconds. A portion of the out put for 20,000 inscribed triangles in 1mm circles :

Run with lualatex:

\documentclass{article}
\usepackage{luamplib}
\usepackage{geometry}
\mplibnumbersystem{double}
\mplibtextextlabel{enable}
\mplibcodeinherit{enable}
\begin{document}
\begin{mplibcode}
vardef triarray(expr r,n)=
save x,tmp,width;
width:=\mpdim{\linewidth} div r;
count:=0;
tot:=n;
for j=0 upto n:
% for the grid
drawoptions(withpen pencircle scaled .1bp shifted ((r+.1)*(j mod width),-(r+.1)*(j div width)));
for i=1 upto 3: x[i]:=uniformdeviate(8); endfor;
% sort vals, probably didn't need to, but made things tidier.
if x1>x2:
tmp:=x1; x1:=x2; x2:=tmp;
fi;
if x2>x3:
tmp:=x2; x2:=x3; x3:=tmp;
if x1>x2:
tmp:=x1; x1:=x2; x2:=tmp;
fi;
fi;
% end sort
% points on a circle in mp are mapped to the interval [0,8] with 0->0 and 8->360
% reflected points rather than rotating arc
if ((x1+4) mod 8>x2) and ((x1+4) mod 8<x3) and ((x3+4) mod 8>x1) and ((x3+4) mod 8<x2):
fill fullcircle scaled r withcolor .2[white,green];
count:=count+1;
else:
fill fullcircle scaled r withcolor .2[white,red];
fi;
% uncomment below for the triangles
draw for i=1 upto 3: point x[i] of (fullcircle scaled r)-- endfor cycle;
endfor;
enddef;
beginfig(0);

triarray(1mm,20000);

endfig;
\end{mplibcode}
\begin{mplibcode}
beginfig(1);
picture p; string s;
s="$\frac{"&decimal(count)&"}{"&decimal(tot)&"}="&decimal(count/tot)&"$";
p= s infont defaultfont scaled defaultscale;
draw p;
endfig;
\end{mplibcode}
\end{document}

• This is really neat! Well done. Mar 31, 2020 at 2:33
• Neat! The relative error goes like 1/sqrt(N), so even this is consistent with theory. ;-)
– user194703
Mar 31, 2020 at 2:44
• @schrodinger'scat. Thanks for pointing out the error, it seemed weird it was taking so long, but that explains it :) Mar 31, 2020 at 3:52