무작위로 생성된 내접삼각형 모델링

Question 1

도서관 을 이용할 수 있고 calc,이 처방전, 이는 귀하의 것과 매우 유사하지만 약간 더 짧을 수도 있습니다. 라이브러리를 사용하면 calc새로운 차원을 도입하는 것을 피할 수도 있습니다. a를 정의하면 picTi를 사용할 수 있다는 장점이 있습니다.케이Z 원하는 방식으로 그림을 배열합니다.

\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}
  \draw (1,0) arc[start angle=0,end angle=\rndB,radius=1] 
    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);
  \draw[blue,thick] (180:rr) arc[start angle=180,end angle=180+\rndB,radius=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}
  \draw  (1,0) arc[start angle=0,end angle=\rndB,radius=1] 
    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);
  \draw[blue,thick] (180:rr) arc[start angle=180,end angle=180+\rndB,radius=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}}%
        \draw (0,0) coordinate (O) circle[radius=\pv{r}];
        \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}

교차점을 기반으로 한 대안 제안입니다. 중심에서 원을 벗어나는 광선을 구성합니다. 삼각형과의 교점 개수가 짝수이면 중심은 삼각형의 외부에 있고, 그렇지 않으면 내부에 있습니다.

\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}}%
        \draw (0,0) coordinate (O) circle[radius=\pv{r}];
        \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}

이 접근 방식은 교차점의 정확성에 의해 제한되며 삼각형이 얇아지면(즉, 기본적으로 선인 경우) 실패할 수 있습니다.

PS 이러한 분포는 실제 확률과 일치합니다.

Answer