画一个中间有平滑曲线的三角形

Question

我需要画一幅这样的画。

我是用 Paint 画的，所以不太清楚。我需要三角形内部的平滑曲线以及连接曲线和圆的直线为虚线。出现两次的符号是\epsilon。

Answer 1

使用xelatex或运行latex->dvips->ps2pdf

\documentclass{article}
\usepackage{pstricks-add}
\begin{document}

\psset{unit = 2}
\begin{pspicture}(-1.7,-1.7)(2,2)
  \psaxes[xlabelsep =-0.02,xlabelOffset = 0.07,
          ylabelsep = -0.02,ylabelOffset = 0.1,
          arrowscale=1.5,linewidth=0.5pt]{->}(0,0)(-1.7,-1.7)(2,2)
  \pscircle(0,0){1}
  \psset{origin={0.15,0.05}}
  \pswedge[fillcolor=black!10,fillstyle=solid](0,0){0.65}{0}{30}
  \pscustom[fillcolor=black!10,fillstyle=solid]{
    \psline(1.05;30)(1.5;30)(1.5,0)
    \psarc(0,0){1.05}{0}{30}}  
  \pcline(0.8;20)(1.2;20)
  \naput[npos=0.2]{\small $\epsilon$}
  \naput[npos=0.8]{\small $\epsilon$}
  \naput[npos=1.5,labelsep=0.2]{\small $\overline{T}$}
\end{pspicture}

\end{document}

Answer 2

这是一个解决方案，其中饼状图的起始角度为 15 度（半径为 1.5 厘米），终止角度为 45 度。对于扇形区域，半径分别为 2.5 厘米和 4 厘米。这些都可以根据需要进行更改。

绘制圆弧的语法可以通过

\draw (x,y) arc (alph:beta:r);     % where x+jy=r\angle alph
\draw (alph:r) arc (alph:beta:r);  % start angle=alpha, end angle=beta, radius=r.

代码

\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{positioning,intersections}


\begin{document}

\begin{tikzpicture}
\draw (-2.5,0)node[left]{-1}--(2.5,0) node[right]{1};
\draw (0,2.5) node[right]{1}--(0,-2.5)node[right]{-1};
\draw (0,0) circle (2cm);
\draw (0,0) -- +(15:1.5cm) arc (15:45:1.5cm) -- cycle;
\draw (15:2.5)--(15:4) arc (15:45:4) --node[left]{$\bar T$} (45:2.5) arc (45:15:2.5);
\draw (25:1.5)--node[pos=0.2,above]{$\epsilon$}
                node[pos=0.7,above]{$\epsilon$}(25:2.5);
\end{tikzpicture}


\end{document}

Answer 3

这只是 PSTricks 的另一种解决方案。

\documentclass[pstricks,border=12pt,12pt]{standalone}
\usepackage{pst-eucl}
\psset{unit=2,PointSymbol=none,linejoin=1}
\begin{document}
\multido{\i=15+5}{15}{%
\begin{pspicture}(-1.5,-1.5)(2,1.5)
    \pstGeonode[PosAngle={-135,-45,-135,135},PointName={-1,1,-1,1}](-1,0){A}(1,0){B}(0,-1){C}(0,1){D}
    \pcline[nodesep=-.5](A)(B)
    \pcline[nodesep=-.5](C)(D)
    \pscircle{1}
    \def\temp{%
        \pnodes(0,0){O}(1.5;10){P}(1.5;\i){Q}%
        \pswedge{.5}{(P)}{(Q)}%
        \pscustom{\psarc(O){1}{(P)}{(Q)}\psline(Q)(P)\closepath}%
        \pstBissectBAC[linestyle=none,PointName=none]{P}{O}{Q}{R}%
        \pnodes([nodesep=.5]{R}O){X}([nodesep=1]{R}O){Y}%
        \pcline(X)(Y)%
        \psset{labelsep=2pt}
        \naput[npos=.2]{$\epsilon$}%
        \naput[npos=.8]{$\epsilon$}%
        \pcline[linestyle=none]([nodesep=1]{Q}O)(Q)%
        \naput{$\overline{T}$}%
    }%
    \rput(.3;10){\temp}
\end{pspicture}}

\end{document}

Answer 4

PSTricks 解决方案：

\documentclass{article}

\usepackage{pstricks-add}
\usepackage{xfp}

\newcommand*\radiusA{\fpeval{0.8-\xOrigin}}
\newcommand*\radiusB{\fpeval{1.2-\xOrigin}}
\newcommand*\radiusC{\fpeval{1.5+\xOrigin}}

\def\xOrigin{0.15}
\def\yOrigin{0.05}


\begin{document}

\psset{
  unit = 2,
  dimen = m
}
\begin{pspicture}(-1.7,-1.7)(2,2)
  \psaxes[
    xlabelsep = -0.02,
    xlabelOffset = 0.07,
    ylabelsep = -0.02,
    ylabelOffset = 0.1
  ]{->}(0,0)(-1.7,-1.7)(2,2)
  \pscircle(0,0){1}
 {\psset{linecap = 2}
  \pswedge(\xOrigin,\yOrigin){\radiusA}{0}{30}
  \psarc(\xOrigin,\yOrigin){1.05}{0}{30}
  \psline(\radiusB;0)(1.5;0)
  \psline(\radiusC,\yOrigin)(\fpeval{1.5*cosd(30)+\xOrigin},\fpeval{1.5*sind(30)+\yOrigin})}
 {\psset{origin = {\xOrigin,\yOrigin}}
  \psline(\radiusB;30)(1.5;30)
  \psline(\radiusB;0)(1.5;0)
  \psline(\radiusA;20)(\radiusB;20)
 {\psset{linestyle = none, offset = 4pt}
  \pcline(0.8;20)(1;20)
  \ncput{\small $\epsilon$}
  \pcline(1;20)(1.2;20)
  \ncput{\small $\epsilon$}
  \pcline(1.2;30)(\radiusC;30)
  \ncput{\small $\overline{T}$}}}
\end{pspicture}

\end{document}