Tikzpicture: termina de dibujar una línea curva para un trozo de pastel

Question

Como se describió anteriormente, quiero dar el toque final dibujando la línea curva azul para que la imagen terminada parezca un trozo de pastel.

Gracias.

Lo que tengo hasta ahora:

\documentclass[]{article}
\usepackage[margin=0.5in]{geometry}
\usepackage{pgfplots}
\renewcommand{\thesection}{\arabic{section}}
\usepackage{mathtools}
\usepackage{cancel}
\usepackage{pgfplots}
\usepackage{amsmath}
\newtheorem{theorem}{THEOREM}
\newtheorem{proof}{PROOF}
\usepackage{tikz}
\usepackage{amssymb}
\usetikzlibrary{patterns}
\usepackage{fancyhdr}
\usepackage{bigints}
\usepackage{color}
\usepackage{tcolorbox}
\usepackage{color,xcolor}
\usepackage{booktabs,array}
\usepackage{hyperref}
\usepackage{graphicx}
\usetikzlibrary{arrows}
\usepackage{polynom}
\usepackage{flexisym}
\usepackage{wallpaper}
\usepackage{blkarray}
\usepackage{caption}
\usepackage{lscape}
\usepgfplotslibrary{fillbetween}
\usepgfplotslibrary{statistics}
\usetikzlibrary{shapes.misc}
\usetikzlibrary{arrows.meta}
\newenvironment{tightcenter}{
\setlength\topsep{0pt}
\setlength\parskip{0pt}
\begin{center}}{\end{center}}
\begin{document}
\begin{tikzpicture}
%\draw[thick] (0,0) circle (4.5cm);
%\draw[fill=black] (0,0) circle (0.3mm);
%\node[below] at (0,0){$O$};
%\node[below] at (1.25,0.9){$2$ cm};
%\node[below] at (-0.9,0.9){$2$ cm};
%
%\draw[thick,color=black,fill=gray!30] (0,0) --  (120:4.5) arc(120:45:4.5) -- cycle;
\draw[thick,color=black,fill=gray!30] (0,0) --  (-20:4.5) arc(-20:-50:4.5) -- cycle;
\draw[thick,color=blue!30] (-35.5:5.2) arc(-35.5:-60:5.2);
\draw[thick,color=black] (4.23,-1.55) -- (4.23,-3.52);
\draw[thick,color=black] (2.9,-3.43) -- (2.9,-4.67);
\draw[thick,color=black] (0,0) -- (0,-1.04)-- (2.9,-4.67);
%
%\draw [thick,<->] (0.28,-0.4) -- node[fill=white] {\small $2$ \text{cm}} (2.12,1.33);
%\draw [thick,<->] (-0.5,-0.3) -- node[fill=white] {\small $2$ \text{cm}} (-1.73,1.83);
%
\end{tikzpicture}
\end{document}

Answer 1

Como esto ?

Para evitar tener que calcular manualmente las coordenadas de los puntos, utilizo el posicionamiento relativo de los puntos con la sintaxis --++. Esta sintaxis indica que para obtener las coordenadas del siguiente punto, sumamos el punto anterior(0,-1.24)

(4.23,-1.55) --++ (0,-1.24)

es equivalente a

(4.23,-1.55) -- (4.23,-2.79)

de hecho 4.23 + 0 = 4.23y-1.55 + (-1.24) = -2.79

\documentclass[tikz,border=5mm]{standalone}

\begin{document}
\begin{tikzpicture}
\draw[thick,color=black,fill=gray!30] (0,0) --  (-20:4.5) arc(-20:-50:4.5) -- cycle;
\draw[thick,color=black,yshift=-1.24cm] (-20:4.5) arc(-20:-50:4.5) -- (0,0);
\draw[thick,color=black] (4.23,-1.55) --++ (0,-1.24);
\draw[thick,color=black] (2.9,-3.43) --++ (0,-1.24);
\draw[thick,color=black] (0,0) --++ (0,-1.24);

\end{tikzpicture}
\end{document}

Answer 2

Si usa tikz-3dplot, no tendrá que adivinar las curvas y podrá ajustar los ángulos de visión a voluntad.

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz-3dplot}
\begin{document}
\tdplotsetmaincoords{75}{60}
\begin{tikzpicture}[tdplot_main_coords]
\begin{scope}[canvas is xy plane at z=0,name prefix=bot-]
 \draw (0,0) coordinate (O) -- (4,0) coordinate (A) arc (0:60:4) coordinate
 (B);
\end{scope} 
\begin{scope}[canvas is xy plane at z=2,name prefix=top-]
 \draw[fill=blue!20] (0,0) coordinate (O) -- (4,0) coordinate (A) arc (0:60:4) coordinate
 (B) -- cycle;
\end{scope} 
 \draw foreach \X in {O,A,B}
 {(bot-\X) -- (top-\X)};
\end{tikzpicture}
\end{document}

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz-3dplot}
\begin{document}
\foreach \X in {89,88,...,60,61,62,...,88}
{\tdplotsetmaincoords{70+10*sin(6*\X)}{\X}
\pgfmathsetmacro{\xmin}{0}
\pgfmathsetmacro{\xmax}{0}
\pgfmathsetmacro{\ymin}{0}
\pgfmathsetmacro{\ymax}{0}
\begin{tikzpicture}[tdplot_main_coords]
    \ifdefined\figbb\relax
    \path \figbb;
    \fi
    \begin{scope}[canvas is xy plane at z=0,name prefix=bot-]
     \draw (0,0) coordinate (O) -- (4,0) coordinate (A) arc (0:60:4) coordinate
     (B);
    \end{scope} 
    \begin{scope}[canvas is xy plane at z=2,name prefix=top-]
     \draw[fill=blue!20] (0,0) coordinate (O) -- (4,0) coordinate (A) arc (0:60:4) coordinate
     (B) -- cycle;
    \end{scope} 
    \draw foreach \X in {O,A,B} {(bot-\X) -- (top-\X)};
    \path let \p1=(current bounding box.south west),
    \p2=(current bounding box.north east)
    in \pgfextra{%
    \pgfmathsetmacro{\xmin}{min(\x1,\xmin)}
    \pgfmathsetmacro{\xmax}{max(\x2,\xmax)}
    \pgfmathsetmacro{\ymin}{min(\y1,\ymin)}
    \pgfmathsetmacro{\ymax}{max(\y2,\ymax)}
    \xdef\xmin{\xmin pt}
    \xdef\xmax{\xmax pt}    
    \xdef\ymin{\ymin pt}
    \xdef\ymax{\ymax pt}    
    }; 
\end{tikzpicture}}
\makeatletter               
\edef\figbb{(\xmin,\ymin) rectangle (\xmax,\ymax)}
\immediate\write\@mainaux{\xdef\string\figbb{\figbb}\relax}
\makeatother
\end{document}

Si desea permitir ángulos de visión arbitrarios, debe distinguir algunos casos como enesta respuestaque te proporciona el resto de la tarta (de queso) (excepto el trozo robado por el ratón ;-).

Answer 3

Un ejemplo más: en el dibujo los ángulos se consideran en orden inverso. Para líneas verticales, la coordenada está definida, de modo que ahora solo es necesaria una coordenada para determinar la altura del corte:

\documentclass[tikz, margin=3mm]{standalone}

\begin{document}
    \begin{tikzpicture}[
every path/.style = {thick, line join=round} % style of lines
                        ]
\draw[fill=gray!30] (0, 0) --   (-50:4.5) coordinate (a1) arc(-50:-20:4.5) coordinate (a2) -- cycle;
\draw      (0,0) -- (0,-2) % determine height of slice
                           -- ++(-50:4.5) coordinate (b1) arc(-50:-20:4.5) coordinate (b2);
\draw      (a1) -- (b1)    (a2) -- (b2);
    \end{tikzpicture}
\end{document}