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编辑：完整的 MWE

tag-icon • tag-icon tikz-pgf floats wrapfigure
编辑：完整的 MWE

我想使用wrapfigure环境将文本环绕在图形周围，在环境adjustwidth内的环境中proof，但是以下代码

\begin{proof}
\begin{adjustwidth}{2em}{2em}
         [...]
[...] arctan a = \frac{\pi}{2} \,$.
%
\begin{wrapfigure}{L}{0.5\textwidth}
\begin{tikzpicture}[scale=3]
      [...]
\end{tikzpicture}

\caption{Angles $ \alpha,\beta,\gamma,\delta $} that correspond to $ a,b,c,d $ respectively.
\label{angles}
\end{wrapfigure}

If we try to ``escape'' the inequality by augmenting the difference between two angle [...]

将图放在最后一页：

not wrapping

figure on separate page

知道为什么吗？这与环境有关吗adjustwidth？

编辑：完整的 MWE

\documentclass[10pt,a4paper,usenames,dvipsnames]{article}

\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{textcomp}
\usepackage[labelfont=bf]{caption}
\usepackage{tikz}
\usepackage{changepage}
\usepackage{wrapfig}
\usepackage{siunitx}
\usepackage{calc}
\usepackage{array}
\usepackage{pythonhighlight}
\usepackage{multirow}
\usepackage{listings}
\usepackage{framed}
\usepackage[symbol, perpage]{footmisc}
\usepackage{lmodern}
\usepackage{hyperref}

\lstset{
    basicstyle=\ttfamily,
    columns=fullflexible,
    frame=single,
    breaklines=true,
    postbreak=\mbox{\textcolor{red}{$\hookrightarrow$}\space},
}

\usepackage{float}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{array}

\newcolumntype{L}[1]{>{\raggedright\let\newline\\\arraybackslash\hspace{0pt}}m{#1}}
\newcolumntype{C}[1]{>{\centering\let\newline\\\arraybackslash\hspace{0pt}}m{#1}}
\newcolumntype{R}[1]{>{\raggedleft\let\newline\\\arraybackslash\hspace{0pt}}m{#1}}

\usepackage{makecell}
\usepackage{amsthm}
\usepackage{thmtools}
\usepackage{ragged2e}
\usepackage{amssymb}
\usepackage{longtable}
\usepackage{graphicx}

\usepackage{xcolor}    


\usepackage{chngcntr}
\counterwithin{figure}{section}
\counterwithin{equation}{section}

\newcommand\ddfrac[2]{\frac{\displaystyle #1}{\displaystyle #2}}


\usetikzlibrary{calc}
\usetikzlibrary{quotes, angles, arrows}

\newcommand{\degre}{\ensuremath{^\circ}}

\definecolor{dbwrru}{rgb}{0.8588235294117647,0.3803921568627451,0.0784313725490196}
\definecolor{dtsfsf}{rgb}{0.8274509803921568,0.1843137254901961,0.1843137254901961}
\definecolor{wrwrwr}{rgb}{0.3803921568627451,0.3803921568627451,0.3803921568627451}
\definecolor{rvwvcq}{rgb}{0.08235294117647059,0.396078431372549,0.7529411764705882}
\definecolor{cqcqcq}{rgb}{0.7529411764705882,0.7529411764705882,0.7529411764705882}
\definecolor{dark blue}{HTML}{002663}
\definecolor{dark green}{HTML}{085e23}
\definecolor{plum}{HTML}{3d085e}

\hypersetup{
    colorlinks=true,
    linkcolor=dark blue,
    filecolor=magenta,      
    urlcolor=cyan,
}


\begin{document}


If we now think of the elements in $ S $ as the \textit{angles} they map to (through the $ \arctan $ function), instead of their actual values, the inequality becomes intuitively evident.

\begin{proof} \ref{ineq}
\begin{adjustwidth}{2em}{2em}
    \ \\
    Let $ a $, $ b $, $ c $, $ d $ be elements in $ S $, where $ a>b>c>d $. Let $ \alpha \equiv \arctan a,\ \beta  \equiv \arctan b,\ \gamma \equiv \arctan c,\ \delta \equiv \arctan d$ (and therefore let $ \alpha>\beta>\gamma>\delta $).
    
    Since $ a,b,c,d $ are positive, the angles $ \alpha, \beta,\gamma,\delta $ belong to the first quadrant (or the third quadrant):
    \begin{equation*}
    \begin{gathered}
        0 < \theta < \frac{\pi}{2} \quad \mod \pi\\
        \forall \theta \in \{\alpha, \beta, \gamma, \delta\}\,.
    \end{gathered}
    \end{equation*}
    Now let us visualize these angles in the first quadrant\footnote{I'm picking just the first quadrant because $\quad \tan (\theta + k\pi) = \tan \theta \ \ \forall k \in \mathbb{Z} \;$.}; see fig. \ref{angles}. If we pick a set $ \{a,b,c,d\} $ such that $ \alpha,\beta,\gamma,\delta  $ are equally spaced, we can space them at most by $ \frac{\pi}{6} $, since there are $ 4-1=3 $ pairs of adjacent angles, and $ \frac{\pi}{2} \div 3 = \frac{\pi}{6}$. In fact, the difference between each adjacent angle can only be \textit{less} than $ \frac{\pi}{6} $, since there is no $ a\in \mathbb{R}^+ $ with finite value such that $\alpha = \arctan a = \frac{\pi}{2} \,$\footnote{Because $ \lim\limits_{\theta\rightarrow \frac{\pi}{2}\,^-} \tan \theta = \infty$.}.
    %
    \begin{wrapfigure}{L}{0.5\textwidth}
    \begin{tikzpicture}[scale=3]
        \coordinate (A) at (1,0);
        \coordinate (B) at (0,0);
        \coordinate (C) at ($ (0,0) +(30:1cm)$);
        \coordinate (D) at ($ (0,0) +(60:1cm)$);
        \coordinate (E) at ($ (0,0) +(90:1cm)$);
        %
        \draw (0,0) node [right=0.2cm, below=0.5mm] {\textcolor{blue}{$ \delta $}};
        \draw (A) -- (B) -- (E)
        pic [draw=red!50!black, fill=red!20, angle radius=2.4cm,
        "\textcolor{red}{$\alpha$}", pic text options={left=5.5mm, above=.3cm}] {angle = A--B--E};
        \draw (A) -- (B) -- (D)
        pic [draw=green!50!black, fill=green!20, angle radius=1.7cm,
        "\textcolor{dark green}{$\beta$}", pic text options={above=.1cm}] {angle = A--B--D};
        \draw (A) -- (B) -- (C)
        pic [draw=purple!50!black, fill=purple!20, angle radius=1cm,
        "\textcolor{plum}{$\gamma$}", pic text options={right=.05mm}] {angle = A--B--C};
        %
        \clip (-0.2, -0.2) rectangle (1.2,1.2);
        \draw (0,0) circle [radius=1cm];
        \fill [white] (0.95, -0.1) rectangle (1.05, 0.1);
        \draw (0, -0.2) -- (0,1.2);
        \draw (-0.2, 0) -- (1.2,0);
        \draw [thick] (A) -- +(90:0.5mm) -- +(270:0.5mm) node [below=.2mm, right=.1 mm] {$ 1 $};
        
        \draw [very thick, blue] (0,0) -- +(0:1cm);
        \draw [very thick, purple] (0,0) -- +(30:1cm);
        \draw [very thick, green] (0,0) -- +(60:1cm);
        \draw [very thick, red] (0,0) -- +(90:1cm);
        
        \draw [thick, dark blue] (C) arc [start angle = 30, end angle = 60, radius = 1cm] node [midway, sloped, above] {\small $ <\frac{\pi}{6} $};
    \end{tikzpicture}
    
    \caption{Angles $ \alpha,\beta,\gamma,\delta $} that correspond to $ a,b,c,d $ respectively.
    \label{angles}
    \end{wrapfigure}
    
    If we try to ``escape'' the inequality by augmenting the difference between two angles---say, for example, that we reduce $ \beta $ so that $ \alpha-\beta>\frac{\pi}{6} $---, we will be forcibly reducing the difference between another pair of adjacent angles---in our example, $ \beta-\gamma $ would be getting smaller---; so there will always be some pair of angles such that $ \theta_2 -\theta_1 < \frac{\pi}{6} $.
\end{adjustwidth}
\end{proof}

答案1

enter image description here

  • 首先我从你的 mwe 中真正地认识到 mwe :-)（从序言中删除所有不相关的包，你真的需要它们全部吗？）
  • 然后我稍微纠正一下你的tikz形象。最重要的是，它不包含空行。
  • 插入图像wrapfigure可以使用tex宏insbox。其语法是：

\InsertBoxR{n}{< content>}[correction]

for box on the right side or

\InsertBoxL{b}{< content>}[correction]

where `n` is number of lines above box and `correction` number of lines needed for correction for box space (especial, when protrude into next paragraph).

用于文本左侧的框。

  • tikzpicture与标题一起封装在minipage环境中
  • 标题使用包\captionof{figure}{...}提供的宏caption（也可以使用capt-of包，但caption提供设计标题的延伸功能）
  • 不幸的是，\adjustwidth来自包的宏changepage不适用于insbox（顺便说一句，该包的正确名称是changepage，chngpage它是弃用的名称）。插入的段落\InsertBoxR{n}{< content>}[correction]不遵循文本宽度的变化。

完整 mwe：

\documentclass[10pt,a4paper,usenames,dvipsnames]{article}

\usepackage{graphicx}
\usepackage[skip=1ex, labelfont=bf, font=footnotesize]{caption}
\usepackage{amsmath, amssymb, amsthm}
\input{insbox}%%%%%%%%%%%%%% TeX macro,
\usepackage{tikz}
\usetikzlibrary{angles, arrows,
                calc,
                quotes,
                }
\usepackage{siunitx}    % to write units. also defines `\ degree`
%\newcommand{\degre}{\ensuremath{^\circ}} % beter use \si{\degree} or \SI{90}{\degree}

%\from defined colors are used only the following
\definecolor{dark blue}{HTML}{002663}
\definecolor{dark green}{HTML}{085e23}
\definecolor{plum}{HTML}{3d085e}

\usepackage{hyperref}
\hypersetup{
    colorlinks=true,
    linkcolor=dark blue,
    filecolor=magenta,
    urlcolor=cyan,
}

\begin{document}
If we now think of the elements in $ S $ as the \textit{angles} they map to (through the $ \arctan $ function), instead of their actual values, the inequality becomes intuitively evident.

\begin{proof} \label{ineq}% <--- \label{...} not \ref{ineq} ?!

    Let $ a $, $ b $, $ c $, $ d $ be elements in $ S $, where $ a>b>c>d $. Let $ \alpha \equiv \arctan a,\ \beta  \equiv \arctan b,\ \gamma \equiv \arctan c,\ \delta \equiv \arctan d$ (and therefore let $ \alpha>\beta>\gamma>\delta $).

    Since $ a,b,c,d $ are positive, the angles $ \alpha, \beta,\gamma,\delta $ belong to the first quadrant (or the third quadrant):
    \begin{gather*}
        0 < \theta < \frac{\pi}{2} \quad \mod \pi\\
        \forall \theta \in \{\alpha, \beta, \gamma, \delta\}\,.
    \end{gather*}

    \InsertBoxR{2}{\begin{minipage}{0.45\linewidth}\centering
    \begin{tikzpicture}[scale=3]
\coordinate[label=below right:1]    (A) at (1,0);
\coordinate[label=below right:
            \textcolor{blue}{$\delta $}] (B) at (0,0);
\coordinate                         (C) at ($(B)+(30:1cm)$);
\coordinate                         (D) at ($(B)+(60:1cm)$);
\coordinate                         (E) at ($(B)+(90:1cm)$);
%
\draw (A) -- (B) -- (E)
    pic [draw=red!50!black, fill=red!20, angle radius=2.4cm,
        "\textcolor{red}{$\alpha$}",
        pic text options={left=5.5mm, above=.3cm}]  {angle = A--B--E};
\draw (A) -- (B) -- (D)
    pic [draw=green!50!black, fill=green!20, angle radius=1.7cm,
        "\textcolor{dark green}{$\beta$}",
        pic text options={above=.1cm}]              {angle = A--B--D};
\draw (A) -- (B) -- (C)
    pic [draw=purple!50!black, fill=purple!20, angle radius=1cm,
        "\textcolor{plum}{$\gamma$}",
        pic text options={right=.05mm}]             {angle = A--B--C};
%
\clip (-0.2, -0.2) rectangle (1.2,1.2);
\draw (B) circle [radius=1cm];
\fill [white] (0.95, -0.1) rectangle (1.05, 0.1);
\draw (0, -0.2) -- (0,1.2);
\draw (-0.2, 0) -- (1.2,0);
\draw [thick] (A) +(0,0.5mm) -- + (0,-0.5mm);
%
\draw [very thick, blue]    (B) -- +(0:1cm);
\draw [very thick, purple]  (B) -- +(30:1cm);
\draw [very thick, green]   (B) -- +(60:1cm);
\draw [very thick, red]     (B) -- +(90:1cm);
%
\draw [thick, dark blue] (C) arc (30:60:1cm) node [midway, sloped, above] {\small $ <\frac{\pi}{6} $};
    \end{tikzpicture}
                   \captionof{figure}{Angles $\alpha,\beta,\gamma,\delta$ that correspond to $a,b,c,d$ respectively.}
                   \label{angles}
                   \end{minipage}%
                   }[5]
    Now let us visualize these angles in the first quadrant\footnote{I'm picking just the first quadrant because $\quad \tan (\theta + k\pi) = \tan \theta \ \ \forall k \in \mathbb{Z} \;$.}; see fig. \ref{angles}. If we pick a set $ \{a,b,c,d\} $ such that $ \alpha,\beta,\gamma,\delta  $ are equally spaced, we can space them at most by $ \frac{\pi}{6} $, since there are $ 4-1=3 $ pairs of adjacent angles, and $ \frac{\pi}{2} \div 3 = \frac{\pi}{6}$. In fact, the difference between each adjacent angle can only be \textit{less} than $ \frac{\pi}{6} $, since there is no $ a\in \mathbb{R}^+ $ with finite value such that $\alpha = \arctan a = \frac{\pi}{2} \,$\footnote{Because $ \lim\limits_{\theta\rightarrow \frac{\pi}{2}\,^-} \tan \theta = \infty$.}.

    If we try to ``escape'' the inequality by augmenting the difference between two angles---say, for example, that we reduce $ \beta $ so that $ \alpha-\beta>\frac{\pi}{6} $---, we will be forcibly reducing the difference between another pair of adjacent angles---in our example, $ \beta-\gamma $ would be getting smaller---; so there will always be some pair of angles such that $ \theta_2 -\theta_1 < \frac{\pi}{6} $.
\end{proof}
\end{document}

注意：包hyperref必须在序言中最后加载（很少有例外）。

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