我如何在 LaTeX 中制作这个框？

Question

正如标题所示，我不知道如何制作这样的框，我需要一个确切的框来记录我的化学笔记。任何帮助都将不胜感激，因为我是 LaTeX 新手。

背景：这张图片来自一位讲师的一份文件

https://castel.dev/notes

他也链接了他的 GitHub 代码，但那里\begin{eq}并没​​有解释他是如何使他的盒子看起来像这样的。

Answer 1

正如大家在评论中所说，代码是在preamble.texgithub 上的文件中定义的。我不确定在这里展示一些代码是否有助于您学习。但是，我决定这样做。这是一个示例，使用来自 github 的完全相同的定义（仅包含示例、证明和定义）。

\documentclass{report}
\usepackage[dvipsnames]{xcolor}
\usepackage{amsthm}
\usepackage{thmtools}
\usepackage[framemethod=TikZ]{mdframed}
\mdfsetup{skipabove=1em,skipbelow=0em}

\declaretheoremstyle[
    headfont=\bfseries\sffamily\color{ForestGreen!70!black}, bodyfont=\normalfont,
    mdframed={
        linewidth=2pt,
        rightline=false, topline=false, bottomline=false,
        linecolor=ForestGreen, backgroundcolor=ForestGreen!5,
    }
]{thmgreenbox}

\declaretheoremstyle[
    headfont=\bfseries\sffamily\color{NavyBlue!70!black}, bodyfont=\normalfont,
    mdframed={
        linewidth=2pt,
        rightline=false, topline=false, bottomline=false,
        linecolor=NavyBlue, backgroundcolor=NavyBlue!5,
    }
]{thmbluebox}

\declaretheoremstyle[
    headfont=\bfseries\sffamily\color{NavyBlue!70!black}, bodyfont=\normalfont,
    numbered=no,
    mdframed={
        linewidth=2pt,
        rightline=false, topline=false, bottomline=false,
        linecolor=NavyBlue, backgroundcolor=NavyBlue!1,
    },
]{thmexplanationbox}

\declaretheorem[style=thmgreenbox, name=Definition]{definition}
\declaretheorem[style=thmbluebox, numbered=no, name=Example]{eg}
\declaretheorem[style=thmexplanationbox, name=Proof]{tmpexplanation}
\newenvironment{explanation}[1][]{\vspace{-10pt}\begin{tmpexplanation}}{\end{tmpexplanation}}

\begin{document}
\begin{eg}
    Suppose we have a functor.
    If $G_X \not\cong G_Y$, then  $X$ and  $Y$ are not homeomorphic.
    If `shadows' are different, then objects themselves are different too.
\end{eg}
\begin{explanation}
    Suppose $X$ and $Y$ are homeomorphic.
    Then $\exists f: X \to  Y$ and $g: Y \to  X$, maps (maps are always continuous in this course), such that $g  \circ f = 1_X$ and $f \circ g = 1_Y$.
    Then $f_*: G_X \to G_Y$ and $g_*: G_Y \to  G_X$ such that $(g \circ f)_* = (1_X)_*$ and  $(f \circ g)_* = (1_Y)_*$. Using the rules discussed previously, we get
    \[
    g_*  \circ f_* = 1_{G_X} \quad f_*  \circ  g_* = 1_{G_Y}
    ,\] 
    which means that $f_* : G_X \to  G_Y$ is an isomorphism.
\end{explanation}
\begin{definition}[Homotopy]
    Let $f,g:X\to Y$ be maps (so continuous). Then a homotopy between $f$ and $g$ is a continuous map $H: X\times I \to Y$ such that
    \begin{itemize}
        \item $H(x, 0) = f(x)$, $H(x, 1) = g(x)$ 
        \item For all $t \in I$, define $f_t: X \to  Y: x \mapsto  H(x, t)$
    \end{itemize}
    We say that $f$ is homotopic with $g$, we write $f \simeq g$.
    If $g$ is a constant map, we say that $f$ is null homotopic.
\end{definition}
\end{document}