TikZ 为方程式制作图形平衡

Question

我希望能够=在三角形内画一个白色，能够旋转天平和方程，并将方程对称放置但又不要太靠近边缘。这对我来说相当困难。

\documentclass{beamer}
\usepackage{tikz}
\begin{document}
\begin{frame}
\begin{center}
$2+2$ \hfill $4$
\begin{tikzpicture}
\draw[red!50!blue!70!white,ultra thick](-5.4,1) -- (5.4,1);
\fill[red!50!blue!70!white] (0,0.98) -- (-0.4,0) -- (0.4,0);
\end{tikzpicture}
\end{center}
\end{frame}
\end{document}

Answer 1

quick generating command请参阅本篇文章末尾的更新。

barycentric利用 提供的坐标系可以轻松找到三角形的中心TikZ。

定义三角形的三个角节点并说

\node[color=white]  at (barycentric cs:A=1,B=1,C=1) {=};

为了使用加权顶点（这里所有顶点都具有相同的权重，这对应于三角形的真实重心）。

\documentclass{beamer}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{frame}
\begin{center}
$2+2$ \hfill $4$
\begin{tikzpicture}
\coordinate (A) at (0,0.98);
\coordinate (B) at (-0.4,0);
\coordinate (C) at (0.4,0);
\draw[red!50!blue!70!white,ultra thick](-5.4,1) -- (5.4,1);
\fill[red!50!blue!70!white] (A) -- (B) -- (C);
\node[color=white]  at (barycentric cs:A=1,B=1,C=1) {=};
\end{tikzpicture}
\end{center}
\end{frame}
\end{document}

更新

\documentclass{beamer}
\usepackage{tikz}
\usetikzlibrary{calc}

\begin{document}
\begin{frame}
\begin{center}
\begin{tikzpicture}
\def\winkel{20}
\coordinate (A) at (0,0.98);
\coordinate (B) at (-0.4,0);
\coordinate (C) at (0.4,0);
\coordinate (rodleft) at (-5.4,1);
\coordinate (rodright) at (5.4,1);
\coordinate (balancetop) at (0,1);
\coordinate (rotatedleft) at ($(balancetop)-(\winkel:5)$);
\coordinate (rotatedright) at ($(balancetop)+(\winkel:5)$);
\draw[red!50!blue!70!white,ultra thick] (balancetop) -- (rotatedleft);
\draw[red!50!blue!70!white,ultra thick] (balancetop) -- (rotatedright);
\node[rotate=\winkel,above right] (LHS) at ($(rotatedleft) + (\winkel:1)$) {$2+2$};
\node[rotate=\winkel,above left] (RHS) at ($(rotatedright) - (\winkel:1)$) {$3$};
\fill[red!50!blue!70!white] (A) -- (B) -- (C);
\node[font=\large,color=white]  at (barycentric cs:A=1,B=1,C=1) {$\neq$};
\end{tikzpicture}
\end{center}
\end{frame}
\end{document}

第二次更新

宏\equationbalance检查参数#3并#4打印=或\neq，但是，仍然必须指定角度。我发现根据#3或的值来计算角度没什么用#4（从物理学的角度来看：如果刻度的两侧的值不相等，角度无论如何都不会是恒定的）

\documentclass{beamer}
\usepackage{tikz}
\usetikzlibrary{calc}

% First opt arg: angle
% 2nd arg: inward shift of the labels 
% 3rd arg: left hand label
% 4th arg: right hand label
\newcommand{\equationbalance}[4][0]{%
  \begin{tikzpicture}
    \def\winkel{#1}
    \coordinate (A) at (0,0.98);
    \coordinate (B) at (-0.4,0);
    \coordinate (C) at (0.4,0);
    \coordinate (rodleft) at (-5.4,1);
    \coordinate (rodright) at (5.4,1);
    \coordinate (balancetop) at (0,1);
    \coordinate (rotatedleft) at ($(balancetop)-(\winkel:5)$);
    \coordinate (rotatedright) at ($(balancetop)+(\winkel:5)$);
    \draw[red!50!blue!70!white,ultra thick] (balancetop) -- (rotatedleft);
    \draw[red!50!blue!70!white,ultra thick] (balancetop) -- (rotatedright);
    \node[rotate=\winkel,above right] (LHS) at ($(rotatedleft) + (\winkel:#2)$) {$#3$};
    \node[rotate=\winkel,above left] (RHS) at ($(rotatedright) - (\winkel:#2)$) {$#4$};
    \fill[red!50!blue!70!white] (A) -- (B) -- (C);
    % Works for simple integer expressions only
    \ifnum\numexpr#3=#4
    \node[font=\large,color=white]  at (barycentric cs:A=1,B=1,C=1) {$=$};
    \else
    \node[font=\large,color=white]  at (barycentric cs:A=1,B=1,C=1) {$\neq$};
    \fi
  \end{tikzpicture}%
}


\begin{document}
\begin{frame}
\begin{center}
\equationbalance[10]{1.5}{2+2}{3}
\end{center}
\end{frame}
\end{document}

Answer 2

有了新的TABcline选项tabstackengine( 2018/03/05 (V2.10))，这变成了一个很好的练习。支持宏\balanceindent和\lever分别控制从杠杆末端到杠杆到支点的文本缩进。杠杆厚度由\fboxrule在主宏内部本地定义的控制。主程序是

\balance[fulcrum inset]{left terms}{right terms}

默认的支点插入是?。如果设置为>或<，杠杆将向适当的方向倾斜。

这是MWE：

\documentclass{article}
\usepackage[TABcline]{tabstackengine}
\usepackage{graphicx}
\newcommand\Leqtriangle[1][=]{%
  \stackengine{2pt}{\Ltriangle}{$\scriptstyle#1\mkern1mu$}{O}{c}{F}{F}{L}}
\newcommand\Ltriangle{\LARGE$\triangle$}
\newcommand\balanceindent{\kern10pt}
\newcommand\lever{\kern7em}
\newcommand\balance[3][?]{{%
  \TABbinary\TABstackMath\setstackgap{S}{0pt}\fboxrule=1pt\relax%
  \def\bangle{4}% TILT ANGLE
  \def\bsign{}%
  \ifx>#1\def\bsign{-}\else\ifx<#1\else\def\bangle{0}\fi\fi%
  \rotatebox[origin=b]{\bsign\numexpr-\bangle}{%
    \tabularShortstack{lcr}{\mathstrut\balanceindent#2 && #3\balanceindent\\%
    \TABcline{1-3}\lever &%
    \rotatebox[origin=t]{\bsign\numexpr\bangle}{\Leqtriangle[#1]} & \lever}}}}
\begin{document}
\balance[>]{2 + 2(4-3)}{6 - 6/2}

\balance[=]{2 + 2}{4}

\balance[<]{2 + 3}{7}

\balance{y}{42}
\end{document}