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围绕某个点旋转路径

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围绕某个点旋转路径

我有一条命名路径（三角形）。如何围绕点 A 以各种角度旋转它而不重新计算点 B 和 C？

这是我的代码。我想通过简单地将黑色三角形旋转 60 度的倍数来创建红色和绿色三角形。

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

\begin{document}
\begin{tikzpicture}
    \providecommand* \angle {30}
    \coordinate[label=above left:A](A) at (2,3);
    \coordinate[label=below:B](B) at (0,0);
    \coordinate[label=below:C](C) at (6,0);
    \draw
        (A) -- (B) -- (C) -- cycle;

    \coordinate[label=B'](B') at ($(A)!1!60:(B)$);
    \coordinate[label=C'](C') at ($(A)!1!60:(C)$);
    \draw[dashed,red]
        (A) -- (B') -- (C') -- cycle;

    \coordinate[label=B''](B'') at ($(A)!1!60:(B')$);
    \coordinate[label=C''](C'') at ($(A)!1!60:(C')$);
    \draw[dashed,green]
        (A) -- (B'') -- (C'') -- cycle;
\end{tikzpicture}
\end{document}

旋转三角形

答案1

您正在覆盖节点，因此这有点棘手，但本质上您可以确定路径范围并使用rotate around转换。

\documentclass[tikz]{standalone}

\begin{document}
\begin{tikzpicture}
    \providecommand* \angle {30}
    \coordinate[label=above left:A](A) at (2,3);
    \coordinate[label=below:B](B) at (0,0);
    \coordinate[label=below:C](C) at (6,0);
    \draw
        (A) -- (B) -- (C) -- cycle;

\begin{scope}[rotate around={60:(A)}]
    \coordinate[label=above left:A](A) at (2,3);
    \coordinate[label=below:B](B) at (0,0);
    \coordinate[label=below:C](C) at (6,0);
    \draw[dashed,red]
        (A) -- (B) -- (C) -- cycle;
\end{scope}

\end{tikzpicture}
\end{document}

在此处输入图片描述

答案2

\documentclass[tikz]{standalone}

\begin{document}
\begin{tikzpicture}
    %\providecommand* \angle {30}
    \coordinate[label=above left:A](A) at (2,3);
    \coordinate[label=below:B](B) at (0,0);
    \coordinate[label=below:C](C) at (6,0);
    \draw
        (A) -- (B) -- (C) -- cycle;

\begin{scope}[rotate around={60:(A)}]
    \coordinate[label=above left:A](A) at (2,3);
    \coordinate[label=below:B](B) at (0,0);
    \coordinate[label=below:C](C) at (6,0);
    \draw[dashed,red]
        (A) -- (B) -- (C) -- cycle;
\end{scope}

\end{tikzpicture}
\end{document}

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