TikZ：获取三角形内心的坐标

Question

我想获得三角形内心. 我想要一个宏

\incenter{name}{a}{b}{c}

它设置了三角形内心的坐标，name该三角形的顶点坐标为a、b、c。

我想用这个计算使用命令中的笛卡尔坐标let，但这不适用于坐标。

Answer 1

这或多或少直接取自文档tkz-euclide（texdoc -s euclide）：

\documentclass[border=2mm]{standalone}

\usepackage{tkz-euclide}
\usetkzobj{all}

\newcommand{\incenter}[4]{%
    \begin{tikzpicture}
        \tkzInit[xmax=5,ymax=4]
        \tkzClip
        \tkzDefPoint(0,0){#2} %% Put the coordinates here
        \tkzDefPoint(5,1){#3} %% for the desired
        \tkzDefPoint(1,4){#4} %% triangle.
        \tkzDrawPolygon[color=red](#2,#3,#4)
        \tkzInCenter(#2,#3,#4)
        \tkzGetPoint{G}
        \tkzDrawPoint(G)
        \node[below] at (G) {#1};
    \end{tikzpicture}
}

\begin{document}

\incenter{name}{a}{b}{c}

\end{document}

还有一件事。如果你想用 指定一个任意三角形\incenter，那么你可以尝试这个：

\documentclass[border=2mm]{standalone}

\usepackage{tkz-euclide}
\usetkzobj{all}

\newcommand{\incenter}[2]{%
    \begin{tikzpicture}
        \foreach \x/\y [count=\i from 1] in {#2}{\tkzDefPoint(\x,\y){n-\i}}
        \tkzInit[xmax=5,ymax=4]
        \tkzClip
        \tkzDrawPolygon[color=red](n-1,n-2,n-3)
        \tkzInCenter(n-1,n-2,n-3)
        \tkzGetPoint{G}
        \tkzDrawPoint(G)
        \node[below] at (G) {#1};
    \end{tikzpicture}
}

\begin{document}

\incenter{Incenter!}{0/0.5,5/3,1/4}% {label}{x_1/y_1,x_2/y_2,x_3/y_3}

\end{document}

Answer 2

如果你不介意按点工作：

\documentclass[tikz,border=5]{standalone}
\usetikzlibrary{math}
\begin{document}
\begin{tikzpicture}[x=1pt, y=1pt]
\draw [thick] (0, 0) coordinate (A) -- (20, 40) coordinate (B)
   -- (40, 10) coordinate (C) -- cycle;
\tikzmath{%
  coordinate \p, \I;
  \p1 = (A); \p2 = (B); \p3 = (C);
  \a = veclen(\px3-\px2, \py3-\py2);
  \b = veclen(\px1-\px3, \py1-\py3);
  \c = veclen(\px2-\px1, \py2-\py1);
  % Get the inradius
  \s = (\a + \b + \c) / 2;
  \K = sqrt(\s) * sqrt(\s - \a) * sqrt(\s - \b) * sqrt(\s - \c);
  \r = \K / \s;
  % Normalize the lengths a bit
  \m = max(\a, \b, \c);
  \a = \a / \m; \b = \b / \m; \c = \c / \m;
  % Get the incenter
  \ix = (\a*\px1 + \b*\px2 + \c*\px3) / (\a + \b + \c);
  \iy = (\a*\py1 + \b*\py2 + \c*\py3) / (\a + \b + \c);
  \I = (\ix, \iy);
}
\fill (\I) circle [radius=1];
\draw [red] (\I) circle [radius=\r];
\end{tikzpicture}
\end{document}