指示 TikZ 计算三角形内接圆的半径

Question

我有一个三角形的代码，并且已经找到了可以内切于该三角形的圆心。我想在不使用三角形内切圆半径公式的情况下绘制内切圆。我该如何TikZ计算这个半径并绘制圆？

\documentclass{amsart}
\usepackage{amsmath}
\usepackage{amsfonts}

\usepackage{tikz}
\usetikzlibrary{calc,intersections}

\begin{tikzpicture}

\path (0,1) coordinate (A) (3.5,1.5) coordinate (B) (-1.5,-0.75) coordinate (C);
\draw (A) -- (B) -- (C) -- cycle;

%The angle bisector at A is drawn.
\coordinate (a_path_for_angle_bisector_at_A) at ($(A)!2cm!(B)$);
\coordinate (another_path_for_angle_bisector_at_A) at ($(A)!2cm!(C)$);
\coordinate (midpoint_on_line_segment_to_locate_angle_bisector_at_A) at ($(a_path_for_angle_bisector_at_A)!0.5!(another_path_for_angle_bisector_at_A)$);
\path[name path=a_path_to_locate_center_of_circle] (A) -- ($(A)!1cm!(midpoint_on_line_segment_to_locate_angle_bisector_at_A)$);

%The angle bisector at C is drawn.
\coordinate (a_path_for_angle_bisector_at_C) at ($(C)!2cm!(A)$);
\coordinate (another_path_for_angle_bisector_at_C) at ($(C)!2cm!(B)$);
\coordinate (midpoint_on_line_segment_to_locate_angle_bisector_at_C) at ($(a_path_for_angle_bisector_at_C)!0.5!(another_path_for_angle_bisector_at_C)$);
\path[name path=another_path_to_locate_center_of_circle] (C) -- ($(C)!3cm!(midpoint_on_line_segment_to_locate_angle_bisector_at_C)$);

%The center of the circle inscribed in the triangle is located.
\coordinate[name intersections={of=a_path_to_locate_center_of_circle and another_path_to_locate_center_of_circle, by={O}}];
\draw[fill] (O) circle (1.5pt);


\end{tikzpicture}

\end{document}

Answer 1

您可以使用

\draw 
  let 
  \p1=( $ ( $ (A)!(O)!(B) $ )-(O) $ )
  in
  (O) circle ({veclen(\x1,\y1)});

（半径是从 到其中一条边O的投影的线段的长度）。O

完整代码：

\documentclass{amsart}
\usepackage{amsmath}
\usepackage{amsfonts}

\usepackage{tikz}
\usetikzlibrary{calc,intersections}

\begin{document}

\begin{tikzpicture}

\path (0,1) coordinate (A) (3.5,1.5) coordinate (B) (-1.5,-0.75) coordinate (C);
\draw (A) -- (B) -- (C) -- cycle;

%The angle bisector at A is drawn.
\coordinate (a_path_for_angle_bisector_at_A) at ($(A)!2cm!(B)$);
\coordinate (another_path_for_angle_bisector_at_A) at ($(A)!2cm!(C)$);
\coordinate (midpoint_on_line_segment_to_locate_angle_bisector_at_A) at ($(a_path_for_angle_bisector_at_A)!0.5!(another_path_for_angle_bisector_at_A)$);
\path[name path=a_path_to_locate_center_of_circle] (A) -- ($(A)!1cm!(midpoint_on_line_segment_to_locate_angle_bisector_at_A)$);

%The angle bisector at C is drawn.
\coordinate (a_path_for_angle_bisector_at_C) at ($(C)!2cm!(A)$);
\coordinate (another_path_for_angle_bisector_at_C) at ($(C)!2cm!(B)$);
\coordinate (midpoint_on_line_segment_to_locate_angle_bisector_at_C) at ($(a_path_for_angle_bisector_at_C)!0.5!(another_path_for_angle_bisector_at_C)$);
\path[name path=another_path_to_locate_center_of_circle] (C) -- ($(C)!3cm!(midpoint_on_line_segment_to_locate_angle_bisector_at_C)$);

%The center of the circle inscribed in the triangle is located.
\coordinate[name intersections={of=a_path_to_locate_center_of_circle and another_path_to_locate_center_of_circle, by={O}}];
\draw[fill] (O) circle (1.5pt);

\draw 
  let 
  \p1=( $ ( $ (A)!(O)!(B) $ )-(O) $ )
  in
  (O) circle ({veclen(\x1,\y1)});

\end{tikzpicture}

\end{document}

Answer 2

这是一个tkz-euclide变体。

\documentclass{amsart}
\usepackage{tkz-euclide}
\usetkzobj{all}
\begin{document}
  \begin{tikzpicture}
     \tkzDefPoint(0,1){A}
     \tkzDefPoint(3.5,1.5){B}
     \tkzDefPoint(-1.5,-0.75){C}
     \tkzDrawPolygon(A,B,C)

     \tkzDefCircle[in](A,B,C)
     \tkzGetPoint{O} \tkzGetLength{rIN}         
     \tkzDrawPoints(A,B,C,O)
     \tkzDrawCircle[R,color=red](O,\rIN pt)         
     \tkzLabelPoints[below](B,C)
     \tkzLabelPoints[above left=0mm and -2mm](A,O)
  \end{tikzpicture}
\end{document}