Tikz：绘制一条与另一条路径长度相同的路径，从特定坐标开始并与给定点相交

Question

我有一些 tikz 代码可以绘制带有标记顶点的三角形。

我想画一条从原点开始、与标记为 P 的点相交、长度等于标记为 A 的边的线。

我认为 calc 库就是我想要的，但我无法弄清楚如何将 A 的长度提供给它...

这是我目前的代码，我对我想要绘制的边缘命令的最佳猜测是：

\documentclass{standalone}

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



\begin{document}

\begin{tikzpicture}[scale=3]


% layout some coordinates for a triangle
\coordinate (P) at ($(0,0) + (rand,rand)$);
\coordinate (Q) at ($(2,-2) + .5*(rand, rand)$);
\coordinate (R) at ($(-2, -2) + .5*(rand, rand)$);
%convienience for finding the centre
\coordinate (PQR) at ($(P)+(Q)+(R)$);
\coordinate (O) at ($1/3*(PQR)$);

%draw triangle
\draw[name path=A] (P) -- (Q);
\draw[name path=B] (Q) -- (R);
\draw[name path=C] (R) -- (P);

%labels
\node at (O) {$O$};
\node at (P) [above=2pt]{$P$};
\node at (Q) [right=2pt]{$Q$};
\node at (R) [left=2pt]{$R$};

\node at ($(P)!.5!(Q)$) [above right=1em]{$\mathbf{A}$};

%draw a path of length (A) from (O), passing through (P)
%\draw[name path=U] (O) -- ($(O)! !(B)$)

\end{tikzpicture}
\end{document}

这给了我这个输出

Answer 1

使用tkz-euclide，你可以使用这个：

\documentclass[tikz,border=10pt]{standalone}

\usepackage{tkz-euclide}

\begin{document}

    \begin{tikzpicture}[scale=3]
    
        % Coordinates for a triangle
        \coordinate (P) at ($(0,0) + (rand,rand)$);
        \coordinate (Q) at ($(2,-2) + .5*(rand, rand)$);
        \coordinate (R) at ($(-2, -2) + .5*(rand, rand)$);
        \coordinate (O) at (barycentric cs:P=1,Q=1,R=1) ;
        
        % Draw triangle
        \draw (P) -- (Q) node[midway,above,sloped] {dist $=A$}  -- (R) -- cycle;
        
        \tkzDrawPoints(P,Q,R,O)
        
        % Labels
        \node at (O) [left=2pt]{$O$};
        \node at (P) [above right]{$P$};
        \node at (Q) [right=2pt]{$Q$};
        \node at (R) [left=2pt]{$R$};
        
        % Calculate length of segment PQ and label as dPQ
        \tkzCalcLength(P,Q)\tkzGetLength{dPQ}
        
        % Create some point on a circle centred on O and with radius = PQ length
        \tkzDefPointOnCircle[angle=90,center=O,radius=\dPQ pt]
        \tkzGetPoint{I}
        
        % Find intersection between line OP and previous circle
        % Two points, but only one is useful
        \tkzInterLC(O,P)(O,I) \tkzGetPoints{D}{E}
        
        % Draw the segment from O, passing through P, the length of PQ
        \draw[dashed,blue] (O) -- (E) node[pos=0.7,left] {dist$=A$};
        
        \tkzDrawPoints(E)
        \tkzLabelPoints[left](E)
    
    \end{tikzpicture}
\end{document}

编辑
利用这个其他聪明的选项（除此之外veclen我绝对不会使用），这里还有另一个更短的版本，用于tikz-euclide计算线段长度和绘制线段的更简单实现。

\documentclass[tikz,border=10pt]{standalone}

\usepackage{tkz-euclide}

\begin{document}

    \begin{tikzpicture}[scale=3]
    
        % Coordinates for a triangle
        \coordinate (P) at ($(0,0) + (rand,rand)$);
        \coordinate (Q) at ($(2,-2) + .5*(rand, rand)$);
        \coordinate (R) at ($(-2, -2) + .5*(rand, rand)$);
        \coordinate (O) at (barycentric cs:P=1,Q=1,R=1) ;
        
        % Draw triangle
        \draw (P) -- (Q) node[midway,above,sloped] {dist $=A$}  -- (R) -- cycle;
        
        \tkzDrawPoints(P,Q,R,O)
        
        % Labels
        \node at (O) [left=2pt]{$O$};
        \node at (P) [above right]{$P$};
        \node at (Q) [right=2pt]{$Q$};
        \node at (R) [left=2pt]{$R$};
        
        % Calculate length of segment PQ
        \tkzCalcLength(P,Q)\tkzGetLength{dPQ}
        
        % Draw the segment from O, passing through P, the length of PQ
        \draw[orange,thick,dashed] (O)--++ ($(O)!\dPQ pt!(P)-(O)$) coordinate[at end] (E);

        \tkzDrawPoints(E)
        \tkzLabelPoints[left](E)

    
    \end{tikzpicture}
\end{document}

Answer 2

(\x,\y)您可以使用计算“向量”的长度sqrt(\x1*\x1+\y1*\y1)。然后使用构造

($(A)!<len>!(B)-(A)$)

创建一个从 (A) 到 (B) – (A) 方向的矢量，其长度为<len>：

这里，蓝线的端点为 (A) = (1,0) 和 (B) = (3,1)。红线从 (C) = (1,1) 到 (D) = (2,2) 绘制，线长从 (A) 到 (B)，使用代码

\draw[red,thick] let\p1=($(B)-(A)$) in (C)--++($(C)!sqrt(\x1*\x1+\y1*\y1)!(D)-(C)$);

该let命令\p1将从 (A) 分配给 (B) 的“向量”，并自动将其 x 和 y 分量分配给\x1和\y1。

如果愿意，您可以定义一个\thru接受 5 个参数的命令，其中一个是可选的：

\newcommand{\thru}[5][]{\draw[#1] let\p1=($#5-#4$) in #2--++($#2!sqrt(\x1*\x1+\y1*\y1)!#3-#2$);}

可选参数可以是任何tikz你想要的属性。第一个必需参数是你的起点，第二个是终点，第三个和第四个是你想要的长度的线段的端点。

因此，\thru[green,thick]{(2,1)}{(3,2)}{(A)}{(B)}将从 (2,1) 到 (3,2) 绘制一条粗绿线，其长度为线段 (A)--(B)。

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{calc}

\newcommand{\thru}[5][]{\draw[#1] let\p1=($#5-#4$) in #2--++($#2!sqrt(\x1*\x1+\y1*\y1)!#3-#2$);}

\begin{document}

\begin{tikzpicture}
\draw[help lines] (0,0) grid (4,3);
\coordinate(A) at (1,0);
\coordinate(B) at (3,1);
\coordinate(C) at (1,1);
\coordinate(D) at (2,2);
\draw[blue,thick] (A)--(B);

\draw[red,thick] let\p1=($(B)-(A)$) in (C)--++($(C)!sqrt(\x1*\x1+\y1*\y1)!(D)-(C)$);
\thru[green,thick]{(2,1)}{(3,2)}{(A)}{(B)}

\end{tikzpicture}

\end{document}