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使用 `let-in` 语法指定角度和长度

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使用 `let-in` 语法指定角度和长度

我有$\triangle{ABC}$一条$\ell$通过三角形两边中点的线。 $\ell$平行于$\overline{BC}$。在代码的最后几行中，我试图绘制另一条k通过A 平行于 $\ell$和 的线$\overline{BC}$。我想k在上面延伸$\overline{AC}$相同的量$\ell$。我尝试使用以下命令。

\draw[dashed, green] let \p1=($(upper_left_arrowhead)-(A)$), \n1={atan(\y1/\x1)} in (A) -- ($(A) +(\n1:{veclen(\x1,\y1)})$);

在我看来，我不能在命令中同时使用\n1={atan(\y1/\x1)}和。veclen(\x1,\y1)\draw

代码可以编译，因为我把它放在了%最后三行的前面。

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

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


\begin{document}
\begin{tikzpicture}


%The vertices of a triangle are located.
\coordinate (A) at (0,0);
\coordinate (B) at (20:7);
\coordinate (C) at ($(B) +(150:3)$);
\draw (A) -- (B) -- (C) -- cycle;

%The midpoint of AB is called P.
\coordinate (P) at ($(A)!0.5!(B)$);
\path[name path=the_part_of_ell_through_AB] let \p1=($(B)-(C)$), \n1={atan(\y1/\x1)} in (P) -- ($(P) +(\n1:2.5)$);
\path[name path=the_part_of_ell_through_AC] let \p1=($(B)-(C)$), \n1={atan(\y1/\x1)} in (P) -- ($(P) +({\n1+180}:4)$);

%The arrowheads of line $\ell$ are to be drawn at the intersections of the horizontal and
%vertical lines through A.
\path[name path=the_vertical_line_through_A] (A) -- ($(A) +(0,3.5)$);
\path[name path=the_horizontal_line_through_A] (A) -- ($(A) +(6,0)$);

\coordinate[name intersections={of=the_part_of_ell_through_AC and the_vertical_line_through_A, by=upper_left_arrowhead}];
\coordinate[name intersections={of=the_part_of_ell_through_AB and the_horizontal_line_through_A, by=lower_right_arrowhead}];
\draw[latex-latex] (upper_left_arrowhead) -- (lower_right_arrowhead);
\draw let \p1=($(B)-(C)$), \n1={atan(\y1/\x1)} in node[anchor={\n1+180}, inner sep=0] at ($(lower_right_arrowhead) + (\n1:0.15)$){$\ell$};

%Sides $\overline{AP}$ and $\overline{BP}$ are marked with "|".
\draw ($($(A)!0.5!(P)$)!3pt!90:(P)$) -- ($($(A)!0.5!(P)$)!3pt!-90:(P)$);
\draw ($($(B)!0.5!(P)$)!3pt!90:(P)$) -- ($($(B)!0.5!(P)$)!3pt!-90:(P)$);


%The labels for the vertices of the triangle are typeset.
\draw let \p1=($(A)-(B)$), \n1={atan(\y1/\x1)}, \p2=($(A)-(C)$), \n2={atan(\y2/\x2)} in node[anchor={0.5*(\n1+\n2)}, inner sep=0] at ($(A) + ({0.5*(\n1+\n2)+180}:0.15)$){$A$};

\draw let \p1=($(B)-(A)$), \n1={atan(\y1/\x1)}, \p2=($(B)-(C)$), \n2={atan(\y2/\x2)} in node[anchor={0.5*(\n1+\n2)+180}, inner sep=0] at ($(B) + ({0.5*(\n1+\n2)}:0.15)$){$B$};

\draw let \p1=($(C)-(B)$), \n1={atan(\y1/\x1)}, \p2=($(C)-(A)$), \n2={atan(\y2/\x2)} in node[anchor={0.5*(\n1-180+\n2)}, inner sep=0] at ($(C) + ({0.5*(\n1-180+\n2)+180}:0.15)$){$C$};

%The label for the midpoint of AB is typeset.
\draw let \p1=($(B)-(A)$), \n1={atan(\y1/\x1)}, \p2=($(B)-(C)$), \n2={atan(\y2/\x2)} in node[anchor={0.5*(\n1-180+\n2)+180}, inner sep=0] at ($(P) + ({0.5*(\n1-180+\n2)}:0.15)$){$P$};


%A line k is drawn through A that is parallel to BC. k is to extend above side AC as much as $\ell$ does,
%and k is to extend below side AB as much as $\ell$ does.
%\draw[dashed, green] let \p1=($(upper_left_arrowhead)-(A)$) in (A) -- ($(A) +({atan(\y1/\x1)}:{veclen(\x1,\y1)})$);
%\draw[dashed, green] let \p1=($(upper_left_arrowhead)-(A)$), \n1={atan(\y1/\x1)} in (A) -- ($(A) +(\n1:{veclen(\x1,\y1)})$);
%\draw[dashed] let \p1=($(lower_right_arrowhead)-(A)$), \n1={atan(\y1/\x1)} in (A) -- ($(A) +({atan(\y1/\x1)}:{veclen(\x1,\y1)})$);

\end{tikzpicture}

\end{document}

答案1

问题是(upper_left_arrowhead)直接位于 上方(A)。因此，\x1等于零，因此\y1/\x1无法计算。TeX 会在控制台上告诉您这一点。它与 无关veclen。编译还没有进行到这一步。

我有点不明白你为什么要用看起来不必要的复杂方式做事。例如，为什么要用这样的方式标记顶点和中点？

为什么不只是，例如

  \coordinate [label=below left:$A$] (A) at (0,0);
  \coordinate [label=right:$B$] (B) at (20:7);
  \coordinate [label=above:$C$] (C) at ($(B) +(150:3)$);

  ...

  \coordinate [label=below:$P$] (P) at ($(A)!0.5!(B)$);
  \path[name path=the_part_of_ell_through_AB] let \p1=($(B)-(C)$), \n1={atan(\y1/\x1)} in (P) -- ($(P) +(\n1:2.5)$) node {$\ell$};

据我所知，它的作用基本上是相同的？

我认为对于新路径，您需要类似以下内容的内容。

改变线条的绘制升这样路径就被命名为upper path：

  \draw[latex-latex,name path=upper path] (upper_left_arrowhead) -- (lower_right_arrowhead);

现在命名一条路径A到C：

  \path [name path=ac] (A) -- (C);

并得到这些线的交点：

  \coordinate [name intersections={of=upper path and ac, by=intersect}];

然后画出相关部分升移动，以便它们相对于A而不是升和AB和交流：

  \draw [dashed, green] ($(upper_left_arrowhead)-(intersect)$) -- ($(lower_right_arrowhead)-(P)$);

l 和 k

完整代码：

\documentclass[tikz,border=10pt]{standalone}
\usetikzlibrary{calc,intersections}
\begin{document}
\begin{tikzpicture}
  %The vertices of a triangle are located.
  \coordinate [label=below left:$A$] (A) at (0,0);
  \coordinate [label=right:$B$] (B) at (20:7);
  \coordinate [label=above:$C$] (C) at ($(B) +(150:3)$);
  \draw (A) -- (B) -- (C) -- cycle;

  %The midpoint of AB is called P.
  \coordinate [label=below:$P$] (P) at ($(A)!0.5!(B)$);
  \path[name path=the_part_of_ell_through_AB] let \p1=($(B)-(C)$), \n1={atan(\y1/\x1)} in (P) -- ($(P) +(\n1:2.5)$) node {$\ell$};
  \path[name path=the_part_of_ell_through_AC] let \p1=($(B)-(C)$), \n1={atan(\y1/\x1)} in (P) -- ($(P) +({\n1+180}:4)$);

  %The arrowheads of line $\ell$ are to be drawn at the intersections of the horizontal and
  %vertical lines through A.
  \path[name path=the_vertical_line_through_A] (A) -- ($(A) +(0,3.5)$);
  \path[name path=the_horizontal_line_through_A] (A) -- ($(A) +(6,0)$);

  \coordinate[name intersections={of=the_part_of_ell_through_AC and the_vertical_line_through_A, by=upper_left_arrowhead}];
  \coordinate[name intersections={of=the_part_of_ell_through_AB and the_horizontal_line_through_A, by=lower_right_arrowhead}];
  \draw[latex-latex,name path=upper path] (upper_left_arrowhead) -- (lower_right_arrowhead);

  %Sides $\overline{AP}$ and $\overline{BP}$ are marked with "|".
  \draw ($($(A)!0.5!(P)$)!3pt!90:(P)$) -- ($($(A)!0.5!(P)$)!3pt!-90:(P)$);
  \draw ($($(B)!0.5!(P)$)!3pt!90:(P)$) -- ($($(B)!0.5!(P)$)!3pt!-90:(P)$);

  \path [name path=ac] (A) -- (C);
  \coordinate [name intersections={of=upper path and ac, by=intersect}];
  \draw [dashed, green] ($(upper_left_arrowhead)-(intersect)$) -- ($(lower_right_arrowhead)-(P)$);

\end{tikzpicture}
\end{document}

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