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在笛卡尔平面上绘制等腰三角形

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在笛卡尔平面上绘制等腰三角形

要在笛卡尔平面上画一个等腰三角形。等长的两条边应位于第一和第二象限。（我手头的代码给出的两条边相对于正 x 轴的角度分别为 41 度和 122 度。）两条边应标记为“1”，第三条边应标记为“c”——我猜是在边缘下方，以免干扰圆。

在以下代码中，以原点为端点的两条边的长度为 1。我如何让 TikZ 进行必要的计算以保持 41 度和 122 度的角度？原点标记为“O”，但它与 y 轴相干扰。我想覆盖它或将其移动到第三象限。（我尝试了命令\node[mydot,label={below,left:$O$}] at (O) {};。）

\documentclass[10pt]{amsart}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{mathtools,array}

\newlength{\widestright}
\newenvironment{horriblesystem}[1]
 {\settowidth{\widestright}{${}=#1$}%
  \left\{\,
  \begin{array}{@{}r@{}>{\raggedright\arraybackslash${}}p{\widestright}<{$}@{}}}
 {\end{array}\right.}

\usepackage{tikz}

\tikzset{
mydot/.style={
  fill,
  circle,
  inner sep=1.5pt
  }
}

\begin{document}

\begin{tikzpicture}[>=latex]


% the coordinates of the vertices
\coordinate (O) at (0,0);
\coordinate (A) at (2,1);
\coordinate (B) at (-3,5);

% the axis
\draw[help lines,->] (-3.5,0) -- (2.5,0);
\draw[help lines,->] (0,-0.5) -- (0,5.5);

% the edges of the triangle
\draw (O) -- (A) -- (B) -- cycle;

% labelling the vertices
\node[mydot,label={right:$A$}] at (A) {};
\node[mydot,label={left:$B$}] at (B) {};
\node[mydot,label={below:$O$}] at (O) {};

% the arcs for the angles
\begin{scope}[gray]
\draw[->]
  (1,0) +(0:0.5cm) arc [radius=1cm,start angle=0,end angle=41] node[midway,right] {$\alpha$};
\draw[->]
  (0.5,0) +(0:0.25cm) arc [radius=0.75cm,start angle=0,end angle=122] node[midway,above] {$\beta$};
\end{scope}

\end{tikzpicture}

\end{document}

答案1

一种使用极坐标的选项（见下文备注）：

在此处输入图片描述

代码：

\documentclass[10pt]{amsart}
\usepackage{tikz}
\usetikzlibrary{angles,quotes}

\tikzset{
mydot/.style={
  fill,
  circle,
  inner sep=1.5pt
  }
}
\def\Side{6cm}
\begin{document}

\begin{tikzpicture}[>=latex]
% the coordinates of the vertices
\coordinate (O) at (0,0);
\path  ++(122:\Side) coordinate (A) ++(-40:\Side)
coordinate (B);

% auxiliary coordinates to draw the marks for the angles
\coordinate (aux) at (3,0);

% the axis
\draw[help lines,->] (-3.5,0) -- (2.5,0);
\draw[help lines,->] (0,-0.5) -- (0,5.5);

% the edges of the triangle
\draw 
  (O) --
    node[auto] {$l$} 
  (A) -- 
    node[auto] {$l$} 
  (B) -- 
    node[auto] {$c$} 
  cycle;

% labelling the vertices
\node[mydot,label={above:$A$}] at (A) {};
\node[mydot,label={right:$B$}] at (B) {};
\node[mydot,label={below left:$O$}] at (O) {};

% the arcs for the angles
\begin{scope}[gray,angle eccentricity=1.2,->]
\path 
  pic[draw,angle radius=1.5cm,"$\alpha$"] {angle = aux--O--B}
  pic[draw,->,angle radius=0.75cm,"$\beta$"] {angle = aux--O--A}
  ;
\end{scope}
\end{tikzpicture}

\end{document}

评论

  1. 可以使用极坐标获得顶点：

    % the coordinates of the vertices
\coordinate (O) at (0,0);
\path  ++(122:\Side) coordinate (A) ++(-40:\Side)
coordinate (B);

  2. 可以使用节点生成侧面的标签：

    % the edges of the triangle
\draw 
  (O) --
    node[auto] {$l$} 
  (A) -- 
    node[auto] {$l$} 
  (B) -- 
    node[auto] {$c$} 
  cycle;

  3. 使用angles和quotes库来生成角度标记更加容易：

    \path 
  pic[draw,angle radius=1.5cm,"$\alpha$"] {angle = aux--O--B}
  pic[draw,->,angle radius=0.75cm,"$\beta$"] {angle = aux--O--A}
  ;

  4. 使用\Side您可以轻松改变长度。

答案2

这个使用intersections库，你必须知道两边的长度。其余的，我采用了 Gonzalo 的代码。

\documentclass[10pt]{amsart}
\usepackage{tikz}
\usetikzlibrary{angles,quotes,intersections}

\tikzset{
mydot/.style={
  fill,
  circle,
  inner sep=1.5pt
  }
}

%% two sides
\def\Side{6cm}  
\def\side{2cm}
\begin{document}

\begin{tikzpicture}[>=latex]
% the coordinates of the two vertices
\coordinate (O) at (0,0);
\coordinate (A) at (122:\Side);
\begin{pgfinterruptboundingbox}
  \path[name path global=circle] (A) circle (\Side);
  \path[name path global=line] (O)circle (\side); 
\end{pgfinterruptboundingbox}
  \path[name intersections={of=line and circle}];
  \coordinate (B) at (intersection-1);   %% third vertex
  \draw (O) -- (B) -- (A) -- cycle;

% auxiliary coordinates to draw the marks for the angles
\coordinate (aux) at (3,0);

% the axis
\draw[help lines,->] (-3.5,0) -- (2.5,0);
\draw[help lines,->] (0,-0.5) -- (0,5.5);

% the edges of the triangle
\draw
  (O) --
    node[auto] {$l$}
  (A) --
    node[auto] {$l$}
  (B) --
    node[auto] {$c$}
  cycle;

% labelling the vertices
\node[mydot,label={above:$A$}] at (A) {};
\node[mydot,label={right:$B$}] at (B) {};
\node[mydot,label={below left:$O$}] at (O) {};

% the arcs for the angles
\begin{scope}[gray,angle eccentricity=1.2,->]
\path
  pic[draw,angle radius=1.5cm,"$\alpha$"] {angle = aux--O--B}
  pic[draw,->,angle radius=0.75cm,"$\beta$"] {angle = aux--O--A}
  ;
\end{scope}
\end{tikzpicture}

\end{document}

在此处输入图片描述

插图

\path[draw,red,name path global=circle] (A) circle (\Side);
\path[draw,red,name path global=line] (O)circle (\side);

给出

在此处输入图片描述

答案3

嗯，从其他答案来看，也许我不明白 OP 的要求，因为我认为这是所需的图像：

在此处输入图片描述

我还使用了和库pic的语法，因此这需要最新版本：anglesquotesPGF

\documentclass[tikz,border=5]{standalone}
\usetikzlibrary{angles,quotes}
\tikzset{mydot/.style={ fill,  circle,  inner sep=1.5pt }}

\begin{document}
\begin{tikzpicture}[>=latex]

% the coordinates of the vertices
\coordinate (O) at (0:0);
\coordinate (A) at (41:5);
\coordinate (B) at (122:5);

% the axis
\draw[help lines,->] (-3,0) -- (4,0) coordinate (X);
\draw[help lines,->] (0,-0.5) -- (0,5);

% the edges of the triangle
\draw (O)
   -- (A)   node [midway, below right] {$l$}
   -- (B)   node [midway, above]       {$c$} 
   -- cycle node [midway, below left]  {$l$};

% labelling the vertices
\node[mydot,label={right:$A$}]      at (A) {};
\node[mydot,label={left:$B$}]       at (B) {};
\node[mydot,label={below left:$O$}] at (O) {};

% the arcs for the angles
\begin{scope}[gray]
\path pic ["$\alpha$", draw, ->, angle eccentricity=1.10, angle radius=1.5cm] 
  {angle=X--O--A};
\path pic ["$\beta$",  draw, ->, angle eccentricity=1.25, angle radius=1.0cm] 
  {angle=X--O--B};
\end{scope}
\end{tikzpicture}
\end{document}

答案4

使用 tkz-euclide v2.42b (beta)这里 但也许罐装版本就还不错。

\documentclass{standalone}
\usepackage{tkz-euclide}
\def\side{6}

\begin{document}

\begin{tikzpicture}[>=latex,scale=2]
    \tkzInit[xmin=-4,xmax=2,ymax=5]
    \tkzDrawXY[noticks]
    \tkzDefPoint(0,0){O}
    \tkzDefPoint(1,0){i}
    \tkzDefPoint(122:\side){A}
    \tkzDefShiftPoint[A](-40:\side){B}

    \tkzDrawPolygon(O,A,B)
    \tkzDrawPoints(O,A,B)

    \tkzLabelPoints(O,B)
    \tkzLabelPoints[above left](A)
    \tkzLabelSegment(A,B){$l$}
    \tkzLabelSegments[below](A,O){$l$}
    \tkzLabelSegments[below](O,B){$c$}

    \tkzMarkAngle[mark=none, size=0.75,->](i,O,A)
    \tkzLabelAngle[pos=.85](i,O,A){$\beta$}
    \tkzMarkAngle[mark=none, size=1.5,->](i,O,B)
    \tkzLabelAngle[pos=1.60](i,O,B){$\alpha$}
\end{tikzpicture}

\end{document}

在此处输入图片描述

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