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在 tkz-euclide 上绘制外圆

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在 tkz-euclide 上绘制外圆

我知道可以在 tkz-euclide 中使用以下命令绘制三角形的内切圆：

\tkzDefCircle[in](A,B,C)\tkzGetPoint{I}\tkzGetLength{r}
\tkzDrawCircle[R](I,r pt)

有没有画三角形外圆的快捷方式？作为参考，这是三角形的三个外圆： 三角形的外接圆

谢谢！

答案1

我实际上正在开发 tkz-euclide 的新版本。现在有 ex 圆圈

% tkz-euclide 的下一个版本发布时附带：

%tkzExCenter(A,B,C)定义以 AB 为圆心的外圆点，

%\tkzDefCircle[ex](A,B,C)表示以 AB 为中心点的圆，你得到一个点和一个半径

% 您还\tkzDrawCircle[ex](A,B,C)

％ 例子

其实你可以这样做

\documentclass{article}
\usepackage{tkz-euclide} % loads  TikZ and tkz-base
\usetkzobj{all}

\makeatletter
%<----------------------------------–>
%                     ExCenter
%<----------------------------------–>
\def\tkzExCenter(#1,#2,#3){%
\begingroup
    \tkzDefBisectorOutLine(#1,#2,#3)\tkzGetPoint{tkz@bex2}
    \tkzDefBisectorOutLine(#2,#1,#3)\tkzGetPoint{tkz@bex1}
    \tkzInterLL(#1,tkz@bex1)(#2,tkz@bex2)
\endgroup
} 

\def\tkzDefExCircle(#1,#2,#3){%
\begingroup
   \tkzExCenter(#1,#2,#3) \tkzGetPoint{tkz@ea}
   \tkzUProjection(#1,#2)(tkz@ea)
   \tkzCalcLength(tkzPointResult,tkz@ea)
   \tkzRenamePoint(tkz@ea){tkzPointResult} 
\endgroup
} 
\makeatother



\begin{document}
    \begin{tikzpicture}[scale=.4]
      \tkzInit[xmin=-10,xmax=10,ymin=-10,ymax=10]
      \tkzClip[space=3]
      \tkzDefPoint(0,0){A}
      \tkzDefPoint(6,1){B}
      \tkzDefPoint(1,4){C}

      \tkzDrawLines[add=1 and 1](A,B A,C B,C)
      \tkzDefExCircle(A,B,C) \tkzGetPoint{I} \tkzGetLength{rI}
      \tkzDefExCircle(C,A,B) \tkzGetPoint{J} \tkzGetLength{rJ}
      \tkzDefExCircle(B,C,A) \tkzGetPoint{K} \tkzGetLength{rK}
      \tkzDrawPoints(I,J,K)
      \tkzLabelPoints(A,B,C,I,J,K)
      \tkzDrawPolygon(A,B,C)
      \tkzDrawPoints(A,B,C)
      \tkzDrawCircle[R](J,\rJ pt)  
      \tkzDrawCircle[R](I,\rI pt)      
      \tkzDrawCircle[R](K,\rK pt)
    \end{tikzpicture}
\end{document}

在此处输入图片描述

答案2

只是为了比较，也可能是为了其他人阅读这个问题的利益，这里有一种在平原中生成外圆（和内圆）的方法元帖子如图所示：

在此处输入图片描述

prologues := 3;
outputtemplate := "%j%c.eps";

vardef incircle(expr A,B,C) = 
  save a,b,c,t; pair a,b,c,t;
  a = A + unitvector(C-A)+unitvector(B-A);
  b = B + unitvector(A-B)+unitvector(C-B);
  c = whatever[A,a] = whatever [B,b]; t = whatever[A,B];
  t-c = whatever * (B-A) rotated 90;
  fullcircle scaled 2 abs(c-t) shifted c
enddef;

vardef excircle(expr A,B,C) = 
  save a,b,c,t; pair a,b,c,t;
  a = A + (unitvector(C-A)+unitvector(B-A)) rotated 90;
  b = B + (unitvector(A-B)+unitvector(C-B)) rotated 90;
  c = whatever[A,a] = whatever [B,b]; t = whatever[A,B];
  t-c = whatever * (B-A) rotated 90;
  fullcircle scaled 2 abs(c-t) shifted c
enddef;

beginfig(1);

pair A, B, C;
A = origin;
B = 89 right rotated 10;
C = 55 right rotated 72;

fill A--B--C--cycle withcolor .87[blue,white];
path c[];
c1 = excircle(B,C,A);
c2 = excircle(C,A,B);
c3 = excircle(A,B,C);

draw c1 withcolor .53 red;
draw c2 withcolor .53 green;
draw c3 withcolor .53 blue;

drawoptions(withcolor .6 white);
draw A -- center c1;
draw B -- center c2;
draw C -- center c3;
draw center c1 -- center c2 -- center c3 -- cycle;
draw incircle(A,B,C);

drawoptions();
draw 2[A,B] -- 2[B,A];
draw 2[B,C] -- 2[C,B];
draw 3[C,A] -- 3[A,C];

label(btex $A$ etex, A - (10,10)); drawdot A withpen pencircle scaled dotlabeldiam;
label(btex $B$ etex, B + (18,-2)); drawdot B withpen pencircle scaled dotlabeldiam;
label(btex $C$ etex, C + (-4,10)); drawdot C withpen pencircle scaled dotlabeldiam;

dotlabel.urt (btex $a$ etex, center c1);
dotlabel.ulft(btex $b$ etex, center c2);
dotlabel.bot (btex $c$ etex, center c3);

endfig;
end.

笔记

该incircle例程的工作原理是找到两个相邻内角平分线 ( ) 的交点和相应边上的c切点 ( )。该例程完全相同，只是平分线旋转了 90°，这样我们就得到了外角平分线。texcircle

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