tkz-euclide作图精度

abababa

如题。我现在做一个反演的题，题目如下：
$\odot O_1,\odot O_2,\odot O_3,\odot O_4$依次两两外切于点$A_1, A_2, A_3, A_4$，求证$A_1,A_2,A_3,A_4$四点共圆。
作图用了tkz-euclide，代码如下（以下代码中tkzGetPoints{}{}的第二个参数，多数都是没用的，只是要有个名字用来占个位置）：

1. \begin{tikzpicture}
2. \tkzDefPoints{0/0/O_1,0/-2.5/O_2,3/0/O_4,1/0/A_4,0/-1/A_1,4/-4/X,4/-3/Y,1/2.5/Z}
3. % 以$\odot(X,Y)$为反演圆，将$\odot O_2,\odot O_4$都反演，取得反形的一条外公切线PT_1'T_2'
4. \tkzDefCircleBy[inversion = center X through Y](O_4,A_4)\tkzGetPoints{O_4''}{A_4''}
5. \tkzDefCircleBy[inversion = center X through Y](O_2,A_1)\tkzGetPoints{O_2''}{A_1''}
6. \tkzInterOCT(O_4'',A_4'')(O_2'',A_1'')\tkzGetPoint{P}
7. \tkzDefTangent[from=P](O_2'',A_1'')\tkzGetPoints{T_1'}{T_1''}
8. \tkzDefTangent[from=P](O_4'',A_4'')\tkzGetPoints{T_2'}{T_2''}
9. % 将直线$PT_1'T_2'$按反演圆$\odot(X,Y)$反演，此圆过反演中心$X$，得到的反形$\odot O_3$与$\odot O_2,\odot O_4$都相切
10. \tkzDefPointBy[inversion = center X through Y](T_1')\tkzGetPoint{T_1}
11. \tkzDefPointBy[inversion = center X through Y](T_2')\tkzGetPoint{T_2}
12. \tkzDefTriangleCenter[circum](X,T_1,T_2)\tkzGetPoint{O_3}
13. \tkzInterCC(O_3,X)(O_2,A_1)\tkzGetPoints{A_2}{A_2'}
14. \tkzInterCC(O_3,X)(O_4,A_4)\tkzGetPoints{A_3}{A_3'}
15. % 以$\odot(A_4,Z)$为反演圆，将原图形反演
16. \tkzDefCircleBy[inversion = center A_4 through Z](O_2,A_2)\tkzGetPoints{O_2'}{B_2'}
17. \tkzDefCircleBy[inversion = center A_4 through Z](O_3,A_3)\tkzGetPoints{O_3'}{B_3'}
18. \tkzDefPointBy[inversion = center A_4 through Z](A_1)\tkzGetPoint{A_1'}
19. \tkzDefPointBy[inversion = center A_4 through Z](A_3)\tkzGetPoint{A_3'}
20. \tkzInterCC(O_2',B_2')(O_3',B_3')\tkzGetPoints{C}{C'}
21. \tkzDefTangent[at=A_1'](O_2')\tkzGetPoint{K_1}
22. \tkzDefTangent[at=C](O_2')\tkzGetPoint{K_2}
23. \tkzInterLL(A_1',K_1)(C,K_2)\tkzGetPoint{M}
24. \tkzDefTangent[at=A_3'](O_3')\tkzGetPoint{K_1}
25. \tkzDefTangent[at=C](O_3')\tkzGetPoint{K_2}
26. \tkzInterLL(A_3',K_1)(C,K_2)\tkzGetPoint{N}
28. \tkzDrawCircles(O_1,A_1 O_2,A_2 O_3,A_3 O_4,A_4)
29. \tkzDrawCircles(O_2',C O_3',C)
30. \tkzDrawSegments(M,N A_1',M A_3',N)
32. \tkzLabelPoints[left](O_1)
33. \tkzLabelPoints[below](O_2)
34. \tkzLabelPoints[below right](O_3)
35. \tkzLabelPoints[below left](O_4)
36. \tkzLabelPoints[left](A_1)
37. \tkzLabelPoints[below](A_2)
38. \tkzLabelPoints[below right](A_3)
39. \tkzLabelPoints[below left](A_4)
40. \tkzLabelPoints[below left](C)
41. \tkzLabelPoints[below left](A_1')
42. \tkzLabelPoints[below left](A_3')
44. \tkzDrawPoints(A_1,A_2,A_3,A_4,O_1,O_2,O_3,O_4)
45. \tkzDrawPoints(A_1',A_3',C)
46. \end{tikzpicture}

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其中tkzInterOCT是用来定义两个圆的两条外公切线的交点，定义如下：

1. \def\tkzInterOCT(#1,#2)(#3,#4){
2. \begingroup
3. % 取得两圆的半径
4. \tkzCalcLength(#1,#2)\tkzGetLength{r}
5. \tkzCalcLength(#3,#4)\tkzGetLength{R}
6. \pgfmathsetmacro{\K}{\r/(\r-\R)}
7. \tkzDefPointWith[linear,K=\K](#1,#3)
8. \endgroup
9. }

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正常的在GeoGebra里画出来，点$A_1'$应该在$\odot O_2'$上，所以后面才能用\tkzDefTangent[at=A_1'](O_2')\tkzGetPoint{K_1}来得到过圆上一点的切线。但在tikz里画出来，$A_1'$就不在$\odot O_2'$上了，差了一点，导致整个图都有问题。这要怎么改才行？

hbghlyj

Precision of \tkzCalcLength

abababa

hbghlyj 发表于 2024-2-28 17:35
Precision of \tkzCalcLength

可能就是那些计算时有问题。主楼的问题我解决了，换了个作法，不用两圆的切点（因为有精度问题，现在这两圆可能不相切而是相交，但两个交点距离也很近），而是用连心线与其中一圆的交点，这样就算两圆是相交于两点的，连心线与其中一圆的交点仍然位于正中间，看上去就没有那么大偏差了。最终的代码为：

1. \begin{tikzpicture}[scale=0.8,label style/.style={font=\scriptsize}]
2. \tkzDefPoints{0/0/O_1,0/-2.5/O_2,3/0/O_4,1/0/A_4,0/-1/A_1,4/-4/X,4/5/Y,1/2.5/Z}
3. % 以$\odot(X,Y)$为反演圆，将$\odot O_2, \odot O_4$ 都反演，取得反形的一条外公切线PT_1'T_2'
4. \tkzDefCircleBy[inversion = center X through Y](O_4,A_4)\tkzGetPoints{O_4''}{A_4''}
5. \tkzDefCircleBy[inversion = center X through Y](O_2,A_1)\tkzGetPoints{O_2''}{A_1''}
6. \tkzInterOCT(O_4'',A_4'')(O_2'',A_1'')\tkzGetPoint{P}
7. \tkzDefTangent[from=P](O_2'',A_1'')\tkzGetPoints{T_1'}{T_1''}
8. \tkzDefTangent[from=P](O_4'',A_4'')\tkzGetPoints{T_2'}{T_2''}
9. % 将直线$PT_1'T_2'$按反演圆$\odot(X,Y)$反演，此圆过反演中心$X$，得到的反形$\odot O_3$与$\odot O_2, \odot O_4$都相切
10. \tkzDefPointBy[inversion = center X through Y](T_1')\tkzGetPoint{T_1}
11. \tkzDefPointBy[inversion = center X through Y](T_2')\tkzGetPoint{T_2}
12. \tkzDefTriangleCenter[circum](X,T_1,T_2)\tkzGetPoint{O_3}
13. \tkzInterLC[near](O_3,O_2)(O_2,A_1)\tkzGetPoints{A_2}{A_2'}
14. \tkzInterLC[near](O_3,O_4)(O_4,A_4)\tkzGetPoints{A_3}{A_3'}
15. %\tkzInterCC(O_3,X)(O_2,A_1)\tkzGetPoints{A_2}{A_2'}
16. %\tkzInterCC(O_3,X)(O_4,A_4)\tkzGetPoints{A_3}{A_3'}
17. % 以$\odot(A_4,Z)$为反演圆，将原图形反演
18. \tkzDefCircleBy[inversion = center A_4 through Z](O_2,A_2)\tkzGetPoints{O_2'}{B_2'}
19. \tkzDefCircleBy[inversion = center A_4 through Z](O_3,A_3)\tkzGetPoints{O_3'}{B_3'}
20. \tkzDefPointBy[inversion = center A_4 through Z](A_1)\tkzGetPoint{A_1'}
21. \tkzDefPointBy[inversion = center A_4 through Z](A_2)\tkzGetPoint{A_2'}
22. \tkzDefPointBy[inversion = center A_4 through Z](A_3)\tkzGetPoint{A_3'}
23. \tkzDefTangent[at=A_1'](O_2')\tkzGetPoint{K_1}
24. \tkzDefTangent[at=A_2'](O_2')\tkzGetPoint{K_2}
25. \tkzInterLL(A_1',K_1)(A_2',K_2)\tkzGetPoint{M}
26. \tkzDefTangent[at=A_3'](O_3')\tkzGetPoint{K_1}
27. \tkzDefTangent[at=A_2'](O_3')\tkzGetPoint{K_2}
28. \tkzInterLL(A_3',K_1)(A_2',K_2)\tkzGetPoint{N}
30. \tkzDrawCircles(O_1,A_1 O_2,A_2 O_3,A_3 O_4,A_4)
31. \tkzDrawCircles(O_2',B_2' O_3',B_3')
32. \tkzDrawLines[add = 0.3 and 0.1](A_1',M)
33. \tkzDrawLines[add = 2.3 and 0.5](A_3',N)
34. \tkzDrawSegments(M,N)
36. \tkzLabelPoints[above](O_1)
37. \tkzLabelPoints[below](O_2)
38. \tkzLabelPoints[right](O_3)
39. \tkzLabelPoints[above](O_4)
40. \tkzLabelPoints[above](A_1)
41. \tkzLabelPoints[shift=(150:7pt)](A_2)
42. \tkzLabelPoints[below](A_3)
43. \tkzLabelPoints[right](A_4)
44. \tkzLabelPoints[below left](A_1')
45. \tkzLabelPoints[shift=(-150:7pt)](A_2')
46. \tkzLabelPoints[left](A_3')
47. \tkzLabelPoints[left](M)
48. \tkzLabelPoints[right](N)
50. \tkzDrawPoints(A_1,A_2,A_3,A_4,O_1,O_2,O_3,O_4)
51. \tkzDrawPoints(A_1',A_2',A_3',M,N)
52. \end{tikzpicture}

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