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[几何] 使三个互相外切的圆轮换的保角变换

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hbghlyj Posted at 2021-12-16 22:24:07 |Read mode
Last edited by hbghlyj at 2022-12-30 16:19:00

设⊙A,B,C互相外切于D,E,F,它们的逆相似圆为⊙G,H,I,变换T是关于⊙G再关于⊙H反演,则T(⊙A)=⊙C,T(⊙C)=⊙B,T(⊙B)=⊙A,也就是说,正好使⊙A,B,C向右轮换.

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 Author| hbghlyj Posted at 2021-12-16 22:38:04
  1. AB = Sqrt[(a - b) Conjugate[a - b]];
  2. BC = Sqrt[(b - c) Conjugate[b - c]];
  3. CA = Sqrt[(c - a) Conjugate[c - a]];
  4. p = (AB + BC + CA)/2;
  5. d = (b (p - AB) + c (p - CA))/(CA + AB);
  6. g = (b (p - AB) - c (p - CA))/(CA - AB);
  7. e = (c (p - BC) + a (p - AB))/(AB + BC);
  8. h = (c (p - BC) - a (p - AB))/(AB - BC);
  9. G[z_] = (g - d) Conjugate[g - d]/Conjugate[(z - g)] + g;
  10. H[z_] = (h - e) Conjugate[h - e]/Conjugate[(z - h)] + h;
  11. Factor[H[G[z]]]
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出来是一个很复杂的式子,就不贴了...

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 Author| hbghlyj Posted at 2021-12-16 22:46:32
Last edited by hbghlyj at 2021-12-16 23:01:00T(D)=F,T(E)=D,T(F)=E.
T(G)=H,T(H)=I,T(I)=G.
⊙G,H,I交于两个点,它们是T的不动点.
⊙J为△ABC的内切圆.
T(⊙J)=⊙J.
设⊙G与⊙J交于D,D',⊙H与⊙J交于E,E',⊙I与⊙J交于F,F',则
T(D')=F',T(E')=D',T(F')=E'.
GeoGebra链接
注:在上面的ggb文件中,K是⊙J上的一点,可以看到K'仍在⊙J上,且K与D重合时K'与F重合,等等.可以将K重新定义为⊙A上的点,那么K'就在⊙C上,等等.

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2025-4-21 14:17 GMT+8

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