■初等幾何の楽しみ(その103)

 (その49)にて,n=8の場合のグレブナー基底をお知らせしたが,作図してみて誤りに気づいた.

  d^16−8d^14r^2+8d^12r^4−8d^14R^2+40d^12r^2R^2+48d^10r^4R^2−128d^8r^6R^2+128d^6r^8R^2+28d^12R^4−72d^10r^2R^4−264d^8r^4R^4+128d^6r^6R^4−56d^10R^6+40d^8r^2R^6+416d^6r^4R^6+128d^4r^6R^6+128d^2r^8R^6+70d^8R^8+40d^6r^2R^8−264d^4r^4R^8−128d^2r^6R^8−56d^6R^10−72d^4r^2R^10+48d^2r^4R^10+28d^4R^12+40d^2r^2R^12+8r^4R^12−8d^2R^14−8r^2R^14+R^16=0

が正解のようである.

 d=0とおくと

  8r^4R^12−8r^2R^14+R^16=0

となるが,

  r/R=cos(π/8)=(2+√2)^1/2/2

  r/R=cos(3π/8)=(2−√2)^1/2/2

はこれを満たす.

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[1]双心三角形

  R^2−2Rr=d^2   (オイラーの定理)

[2]双心四角形

  2r^2(R^2+d^2)=(R^2−d^2)^2    (フースの定理)

[3]双心五角形

  d^6−2d^4rR+8d^2r^3R−3d^4R^2−4d^2r^2R^2+4d^2rR^3+3d^2R^4+4r^2R^4−2rR^5−R^6=0

[4]双心六角形

  3d^8−4d^6r^2−12d^6R^2+4d^4r^2R^2−16d^2r^4R^2+18d^4R^4+4d^2r^2R^4−12d^2R^6−4r^2R^6+3R^8=0

[5]双心七角形

  d^12+4d^10rR−24d^8r^3R+32d^6r^5R−6d^10R^2−4d^8r^2R^2−16d^6r^4R^2−20d^8rR^3+64d^6r^3R^3+15d^8R^4+16d^6r^2R^4+32d^4r^4R^4+64d^2r^6R^4+40d^6rR^5−48d^4r^3R^5−32d^2r^5R^5−20d^6R^6−24d^4r^2R^6−16d^2r^4R^6−40d^4rR^7+15d^4R^8+16d^2r^2R^8+20d^2rR^9+8r^3R^9−6d^2R^10−4r^2R^10−4rR^11+R^12=0

[6]双心八角形

  d^16−8d^14r^2+8d^12r^4−8d^14R^2+40d^12r^2R^2+48d^10r^4R^2−128d^8r^6R^2+128d^6r^8R^2+28d^12R^4−72d^10r^2R^4−264d^8r^4R^4+128d^6r^6R^4−56d^10R^6+40d^8r^2R^6+416d^6r^4R^6+128d^4r^6R^6+128d^2r^8R^6+70d^8R^8+40d^6r^2R^8−264d^4r^4R^8−128d^2r^6R^8−56d^6R^10−72d^4r^2R^10+48d^2r^4R^10+28d^4R^12+40d^2r^2R^12+8r^4R^12−8d^2R^14−8r^2R^14+R^16=0

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