變態幾何問題…

首先, 將各自的點設如下:

圖片參考：http://i388.photobucket.com/albums/oo325/loyitak1990/Jul10/Crazygeom1.jpg


如未能見, 請瀏覽:

http://i388.photobucket.com/albums/oo325/loyitak1990/Jul10/Crazygeom1.jpg

即:

D, F, H 和 J 為各自圓心.

E, G, I 和 K 為 AC 與各圓的接觸點.

另設: r1 為第二圓半徑, r2 為第三圓半徑, ∠EAD = θ 和 AD = x

由切線性質可知:

△ADE ~ △AGF ~ △AIH ~ △AKJ ~ △ALC

由於 △AED ~ △AGF:

AF/AD = GF/ED

(x + 8 + r1)/x = r1/8

(x + 8)/x + r1/x = r1/8

r1(1/8 - 1/x) = (x + 8)/x

r1(x - 8)/(8x) = (x + 8)/x

r1 = 8(x + 8)/(x - 8)

由於 △AED ~ △AIH:

AH/AD = IH/ED

(x + 2r1 + r2)/x = r2/8

(x + 2r1 + 8)/x + r2/x = r2/8

r2(1/8 - 1/x) = (x + 2r1 + 8)/x

r2(x - 8)/(8x) = (x + 2r1 + 8)/x

r2 = 8(x + 2r1 + 8)/(x - 8)

= 8{x + 8 + 2[8(x + 8)/(x - 8)]}/(x - 8)

= 8[(x + 8)(x - 8) + 16(x + 8)]/(x - 8)2

= 8(x2 + 16x + 64)/(x - 8)2

= 8(x + 8)2/(x - 8)2

由於 △AED ~ △AKL, 以相同手法可得出:

27 = 8(x + 8)3/(x - 8)3

3 = 2(x + 8)/(x - 8)

3x - 24 = 2x + 16

x = 40

所以, sin θ = 8/40 = 1/5

AJ sin θ = 27

AJ = 135

AL = 135 + 27 = 162

cos θ = (2√6)/5

tan θ = 1/(2√6)

LC = AL tan θ = 162/(2√6) = 81/√6

BC = 162/√6

所以 △ABC 面積為:

BC x AL/2 = 81 x 162/√6

= 81 x 27√6

= 2187√6

即 p 最大可能值為 2187

半徑成等比, 則 r^3=8/27, r= 2/3
故半徑為27, 18, 12, 8, sin(A/2)=(27-18)/(27+18)= 1/5
高=135+27=162, 底邊長=162/(2√6)

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