凸四邊形ABCD為圓O的內接四邊形,其中AC為直徑,BD交AC於E,且CE:EA=1:5,又AB=BD,圓O的半徑12,則BC的長度最接近哪一個整數?

簡圖請參考 :
http://imgur.com/a/XH7pD

令 θ = ∠BAC , α = ∠DAC
因為 OA = OD = 圓半徑 , 所以 ∠ODA = α
因為 ∠CBD 與 ∠DAC 皆對應 CD弧 , 因此 ∠CBD = α
因為 OA = OB , 所以 ∠OBA = θ

因為 AB = BD
所以 ∠BAD = ∠BDA
θ + α = ∠BDO + α
∠BDO = θ

因為 OB = OD , 所以 ∠OBD = θ

ΔABD 的內角和 = 4θ + 2α = π
α = ( π - 4θ ) / 2 = π/2 - 2θ
sin α = sin ( π/2 - 2θ ) = cos 2θ

因為 CE : EA = 1 : 5
所以 ΔABE面積 / ΔCBE面積 = 5
(1/2)AB*BE*sin 2θ / [ (1/2)CB*BE*sin α ] = 5

5
= AB*sin 2θ / [ CB*sin α ]
= 24*cos θ*sin 2θ / [ 24*sin θ*cos 2θ ]
= tan 2θ / tan θ

tan 2θ = 5 * tan θ
2 * tan θ / ( 1 - tan² θ ) = 5 * tan θ
令 x = tan θ
2x / ( 1 - x² ) = 5x
2x = 5x - 5x³
5x³ - 3x = 0
x ( 5x² - 3 ) = 0
因為 x = tan θ > 0
所以 x = tan θ = √3 / √5
sin θ = √3 / √8

BC = 24*sin θ = 24 * ( √3 / √8 ) = 6√6 ≒ 14.7
Ans: 15