幾何問題~求OA+OB+OC

圖片參考：http://i778.photobucket.com/albums/yy68/CH_/ac6caaf9.jpg

O是該三角形的外心, 所以 OA = OB = OC = y,
AB^2 = BD^2 + AD^2
18^2 = 9^2 + AD^2
AD = 9√3
BO^2 = BD^2 + OD^2
BO^2 = BD^2 + (AD -AO)^2
y^2 = 9^2 + (9√3 - y)^2
y^2 = 81 + (243 + y^2 - 18√3 * y)
18√3 * y = 324
y = 18√3
OA + OB + OC = 3y = 54√3