>>478
A(0,0)，B(2,0)
Cが直線y=1上にあればいいのでC(x,1)とおくと
AC^2 = (x-0)^2 + (1-0)^2 = x^2 + 1
BC^2 = (x-2)^2 + (1-0)^2 = x^2 -4x + 5
L = BC^2+(2√3-1)AC^2 とおくと
dL/dx = 2x - 4 + (2√3-1)*2x = 4√3x - 4 = 0
x = 1/√3
∠BAC = π/3