设:∠BOA=t,则:∠COA=(2π/3)+t,则:
B(cost,sint),C(cos(2π/3+t),sin(2π/3+t)),A(2,0)
设三角形ABC重心为G(x,y),则:
x=(1/3)[cost+cos(2π/3+t)+2]
y=(1/3)[sint+sin(2π/3+t)]
化简下,得:
3x-2=cos(π/3+t)
3y=sin(π/3+t)
消去参数t,得:(3x-2)²+(3y)²=1
解答:设:∠BOA=t,则:∠COA=(2π/3)+t,则:
B(2cost,2sint),C(2cos(2π/3+t),2sin(2π/3+t)),A(2,0)
设三角形ABC重心为G(x,y),则:
x=(1/3)[2cost+2cos(2π/3+t)+2]
y=(1/3)[2sint+2sin(2π/3+t)]
化简下,得:
(3x-2)/2=cos(π/3+t)
(3y)/2=sin(π/3+t)
消去参数t,得:(3x-2)²+(3y)²=4
是抛物线y=(3的负二分之一次方)x^2