设直线系M:xcosФ+(y-2)sinФ=1(0≤Ф≤2π),对于下面4个命题:

不必从从内切圆还是外接圆考虑。可从角度关系入手。取φ=t,t+π/3，t-π/3,得夹角都是π/3的三条直线：l1:xcost+(y-2)sint=1,(1）l2:xcos(t+π/3)+(y-2)sin(t+π/3)=1,(2)l3:xcos(t-π/3)+(y-2)sin(t-π/3)=1.(3）由(1),(2)得l1,l2的交点A(2[sin(t+π/3)-sint]/√3，2[cost-cos(t+π/3)]/√3+2），由(1),(3)得l1,l3的交点B(2[sin(t-π/3)-sint]/√3，2[cost-cos(t-π/3)]/√3+2），∴AB^2=4/3*{2-2[sin(t+π/3)*sin(t-π/3)+cos(t+π/3)*cos(t-π/3)]=4/3*{2-2cos(2π/3)}=4,为定值。∴M中的直线所能围成的正三角形的面积都相等。 收起