Question

已知三角形ABC是边长为1的正三角形,M,N分别是边AB,AC上的点,线段MN经过三角形ABC的中心G,设角MGA=a（

Answer 1

(Ⅰ)在三角形AGM中,由正弦定理：

sin∠AMG/AG=sin∠MAG/GM

其中∠MAG=30°,

∠AMG=180°-(30°+α),

AG=2/3*AD=2/3*sin60°*AB=根号3/3,

GM=sin∠MAG*AG/sin∠AMG=根号3/6sin(30°+α)

同理,在三角形AGN中,

GN=根号3/6sin(a-30°)

S1=1/2AG·GMsinα=1/2*根号3/3*根号3/6sin(30°+α)*sinα=sinαsin(30°+α)/12

S2=1/2AG·GNsin(180°-α)=1/2*根号3/3*根号3/6sin(a-30°)*sinα=sinαsin(a-30°)/12

(Ⅱ)y=1/(S1^2)+ 1/(S2^2)

=144/[(sinα)^2*sin^2(a-30°)]+144/[(sinα)^2*sin^2(30°+α]

=72(3+cot^2α)

∵π/3