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由sinA/a=sinB/b=sinC/c(其中a,b,c为角A,B,C对应的三条边)设sinA/a=sinB/b=sinC/c=k则a=sinA/k,b=sinB/k,c=sinC/k带入(sinB-sinA)x^2 +（sinA-sinC)x +(sinC-sinB)=0得(b-a)x^2+(a-c)x+(c-b)=0由有两相等实根得:Δ=(a-c)^2-4(b-a)(c-b)=(2b-a-c)^2=0所以2b=a+c即b=(a+c)/2(1)a+b>c(2)c+b>a(3)(1)带入(2)得a>c/3(4)(1)带入(3)得a<3c(5)设a/c=y则(4),(5)变为1/3<y<3(6)(1)... |
