cosC=(a^2＋b^2－c^2)/2ab 设AB=x,BC=y 则0.5=(y^2+(x+0.5)^2-x^2)/(2y(x+0.5)) => y^2+x+0.25=xy+0.5y x=(y^2-0.5y+0.25)/(y-1) =((y-1)^2+1.5(y-1)+0.75)/(y-1) 因为y>1,所以y-1>0 x=(y-1)+1.5+0.75/(y-1)>=2(0.75^0.5)+1.5=1.5+3^0.5 当且仅当y=1+(3)^0.5/2时成立所以AC最短为2+根号3 BC为1+0.5根号3
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