△ABC中,三边长为连续的自然数,且最大角是最小角的2倍,求此三角形的三边长.

设三角形最小角为a,三边长分别为k-1,k,k+1
则根据正弦定理和已知有
（k-1)/sina=(k+1)/sin2a=(k+1)/2sinacosa
∴cosa=(k+1)/(2k-2)
又∵cosa=[k²+(k+1)²-(k-1)²]/[2k(k+1)]
=(k²+k²+2k+1-k²+2k-1)/(2k²+2k)
=(k²+4k)/(2k²+2k)
=(k+4)/(2k+2)
∴(k+1)/(2k-2)=(k+4)/(2k+2)
(k+1)/(k-1)=(k+4)/(k+1)
(k+1)²=(k+4)(k-1)
k²+2k+1=k²+3k-4
∴k=5
∴△ABC的三边长分别为4,5,6.