ABC中三条边为abc a>b>c,a,b,c整数． a分之1＋b分之1＋c分之1＝1．三角形ABC是否存在,为什么

不存在。假设存在∵ a>b>c∴ 1/a < 1/c ,1/b < 1/c∴ 1/a + 1/b + 1/c < 1/c + 1/c +1/c = 3/c又∵ 1/a + 1/b + 1/c = 1∴ 3/c>1 ∴ c0∴ c=1或2当c=1时,代入1/a + 1/b + 1/c = 1得,1/a +1/b=0,无解,∴c≠1,∴c=2。∴1/a + 1/b =1-1/c =1-1/2=1/2又a>b ,∴1/a