09/08/03 07:36:25
>>309 (上)
〔類題〕
三角形の辺長をa,b,c 中線の長さを AA', BB', CC' とおくと
1 < (a+b+c)/(AA'+BB'+CC') < 4/3.
(略証)
・右側は >>313
・左側を示す。
B'A' = AC' = C'B = (1/2)AB = c/2,
C'B' = BA' = A'C = (1/2)BC = a/2,
A'C' = CB' = B'A = (1/2)CA = b/2,
より
AA' < AB' + B'A' = (c+b)/2,
BB' < BC' + C'B' = (a+c)/2,
CC' < CA' + A'C' = (b+a)/2,
辺々たすと
AA' + BB' + CC' < a+b+c,