本章讨论，当0<n<1时什么情况会存在。（由于有可能不存在）
据题意，设B的长和宽为x和y，A的长宽分别为a、b，得方程组：
①x + y = n ( a + b )
②x y = a b n
由①得：③y = a n + b n – x
将③带入②得：
x ( a n + b n – x ) = a b n
x2 – a n x – b n x + a b n = 0
x2 – ( a + b ) n x + a b n = 0
要使其有解，则Δ≥0
Δ = ( a + b )2 n2 – 4 a b n ≥ 0
( a + b )2 n – 4 a b ≥ 0
( a + b )2 n ≥ 4 a b n
( a2 + 2 a b + b2 ) n ≥ 4 a b
( a2 + 2 a b + b2 ) / 4 a b * n ≥ 1
( a / 4 b + 1 / 2 + b / 4 a ) n ≥ 1
( a2 + b2 ) / 4 a b ≥ ( 2 – n ) / 2 n
设b = a m
( a2 + a2 m2 ) / ( 4 a2 m ) ≥ ( 2 – n ) / 2 n
( m2 + 1 ) / 2 m ≥ ( 2 – n ) / 2 n
( m + 1 )2 / 2 m - 1 ≥ ( 2 – n ) / 2 n
( m + 1 )2 ≥ 4 m / n
m2 + ( 2 – 4 / n ) m + 1 ≥ 0
解得：
m1 ≥ [ 2 + sqrt( 1 – n ) ] / n – 1
m2 ≤ [ 2 – sqrt( 1 – n ) ] / n - 1
所以，当a与b的比值m ≥ [ 2 + sqrt( 1 – n ) ] / n – 1或0 < m ≤ [ 2 – sqrt( 1 – n ) ] / n - 1时可以找到
