Problem: S S ( x , y , z ) ( x , y , z )
10 ( x + y ) = z and 10 ( x 2 + y 2 ) = z + 1 log 1 0 ​ ( x + y ) = z and log 1 0 ​ ( x 2 + y 2 ) = z + 1
There are real numbers a a b b ( x , y , z ) ( x , y , z ) S S x 3 + y 3 = a â‹… 1 0 3 z + b â‹… 1 0 2 z x 3 + y 3 = a â‹… 1 0 3 z + b â‹… 1 0 2 z a + b a + b
Answer Choices:
A. 15 2 2 1 5 ​ 29 2 2 2 9 ​ 15 1 5 39 2 2 3 9 ​ 24 2 4 Solution:
From the given conditions it follows that
x + y = 1 0 z , x 2 + y 2 = 10 â‹… 1 0 z and 1 0 2 z = ( x + y ) 2 = x 2 + 2 x y + y 2 x + y = 1 0 z , x 2 + y 2 = 1 0 â‹… 1 0 z and 1 0 2 z = ( x + y ) 2 = x 2 + 2 x y + y 2
Thus
x y = 1 2 ( 1 0 2 z − 10 ⋅ 1 0 z ) x y = 2 1 ​ ( 1 0 2 z − 1 0 ⋅ 1 0 z )
Also
( x + y ) 3 = 1 0 3 z and x 3 + y 3 = ( x + y ) 3 − 3 x y ( x + y ) ( x + y ) 3 = 1 0 3 z and x 3 + y 3 = ( x + y ) 3 − 3 x y ( x + y )
which yields
x 3 + y 3 = 1 0 3 z − 3 2 ( 1 0 2 z − 10 ⋅ 1 0 z ) ( 1 0 z ) = 1 0 3 z − 3 2 ( 1 0 3 z − 10 ⋅ 1 0 2 z ) = − 1 2 1 0 3 z + 15 ⋅ 1 0 2 z x 3 + y 3 = 1 0 3 z − 2 3 ​ ( 1 0 2 z − 1 0 ⋅ 1 0 z ) ( 1 0 z ) = 1 0 3 z − 2 3 ​ ( 1 0 3 z − 1 0 ⋅ 1 0 2 z ) = − 2 1 ​ 1 0 3 z + 1 5 ⋅ 1 0 2 z ​
and a + b = − 1 2 + 15 = 29 / 2 a + b = − 2 1 ​ + 1 5 = 2 9 / 2
No other value of a + b a + b S S ( 1 2 ( 1 + 19 ) , 1 2 ( 1 − 19 ) , 0 ) ( 2 1 ​ ( 1 + 1 9 ​ ) , 2 1 ​ ( 1 − 1 9 ​ ) , 0 ) S S x 3 + y 3 = a ⋅ 1 0 3 z + b ⋅ 1 0 2 z x 3 + y 3 = a ⋅ 1 0 3 z + b ⋅ 1 0 2 z a + b = 29 / 2 a + b = 2 9 / 2 ​
The problems on this page are the property of the MAA's American Mathematics Competitions