Problem: x , y x , y z z
2 x β x y + 2 y β x y = 1 2 y β y z + 2 z β y z = 2 2 z β z x + 2 x β z x = 3 β 2 x β x y β + 2 y β x y β = 1 2 y β y z β + 2 z β y z β = 2 β 2 z β z x β + 2 x β z x β = 3 β β
Then [ ( 1 β x ) ( 1 β y ) ( 1 β z ) ] 2 [ ( 1 β x ) ( 1 β y ) ( 1 β z ) ] 2 m n n m β m m n n m + n m + n
Solution:
Answer (033):
The square root of any real number is either real or pure imaginary, while the right side of each of the given equations is a nonzero real number. Thus in each of the equations both square roots evaluate to a real number and each of the six expressions under the radicals is nonnegative. This in turn implies that 0 β€ x , y , z β€ 2 0 β€ x , y , z β€ 2 Ξ± , Ξ² Ξ± , Ξ² Ξ³ Ξ³ 0 β€ Ξ± , Ξ² , Ξ³ β€ 9 0 β 0 β€ Ξ± , Ξ² , Ξ³ β€ 9 0 β x = 2 2 Ξ± , y = 2 2 Ξ² x = 2 sin 2 Ξ± , y = 2 sin 2 Ξ² z = 2 2 Ξ³ z = 2 sin 2 Ξ³
2 x β x y + 2 y β x y = 2 2 Ξ± 2 Ξ² + 2 2 Ξ± 2 Ξ² = 2 sin β‘ ( Ξ± + Ξ² ) 2 x β x y β + 2 y β x y β = 2 sin 2 Ξ± cos 2 Ξ² β + 2 cos 2 Ξ± sin 2 Ξ² β = 2 sin ( Ξ± + Ξ² )
After a similar substitution into the second and third equations, the system becomes
sin β‘ ( Ξ± + Ξ² ) = 1 2 sin β‘ ( Ξ² + Ξ³ ) = 2 2 sin β‘ ( Ξ± + Ξ³ ) = 3 2 β sin ( Ξ± + Ξ² ) = 2 1 β sin ( Ξ² + Ξ³ ) = 2 2 β β sin ( Ξ± + Ξ³ ) = 2 3 β β β
Hence Ξ± + Ξ² β { 3 0 β , 15 0 β } , Ξ² + Ξ³ β { 4 5 β , 13 5 β } Ξ± + Ξ² β { 3 0 β , 1 5 0 β } , Ξ² + Ξ³ β { 4 5 β , 1 3 5 β } Ξ± + Ξ³ β { 6 0 β , 12 0 β } Ξ± + Ξ³ β { 6 0 β , 1 2 0 β }
Note that if Ξ± , Ξ² Ξ± , Ξ² Ξ³ Ξ³ 9 0 β β Ξ± , 9 0 β β Ξ² 9 0 β β Ξ± , 9 0 β β Ξ² 9 0 β β Ξ³ 9 0 β β Ξ³ Ξ± + Ξ² = 3 0 β Ξ± + Ξ² = 3 0 β Ξ³ β€ 9 0 β Ξ³ β€ 9 0 β Ξ± + Ξ³ = 6 0 β Ξ± + Ξ³ = 6 0 β Ξ² + Ξ³ = 4 5 β Ξ² + Ξ³ = 4 5 β 0 β€ Ξ± , Ξ² , Ξ³ β€ 9 0 β 0 β€ Ξ± , Ξ² , Ξ³ β€ 9 0 β
Ξ± + Ξ² = 3 0 β Ξ² + Ξ³ = 4 5 β Ξ± + Ξ³ = 6 0 β and Ξ± + Ξ² = 15 0 β Ξ² + Ξ³ = 13 5 β Ξ± + Ξ³ = 12 0 β Ξ± + Ξ² Ξ² + Ξ³ Ξ± + Ξ³ β = 3 0 β = 4 5 β = 6 0 β β and Ξ± + Ξ² Ξ² + Ξ³ Ξ± + Ξ³ β = 1 5 0 β = 1 3 5 β = 1 2 0 β β
Note that
( 1 β x ) ( 1 β y ) ( 1 β z ) = ( 1 β 2 2 Ξ± ) ( 1 β 2 2 Ξ² ) ( 1 β 2 2 Ξ³ ) = cos β‘ ( 2 Ξ± ) cos β‘ ( 2 Ξ² ) cos β‘ ( 2 Ξ³ ) ( 1 β x ) ( 1 β y ) ( 1 β z ) β = ( 1 β 2 sin 2 Ξ± ) ( 1 β 2 sin 2 Ξ² ) ( 1 β 2 sin 2 Ξ³ ) = cos ( 2 Ξ± ) cos ( 2 Ξ² ) cos ( 2 Ξ³ ) β
In the first case, Ξ± = 22. 5 β , Ξ² = 7. 5 β , Ξ³ = 37. 5 β Ξ± = 2 2 . 5 β , Ξ² = 7 . 5 β , Ξ³ = 3 7 . 5 β
( 1 β x ) ( 1 β y ) ( 1 β z ) = cos β‘ ( 4 5 β ) β
cos β‘ ( 1 5 β ) β
cos β‘ ( 7 5 β ) = cos β‘ ( 4 5 β ) β
cos β‘ ( 1 5 β ) β
sin β‘ ( 1 5 β ) = cos β‘ ( 4 5 β ) β
sin β‘ ( 3 0 β ) 2 = 2 8 ( 1 β x ) ( 1 β y ) ( 1 β z ) β = cos ( 4 5 β ) β
cos ( 1 5 β ) β
cos ( 7 5 β ) = cos ( 4 5 β ) β
cos ( 1 5 β ) β
sin ( 1 5 β ) = cos ( 4 5 β ) β
2 sin ( 3 0 β ) β = 8 2 β β β
In the second case, Ξ± = 67. 5 β , Ξ² = 82. 5 β , Ξ³ = 52. 5 β Ξ± = 6 7 . 5 β , Ξ² = 8 2 . 5 β , Ξ³ = 5 2 . 5 β
( 1 β x ) ( 1 β y ) ( 1 β z ) = cos β‘ ( 13 5 β ) β
cos β‘ ( 16 5 β ) β
cos β‘ ( 10 5 β ) = β cos β‘ ( 4 5 β ) β
cos β‘ ( 1 5 β ) β
cos β‘ ( 7 5 β ) = β 2 8 ( 1 β x ) ( 1 β y ) ( 1 β z ) β = cos ( 1 3 5 β ) β
cos ( 1 6 5 β ) β
cos ( 1 0 5 β ) = β cos ( 4 5 β ) β
cos ( 1 5 β ) β
cos ( 7 5 β ) = β 8 2 β β β
Hence in both cases,
[ ( 1 β x ) ( 1 β y ) ( 1 β z ) ] 2 = 1 32 [ ( 1 β x ) ( 1 β y ) ( 1 β z ) ] 2 = 3 2 1 β
The requested sum is 1 + 32 = 33 1 + 3 2 = 3 3
OR
As above, deduce that 0 < x , y , z β€ 2 0 < x , y , z β€ 2 a = 1 β x , b = 1 β y , c = 1 β z a = 1 β x , b = 1 β y , c = 1 β z β 1 β€ a , b , c β€ 1 β 1 β€ a , b , c β€ 1 ( a b c ) 2 ( a b c ) 2
( 1 β a ) ( 1 + b ) + ( 1 + a ) ( 1 β b ) = 1 ( 1 β b ) ( 1 + c ) + ( 1 + b ) ( 1 β c ) = 2 ( 1 β a ) ( 1 + c ) + ( 1 + a ) ( 1 β c ) = 3 β ( 1 β a ) ( 1 + b ) β + ( 1 + a ) ( 1 β b ) β = 1 ( 1 β b ) ( 1 + c ) β + ( 1 + b ) ( 1 β c ) β = 2 β ( 1 β a ) ( 1 + c ) β + ( 1 + a ) ( 1 β c ) β = 3 β β
Squaring the equations and simplifying gives
2 ( 1 β a 2 ) ( 1 β b 2 ) = 2 a b β 1 ( 1 β b 2 ) ( 1 β c 2 ) = b c 2 ( 1 β a 2 ) ( 1 β c 2 ) = 2 a c + 1 2 ( 1 β a 2 ) ( 1 β b 2 ) β ( 1 β b 2 ) ( 1 β c 2 ) β 2 ( 1 β a 2 ) ( 1 β c 2 ) β β = 2 a b β 1 = b c = 2 a c + 1 β
Note that it is necessary that 2 a b β 1 , b c 2 a b β 1 , b c 2 a c + 1 2 a c + 1
4 a 2 + 4 b 2 β 4 a b = 3 b 2 + c 2 = 1 4 a 2 + 4 c 2 + 4 a c = 3 4 a 2 + 4 b 2 β 4 a b = 3 b 2 + c 2 = 1 4 a 2 + 4 c 2 + 4 a c = 3 β
Subtracting the third equation from the first yields
b 2 β c 2 β a b β a c = ( b + c ) ( b β a β c ) = 0 b 2 β c 2 β a b β a c = ( b + c ) ( b β a β c ) = 0
The case b = β c b = β c ( 1 β b 2 ) ( 1 β c 2 ) = b c ( 1 β b 2 ) ( 1 β c 2 ) β = b c
a + c = b b 2 + c 2 = 1 a 2 + c 2 + a c = 3 4 a + c b 2 + c 2 a 2 + c 2 + a c β = b = 1 = 4 3 β β
Substituting the first equation into the second gives
a 2 + 2 c 2 + 2 a c = 1 a 2 + 2 c 2 + 2 a c = 1
which, combined with the third equation yields a 2 = 1 2 a 2 = 2 1 β
The condition b = a + c b = a + c
( b β c ) 2 = b 2 + c 2 β 2 b c = a 2 = 1 2 ( b β c ) 2 = b 2 + c 2 β 2 b c = a 2 = 2 1 β
from which b c = 1 4 b c = 4 1 β ( a b c ) 2 = ( b c ) 2 β
a 2 = 1 32 ( a b c ) 2 = ( b c ) 2 β
a 2 = 3 2 1 β
( a , b , c ) = ( 2 2 , 2 2 + 1 2 , 1 2 2 2 + 1 ) . ( a , b , c ) = ( 2 2 β β , 2 2 2 β + 1 β β , 2 2 2 β + 1 β 1 β ) .
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions