Problem: P ( x ) = x 3 + a x 2 + b x + c P ( x ) = x 3 + a x 2 + b x + c cos β‘ 2 Ο 7 , cos β‘ 4 Ο 7 cos 7 2 Ο β , cos 7 4 Ο β cos β‘ 6 Ο 7 cos 7 6 Ο β a b c a b c
Answer Choices:
A. β 3 49 β 4 9 3 β β 1 28 β 2 8 1 β 3 7 64 6 4 3 7 β β 1 32 3 2 1 β 1 28 2 8 1 β Solution:
By Vieta's formulas,
a = β ( cos β‘ 2 Ο 7 + cos β‘ 4 Ο 7 + cos β‘ 6 Ο 7 ) b = cos β‘ 2 Ο 7 β
cos β‘ 4 Ο 7 + cos β‘ 4 Ο 7 β
cos β‘ 6 Ο 7 + cos β‘ 6 Ο 7 β
cos β‘ 2 Ο 7 c = β cos β‘ 2 Ο 7 β
cos β‘ 4 Ο 7 β
cos β‘ 6 Ο 7 . a = β ( cos 7 2 Ο β + cos 7 4 Ο β + cos 7 6 Ο β ) b = cos 7 2 Ο β β
cos 7 4 Ο β + cos 7 4 Ο β β
cos 7 6 Ο β + cos 7 6 Ο β β
cos 7 2 Ο β c = β cos 7 2 Ο β β
cos 7 4 Ο β β
cos 7 6 Ο β . β
The roots of the polynomial Q ( x ) = x 6 + x 5 + β― + x + 1 Q ( x ) = x 6 + x 5 + β― + x + 1 cos β‘ 2 k Ο 7 + i sin β‘ 2 k Ο 7 cos 7 2 k Ο β + i sin 7 2 k Ο β k = 1 , 2 , 3 , 4 , 5 , 6 k = 1 , 2 , 3 , 4 , 5 , 6 Q ( x ) Q ( x )
Q ( x ) = ( x β ( cos β‘ 2 Ο 7 + i sin β‘ 2 Ο 7 ) ) ( x β ( cos β‘ 2 Ο 7 β i sin β‘ 2 Ο 7 ) ) β
( x β ( cos β‘ 4 Ο 7 + i sin β‘ 4 Ο 7 ) ) ( x β ( cos β‘ 4 Ο 7 β i sin β‘ 4 Ο 7 ) ) β
( x β ( cos β‘ 6 Ο 7 + i sin β‘ 6 Ο 7 ) ) ( x β ( cos β‘ 6 Ο 7 β i sin β‘ 6 Ο 7 ) ) = ( x 2 β ( 2 cos β‘ 2 Ο 7 ) x + 1 ) β
( x 2 β ( 2 cos β‘ 4 Ο 7 ) x + 1 ) β
( x 2 β ( 2 cos β‘ 6 Ο 7 ) x + 1 ) . Q ( x ) = ( x β ( cos 7 2 Ο β + i sin 7 2 Ο β ) ) ( x β ( cos 7 2 Ο β β i sin 7 2 Ο β ) ) β
( x β ( cos 7 4 Ο β + i sin 7 4 Ο β ) ) ( x β ( cos 7 4 Ο β β i sin 7 4 Ο β ) ) β
( x β ( cos 7 6 Ο β + i sin 7 6 Ο β ) ) ( x β ( cos 7 6 Ο β β i sin 7 6 Ο β ) ) = ( x 2 β ( 2 cos 7 2 Ο β ) x + 1 ) β
( x 2 β ( 2 cos 7 4 Ο β ) x + 1 ) β
( x 2 β ( 2 cos 7 6 Ο β ) x + 1 ) . β
By expanding the last expression above for Q ( x ) Q ( x ) x 5 x 5 Q ( x ) Q ( x )
1 = β 2 cos β‘ 2 Ο 7 β 2 cos β‘ 4 Ο 7 β 2 cos β‘ 6 Ο 7 . 1 = β 2 cos 7 2 Ο β β 2 cos 7 4 Ο β β 2 cos 7 6 Ο β .
It then follows from (1) that a = 1 2 a = 2 1 β x 4 x 4 Q ( x ) Q ( x )
1 = 3 + 4 ( cos β‘ 2 Ο 7 cos β‘ 4 Ο 7 + cos β‘ 4 Ο 7 cos β‘ 6 Ο 7 + cos β‘ 6 Ο 7 cos β‘ 2 Ο 7 ) 1 = 3 + 4 ( cos 7 2 Ο β cos 7 4 Ο β + cos 7 4 Ο β cos 7 6 Ο β + cos 7 6 Ο β cos 7 2 Ο β )
and it follows from (2) that b = β 1 2 b = β 2 1 β x 3 x 3 Q ( x ) Q ( x )
1 = β 4 ( cos β‘ 2 Ο 7 + cos β‘ 4 Ο 7 + cos β‘ 6 Ο 7 ) β 8 cos β‘ 2 Ο 7 cos β‘ 4 Ο 7 cos β‘ 6 Ο 7 = 4 a + 8 c . 1 = β 4 ( cos 7 2 Ο β + cos 7 4 Ο β + cos 7 6 Ο β ) β 8 cos 7 2 Ο β cos 7 4 Ο β cos 7 6 Ο β = 4 a + 8 c .
Thus c = β 1 8 c = β 8 1 β a b c = 1 2 ( β 1 2 ) ( β 1 8 ) = 1 32 a b c = 2 1 β ( β 2 1 β ) ( β 8 1 β ) = 3 2 1 β β
OR
Applying the identity cos β‘ A cos β‘ B = 1 2 ( cos β‘ ( A + B ) + cos β‘ ( A β B ) ) cos A cos B = 2 1 β ( cos ( A + B ) + cos ( A β B ) ) a , b a , b c c
b = cos β‘ 2 Ο 7 cos β‘ 4 Ο 7 + cos β‘ 2 Ο 7 cos β‘ 6 Ο 7 + cos β‘ 4 Ο 7 cos β‘ 6 Ο 7 = 1 2 ( cos β‘ 6 Ο 7 + cos β‘ 2 Ο 7 ) + 1 2 ( cos β‘ 8 Ο 7 + cos β‘ 4 Ο 7 ) + 1 2 ( cos β‘ 10 Ο 7 + cos β‘ 2 Ο 7 ) = 1 2 ( cos β‘ 6 Ο 7 + cos β‘ 2 Ο 7 ) + 1 2 ( cos β‘ 6 Ο 7 + cos β‘ 4 Ο 7 ) + 1 2 ( cos β‘ 4 Ο 7 + cos β‘ 2 Ο 7 ) = β a b = cos 7 2 Ο β cos 7 4 Ο β + cos 7 2 Ο β cos 7 6 Ο β + cos 7 4 Ο β cos 7 6 Ο β = 2 1 β ( cos 7 6 Ο β + cos 7 2 Ο β ) + 2 1 β ( cos 7 8 Ο β + cos 7 4 Ο β ) + 2 1 β ( cos 7 1 0 Ο β + cos 7 2 Ο β ) = 2 1 β ( cos 7 6 Ο β + cos 7 2 Ο β ) + 2 1 β ( cos 7 6 Ο β + cos 7 4 Ο β ) + 2 1 β ( cos 7 4 Ο β + cos 7 2 Ο β ) = β a β
and
c = β cos β‘ 2 Ο 7 cos β‘ 4 Ο 7 cos β‘ 6 Ο 7 = β 1 2 ( cos β‘ 6 Ο 7 + cos β‘ 2 Ο 7 ) cos β‘ 6 Ο 7 = β 1 2 2 6 Ο 7 β 1 4 ( cos β‘ 8 Ο 7 + cos β‘ 4 Ο 7 ) = β 1 4 ( 1 + cos β‘ 12 Ο 7 ) β 1 4 ( cos β‘ 6 Ο 7 + cos β‘ 4 Ο 7 ) = 1 4 ( a β 1 ) . c = β cos 7 2 Ο β cos 7 4 Ο β cos 7 6 Ο β = β 2 1 β ( cos 7 6 Ο β + cos 7 2 Ο β ) cos 7 6 Ο β = β 2 1 β cos 2 7 6 Ο β β 4 1 β ( cos 7 8 Ο β + cos 7 4 Ο β ) = β 4 1 β ( 1 + cos 7 1 2 Ο β ) β 4 1 β ( cos 7 6 Ο β + cos 7 4 Ο β ) = 4 1 β ( a β 1 ) . β
To compute a a
cos β‘ 6 Ο 7 = cos β‘ 8 Ο 7 = 2 2 4 Ο 7 β 1 cos β‘ 2 Ο 7 = cos β‘ 16 Ο 7 = 2 2 8 Ο 7 β 1 = 2 2 6 Ο 7 β 1 cos β‘ 4 Ο 7 = 2 2 2 Ο 7 β 1 cos 7 6 Ο β = cos 7 8 Ο β = 2 cos 2 7 4 Ο β β 1 cos 7 2 Ο β = cos 7 1 6 Ο β = 2 cos 2 7 8 Ο β β 1 = 2 cos 2 7 6 Ο β β 1 cos 7 4 Ο β = 2 cos 2 7 2 Ο β β 1 β
Then
β a = 2 ( 2 4 Ο 7 + 2 6 Ο 7 + 2 2 Ο 7 ) β 3 = 2 ( a 2 β 2 b ) β 3 = 2 ( a 2 + 2 a ) β 3. β a = 2 ( cos 2 7 4 Ο β + cos 2 7 6 Ο β + cos 2 7 2 Ο β ) β 3 = 2 ( a 2 β 2 b ) β 3 = 2 ( a 2 + 2 a ) β 3 . β
So 2 a 2 + 5 a β 3 = 0 2 a 2 + 5 a β 3 = 0 a = 1 2 a = 2 1 β a = β 3 a = β 3 β 3 < a < 3 β 3 < a < 3 a = 1 2 , b = β 1 2 a = 2 1 β , b = β 2 1 β c = β 1 8 c = β 8 1 β a b c = 1 32 a b c = 3 2 1 β β
The problems on this page are the property of the MAA's American Mathematics Competitions