Problem: z 1 , z 2 , z 3 , β¦ , z 12 z 1 β , z 2 β , z 3 β , β¦ , z 1 2 β z 12 β 2 36 z 1 2 β 2 3 6 j j w j w j β z j z j β i z j i z j β β j = 1 12 w j β j = 1 1 2 β w j β m + n m + n β m m n n m + n m + n
Solution:
Without loss of generality, let z j = 8 ( cos β‘ j Ο 6 + i sin β‘ j Ο 6 ) z j β = 8 ( cos 6 j Ο β + i sin 6 j Ο β ) i z j = i z j β = 8 ( β sin β‘ j Ο 6 + i cos β‘ j Ο 6 ) 8 ( β sin 6 j Ο β + i cos 6 j Ο β ) β ( β j = 1 12 w j ) = β j = 1 12 β ( w j ) β ( β j = 1 1 2 β w j β ) = β j = 1 1 2 β β ( w j β ) β ( w j ) = max β‘ ( 8 cos β‘ j Ο 6 , β 8 sin β‘ j Ο 6 ) β ( w j β ) = max ( 8 cos 6 j Ο β , β 8 sin 6 j Ο β ) 1 β€ j β€ 12 1 β€ j β€ 1 2 cos β‘ j Ο 6 < β sin β‘ j Ο 6 cos 6 j Ο β < β sin 6 j Ο β 5 β€ j β€ 10 5 β€ j β€ 1 0
8 ( cos β‘ 0 Ο 6 + cos β‘ 1 Ο 6 + cos β‘ 2 Ο 6 + cos β‘ 3 Ο 6 + cos β‘ 4 Ο 6 + cos β‘ 11 Ο 6 ) β 8 ( sin β‘ 5 Ο 6 + sin β‘ 6 Ο 6 + sin β‘ 7 Ο 6 + sin β‘ 8 Ο 6 + sin β‘ 9 Ο 6 + sin β‘ 10 Ο 6 ) = 8 ( 1 + 3 2 + 1 2 + 0 β 1 2 + 3 2 ) β 8 ( 1 2 + 0 β 1 2 β 3 2 β 1 β 3 2 ) = 16 + 16 3 = 16 + 768 , β 8 ( cos 6 0 Ο β + cos 6 1 Ο β + cos 6 2 Ο β + cos 6 3 Ο β + cos 6 4 Ο β + cos 6 1 1 Ο β ) β 8 ( sin 6 5 Ο β + sin 6 6 Ο β + sin 6 7 Ο β + sin 6 8 Ο β + sin 6 9 Ο β + sin 6 1 0 Ο β ) = 8 ( 1 + 2 3 β β + 2 1 β + 0 β 2 1 β + 2 3 β β ) β 8 ( 2 1 β + 0 β 2 1 β β 2 3 β β β 1 β 2 3 β β ) = 1 6 + 1 6 3 β = 1 6 + 7 6 8 β , β
and the requested sum is 16 + 768 = 784 1 6 + 7 6 8 = 7 8 4 β
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions