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