Problem: P 0 P 0 β P 1 P 1 β j β₯ 1 j β₯ 1 P j P j β 3 0 β 3 0 β j + 1 j + 1 P j + 1 P j + 1 β P 2015 P 2 0 1 5 β a b + c d a b β + c d β P 0 P 0 β a , b , c a , b , c d d b b d d a + b + c + d a + b + c + d
Answer Choices:
A. 2016 2 0 1 6 2024 2 0 2 4 2032 2 0 3 2 2040 2 0 4 0 2048 2 0 4 8 Solution:
Modeling the bee's path with complex numbers, set P 0 = 0 P 0 β = 0 z = e Ο i / 6 z = e Ο i / 6 j β₯ 1 j β₯ 1
P j = β k = 1 j k z k β 1 P j β = k = 1 β j β k z k β 1
Thus
P 2015 = β k = 0 2015 k z k β 1 = β k = 0 2014 ( k + 1 ) z k = β k = 0 2014 β j = 0 k z k P 2 0 1 5 β = k = 0 β 2 0 1 5 β k z k β 1 = k = 0 β 2 0 1 4 β ( k + 1 ) z k = k = 0 β 2 0 1 4 β j = 0 β k β z k
Interchanging the order of summation and summing the geometric series gives
P 2015 = β j = 0 2014 β k = j 2014 z k = β j = 0 2014 z j β k = 0 2014 β j z k = β j = 0 2014 z j ( z 2015 β j β 1 ) z β 1 = β j = 0 2014 z 2015 β z j z β 1 = 1 z β 1 β j = 0 2014 ( z 2015 β z j ) = 1 z β 1 ( 2015 z 2015 β β j = 0 2014 z j ) = 1 z β 1 ( 2015 z 2015 β z 2015 β 1 z β 1 ) = 1 ( z β 1 ) 2 ( 2015 z 2015 ( z β 1 ) β z 2015 + 1 ) = 1 ( z β 1 ) 2 ( 2015 z 2016 β 2016 z 2015 + 1 ) P 2 0 1 5 β = j = 0 β 2 0 1 4 β k = j β 2 0 1 4 β z k = j = 0 β 2 0 1 4 β z j k = 0 β 2 0 1 4 β j β z k = j = 0 β 2 0 1 4 β z β 1 z j ( z 2 0 1 5 β j β 1 ) β = j = 0 β 2 0 1 4 β z β 1 z 2 0 1 5 β z j β = z β 1 1 β j = 0 β 2 0 1 4 β ( z 2 0 1 5 β z j ) = z β 1 1 β ( 2 0 1 5 z 2 0 1 5 β j = 0 β 2 0 1 4 β z j ) = z β 1 1 β ( 2 0 1 5 z 2 0 1 5 β z β 1 z 2 0 1 5 β 1 β ) = ( z β 1 ) 2 1 β ( 2 0 1 5 z 2 0 1 5 ( z β 1 ) β z 2 0 1 5 + 1 ) = ( z β 1 ) 2 1 β ( 2 0 1 5 z 2 0 1 6 β 2 0 1 6 z 2 0 1 5 + 1 ) β
Note that z 12 = 1 z 1 2 = 1 z 2016 = ( z 12 ) 168 = 1 z 2 0 1 6 = ( z 1 2 ) 1 6 8 = 1 z 2015 = 1 z z 2 0 1 5 = z 1 β
P 2015 = 2016 ( z β 1 ) 2 ( 1 β 1 z ) = 2016 z ( z β 1 ) P 2 0 1 5 β = ( z β 1 ) 2 2 0 1 6 β ( 1 β z 1 β ) = z ( z β 1 ) 2 0 1 6 β
Finally,
β£ z β 1 β£ 2 = β£ cos β‘ ( Ο 6 ) β 1 + i sin β‘ ( Ο 6 ) β£ 2 = β£ 3 2 β 1 + i 2 β£ 2 = 2 β 3 = ( 3 β 1 ) 2 2 β£ z β 1 β£ 2 = β£ β£ β£ β£ β cos ( 6 Ο β ) β 1 + i sin ( 6 Ο β ) β£ β£ β£ β£ β 2 = β£ β£ β£ β£ β£ β£ β 2 3 β β β 1 + 2 i β β£ β£ β£ β£ β£ β£ β 2 = 2 β 3 β = 2 ( 3 β β 1 ) 2 β
and thus
β£ P 2015 β£ = β£ 2016 z ( z β 1 ) β£ = 2016 β£ z β 1 β£ = 2016 2 3 β 1 = 1008 2 ( 3 + 1 ) = 1008 6 + 1008 2 β£ P 2 0 1 5 β β£ = β£ β£ β£ β£ β£ β z ( z β 1 ) 2 0 1 6 β β£ β£ β£ β£ β£ β = β£ z β 1 β£ 2 0 1 6 β = 3 β β 1 2 0 1 6 2 β β = 1 0 0 8 2 β ( 3 β + 1 ) = 1 0 0 8 6 β + 1 0 0 8 2 β β
The requested sum is 1008 + 6 + 1008 + 2 = 2024 1 0 0 8 + 6 + 1 0 0 8 + 2 = 2 0 2 4 β
The problems on this page are the property of the MAA's American Mathematics Competitions