Problem: P 0 P 0 P 1 P 1 j ≥ 1 j ≥ 1 P j P j 3 0 ∘ 3 0 ∘ j + 1 j + 1 P j + 1 P j + 1 P 2015 P 2 0 1 5 a b + c d a b + c d P 0 P 0 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 = 0 z = e π i / 6 z = e π i / 6 j ≥ 1 j ≥ 1
P j = ∑ k = 1 j k z k − 1 P j = k = 1 ∑ j k z k − 1
Thus
P 2015 = ∑ k = 0 2015 k z k − 1 = ∑ k = 0 2014 ( k + 1 ) z k = ∑ k = 0 2014 ∑ j = 0 k z k P 2 0 1 5 = k = 0 ∑ 2 0 1 5 k z k − 1 = k = 0 ∑ 2 0 1 4 ( k + 1 ) z k = k = 0 ∑ 2 0 1 4 j = 0 ∑ k z k
Interchanging the order of summation and summing the geometric series gives
P 2015 = ∑ j = 0 2014 ∑ k = j 2014 z k = ∑ j = 0 2014 z j ∑ k = 0 2014 − j z k = ∑ j = 0 2014 z j ( z 2015 − j − 1 ) z − 1 = ∑ j = 0 2014 z 2015 − z j z − 1 = 1 z − 1 ∑ j = 0 2014 ( z 2015 − z j ) = 1 z − 1 ( 2015 z 2015 − ∑ j = 0 2014 z j ) = 1 z − 1 ( 2015 z 2015 − z 2015 − 1 z − 1 ) = 1 ( z − 1 ) 2 ( 2015 z 2015 ( z − 1 ) − z 2015 + 1 ) = 1 ( z − 1 ) 2 ( 2015 z 2016 − 2016 z 2015 + 1 ) P 2 0 1 5 = j = 0 ∑ 2 0 1 4 k = j ∑ 2 0 1 4 z k = j = 0 ∑ 2 0 1 4 z j k = 0 ∑ 2 0 1 4 − j z k = j = 0 ∑ 2 0 1 4 z − 1 z j ( z 2 0 1 5 − j − 1 ) = j = 0 ∑ 2 0 1 4 z − 1 z 2 0 1 5 − z j = z − 1 1 j = 0 ∑ 2 0 1 4 ( z 2 0 1 5 − z j ) = z − 1 1 ( 2 0 1 5 z 2 0 1 5 − j = 0 ∑ 2 0 1 4 z j ) = z − 1 1 ( 2 0 1 5 z 2 0 1 5 − z − 1 z 2 0 1 5 − 1 ) = ( z − 1 ) 2 1 ( 2 0 1 5 z 2 0 1 5 ( z − 1 ) − z 2 0 1 5 + 1 ) = ( z − 1 ) 2 1 ( 2 0 1 5 z 2 0 1 6 − 2 0 1 6 z 2 0 1 5 + 1 )
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
P 2015 = 2016 ( z − 1 ) 2 ( 1 − 1 z ) = 2016 z ( z − 1 ) P 2 0 1 5 = ( z − 1 ) 2 2 0 1 6 ( 1 − z 1 ) = z ( z − 1 ) 2 0 1 6
Finally,
∣ z − 1 ∣ 2 = ∣ cos ( π 6 ) − 1 + i sin ( π 6 ) ∣ 2 = ∣ 3 2 − 1 + i 2 ∣ 2 = 2 − 3 = ( 3 − 1 ) 2 2 ∣ z − 1 ∣ 2 = ∣ ∣ ∣ ∣ cos ( 6 π ) − 1 + i sin ( 6 π ) ∣ ∣ ∣ ∣ 2 = ∣ ∣ ∣ ∣ ∣ ∣ 2 3 − 1 + 2 i ∣ ∣ ∣ ∣ ∣ ∣ 2 = 2 − 3 = 2 ( 3 − 1 ) 2
and thus
∣ P 2015 ∣ = ∣ 2016 z ( z − 1 ) ∣ = 2016 ∣ z − 1 ∣ = 2016 2 3 − 1 = 1008 2 ( 3 + 1 ) = 1008 6 + 1008 2 ∣ P 2 0 1 5 ∣ = ∣ ∣ ∣ ∣ ∣ z ( z − 1 ) 2 0 1 6 ∣ ∣ ∣ ∣ ∣ = ∣ z − 1 ∣ 2 0 1 6 = 3 − 1 2 0 1 6 2 = 1 0 0 8 2 ( 3 + 1 ) = 1 0 0 8 6 + 1 0 0 8 2
The requested sum is 1008 + 6 + 1008 + 2 = 2024 1 0 0 8 + 6 + 1 0 0 8 + 2 = 2 0 2 4
The problems on this page are the property of the MAA's American Mathematics Competitions