Problem:
A sequence (a1β,b1β),(a2β,b2β),(a3β,b3β),β¦ of points in the coordinate plane satisfies
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(an+1β,bn+1β)=(3βanββbnβ,3βbnβ+anβ) for n=1,2,3,β¦
Suppose that (a100β,b100β)=(2,4). What is a1β+b1β?
Answer Choices:
A. β2971β
B. β2991β
C. 0
D. 2981β
E. 2961β
Solution:
Let znβ=anβ+bnβi. Then
zn+1β=(3βanββbnβ)+(3βbnβ+anβ)i=(anβ+bnβi)(3β+i)=znβ(3β+i)=z1β(3β+i)nβ
Noting that 3β+i=2(cos6Οβ+isin6Οβ) and applying DeMoivre's formula gives
2+4i=z100β=z1β(2(cos(6Οβ)+isin(6Οβ)))99=z1ββ
299(cos(699Οβ)+isin(699Οβ))=(a1β+b1βi)β
299β
i=β299b1β+299a1βiβ
So 2=β299b1β,4=299a1β, and
a1β+b1β=2994ββ2992β=2981ββ
Note that
(an+2β,bn+2β)=(3β(3βanββbnβ)β(3βbnβ+anβ)3β(3βbnβ+anβ)+(3βanββbnβ))=(β23βbnβ+2anβ,23βanβ+2bnβ)(an+3β,bn+3β)=(3β(β23βbnβ+2anβ)β(23βanβ+2bnβ)3β(23βanβ+2bnβ)+(β23βbnβ+2anβ))=8(βbnβ,anβ)β
and (an+6β,bn+6β)=8(βbn+3β,an+3β)=β64(anβ,bnβ). Because 97=1+16β
6, we have
(a97β,b97β)=(β64)16(a1β,b1β)=296(a1β,b1β)
and
(2,4)=(a100β,b100β)=23(βb97β,a97β)=299(βb1β,a1β)
The conclusion follows as in the first solution.
The problems on this page are the property of the MAA's American Mathematics Competitions