Problem:
How many integers nβ₯2 are there such that whenever z1β,z2β,β¦,znβ are complex numbers such that
β£z1ββ£=β£z2ββ£=β¦=β£znββ£=1 and z1β+z2β+β¦+znβ=0
then the numbers z1β,z2β,β¦,znβ are equally spaced on the unit circle in the complex plane?
Answer Choices:
A. 1
B. 2
C. 3
D. 4
E. 5
Solution:
For n=2, we see that if z1β+z2β=0, then z1β=βz2β, so they are evenly spaced along the unit circle.
For n=3, WLOG, we can set z1β=1. Notice that now β(z2β+z3β)=β1 and β{z2β}=ββ{z3β}. This forces z2β and z3β to be equal to ei32Οβ and eβi32Οβ, meaning that all three are equally spaced along the unit circle.
We can now show that we can construct complex numbers when nβ₯4 that do not satisfy the conditions in the problem.
Suppose that the condition in the problem holds for some n=k. We can now add two points zk+1β and zk+2β anywhere on the unit circle such that zk+1β=βzk+2β, which will break the condition. Now that we have shown that n=2 and n=3 works, by this construction, any nβ₯4 does not work, making the answer (B)2β .
The problems on this page are the property of the MAA's American Mathematics Competitions