Problem:
Let z1β,z2β,z3β,β¦,z12β be the 12 zeros of the polynomial z12β236. For each j, let wjβ be one of zjβ or izjβ. Then the maximum possible value of the real part of βj=112βwjβ can be written as m+nβ, where m and n are positive integers. Find m+n.
Solution:
Without loss of generality, let zjβ=8(cos6jΟβ+isin6jΟβ), so that izjβ=8(βsin6jΟβ+icos6jΟβ). Because β(βj=112βwjβ)=βj=112ββ(wjβ), the sum is maximized when β(wjβ)=max(8cos6jΟβ,β8sin6jΟβ) for each 1β€jβ€12. Because cos6jΟβ<βsin6jΟβ for 5β€jβ€10, the maximum possible sum is