Problem 12202
(American Mathematical Monthly, Vol.127, October 2020) Proposed by Koopa Tak Lun Koo (China).
LetV be a finite set of vectors in R2 such thatP
v∈V |v| = π. Prove that there exists a subset U of V such that
P
v∈Uv ≥ 1.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. For any unit vector w, let
Pw(v) =
(v · w = |v| cos(θ) if cos(θ) > 0,
0 otherwise.
where θ is the angle between v and w. Then |v| ≥ Pw(v) and by letting Uw= {v ∈ V : Pw(v) > 0}, we find that
X
v∈Uw
v
≥ Pw
X
v∈Uw
v
!
= X
v∈Uw
Pw(v) =X
v∈V
Pw(v) .
Hence there exists a unit vector w such that P
v∈Uwv
is greater or equal the average value of P
v∈V Pw(v) given by 1 2π
Z π
−π
X
v∈V
Pw(v) dθ = 1 2π
X
v∈V
|v|
Z π/2
−π/2
cos(θ) dθ = 1 π
X
v∈V
|v| = 1.
Remark. The lower bound 1 is the best possible one. Consider this set of evenly distributed two- dimensional vectors (with complex number notation)
Vn= {vk =π
nei2πkn : k = 0, . . . , n − 1}
where n is an even positive integer. Let Un = {vk : k = 0, . . . ,n2 − 1} ⊂ Vn. ThenP
v∈Vn|v| = π and
X
v∈Un
|v| = π n
n 2−1
X
k=0
ei2πkn
=π n
1 − eiπ 1 − ei2πn = π
n 2
|e−iπn− eiπn| = π/n
sin(π/n) → 1 as n → ∞.
