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IMC / 1999 / Problems / Day 2, P12

IMC 1999 · Day 2 · P12

killer

Let AA be a subset of Zn=Z/nZ\mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z} containing at most 1100lnn\frac{1}{100} \ln n elements. Define the rrth Fourier coefficient of AA for rZnr \in \mathbb{Z}_n by f(r)=sAexp(2πinsr).f(r) = \sum_{s \in A} \exp \left( \frac{2\pi i}{n} sr \right). Prove that there exists an r0r \ne 0, such that f(r)A2\bigl| f(r) \bigr| \ge \frac{|A|}{2}.

Solution (official)

Let A={a1,,ak}A = \{a_1, \dots, a_k\}. Consider the kk-tuples (exp2πia1tn,,exp2πiaktn)Ck,t=0,1,,n1.\left( \exp \frac{2\pi i a_1 t}{n}, \dots, \exp \frac{2\pi i a_k t}{n} \right) \in \mathbb{C}^k, \quad t = 0, 1, \dots, n-1. Each component is in the unit circle z=1|z| = 1. Split the circle into 6 equal arcs. This induces a decomposition of the kk-tuples into 6k6^k classes. By the condition k1100lnnk \le \frac{1}{100} \ln n we have n>6kn > 6^k, so there are two kk-tuples in the same class say for t1<t2t_1 < t_2. Set r=t2t1r = t_2 - t_1. Then Reexp2πiajrn=cos(2πajt2n2πajt1n)cosπ3=12\operatorname{Re} \exp \frac{2\pi i a_j r}{n} = \cos \left( \frac{2\pi a_j t_2}{n} - \frac{2\pi a_j t_1}{n} \right) \ge \cos \frac{\pi}{3} = \frac{1}{2} for all jj, so f(r)Ref(r)k2.|f(r)| \ge \operatorname{Re} f(r) \ge \frac{k}{2}.

How the field did

contestants scored
87
average (of 20)
0.78
solved (≥ 80%)
2.3%
near-0 (≤ 10%)
90.8%
discrimination
0.30

Score distribution (field cohort)

Computed on contestants with a meaningful total (field cohort); discrimination is the corrected item–total correlation.

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