see http://imomath.com/index.php?options=oth|Bul|Ne|Ne&ttn=Bulgaria&p=0

enndb0x wrote: see http://imomath.com/index.php?options=oth|Bul|Ne|Ne&ttn=Bulgaria&p=0 It is problem of Bulgari MO 1960-2008 not have problem Bul MO,TST 2009 and Bul TST 2008 if you have them,you can post them?

I and a friend of mine have posted lots of Bulgarian problems at imomath. I have the missing problems. When I've time I'll post them.

I think $ b_i \ge 0$ for $ i = 1,...n$ Otherwise let $ b_1 = 1,b_2 = - 1,a_1 = a_2$ which gives a cunterexample :

A solution based on $ b_i\ge 0$ The idea is using approximation Very sorry for written in Chinese,now I restate the lemma(the proof is not difficult to come up with) Lemma:$ x_1,x_2,...,x_n \in R$ we can find infinitely many $ k,k\in N$ and $ c_1,c_2,...,c_n \in Z$ so that $ |kx_i-c_i| \le k^{-\frac{1}{n}}$

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For those interested in a proof the lemma above, a special case which is easily generalized is outlined here Here's a proof of the slightly stronger claim that given unrestricted $b_i$ with the property below, they must sum to $0$ (of course we recover the problem statement when we impose $b_i \ge 0$) Also, it's worth noting that it is critical we work with $\cos$ and not, say, $\sin$ since then Lemma 2 (and the conclusion) fails to hold. Lemma 1: Given real numbers $a_1, a_2, \dots, a_n$ and $\delta > 0$, there exists $k \in \mathbb{N}$ such that \[-\delta < ka_i - ka_1 \pmod{2\pi} < \delta\] for $1 \le i \le n$. Proof: We prove by induction on $n$. When $n = 1$ the statement is obvious. Then suppose for some $m \ge 1$ and all $\delta > 0$ there exists $k \in \mathbb{N}$ such that \[-\delta < kb_i - kb_1 \pmod{2\pi} < \delta \; \; \; \; \; \; \; (1)\] for all $1 \le i \le m$. Let $b_{m + 1}$ and $\delta > 0$ be given. Taking $\delta$ in $(1)$ to be $\frac{\delta}{\lceil \frac{2\pi}{\delta} \rceil + 1}$ we find $k \in \mathbb{N}$ for which \[-\frac{\delta}{\lceil \frac{2\pi}{\delta} \rceil + 1} < k(b_i - b_1) < \frac{\delta}{\lceil \frac{2\pi}{\delta} \rceil + 1}\] for all $1 \le i \le m$. By Pigeonhole Principle, for some $a \in \mathbb{N}, a < \lceil \frac{2\pi}{\delta} \rceil + 1$ we have \[-\delta < ka(b_{m + 1} - b_1) \pmod{2\pi} < \delta\] where $k$ is the $k$ chosen above. But \[-\delta < ka(b_i - b_1) \pmod{2\pi} < \delta\] for $1 \le i \le m$ since $a < \lceil \frac{2\pi}{\delta} \rceil + 1$, so we are done since $(1)$ is satisfied for all $1 \le i \le m + 1$ as desired. $\blacksquare$ Then by the continuity of $\cos(x)$ and its periodicity modulo $2\pi$, for all $\epsilon > 0$ there exists $k \in \mathbb{N}$ such that \[\cos(ka_1) - \epsilon < \cos(ka_1), \cos(ka_2), \dots, \cos(ka_n) < \cos(ka_1) + \epsilon \; \; \; \; \; \; \; \; \; (2)\] Furthermore, there exist such $k$ arbitrarily large; given any $\epsilon > 0$, there is some $\delta_0 > 0$ for which $(1)$ implies $(2)$. To find a $k$ value larger than $N$, simply take $\delta = \frac{\delta_0}{N}$ in $(1)$; since \[-\frac{\delta_0}{N} < ka_i - ka_1 \pmod{2\pi} < \frac{\delta_0}{N}\] holds for some $k \in \mathbb{N}$, \[-\delta_0 < kNa_i - kNa_1 \pmod{2\pi} < \delta_0\] so $(2)$ holds for $kN$ which is at least $N$, as claimed. We need just one more lemma to finish: Lemma 2: For any $\theta$, \[\max(|\cos(\theta)|, |\cos(2\theta)|) \ge \frac{1}{2}\] Proof: If $|\cos(\theta)| \ge \frac{1}{2}$, we're done. Otherwise, \[|\cos(2\theta)| = |2\cos^2(\theta) - 1| = |1 - 2\cos^2(\theta)| \ge |1 - 2(\frac{1}{2})^2| = \frac{1}{2}. \blacksquare\] Suppose $\sum{b_i} \ne 0$. Using $(2)$ and the construction for arbitrarily large $k$, we can find $k \in \mathbb{N}$ arbitrarily large such that \[\cos(ka_1) - \epsilon < \cos(ka_1), \cos(ka_2), \dots, \cos(ka_n) < \cos(ka_1) + \epsilon\] \[\cos(2ka_1) - \epsilon < \cos(2ka_1), \cos(2ka_2), \dots, \cos(2ka_n) < \cos(2ka_1) + \epsilon \] Then we can choose $\epsilon = \frac{|\sum{b_i}|}{3\sum{|b_i|}}$ and (using Lemma 2) find $k$ arbitrarily large for which either \[|(\sum{b_i})|(\frac{1}{2}) - \frac{|\sum{b_i}|}{3\sum{|b_i|}} \sum{|b_i|} < |\sum{b_i\cos(ka_i)}| < |(\sum{b_i})|(\frac{1}{2} + \frac{1}{3})\] or \[|(\sum{b_i})|(\frac{1}{2}) - \frac{|\sum{b_i}|}{3\sum{|b_i|}}\[ \sum{|b_i|} < |\sum{b_i\cos(2ka_i)}| < |(\sum{b_i})|(\frac{1}{2} + \frac{1}{3})\] where the lower bound for each is $\frac{|\sum{b_i}|}{6} > 0$, contradicting the fact that all $k$ satisfy the problem statement.

The above solutions all use too much thinking! Here is a troll solution. Squaring both sides of the inequality given, we have that $$\sum b_ib_j \cos (ka_i) \cos (ka_j) < \frac{1}{k^2}, \qquad (1)$$where the sum is over all ordered pairs $(i, j) \in [n]^2.$ Observe that $\cos(ka_i) \cos(ka_j) = \frac12 (\cos (k(a_i+a_j)) + \cos (k(a_i-a_j)))$ from the Product-to-Sum Identity. Lemma. The sequence given by $x_m= \sum_{i=1}^{m} \cos (i\alpha)$ is unbounded whenever $\alpha$ is not a multiple of $2 \pi$. Proof. Let $\omega = e^{i \alpha}.$ Then $x_m$ is the real part of the complex number $$\frac{\omega^{m+1} - \omega}{\omega - 1}.$$This is bounded by $|\frac{2}{\omega-1}|$ in magnitude, so we're done. $\blacksquare$ Using the lemma and the Product-to-Sum Identity, we can sum $(1)$ from $1$ to $N$ for a large positive integer $N$, and rewrite it as $$(stuff) + \sum_{i=1}^{n} \frac{Nb_i^2}{2}+\sum_{i \neq j, a_i = a_j} \frac{Nb_ib_j}{2} + (trash)_N,$$where $(trash)_N$ is a quantity dependent on $N$ that is bounded. As for $(stuff)$, it denotes the potential increase of $\frac{Nb_1^2}{2}$ if $a_1 = 0$ and of $\frac{Nb_n^2}{2}$ if $a_n = \pi.$ Since it's easy to see that $\frac2N (stuff) + \sum_{i=1}^{n} b_i^2 + \sum_{i \neq j, a_i = a_j} b_i b_j > 0$ whenever not all $b_i$'s are zero, the above sum is unbounded whenever not all $b_i$'s are zero. However, $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{N^2}$ is bounded above by $\frac{\pi^2}{6}$. Hence, we must have that $\sum_{i=1}^{n} b_i^2 + \sum_{i \neq j, a_i = a_j} b_ib_j = 0$ which implies that $b_1 = b_2 = \cdots = b_n = 0,$ as desired. $\square$