Determine the length of $BC$ in an acute triangle $ABC$ with $\angle ABC = 45^{\circ}$, $OG = 1$ and $OG \parallel BC$. (As usual $O$ is the circumcenter and $G$ is the centroid.)

The sequence of real numbers $(a_n)_{n\geq 0}$ is such that $a_0 = 1$, $a_1 = a > 2$ and $\displaystyle a_{n+1} = \left(\left(\frac{a_n}{a_{n-1}}\right)^2 -2\right)a_n$ for every positive integer $n$. Prove that $\displaystyle \sum_{i=0}^k \frac{1}{a_i} < \frac{2+a-\sqrt{a^2-4}}{2}$ for every positive integer $k$.

$A$ and $B$ play a game, given an integer $N$, $A$ writes down $1$ first, then every player sees the last number written and if it is $n$ then in his turn he writes $n+1$ or $2n$, but his number cannot be bigger than $N$. The player who writes $N$ wins. For which values of $N$ does $B$ win? Proposed by A. Slinko & S. Marshall, New Zealand

Let $x_1,\ldots,x_n$ be a sequence with each term equal to $0$ or $1$. Form a triangle as follows: its first row is $x_1,\ldots,x_n$ and if a row if $a_1, a_2, \ldots, a_m$, then the next row is $a_1 + a_2, a_2 + a_3, \ldots, a_{m-1} + a_m$ where the addition is performed modulo $2$ (so $1+1=0$). For example, starting from $1$, $0$, $1$, $0$, the second row is $1$, $1$, $1$, the third one is $0$, $0$ and the fourth one is $0$. A sequence is called good it is the same as the sequence formed by taking the last element of each row, starting from the last row (so in the above example, the sequence is $1010$ and the corresponding sequence from last terms is $0010$ and they are not equal in this case). How many possibilities are there for the sequence formed by taking the first element of each row, starting from the last row, which arise from a good sequence?

Let $ABCD$ be a cyclic quadrilateral with circumcircle $\omega$ centered at $O$, whose diagonals intersect at $H$. Let $O_1$ and $O_2$ be the circumcenters of triangles $AHD$ and $BHC$. A line through $H$ intersects $\omega$ at $M_1$ and $M_2$ and intersects the circumcircles of triangles $O_1HO$ and $O_2HO$ at $N_1$ and $N_2$, respectively, so that $N_1$ and $N_2$ lie inside $\omega$. Prove that $M_1N_1 = M_2N_2$.

In terms of the fixed non-negative integers $\alpha$ and $\beta$ determine the least upper bound of the ratio (or show that it is unbounded) \[ \frac{S(n)}{S(2^{\alpha}5^{\beta}n)} \]as $n$ varies through the positive integers, where $S(\cdot)$ denotes sum of digits in decimal representation.