Find $k \in \mathbb{N}$ such that a.) For any $n \in \mathbb{N}$, there does not exist $j \in \mathbb{Z}$ which satisfies the conditions $0 \leq j \leq n - k + 1$ and $\left( \begin{array}{c} n\\ j\end{array} \right), \left( \begin{array}{c} n\\ j + 1\end{array} \right), \ldots, \left( \begin{array}{c} n\\ j + k - 1\end{array} \right)$ forms an arithmetic progression. b.) There exists $n \in \mathbb{N}$ such that there exists $j$ which satisfies $0 \leq j \leq n - k + 2$, and $\left( \begin{array}{c} n\\ j\end{array} \right), \left( \begin{array}{c} n\\ j + 1\end{array} \right), \ldots , \left( \begin{array}{c} n\\ j + k - 2\end{array} \right)$ forms an arithmetic progression. Find all $n$ which satisfies part b.)

$n \geq 5$ football teams participate in a round-robin tournament. For every game played, the winner receives 3 points, the loser receives 0 points, and in the event of a draw, both teams receive 1 point. The third-from-bottom team has fewer points than all the teams ranked before it, and more points than the last 2 teams; it won more games than all the teams before it, but fewer games than the 2 teams behind it. Find the smallest possible $n$.

For a fixed $\theta \in \lbrack 0, \frac{\pi}{2} \rbrack$, find the smallest $a \in \mathbb{R}^{+}$ which satisfies the following conditions: I. $\frac{\sqrt a}{\cos \theta} + \frac{\sqrt a}{\sin \theta} > 1$. II. There exists $x \in \lbrack 1 - \frac{\sqrt a}{\sin \theta}, \frac{\sqrt a}{\cos \theta} \rbrack$ such that $\lbrack (1 - x)\sin \theta - \sqrt{a - x^2 \cos^{2} \theta} \rbrack^{2} + \lbrack x\cos \theta - \sqrt{a - (1 - x)^2 \sin^{2} \theta} \rbrack^{2} \leq a$.

In acute-angled $\bigtriangleup ABC$, $H$ is the orthocenter, $O$ is the circumcenter and $I$ is the incenter. Given that $\angle C > \angle B > \angle A$, prove that $I$ lies within $\bigtriangleup BOH$.

Click for solution guess no one bothered with this one, but I'll give an approach (I'll leave it for those that care to finish... ) Two steps: first show that angleCBH < angleCBI < angleCBO (I lies within angle HBO), this is really easy once you simplify everything. Second, prove that IY (Y being the perpendicular foot of I on AB) is not long enough to reach line segment OH. This is proven by assigning X and Z to be the perpendicular feet of H and O on AB, respectively. Then, showing that (|HX|-|OZ|)*(|YZ|/|XZ|) > |IY|-|OZ|. I give fair warning that this is a nasty and commutative trig thing, it's really not a pleasure doing it. But it works!

Let $n$ be a natural number greater than 2. $l$ is a line on a plane. There are $n$ distinct points $P_1$, $P_2$, …, $P_n$ on $l$. Let the product of distances between $P_i$ and the other $n-1$ points be $d_i$ ($i = 1, 2,$ …, $n$). There exists a point $Q$, which does not lie on $l$, on the plane. Let the distance from $Q$ to $P_i$ be $C_i$ ($i = 1, 2,$ …, $n$). Find $S_n = \sum_{i = 1}^{n} (-1)^{n-i} \frac{c_i^2}{d_i}$.

For any $h = 2^{r}$ ($r$ is a non-negative integer), find all $k \in \mathbb{N}$ which satisfy the following condition: There exists an odd natural number $m > 1$ and $n \in \mathbb{N}$, such that $k \mid m^{h} - 1, m \mid n^{\frac{m^{h}-1}{k}} + 1$.

Click for solution Let $2^a||m^h-1$ and $2^b||k$. Then $2^{a-b}||\frac{m^h-1}{k}$, so since $m$ divides $n^{\frac{m^h-1}{k}}+1$ there is a residue modulo $m$ of order $2^{a-b+1}$, it easily follows that $2^{a-b+1}$ divides $p-1$ for every prime divisor of $m$, and hence $2^{a-b+2}|m^2-1$. By induction on $r$, we can prove that $2^t||m^h-1$ iff and only iff $2^t||(m^2-1)2^{r-1}$ so we get that: $a-b+2+r-1$ is at most $a$, so from here it follows that $b\geq r+1$. We prove this is a sufficient condition. Let $k=2^{r+1+t}h$ with $h$ odd. Iff t is not $0$, then choose an $m$ such that $2^{t+2}$ is the max power of 2 dividing $m^2-1$ and also such that $h|m-1$, and put $n=m+1$, in this case everything works. iff $t=0$, let $p$ be a prime of the form $8j+5$, and such that $h|p-1$, then there exist an $n$, such that $n^2+1$ is a multiple of $p$, so this $n$ works!