We are given two point sets $A$ and $B$ which are both composed of finite disjoint arcs on the unit circle. Moreover, the length of each arc in $B$ is equal to $\dfrac{\pi}{m}$ ($m \in \mathbb{N}$). We denote by $A^j$ the set obtained by a counterclockwise rotation of $A$ about the center of the unit circle for $\dfrac{j\pi}{m}$ ($j=1,2,3,\dots$). Show that there exists a natural number $k$ such that $l(A^k\cap B)\ge \dfrac{1}{2\pi}l(A)l(B)$.(Here $l(X)$ denotes the sum of lengths of all disjoint arcs in the point set $X$)

Let $x_1, x_2, \dots ,x_n$ ($n\ge 2$) be positive real numbers satisfying $\sum^{n}_{i=1}x_i=1$. Prove that:\[\sum^{n}_{i=1}\dfrac{x_i}{\sqrt{1-x_i}}\ge \dfrac{\sum_{i=1}^{n}\sqrt{x_i}}{\sqrt{n-1}}.\]

Let $S$ be the unit circle in the complex plane (i.e. the set of all complex numbers with their moduli equal to $1$). We define function $f:S\rightarrow S$ as follow: $\forall z\in S$, $ f^{(1)}(z)=f(z), f^{(2)}(z)=f(f(z)), \dots,$ $f^{(k)}(z)=f(f^{(k-1)}(z)) (k>1,k\in \mathbb{N}), \dots$ We call $c$ an $n$-period-point of $f$ if $c$ ($c\in S$) and $n$ ($n\in\mathbb{N}$) satisfy: $f^{(1)}(c) \not=c, f^{(2)}(c) \not=c, f^{(3)}(c) \not=c, \dots, f^{(n-1)}(c) \not=c, f^{(n)}(c)=c$. Suppose that $f(z)=z^m$ ($z\in S; m>1, m\in \mathbb{N}$), find the number of $1989$-period-point of $f$.

Given a triangle $ABC$, points $D,E,F$ lie on sides $BC,CA,AB$ respectively. Moreover, the radii of incircles of $\triangle AEF, \triangle BFD, \triangle CDE$ are equal to $r$. Denote by $r_0$ and $R$ the radii of incircles of $\triangle DEF$ and $\triangle ABC$ respectively. Prove that $r+r_0=R$.

Given $1989$ points in the space, any three of them are not collinear. We divide these points into $30$ groups such that the numbers of points in these groups are different from each other. Consider those triangles whose vertices are points belong to three different groups among the $30$. Determine the numbers of points of each group such that the number of such triangles attains a maximum.

Find all functions $f:(1,+\infty) \rightarrow (1,+\infty)$ that satisfy the following condition: for arbitrary $x,y>1$ and $u,v>0$, inequality $f(x^uy^v)\le f(x)^{\dfrac{1}{4u}}f(y)^{\dfrac{1}{4v}}$ holds.