Given: \[ S = 1 + \frac{1}{1 + \frac{1}{3}} + \frac{1}{1 + \frac{1}{3} + \frac{1} {6}} + \cdots + \frac{1}{1 + \frac{1}{3} + \frac{1}{6} + \cdots + \frac{1} {1993006}} \] where the denominators contain partial sums of the sequence of reciprocals of triangular numbers (i.e. $k=\frac{n(n+1)}{2}$ for $n = 1$, $2$, $\ldots$,$1996$). Prove that $S>1001$.

Find an integer $n$, where $100 \leq n \leq 1997$, such that \[ \frac{2^n+2}{n} \] is also an integer.

Let $ABC$ be a triangle inscribed in a circle and let \[ l_a = \frac{m_a}{M_a} \ , \ \ l_b = \frac{m_b}{M_b} \ , \ \ l_c = \frac{m_c}{M_c} \ , \] where $m_a$,$m_b$, $m_c$ are the lengths of the angle bisectors (internal to the triangle) and $M_a$, $M_b$, $M_c$ are the lengths of the angle bisectors extended until they meet the circle. Prove that \[ \frac{l_a}{\sin^2 A} + \frac{l_b}{\sin^2 B} + \frac{l_c}{\sin^2 C} \geq 3 \] and that equality holds iff $ABC$ is an equilateral triangle.

Triangle $A_1 A_2 A_3$ has a right angle at $A_3$. A sequence of points is now defined by the following iterative process, where $n$ is a positive integer. From $A_n$ ($n \geq 3$), a perpendicular line is drawn to meet $A_{n-2}A_{n-1}$ at $A_{n+1}$. (a) Prove that if this process is continued indefinitely, then one and only one point $P$ is interior to every triangle $A_{n-2} A_{n-1} A_{n}$, $n \geq 3$. (b) Let $A_1$ and $A_3$ be fixed points. By considering all possible locations of $A_2$ on the plane, find the locus of $P$.

Suppose that $n$ people $A_1$, $A_2$, $\ldots$, $A_n$, ($n \geq 3$) are seated in a circle and that $A_i$ has $a_i$ objects such that \[ a_1 + a_2 + \cdots + a_n = nN \] where $N$ is a positive integer. In order that each person has the same number of objects, each person $A_i$ is to give or to receive a certain number of objects to or from its two neighbours $A_{i-1}$ and $A_{i+1}$. (Here $A_{n+1}$ means $A_1$ and $A_n$ means $A_0$.) How should this redistribution be performed so that the total number of objects transferred is minimum?