 Given S = 1 + 1/(1 + 1/3)+1/(1 + 1/3 + 1/6) + ... + 1/ (1 + 1/3 + 1/6 + ... + 1/1993006),
where the denominators contain partial sums of the sequence of reciprocals of triangular
numbers (i.e. k = n(n + 1)=2 for n = 1, 2, ... , 1996). Prove that S > 1001.
 Find an integer n, where 100 ≤ n ≤ 1997, such that (2^n + 2)/n is also an integer.
 Let ABC be a triangle inscribed in a circle and let l_{a} =m_{a}/M_{a}, l_{b} =m_{b}/M_{b}, l_{c} =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
l_{a}/sin^{2} A + l_{b}/sin^{2} B + l_{c}sin^{2} C ≥ 3 .
 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 ≥ 3), a perpendicular line is drawn to meet A_{n2}A_{n1} 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_{n2}A_{n1}A_{n}, n ≥ 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}, ... , A_{n}, (n ≥ 3) are seated in a circle and that A_{i} has a_{i }objects such that
a_{1} + a_{2} + ... + 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_{i1} 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 ?
