36th Vietnamese Mathematical Olympiad 1998 Problems

36th Vietnamese Mathematical Olympiad 1998 Problems

A1.  Define the sequence x₁, x₂, x₃, ... by x₁ = a ≥ 1, x_n+1 = 1 + ln(x_n(x_n²+3)/(1 + 3x_n²) ). Show that the sequence converges and find the limit.

A2.  Let O be the circumcenter of the tetrahedron ABCD. Let A', B', C', D' be points on the circumsphere such that AA', BB', CC' and DD' are diameters. Let A" be the centroid of the triangle BCD. Define B", C", D" similarly. Show that the lines A'A", B'B", C'C", D'D" are concurrent. Suppose they meet at X. Show that the line through X and the midpoint of AB is perpendicular to CD.

A3.  The sequence a₀, a₁, a₂, ... is defined by a₀= 20, a₁ = 100, a_n+2 = 4a_n+1 + 5a_n + 20. Find the smallest m such that a_m - a₀, a_m+1 - a₁, a_m+2 - a₂, ... are all divisible by 1998.

B1.  Does there exist an infinite real sequence x₁, x₂, x₃, ... such that | x_n | ≤ 0.666, and | x_m - x_n | ≥ 1/(n² + n + m² + m) for all distinct m, n?

B2.  What is the minimum value of √( (x+1)² + (y-1)²) + √( (x-1)² + (y+1)²) + √( (x+2)² + (y+2)²)?

B3.  Find all positive integers n for which there is a polynomial p(x) with real coefficients such that p(x¹⁹⁹⁸ - x^-1998) = (xⁿ - x^-n) for all x.

Solution

36th VMO 1998

Problem A1 Define the sequence x₁, x₂, x₃, ... by x₁ = a ≥ 1, x_n+1 = 1 + ln(x_n(x_n²+3)/(1 + 3x_n²) ). Show that the sequence converges and find the limit.

Solution

(x-1)³ ≥ 0, so x(x²+3)/(1+3x²) ≥ 1. So the ln term is well-defined and non-negative and 1 + ln(x(x²+3)/(1+3x²)) ≥ 1. So by a trivial induction all x_n ≥ 1.

Also x² ≥ 1 implies (x²+3)/(1+3x²) ≤ 1, so x(x²+3)/(1+3x²) ≤ x and hence 1 + ln(x(x²+3)/(1+3x²)) ≤ 1 + ln x ≤ x (*). So the sequence is monotonically decreasing and bounded below by 1. So it must converge. Suppose the limit is L. Then L = 1 + ln(L(L²+3)/(1+Lx²)). But we only have equality in (*) at 1. Hence L = 1.