28th Vietnamese Mathematical Olympiad 1990 Problems

28th Vietnamese Mathematical Olympiad 1990 Problems

A1.  -1 < a < 1. The sequence x₁, x₂, x₃, ... is defined by x₁ = a, x_n+1 = ( √(3 - 3x_n²) - x_n)/2. Find necessary and sufficient conditions on a for all members of the sequence to be positive. Show that the sequence is always periodic.

A2.  n-1 or more numbers are removed from {1, 2, ... , 2n-1} so that if a is removed, so is 2a and if a and b are removed, so is a + b. What is the largest possible sum for the remaining numbers?

A3.  ABCD is a tetrahedron with volume V. We wish to make three plane cuts to give a parallelepiped three of whose faces and all of whose vertices belong to the surface of the tetrahedron. Find the intersection of the three planes if the volume of the parallelepiped is 11V/50. Can it be done so that the volume of the parallelepiped is 9V/40?

B1.  ABC is a triangle. P is a variable point. The feet of the perpendiculars from P to the lines BC, CA, AB are A', B', C' respectively. Find the locus of P such that PA PA' = PB PB' = PC PC'.

B2.  The polynomial p(x) with degree at least 1 satisfies p(x) p(2x²) = p(3x³ + x). Show that it does not have any real roots.  

B3.  Some children are sitting in a circle. Each has an even number of tokens (possibly zero). A child gives half his tokens to the child on his right. Then the child on his right does the same and so on. If a child about to give tokens has an odd number, then the teacher gives him an extra token. Show that after several steps, all the children will have the same number of tokens, except one who has twice the number.

Solution

28th VMO 1990

Problem A1

-1 < a < 1. The sequence x₁, x₂, x₃, ... is defined by x₁ = a, x_n+1 = ( √(3 - 3x_n²) - x_n)/2. Find necessary and sufficient conditions on a for all members of the sequence to be positive. Show that the sequence is always periodic.

Answer

0 < a < (√3)/2
period 2

Solution

Put x₁ = sin k. Then x₂ = (√3)/2 cos k - (1/2) sin k = sin 60^o cos k - cos 60^o sin k = sin(60^o-k). Hence x₃ = sin(60^o - 60^o + k) = x₁.

For sin k and sin(60^o-k) to be positive we need 0 < k < 60^o and hence 0 < a < (√3)/2.

Thanks to Suat Namli

28th VMO 1990

Problem B2

The polynomial p(x) with degree at least 1 satisfies p(x) p(2x²) = p(3x³ + x). Show that it does not have any real roots.

Solution

Suppose it has a positive root α, then 3α3+α is another root, which is larger than α. Proceeding, we get infinitely many roots. Contradiction. So there are no positive roots. Similarly, there are no negative roots. So the only possibility is 0. Suppose bx^m is the lowest power of x in p(x). Then the lowest power of x in p(x)p(2x²) is x^3m, but the lowest power of x in p(3x³ + x) is x^m, so m = 0 and hence p(0) = b ≠ 0, so 0 is not a root.

Thanks to Suat Namli