3rd Vietnam Mathematical Olympiad 1964 Problems

3rd Vietnam Mathematical Olympiad 1964 Problems

1.  Find cos x + cos(x + 2π/3) + cos(x + 4π/3) and sin x + sin(x + 2π/3) + sin(x + 4π/3).
2.  Draw the graph of the functions y = | x² - 1 | and y = x + | x² - 1 |. Find the number of roots of the equation x + | x² - 1 | = k, where k is a real constant.

3.  Let O be a point not in the plane p and A a point in p. For each line in p through A, let H be the foot of the perpendicular from O to the line. Find the locus of H.

4.  Define the sequence of positive integers f_n by f₀ = 1, f₁ = 1, f_n+2 = f_n+1 + f_n. Show that f_n = (aⁿ⁺¹ - bⁿ⁺¹)/√5, where a, b are real numbers such that a + b = 1, ab = -1 and a > b. 

Solutions


1. Using cos(A+B) = cos A cos B - sin A sin B, we have cos(x + 2π/3) = -(1/2) cos x + (√3)/2 sin x, cos(x + 4π/3) = -(1/2) cos x - (√3)/2 sin x. Hence cos x + cos(x + 2π/3) + cos(x + 4π/3) = 0. Similarly, sin(x + 2π/3) = -1/2 sin x + (√3)/2 cos x, sin(x + 4π/3) = -1/2 sin x - (√3)/2 cos x, so sin x + sin(x + 2π/3) + sin(x + 4π/3) = 0.

2. Answer
0 for k < -1
1 for k = -1
2 for -1 < k < 1
3 for k = 1
4 for 1 < k < 5/4
3 for k = 5/4
2 for k > 5/4




It is clear from the graph that there are no roots for k < -1, and one root for k = -1 (namely x = -1). Then for k > -1 there are two roots except for a small interval [1, 1+h]. At k = 1, there are 3 roots (x = -2, 0, 1). The upper bound is at the local maximum between 0 and 1. For such x, y = x + 1 - x2 = 5/4 - (x - 1/2)2, so the local maximum is at 5/4. Thus there are 3 roots at k = 5/4 and 4 roots for k ∈ (1, 5/4).


3. Answer: circle diameter AB, where OB is the normal to p


Let B be the foot of the perpendicular from O to p. We claim that the locus is the circle diameter AB. Any line in p through A meets this circle at one other point K (except for the tangent to the circle at A, but in that case A is obviously the foot of the perpendicular from O to the line). Now BK is perpendicular to AK, so OK is also perpendicular to AK, and hence K must be the foot of the perpendicular from O to the line.

4. Put a = (1+√5)/2, b = (1-√5)/2. Then a, b are the roots of x² - x - 1 = 0 and satisfy a + b = 1, ab = -1. We show by induction that f_n = (aⁿ⁺¹ - bⁿ⁺¹)/√5. We have f₀ = (a-b)/√5 = 1, f₁ = (a²-b²)/√5 = (a+1 - b-1)/√5 = 1, so the result is true for n = 0, 1. Finally, suppose f_n = (aⁿ⁺¹ - bⁿ⁺¹)/√5 and f_n+1 = (aⁿ⁺² - bⁿ⁺²)/√5. Then f_n+2 = f_n+1 + f_n = (1/√5)(aⁿ⁺¹(a+1) - bⁿ⁺¹(b+1) ) = (aⁿ⁺¹a² - bⁿ⁺¹b²)/√5, so the result is true for n+1.

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