17th Vietnamese Mathematical Olympiad 1979 Problems

17th Vietnamese Mathematical Olympiad 1979 Problems

A1.  Show that for all x > 1 there is a triangle with sides, x⁴ + x³ + 2x² + x + 1, 2x³ + x² + 2x + 1, x⁴ - 1.

A2.  Find all real numbers a, b, c such that x³ + ax² + bx + c has three real roots α,β,γ (not necessarily all distinct) and the equation x³ + a³x² + b³x + c³ has roots α³, β³, γ³. 

A3.  ABC is a triangle. Find a point X on BC such that area ABX/area ACX = perimeter ABX/perimeter ACX.

B1.  For each integer n > 0 show that there is a polynomial p(x) such that p(2 cos x) = 2 cos nx. 

B2.  Find all real numbers k such that x² - 2 x [x] + x - k = 0 has at least two non-negative roots. B3.  ABCD is a rectangle with BC/AB = √2. ABEF is a congruent rectangle in a different plane. Find the angle DAF such that the lines CA and BF are perpendicular. In this configuration, find two points on the line CA and two points on the line BF so that the four points form a regular tetrahedron. 

Solution

17th VMO 1979

Problem A1

Show that for all x > 1 there is a triangle with sides, x⁴ + x³ + 2x² + x + 1, 2x³ + x² + 2x + 1, x⁴ - 1.

Solution

Put a = x⁴ + x³ + 2x² + x + 1, b = 2x³ + x² + 2x + 1, c = x⁴ - 1. Then obviously a > c. Also a - b = (x⁴ - x³) + (x² - x) > 0. But a - b = x⁴ - (x³ - x²) - x < c. So a is the longest side but shorter than b + c. That is sufficient to ensure that there is a triangle with sides a, b, c.

Thanks to Suat Namli 

17th VMO 1979

Problem A2

Find all real numbers a, b, c such that x³ + ax² + bx + c has three real roots α,β,γ (not necessarily all distinct) and the equation x³ + a³x² + b³x + c³ has roots α³, β³, γ³.

Answer

(a,b,c) = (h,k,hk), where k ≤ 0

Solution

We have -a = α + β + γ, b = αβ + βγ + γα, -c = αβγ. We need -a³ = α³ + β³ + γ³. Since -a³ = (α + β + γ)³ = α³ + β³ + γ³ + 3(αβ² + ... ) + 6αβγ = -a³ - 3ab + 3c, we require c = ab.

Now if c = ab, the cubic factorises as (x+a)(x²+b), so we get three real roots iff b ≤ 0. But if b ≤ 0 and c = ab, then the roots are -a, ±√(-b). So the equation with roots α³, β³, γ³ is (x + a³)(x² + b³) = x³ + a³x² + b³x + a³b³. Thus c = ab, b ≤ 0 is both a necessary and a sufficient condition. 

17th VMO 1979

Problem B1

For each integer n > 0 show that there is a polynomial p(x) such that p(2 cos x) = 2 cos nx.

Solution

Induction on n. Obvious for n = 1. For n = 2, we have 2 cos 2x = 4 cos²x - 2 = (2 cos x)² - 2, so it is true for n = 2. Now suppose it is true for n and n+1. Then 2 cos(n+1) cos x = cos(n+2)x + cos nx, so 2 cos(n+2) x = (2 cos(n+1)x)(2 cos x) - (2 cos nx), and so it is true for n+2.