23rd Vietnamese Mathematical Olympiad 1985 Problems

23rd Vietnamese Mathematical Olympiad 1985 Problems

A1.  Find all integer solutions to m³ - n³ = 2mn + 8.

A2.  Find all real-valued functions f(n) on the integers such that f(1) = 5/2, f(0) is not 0, and f(m) f(n) = f(m+n) + f(m-n) for all m, n.

A3.  A parallelepiped has side lengths a, b, c. Its center is O. The plane p passes through O and is perpendicular to one of the diagonals. Find the area of its intersection with the parallelepiped.

B1.  a, b, m are positive integers. Show that there is a positive integer n such that (aⁿ - 1)b is divisible by m iff the greatest common divisor of ab and m is also the greatest common divisor of b and m.

B2.  Find all real values a such that the roots of 16x⁴ - ax³ + (2a + 17)x² - ax + 16 are all real and form an arithmetic progression.

B3.  ABCD is a tetrahedron. The base BCD has area S. The altitude from B is at least (AC + AD)/2, the altitude from C is at least (AD + AB)/2, and the altitude from D is at least (AB + AC)/2. Find the volume of the tetrahedron.

Solution

23rd VMO 1985

Problem A1

Find all integer solutions to m³ - n³ = 2mn + 8.

Answer

(m,n) = (2,0), (0,-2)

Solution

Put m = n+k. Then 3n²k+3nk²+k³ = 2n²+2nk+8, so (3k-2)n²+(3k²-2k)n+k³-8 = 0. For real solutions we require (3k²-2k)² ≥ 4(3k-2)(k³-8) or (3k-2)(32-2k²-k³) ≥ 0. The first bracket is +ve for k ≥ 1, -ve for k ≤ 0, the second is +ve for k ≤ 2, -ve for k ≥ 3. Hence k = 1 or 2.

If k = 1, then n²+n-7 = 0, which has no integer solutions. If k = 2, then 4n²+8n = 0, so n = 0 or -2.

Thanks to Suat Namli

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23rd VMO 1985

Problem A2

Find all real-valued functions f(n) on the integers such that f(1) = 5/2, f(0) is not 0, and f(m) f(n) = f(m+n) + f(m-n) for all m, n.

Answer

f(n) = 2ⁿ + 1/2ⁿ

Solution

Putting m = n = 0, we get f(0)² = 2f(0), so f(0) = 2. Putting n = 1, we get f(m+1) = 5/2 f(m) + f(m-1). That is a standard linear recurrence relation. The associated quadratic has roots 2, 1/2, so the general solution is f(n) = A 2ⁿ + B/2ⁿ. f(0) = 1 gives A + B = 2, f(1) = 5/2 gives 2A + B/2 = 5/2, so A = B = 1. It is now easy to check that this solutions satisfies the conditions.

Thanks to Suat Namli