13th Vietnamese Mathematical Olympiad 1975 Problems

13th Vietnamese Mathematical Olympiad 1975 Problems

A1.  The roots of the equation x³ - x + 1 = 0 are a, b, c. Find a⁸ + b⁸ + c⁸.


A2.  Find all real x which satisfy (x³ + a³)/(x + a)³ + (x³ + b³)/(x + b)³ + (x³ + c³)/(x + c)³ + 3(x - a)(x - b)(x - c)/( 2(x + a)(x + b)(x + c) ) = 3/2.

A3.  ABCD is a tetrahedron. The three edges at B are mutually perpendicular. O is the midpoint of AB and K is the foot of the perpendicular from O to CD. Show that vol KOAC/vol KOBD = AC/BD iff 2·AC·BD = AB².


B1.  Find all terms of the arithmetic progression -1, 18, 37, 56, ... whose only digit is 5.

B2.  Show that the sum of the maximum and minimum values of the function tan(3x)/tan³x on the interval (0, π/2) is rational.


B3.  L is a fixed line and A a fixed point not on L. L' is a variable line (in space) through A. Let M be the point on L and N the point on L' such that MN is perpendicular to L and L'. Find the locus of M and the locus of the midpoint of MN.

Solution

13th VMO 1975

Problem A1 The roots of the equation x³ - x + 1 = 0 are a, b, c. Find a⁸ + b⁸ + c⁸.
Answer
10
Solution
We have a+b+c = 0, ab+bc+ca = -1, abc = -1.
a²+b²+c² = (a+b+c)² - 2(ab+bc+ca) = 2. Similarly, a²b²+b²c²+c²a² = (ab+bc+ca)² - 2abc(a+b+c) = 1. Hence a⁴+b⁴+c⁴ = (a²+b²+c²)² - 2(a²b²+b²c²+c²a²) = 2. Similarly, a⁴b⁴+b⁴c⁴+c⁴a⁴ = (a²b²+b²c²+c²a²)² - 2a²b²c²(a²+b²+c²) = 1 - 4 = -3. Finally a⁸+b⁸+c⁸ = (a⁴+b⁴+c⁴)² - 2(a⁴b⁴+b⁴c⁴+c⁴a⁴) = 4 + 6 = 10.
Thanks to Suat Namli

13th VMO 1975

Problem B1 Find all terms of the arithmetic progression -1, 18, 37, 56, ... whose only digit is 5.
Answer
5...5 with 18k+5 digits for k = 0, 1, 2, 3, ...
Solution
The general term is 19k - 1 and this must equal 5(10ⁿ - 1)/9 for some n. Hence 5·10ⁿ = 171k - 4 = -4 mod 19. So we certainly require 10ⁿ = 3 mod 19 and hence n = 5 mod 18. Conversely if n = 5 mod 18, then 10ⁿ = 3 mod 19, so 5·10ⁿ = -4 mod 19, so 5·10ⁿ = 19h - 4 for some h. So 5(10ⁿ - 1) = 19h - 9. Now lhs and 9 are divisible by 9, so 19h must be divisible by 9, so h must be divisible by 9. Put h = 9k and we get 19k - 1 = 5(10ⁿ - 1)/9.
Thanks to Suat Namli