29th Vietnamese Mathematical Olympiad 1991 Problems
A1. Find all real-valued functions f(x) on the reals such that f(xy)/2 + f(xz)/2 - f(x) f(yz) ≥ 1/4 for all x, y, z.
A2. For each positive integer n and odd k > 1, find the largest number N such that 2N divides kn - 1.
A3. The lines L, M, N in space are mutually perpendicular. A variable sphere passes through three fixed points A on L, B on M, C on N and meets the lines again at A', B', C'. Find the locus of the midpoint of the line joining the centroids of ABC and A'B'C'.
B1. 1991 students sit in a circle. Starting from student A and counting clockwise round the remaining students, every second and third student is asked to leave the circle until only one remains. (So if the students clockwise from A are A, B, C, D, E, F, ... , then B, C, E, F are the first students to leave.) Where was the surviving student originally sitting relative to A?
B2. The triangle ABC has centroid G. The lines GA, GB, GC meet the circumcircle again at D, E, F. Show that 3/R ≤ 1/GD + 1/GE + 1/GF ≤ √3 (1/AB + 1/BC + 1/CA), where R is the circumradius.
B3. Show that x2y/z + y2z/x + z2x/y ≥ x2 + y2 + z2 for any non-negative reals x, y, z. [This is false, (1,2,3), (1,1,1), (1,2,8) give >, =, < . Does anyone know the correct question?]
Answer
f(x) = 1/2 for all x
Solution
Put x = y = z = 0, then (1/2 - f(0))2 ≤ 0, so f(0) = 1/2. Put z = 0, then f(xy) ≥ f(x) for all x,y. Taking x = 1 we get f(x) ≥ f(1) for all x. Taking y = 1/x we get f(1) ≥ f(x) for all x except possibly x = 0, so f(x) = f(1) for all x except possibly x = 0. But putting x = y = z = 1 we get (1/2 - f(1))2 ≤ 0, so f(1) = 1/2. Hence f(x) = 1/2 for all x.
Thanks to Suat Namli
B1. 1991 students sit in a circle. Starting from student A and counting clockwise round the remaining students, every second and third student is asked to leave the circle until only one remains. (So if the students clockwise from A are A, B, C, D, E, F, ... , then B, C, E, F are the first students to leave.) Where was the surviving student originally sitting relative to A?
B2. The triangle ABC has centroid G. The lines GA, GB, GC meet the circumcircle again at D, E, F. Show that 3/R ≤ 1/GD + 1/GE + 1/GF ≤ √3 (1/AB + 1/BC + 1/CA), where R is the circumradius.
B3. Show that x2y/z + y2z/x + z2x/y ≥ x2 + y2 + z2 for any non-negative reals x, y, z. [This is false, (1,2,3), (1,1,1), (1,2,8) give >, =, < . Does anyone know the correct question?]
Solution
29th VMO 1991
Problem A1 Find all real-valued functions f(x) on the reals such that f(xy)/2 + f(xz)/2 - f(x) f(yz) ≥ 1/4 for all x, y, z.Answer
f(x) = 1/2 for all x
Solution
Put x = y = z = 0, then (1/2 - f(0))2 ≤ 0, so f(0) = 1/2. Put z = 0, then f(xy) ≥ f(x) for all x,y. Taking x = 1 we get f(x) ≥ f(1) for all x. Taking y = 1/x we get f(1) ≥ f(x) for all x except possibly x = 0, so f(x) = f(1) for all x except possibly x = 0. But putting x = y = z = 1 we get (1/2 - f(1))2 ≤ 0, so f(1) = 1/2. Hence f(x) = 1/2 for all x.
Thanks to Suat Namli
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Vietnam Mathematical Olympiad
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