IMO 2000 SL A2

Origin: GBR | Category: Algebra

Problem

Let a, b, c be positive integers satisfying the conditions b > 2a and c > 2b. Show that there exists a real number t with the property that all the three numbers ta, tb, tc have their fractional parts lying in the interval (1/3, 2/3].

Solution

We note that {ta} lies in
1 3, 2 if and only if there is an integer k such that k + 1 3 < ta \leqk + 2 3, i.e., if and only if t \inIk =  k+1/3 a , k+2/3 a  for some k. Similarly, t should belong to the sets Jm =  m+1/3 b , m+2/3 b  and Kn =  n+1/3 c , n+2/3 c  for some m, n. We have to show that Ik \capJm \capKn is nonempty for some integers k, m, n. The intervals Kn are separated by a distance 3c, and since 3c < 3b, each of the intervals Jm intersects at least one of the Kn’s. Hence it is enough to prove that Jm \subsetIk for some k, m. Let um and vm be the left and right endpoints of Jm. Since avm = aum + a 3b < aum + 1 6, it will suffice to show that there is an integer m such that the fractional part of aum lies in / 1 3, 1 . Let a = d\alpha, b = d\beta, gcd(\alpha, \beta) = 1. Setting m = dµ we obtain that aum = a m+1/3 b = \alpham d\beta + \alpha 3\beta = \alphaµ \beta + \alpha 3\beta . Since \alphaµ gives all possible residues modulo \beta, every term of the arithmetic progression j \beta + \alpha 3\beta (j \inZ) has its fractional part equal to the fractional part of some aum. Now for \beta \geq6 the progression step is 1 \beta \leq1 6, so at least one of the aum has its fractional part in [1/3, 1/2]. If otherwise \beta \leq5, the only irreducible fractions \alpha \beta that satisfy 2\alpha < \beta are 1 3, 4, 5, 5; hence one can take m to be 1, 1, 2, 3 respectively. This justifies our claim.