6th Junior Balkan Mathematical Olympiad Problems 2002

1.  The triangle ABC has CA = CB. P is a point on the circumcircle between A and B (and on the opposite side of the line AB to C). D is the foot of the perpendicular from C to PB. Show that PA + PB = 2·PD.

2.  The circles center O₁ and O₂ meet at A and B with the centers on opposite sides of AB. The lines BO₁ and BO₂ meet their respective circles again at B₁ and B₂. M is the midpoint of B₁B₂. M₁, M₂ are points on the circles center O₁ and O₂ respectively such that angle AO₁M₁ = angle AO₂M₂, and B₁ lies on the minor arc AM₁ and B lies on the minor arc AM₂. Show that angle MM₁B = angle MM₂B.

3.  Find all positive integers which have exactly 16 positive divisors 1 = d₁ < d₂ < ... < d₁₆ such that the divisor d_k, where k = d₅, equals (d₂ + d₄) d₆.

4.  Show that 2/( b(a+b) ) + 2/( c(b+c) ) + 2/( a(c+a) ) ≥ 27/(a+b+c)² for positive reals a, b, c. 

Solutions

Problem 1

The triangle ABC has CA = CB. P is a point on the circumcircle between A and B (and on the opposite side of the line AB to C). D is the foot of the perpendicular from C to PB. Show that PA + PB = 2·PD.

Solution

Take Q on the ray PB so that PD = DQ. Then CA = CB, CP = CQ, ∠CAP = ∠CBQ, so triangles CAP and CBQ are congruent, so AP = BQ. Hence 2·PD = PQ = PB + PA.

Thanks to Cristian Ilac

Problem 3

Find all positive integers which have exactly 16 positive divisors 1 = d₁ < d₂ < ... < d₁₆ such that the divisor d_k, where k = d₅, equals (d₂ + d₄) d₆.

Answer

2002

Solution

If d₂ > 2, then all divisors are odd. But (d₂ + d₄) is a divisor and must be even. Contradiction. So d₂ = 2.

If d₃ = 3, then 6 divides N (where we write the number as N), so either d₄ = 6, or d₅ = 6 or d₆ = 6. If d₄ = 6, then (d₂ + d₄) = 8 is a divisor, so 4 is a divisor, so d₄ = 4. Contradiction. If d₅ = 6, then we are given that d₆ = (d₂ + d₄) d₆ > d₆. Contradiction. If d₆ = 6, then d₄ = 4, d₅ = 5. But we are given that d₅ = (d₂ + d₄) d₆ > d₅. Contradiction. So d₃ > 3. So 3 is not a factor of N.

We are told that (d₂ + d₄) d₆ is a factor, so (d₂ + d₄) = d₄ + 2 must be a factor. If d₄ + 1 is also a factor, then we have three consecutive factors. One of them must be a multiple of 3. But we have just shown that 3 does not divide N. So d₄ + 1 is not a factor. Hence d₅ = d₄ + 2. Neither d₄ nor d₅ are multiples of 3. So d₅ must be 1 greater than a multiple of 3.

We have d₅ = k and that d_k is a factor. So in particular k ≤ 16 (since there are only 16 factors). Thus d₅ = 1, 4, 7, 10, 13 or 16. Obviously d₅ ≥ 5, so 1 and 4 are too small.

If d₅ = 7, then since 3 and 6 are not factors we must have d₃ = 4, d₄ = 5. So 10 and 14 are factors. So d₇ ≤ 10. But we are given that d₇ = (d₂ + d₄) d₆ > (2 + 5) 7 = 49. Contradiction.

If d₅ = 10, then d₄ = 8. So 4 and 5 are also factors, but only d₃ is available. Contradiction.

If d₅ = 16, then d₄ = 14. So 4, 7 and 8 are also factors, but only d₃ is available. Contradiction.

The only remaining possibility is d₅ = 13. That implies d₄ = 11. So d₃ is the only factor between 4 and 10. So it cannot be 5, 6, 8, 9 or 10. So it must be 4 or 7.

Suppose it is 4. N has prime factors 2, 11, 13. If it has another prime factor p, then it would have 16 factors, leaving aside multiples of 4 (because each of 2, 11, 13 and p could be in or out). Contradiction. So 2, 11, 13 are its only prime factors. If 11² or 13² divides N, then N has at least 3·3·2 = 18 factors (there are 3 possibilities for 2, 3 for one of 11/13 and 2 for the other). Contradiction. So 2³must divide N. Contradiction (we are supposed to have exhausted the factors under 10).

So the only remaining possibility is d₃ = 7. That gives N = 2002 and it is easy to check that d₁₃ = 182 = (d₂ + d₄) d₆.

Comment. This all seems rather complicated (although it did not take me long). Does anyone have a simpler solution?

Problem 4

Show that 2/( b(a+b) ) + 2/( c(b+c) ) + 2/( a(c+a) ) ≥ 27/(a+b+c)² for positive reals a, b, c.

Solution

Thanks to Michael Lipnowski

By AM/GM 1/( b(a+b) ) + 1/( c(b+c) ) + 1/( a(c+a) ) ≥ 3/(XY), where X = (abc)^1/3, Y = ((a+b)(b+c)(c+a))^1/3. Using AM/GM again gives X ≤ (a+b+c)/3 and Y ≤ 2(a+b+c)/3, so 3/XY ≥ (27/2) 1/(a+b+c)².