3rd Junior Balkan Mathematical Olympiad Problems 1999

1.  a, b, c are distinct reals and there are reals x, y such that a³ + ax + y = 0, b³ + bx + y = 0 and c³ + cx + y = 0. Show that a + b + c = 0.

2.  Let a_n = 2³ⁿ + 3⁶ⁿ⁺² + 5⁶ⁿ⁺². Find gcd(a₀, a₁, a₂, ... , a₁₉₉₉).

3.  A square has side 20. S is a set of 1999 points inside the square and the 4 vertices. Show that we can find three points in S which form a triangle with area ≤ 1/10.

4.  The triangle ABC has AB = AC. D is a point on the side BC. BB' is a diameter of the circumcircle of ABD, and CC' is a diameter of the circumcircle of ACD. M is the midpoint of B'C'. Show that the area of BCM is independent of the position of D. 

Solutions

Problem 1

a, b, c are distinct reals and there are reals x, y such that a³ + ax + y = 0, b³ + bx + y = 0 and c³ + cx + y = 0. Show that a + b + c = 0.

Solution

Subtracting the first two equations and dividing by (a - b) we get (a² + ab + b²) + x = 0. Similarly we get (b² + bc + c²) + x = 0. Subtracting we get (a - c)(a + b + c) = 0. Hence a + b + c = 0.

or more elegantly, thanks to Suat Namli

a,b,c are the roots of the polynomial p(z) = z³ + xz + y, so their sum is the coefficient of z2, which is 0. 

Problem 2

Let a_n = 2³ⁿ + 3⁶ⁿ⁺² + 5⁶ⁿ⁺². Find gcd(a₀, a₁, a₂, ... , a₁₉₉₉).

Answer

7

Solution

a₀ = 35, so we need only consider whether 5 and 7 must be divisors. Consider first 5. Consider a₁. 2³ⁿ = 8 = 3 mod 5. 3⁶ⁿ⁺² = 3⁸ = 1 mod 5, and 5⁶ⁿ⁺² = 0 mod 5, so a₁ = 4 mod 5, and is not divisible by 5.

Now consider 7. We have 2³ⁿ = 1ⁿ = 1 mod 7. Since 3⁶ = 5⁶ = 1 mod 7 (Fermat's little theorem), we have 3⁶ⁿ⁺² = 2 mod 7, and 5⁶ⁿ⁺² = 4 mod 7, so a_n = 0 mod 7.

Thanks to Suat Namli

Problem 4

The triangle ABC has AB = AC. D is a point on the side BC. BB' is a diameter of the circumcircle of ABD, and CC' is a diameter of the circumcircle of ACD. M is the midpoint of B'C'. Show that the area of BCM is independent of the position of D.

Solution

We show that M must lie on the line through A parallel to BC.

Let O be the center of the circumcircle of ABD and O' the center of the other circumcircle. Let X be the midpoint of AD. Take coordinates with origin the midpoint of BC, x-axis along BC and y-axis through A. It is sufficient to show that the y-coordinates of O and O' sum to that of A, or equivalently that their mean is the same as the y-coordinate of X. But it is easy to see that the x-coordinate of X less that of O equals the x-coordinate of O' less that of X. Hence the y-coordinate differences must also be equal.