7th Junior Balkan Mathematical Olympiad Problems 2003

1.  Let A = 44...4 (2n digits) and B = 88...8 (n digits). Show that A + 2B + 4 is a square.

2.  A₁, A₂, ... , A_n are points in the plane, so that if we take the points in any order B₁, B₂, ... , B_n, then the broken line B₁B₂...B_n does not intersect itself. What is the largest possible value of n?

3.  ABC is a triangle. D is the midpoint of the arc BC not containing A. Similarly E, F. DE meets BC at G and AC at H. M is the midpoint of GH. DF meets BC at I and AB at J, and N is the midpoint of IJ. Find the angles of DMN in terms of the angles of ABC. AD meets EF at P. Show that the circumcenter of DMN lies on the circumcircle of PMN.

4.  Show that (1+x²)/(1+y+z²) + (1+y²)/(1+z+x²) + (1+z²)/(1+x+y²) ≥ 2 for reals x, y, z > -1. 

Solutions

Problem 1

Let A = 44...4 (2n digits) and B = 88...8 (n digits). Show that A + 2B + 4 is a square.

Solution

We show that the square root is D = 6...68 (with n-1 6s). We have D = 6(10ⁿ - 1)/9 + 2 = (2/3)(10ⁿ + 2), so D² = (4/9)(10²ⁿ + 4·10ⁿ + 4) = (4/9)(10²ⁿ - 1) + 2(8/9)(10ⁿ - 1) + 4 = A + 2B + 4. 

Problem 2

A₁, A₂, ... , A_n are points in the plane, so that if we take the points in any order B₁, B₂, ... , B_n, then the broken line B₁B₂...B_n does not intersect itself. What is the largest possible value of n?

Answer

4

Solution

We cannot have three points collinear, because if B lies on the segment AC, then the broken line ...ACB... intersects itself. If we take any four points, then their convex hull must be a triangle. This works for 4 points. Suppose we have 5 points A, B, C, D, E. Then we can take D to be inside the triangle ABC.

But there is now nowhere to put E. If it is in 1, then ABDE is self-intersecting. Similarly, in 2, ACDE is self-intersecting, in 3 BCDE is self-intersecting, in 4 ADCE is self intersecting and so on.

Thanks to Suat Namli

Problem 3

ABC is a triangle. D is the midpoint of the arc BC not containing A. Similarly E, F. DE meets BC at G and AC at H. M is the midpoint of GH. DF meets BC at I and AB at J, and N is the midpoint of IJ. Find the angles of DMN in terms of the angles of ABC. AD meets EF at P. Show that the circumcenter of DMN lies on the circumcircle of PMN.

Answer

D = B/2 + C/2, M = A/2 + B/2, N = C/2 + A/2

Solution

∠BFD = A/2, ∠FBC = B + ∠FBA = B + C/2, so ∠BIJ = 180^o - A/2 - (B + C/2) = A/2 + C/2. Similarly, ∠BJI = A/2 + C/2. So BIJ is isosceles. Hence BN is the angle bisector of B. Similarly, CM is the angle bisector of C. So they meet on AD (the angle bisector of A) at the incenter I. Then ∠DNI = ∠DMI = 90^o, so DI is the diameter of the circumcircle of DMN. Hence ∠DNM = ∠DIM = 90^o - ∠IDM = 90^o - B/2 = A/2 + C/2. Similarly, ∠DMN = A/2 + B/2. Hence ∠MDN = B/2 + C/2.

Let O be the midpoint of DI, so that O is the circumcenter of DMN and we have to show that it lies on the circumcircle of PMN. I is the orthocenter of DEF and it is a well-known result that the midpoints of the segments joining the orthocenter to the vertices lie on the circumcircle of the orthic triangle (known as the nine-point circle, because it also contains the midpoints of the sides of DEF).

Thanks to Cristian Ilac

Problem 3

Show that (1+x²)/(1+y+z²) + (1+y²)/(1+z+x²) + (1+z²)/(1+x+y²) ≥ 2 for reals x, y, z > -1.

Solution

Thanks to Michael Bolton

If x < 0, then replacing x by -x, does not change the first two terms and reduces the third term, so we can assume x, y, z ≥ 0.

Since (x-1)2 ≥ 0 with equality iff x = 1, 3(x² + 1) ≥ 2(x² + x + 1) with equality iff x = 1.