3rd Balkan Mathematical Olympiad Problems 1986

3rd Balkan Mathematical Olympiad Problems 1986

A1.  A line through the incenter of a triangle meets the circumcircle and incircle in the points A, B, C, D (in that order). Show that AB·CD ≥ BC²/4. When do you have equality?

A2.  A point is chosen on each edge of a tetrahedron so that the product of the distances from the point to each end of the edge is the same for each of the 6 points. Show that the 6 points lie on a sphere.

A3.  The integers r, s are non-zero and k is a positive real. The sequence a_n is defined by a₁ = r, a₂ = s, a_n+2 = (a_n+1² + k)/a_n. Show that all terms of the sequence are integers iff (r² + s² + k)/(rs) is an integer.

A4.  A point P lies inside the triangle ABC and the triangles PAB, PBC, PCA all have the same area and the same perimeter. Show that the triangle is equilateral. If P lies outside the triangle, show that the triangle is right-angled.

Solutions

3rd Balkan 1986 Problem 1

A line through the incenter of a triangle meets the circumcircle and incircle in the points A, B, C, D (in that order). Show that AB·CD ≥ BC²/4. When do you have equality?

Solution: Answer: equality when the line passes through the circumcenter.

Let M be the midpoint of AD, r the inradius, O the circumcenter and R the circumradius. We have AB.CD - r² = (MA + MI - r)(MA - MI - r) - r² = (MA - r)² - MI² - r² = MA² - 2r MA - MI². Now MI² = OI² - OM². By Euler's formula (a well-known result, see IMO 62/6 for a proof) we have OI² = R² - 2rR. Also OM² = R² - MA². Hence AB.CD - r² = (R² - OM²) - 2r MA - (R² - 2rR - OM²) = 2r(R - MA) ≥ 0, with equality iff R = MA and hence the line passes through O.

3rd Balkan 1986 Problem 2

A point is chosen on each edge of a tetrahedron so that the product of the distances from the point to each end of the edge is the same for each of the 6 points. Show that the 6 points lie on a sphere.

Solution

Let the center of the circumsphere be O and its radius R. Let an edge be AB and a point on it P. Then it is well-known that PA·PB = R² - OP² (see below). So the six points given must all be the same distance from O. Hence they lie on a sphere center O.

[If chords AB, CD of a circle center O intersect at P, then triangles APC, BPD are similar, so PA/PC = PD/PB and hence PA·PB = PC·PD. If we take CD to be the chord perpendicular to OP, then angle OPD = 90^o, so PC = PD and PC² = OC² - OP². Hence PA·PB = OA² - OP².] 

3rd Balkan 1986 Problem 3

The integers r, s are non-zero and k is a positive real. The sequence a_n is defined by a₁ = r, a₂ = s, a_n+2 = (a_n+1² + k)/a_n. Show that all terms of the sequence are integers iff (r² + s² + k)/(rs) is an integer.

Solution

Let t = (r² + s² + k)/(rs). We show by induction that a_n+2 = t a_n+1 - a_n. We have a₃ = (s² + k)/r = st - r = t a₂ - a₁, so the result is true for n = 1. Suppose it is true for n - 1. Then a_n+2 = (a_n+1² + k)/a_n = a_n+1(t a_n - a_n-1)/a_n + k/a_n = t a_n+1 - a_n+1a_n-1/a_n + k/a_n = t a_n+1 - (a_n² + k)/a_n + k/a_n = t a_n+1 - a_n. So the result is true for n. We are given that r and s are integers. So if t is also an integer, then a trivial induction shows that all a_n are integers.

Now suppose that all the a_n are integers but that t is not. So t = u/v with u and v > 1 relatively prime. Since a₃ = u/v a₂ - a₁ and the a_i are integral, v must divide a₂. A simple induction shows that vⁿ must divide a_n+1. Now take N sufficiently large that v^N > k. Then k = a_N+2a_N - a_N+1². But the rhs is divisible by v^N and the lhs is not. Contradiction. So t must be an integer. 

3rd Balkan 1986 Problem 4

A point P lies inside the triangle ABC and the triangles PAB, PBC, PCA all have the same area and the same perimeter. Show that the triangle is equilateral. If P lies outside the triangle, show that the triangle is right-angled.

Solution

Suppose P is inside the triangle. Since area PAB = area PBC, BP must be the median (for if M is the midpoint of AC, then area PAB = 1/2 PB·AM sin AMB, area PBC = 1/2 PB·CM sin AMB). Similarly AP must be the median, so P must be the centroid. Now suppose ∠ABC < ∠ACB. Then AB > AC and PB > PC, so perimeter PAB > perimeter PAC. Hence ∠ABC = ∠ACB. Similarly, the other angles are equal, so the triangle must be equilateral. It is obvious that the centroid then satisfies the conditions.

Suppose P is outside the triangle. Relabeling if necessary, we may assume that A and B lie on opposite sides of PC. Hence A and C lie on the same side of PB (this is where we need P outside the triangle). Since area PBC = area PBA, A and C must be the same distance from PB, so AC is parallel to PB. Similarly PA is parallel to BC. So PACB is a parallelogram. But now the perimeter requirement implies that PC = AB and hence the parallelogram is a rectangle and angle ACB = 90^o.