16th Balkan Mathematical Olympiad Problems 1999

16th Balkan Mathematical Olympiad Problems 1999

A1.  O is the circumcenter of the triangle ABC. XY is the diameter of the circumcircle perpendicular to BC. It meets BC at M. X is closer to M than Y. Z is the point on MY such that MZ = MX. W is the midpoint of AZ. Show that W lies on the circle through the midpoints of the sides of ABC. Show that MW is perpendicular to AY.

A2.  p is an odd prime congruent to 2 mod 3. Prove that at most p-1 members of the set {m² - n³ - 1: 0 < m, n < p} are divisible by p.

A3.  ABC is an acute-angled triangle area 1. Show that the triangle whose vertices are the feet of the perpendiculars from the centroid to AB, BC, CA has area between 4/27 and 1/4.

A4.  0 = a₁, a₂, a₃, ... is a non-decreasing, unbounded sequence of non-negative integers. Let the number of members of the sequence not exceeding n be b_n. Prove that (x₀ + x₁ + ... + x_m)(y₀ + y₁ + ... + y_n) ≥ (m + 1)(n + 1). 

Solutions

16th Balkan 1999 Problem 1

O is the circumcenter of the triangle ABC. XY is the diameter of the circumcircle perpendicular to BC. It meets BC at M. X is closer to M than Y. Z is the point on MY such that MZ = MX. W is the midpoint of AZ. Show that W lies on the circle through the midpoints of the sides of ABC. Show that MW is perpendicular to AY.

Solution

XW and AM are medians of AZX, so they meet at the centroid G of AZX and AG = 2/3 AM. But AM is also a median of ABC, so G is also the centroid of ABC. Now consider the similarity, center G, which takes X to W. It takes A, B, C to the midpoints of the opposite sides and hence the circumcircle of ABC to the circle through the midpoints. But X lies on the circumcircle of ABC, so W lies on the circle through the midpoints.

The similarity takes MW to AX, but the similarity takes lines to lines which are parallel. AX is perpendicular to AY (because XY is a diameter of the circumcircle through AXY). Hence MW is also perpendicular to AY. 

16th Balkan 1999 Problem 2

p is an odd prime congruent to 2 mod 3. Prove that at most p-1 members of the set {m² - n³ - 1: 0 < m, n < p} are divisible by p.

Solution

We show that p has p distinct cubic residues if it is an odd prime congruent to 2 mod 3. Let k be a primitive root of p. So any non-zero residue of p can be written as kⁿ for some n in the range 1, 2, ... , p-1, and kⁿ = k^m mod p iff p-1 divides m-n. Now we claim that, for distinct m, n in the range 1, 2, ... , p-1, we have k³ⁿ and k^3m distinct mod p. For if they are equal then p-1 divides 3(m-n). But p-1 is not a multiple of 3, so p-1 must divide m-n. But m-n is in the range 1, ... , p-2 so it cannot be divisible by p-1.

So the values -1³, -2³, ... , -(p-1)³ are all incongruent mod p. Thus for each m, the p-1 values m² - 1³ - 1, m² - 2³ - 1, ... , m² - (p-1)³ - 1 are all incongruent mod p. Hence at most one of them is divisible by p. So the set {m² - n³ - 1: 0 < m, n < p} has at most one member divisible by p for each m and hence at most p-1 in all. 

16th Balkan 1999 Problem 3

ABC is an acute-angled triangle area 1. Show that the triangle whose vertices are the feet of the perpendiculars from the centroid to AB, BC, CA has area between 4/27 and 1/4.

Solution

Let the centroid be G and the feet of the perpendiculars to BC, CA, AB be D, E, F respectively. Since G is 1/3 of the way along the median from the midpoint, area BGC = 1/3 area ABC = 1/3. But area BGC = a.GD/2, hence GD = 2/(3a). Similarly, GE = 2/(3b) and GF = 2/(3c). Now area GEF = (GE.GF.sin EFG)/2 = (2 sin A)/(9bc). But area ABC = (bc sin A)/2, so bc = 2/sin A. So area GEF = (sin²A)/9. Hence Area DEF = (sin²A + sin²B + sin²C)/9. So we need to show that 4/3 ≤ (sin²A + sin²B + sin²C) ≤ 9/4.

Take A ≤ B ≤ C. We have C ≤ π/2, so A + B ≥ π/2, so B ≥ π/4. Now sin²x has decreasing derivative between π/4 and π/2, so sin²B + sin²C ≤ 2 sin²((B+C)/2) = 2 cos²A/2 = cos A + 1. Hence (sin²A + sin²B + sin²C) ≤ sin²A + cos A + 1 = 2 - cos²A + cos A = 9/4 - (cos A - 1/2)² ≤ 9/4. Note that we can only have equality at the last step if cos A = 1/2 or A = π/3, which implies the triangle is equilateral. It is easy to check that in this case (sin²A + sin²B + sin²C) = 9/4.

In the other direction we have in fact that sin²A + sin²B + sin²C >= 2, which is significantly stronger than what we need. This is slightly awkward to prove. Take C = π/2 - 2y, B = π/4 + y + x, A = π/4 + y - x. So we are assuming that x, y >= 0, and x + 3y ≤ π/4. Now sin²A + sin²B + sin²C = (1/√2 cos(x - y) - 1/√2 sin(x - y) )² + (1/√2 cos(x + y) + 1/√2 sin(x + y) )² + cos²2y = 1 - cos(x - y)sin(x - y) + cos(x + y)sin(x + y) + cos²2y = 1 + cos²2y - 1/2 sin2(x - y) + 1/2 sin2(x + y) = 1 + cos²2y + cos 2x sin 2y = 2 + sin 2y (cos 2x - sin 2y). But sin 2y ≥ 0 and sin 2y < sin 6y ≤ sin(π/2 - 2x) = cos 2x. Hence sin 2y (cos 2x - sin 2y) ≥ 0 (with equality iff y = 0, so that the triangle is right-angled).

Comment. It is curious that the question made the inequality weaker than necessary. It should have been "between 2/9 and 1/4".

16th Balkan 1999 Problem 4

0 = a₁, a₂, a₃, ... is a non-decreasing, unbounded sequence of non-negative integers. Let the number of members of the sequence not exceeding n be b_n. Prove that (x₀ + x₁ + ... + x_m)(y₀ + y₁ + ... + y_n) ≥ (m + 1)(n + 1).

Solution

If x_m = 0, then y₀ ≥ m+1 and so all of y₀, y₁, ... , y_n ≥ m+1, so the inequality holds. If x_m > 0, then let x_k be the first non-zero term and suppose x_k = h > 0. Reduce x_k to 0. Then ∑x_i is reduced by h. Each of y₀, y₁, ... , y_h-1 is increased by 1 and the other y_j are unchanged, so ∑y_j is increased by at most h. Hence ∑x_i + ∑y_j is not increased. Repeat until we reach x_m = 0. Then the resulting inequality is true. We have not increased the lhs and the rhs is unchanged, so the orignal inequality was true.