4th Balkan Mathematical Olympiad Problems 1987

4th Balkan Mathematical Olympiad Problems 1987

A1.  f is a real valued function on the reals satisfying (1) f(0) = 1/2, (2) for some real a we have f(x+y) = f(x) f(a-y) + f(y) f(a-x) for all x, y. Prove that f is constant.

A2.  Find all real numbers x ≥ y ≥ 1 such that √(x - 1) + √(y - 1) and √(x + 1) + √(y + 1) are consecutive integers.

A3.  ABC is a triangle with BC = 1. We have (sin A/2)²³/(cos A/2)⁴⁸ = (sin B/2)²³/(cos B/2)⁴⁸. Find AC.

A4.  Two circles have centers a distance 2 apart and radii 1 and √2. X is one of the points on both circles. M lies on the smaller circle, Y lies on the larger circle and M is the midpoint of XY. Find the distance XY. 

Solutions

4th Balkan 1987 Problem 1

f is a real valued function on the reals satisfying (1) f(0) = 1/2, (2) for some real a we have f(x+y) = f(x) f(a-y) + f(y) f(a-x) for all x, y. Prove that f is constant.

Solution

Put x = y = 0. We get f(0) = 2 f(0) f(a), so f(a) = 1/2. Put y = 0, we get f(x) = f(x) f(a) + f(0) f(a-x), so f(x) = f(a-x). Put y = a-x, we get f(0) = f(x)² + f(a-x)², so f(x) = 1/2 or -1/2.

Now take any x. We have f(x/2) = 1/2 or -1/2 and f(a - x/2) = f(x/2). Hence f(x) = f(x/2 + x/2) = 2 f(x) f(a - x/2) = 1/2. 

4th Balkan 1987 Problem 2

Find all real numbers x ≥ y ≥ 1 such that √(x - 1) + √(y - 1) and √(x + 1) + √(y + 1) are consecutive integers.

Solution

Answer: there is a unique solution for each n > 3. Namely, x = √(1+ ( (7+4k-4k²)/(4-8k) )²), y = √(1+ ( (7-4k-4k²)/(4+8k) )²), where k = √( (2n-7)/(8n+4) ). This gives the integers n and n+1.

Put f(x) = √(x+1) - √(x-1). Then f(x) is strictly monotonic decreasing. f(4 1/16) = 1/2 and f(x) tends to 0 as x tends to infinity. So since x ≥ y, we have f(x) ≤ 1/2 and f(y) ≥ 1/2. For any k in the range 0 ≤ k ≤ 1/2, there is a unique x such that f(x) = 1/2 - k. Indeed we may find this explicitly. Squaring etc gives x = √(1+ ( (7+4k-4k²)/(4-8k) )²). Then since √(x - 1) + √(y - 1) and √(y - 1) + √(y + 1) are consecutive integers we must have f(y) = 1/2 + k. Solving we get y = √(1+ ( (7-4k-4k²)/(4+8k) )²). Now √(x - 1) + √(y - 1) = (4k² + 7)/(2 - 8k²). This must be some integer n, so k² = (2n - 7)/(8n + 4). Clearly k² must be non-negative, so n > 3.

4th Balkan 1987 Problem 3

ABC is a triangle with BC = 1. Put sA = sin(A/2), cA = cos(A/2), sB = sin(B/2), cB = cos(B/2). We have sA²³cB⁴⁸ = sB²³cA⁴⁸. Find AC.

Solution: Answer: 1.

Since ABC is a triangle, both A/2 and B/2 lie strictly between 0 and π/2. Over this range sin x is strictly increasing and cos x is strictly decreasing, so if A/2 > B/2, then sin A/2 > sin B/2 and 1/cos A/2 > 1/cos B/2, so the equality cannot hold. Similarly if A/2 < B/2. Hence A/2 = B/2 and the triangle is isosceles.

4th Balkan 1987 Problem 4

Two circles have centers a distance 2 apart and radii 1 and √2. X is one of the points on both circles. M lies on the smaller circle, Y lies on the larger circle and M is the midpoint of XY. Find the distance XY.

Solution: Answer: √(7/2).

Use coordinates with origin at the center of the small circle. So X is (3/4, (√7)/4 ). Take M to be (cos k, sin k). Then Y is (2 cos k - 3/4, 2 sin k - (√7)/4). Y lies on the circle center (2, 0) radius √2, so (2 cos k - 11/4)² + (2 sin k - (√7)/4)² = 2. Hence 11 cos k + √7 sin k = 10.

We have XY² = (2 cos k - 3/2)² + (2 sin k - (√7)/2 )² = 8 - 6 cos k - 2√7 sin k = 8 - 6 cos k - 2(10 - 11 cos k) = 16 cos k - 12.

So we need to find cos k. We have (10 - 11 cos k)² = 7 (1 - cos²k), so 128 cos²k - 220 cos k + 93 = 0. Factorising: (32 cos k - 31)(4 cos k - 3) = 0. The root cos k = 3/4 corresponds to the point X, so we want the other root, cos k = 31/32. Hence XY² = 31/2 - 12 = 7/2.