1st Balkan Mathematical Olympiad Problems 1984

1st Balkan Mathematical Olympiad Problems 1984

A1.  Let x₁, x₂, ... , x_n be positive reals with sum 1. Prove that x₁/(2 - x₁) + x₂/(2 - x₂) + ... + x_n/(2 - x_n) ≥ n/(2n - 1).

A2.  ABCD is a cyclic quadrilateral. A' is the orthocenter (point where the altitudes meet) of BCD, B' is the orthocenter of ACD, C' is the orthocenter of ABD, and D' is the orthocenter of ABC. Prove that ABCD and A'B'C'D' are congruent.

A3.  Prove that given any positive integer n we can find a larger integer m such that the decimal expansion of 5^m can be obtained from that for 5ⁿ by adding additional digits on the left.

A4.  Given positive reals a, b, c find all real solutions (x, y, z) to the equations ax + by = (x - y)², by + cz = (y - z)², cz + ax = (z - x)². 

Solutions

1st Balkan 1984 Problem 1

Let x₁, x₂, ... , x_n be positive reals with sum 1. Prove that x₁/(2 - x₁) + x₂/(2 - x₂) + ... + x_n/(2 - x_n) ≥ n/(2n - 1).

Solution

Assume x₁ ≤ x₂ ≤ ... ≤ x_n. Then also 1/(2 - x₁) ≤ 1/(2 - x₂) ≤ ... ≤ 1/(2 - x_n). Hence we may apply Chebyshev's inequality to get x₁/(2 - x₁) + x₂/(2 - x₂) + ... + x_n/(2 - x_n) ≥ (x₁ + ... + x_n)(1/(2 - x₁) + ... + 1/(2 - x_n) )/n. Now the harmonic mean inequality gives (1/(2 - x₁) + ... + 1/(2 - x_n) ) ≥ n²/( (2 - x₁) + ... + (2 - x_n) ) = n²/(2n - 1). Hence result.

1st Balkan 1984 Problem 2

ABCD is a cyclic quadrilateral. A' is the orthocenter (point where the altitudes meet) of BCD, B' is the orthocenter of ACD, C' is the orthocenter of ABD, and D' is the orthocenter of ABC. Prove that ABCD and A'B'C'D' are congruent.

Solution

Let A" be the centroid of BCD, B" the centroid of ACD, C" the centroid of ABD and D" the centroid of ABC. We claim that A"B"C"D" is similar to ABCD and 1/3 the size.

Take any two points of A, B, C, D, wlog A and B. Let M the midpoint of the other two. Then A" is 1/3 the way along MB and B" is 1/3 the way along MA. Hence A"B" = AB/3. Similarly, all the other distances in A"B"C"D" are 1/3 the corresponding distances in ABCD, which establishes the claim.

Now consider the Euler line of the triangle BCD. Its circumcenter is O, its centroid is A'' and its altitudes meet at A'. So these three points lie on the Euler line and OA' = 3OA". Similarly, OB' = 3OB", OC' = 3OC" and OD' = 3OD". So A'B'C'D' is similar to A"B"C"D" and 3 times the size. Hence it is congruent to ABCD. 

1st Balkan 1984 Problem 3

Prove that given any positive integer n we can find a larger integer m such that the decimal expansion of 5^m can be obtained from that for 5ⁿ by adding additional digits on the left.

Solution

We prove first that 5^N - 1 is divisible by 2ⁿ⁺² for N = 2ⁿ. This is an easy induction. It is true for n = 1. Suppose it is true for n. Then 5^N - 1 is divisible by 2ⁿ⁺². But 5^N + 1 is certainly even, so it is divisible by 2. Hence (5^N + 1)(5^N - 1) = 5^2N - 1 is divisible by 2ⁿ⁺³, so the result is true for all n.

So suppose 5ⁿ has k decimal digits. Clearly k ≤ n. We can find N such that 5^N - 1 is divisible by 2^k. Hence (5^N - 1)5ⁿ is divisible by 2^k5^k = 10^k. In other words it ends in at least k zeros. So 5^N+n = (5^N - 1)5ⁿ + 5ⁿ ends in the same digits as 5ⁿ. In other words we may take m = n + N.

1st Balkan 1984 Problem 4

Given positive reals a, b, c find all real solutions (x, y, z) to the equations ax + by = (x - y)², by + cz = (y - z)², cz + ax = (z - x)².

Solution: Answer: (x, y, z) = (0, 0, 0), (a, 0, 0), (0, b, 0), (0, c, 0).

We have 2ax = (ax + by) - (by + cz) + (cz + ax) = 2(x² - xy + yz - xz) = 2(y - x)(z - x). Similarly, 2by = (ax + by) + (by + cz) - (cz + ax) = 2(y² + xz - xy - yz) = (z - y)(x - y) and 2cz = -(ax + by) + (by + cz) + (cz + ax) = 2(z² + xy - yz - zx) = 2(x - z)(y - z).

Suppose z ≥ x, y. Then if x >= y, ax = (z - x)(y - x) ≤ 0 , and by = (z - y)(x - y) ≥ 0. But a and are positive, so x ≤ 0 and y ≥ 0. Hence x = y = 0. On the other hand, if x ≤ y, then ax ≥ 0 and by ≤ 0, so again x = y = 0. Hence cz = z², so z = 0 or c. This gives the two solutions (x, y, z) = (0, 0, 0) and (0, 0, c).

Similarly, if y ≥ x, z, then we find x = z = 0 and hence y = 0 or b. If x ≥ y, z, then we find y = z = 0 and x = 0 or a.