1st Junior Balkan Mathematical Olympiad Problems 1997

1.  Show that given any 9 points inside a square side 1 we can always find three which form a triangle with area < 1/8.

2.  Given reals x, y with (x² + y²)/(x² - y²) + (x² - y²)/(x² + y²) = k, find (x⁸ + y⁸)/(x⁸ - y⁸) + (x⁸ - y⁸)/(x⁸ + y⁸) in terms of k.

3.  I is the incenter of ABC. N, M are the midpoints of sides AB, CA. The lines BI, CI meet MN at K, L respectively. Prove that AI + BI + CI > BC + KL.

4.  A triangle has circumradius R and sides a, b, c with R(b+c) = a √(bc). Find its angles.

5.  n₁, n₂, ... , n₁₉₉₈ are positive integers such that n₁² + n₂² + ... + n₁₉₉₇² = n₁₉₉₈². Show that at least two of the numbers are even. 

Solutions

Problem 2

Given reals x, y with (x² + y²)/(x² - y²) + (x² - y²)/(x² + y²) = k, find (x⁸ + y⁸)/(x⁸ - y⁸) + (x⁸ - y⁸)/(x⁸ + y⁸) in terms of k.

Solution

k/2 = (x⁴+y⁴)/(x⁴-y⁴), so k/2 + 2/k = 2(x⁸+y⁸)/(x⁸-y⁸). Hence desired expresson = (k²+4)/4k + 4k/(k²+4) = (k⁴+24k²+16)/(4k³+14k).

Thanks to Cristian Ilac

Problem 4

A triangle has circumradius R and sides a, b, c with R(b+c) = a √(bc). Find its angles.

Answer

A = 90^o, B = 45^o, C = 45^o.

Solution

We have a = 2R sin A, so b+c = 2 sin A √(bc). But by AM/GM (b+c)/2 ≥ √(bc) with equality iff b = c. So we must have b = c and A = 90^o.

Thanks to Suat Namli

Problem 5

n₁, n₂, ... , n₁₉₉₈ are positive integers such that n₁² + n₂² + ... + n₁₉₉₇² = n₁₉₉₈². Show that at least two of the numbers are even.

Solution

If two of n₁, n₂, ... , n₁₉₉₇ are even, then there is nothing to prove. If just one of n₁, n₂, ... , n₁₉₉₇ is even, then an even number of them are odd, so n₁₉₉₈ is also even. So it remains to show that n₁, n₂, ... , n₁₉₉₇ cannot all be odd.

Odd squares are all = 1 mod 8 and 1997 = 5 mod 8. So if all of n₁, n₂, ... , n₁₉₉₇ are odd, then n₁₉₉₈² = 5 mod 8. Contradiction.

Thanks to Suat Namli