5th Balkan Mathematical Olympiad Problems 1988

5th Balkan Mathematical Olympiad Problems 1988

A1.  ABC is a triangle area 1. AH is an altitude, M is the midpoint of BC and K is the point where the angle bisector at A meets the segment BC. The area of the triangle AHM is 1/4 and the area of AKM is 1 - (√3)/2. Find the angles of the triangle.

A2.  Find all real polynomials p(x, y) such that p(x, y) p(u, v) = p(xu + yv, xv + yu) for all x, y, u, v.

A3.  The sum of the squares of the edges of a tetrahedron is S. Prove that the tetrahedron can be fitted between two parallel planes a distance √(S/12) apart.

A4.  x_n is the sequence 51, 53, 57, 65, ... , 2ⁿ + 49, ... Find all n such that x_n and x_n+1 are each the product of just two distinct primes with the same difference. 

Solutions

5th Balkan 1988 Problem 1

ABC is a triangle area 1. AH is an altitude, M is the midpoint of BC and K is the point where the angle bisector at A meets the segment BC. The area of the triangle AHM is 1/4 and the area of AKM is 1 - (√3)/2. Find the angles of the triangle.

Solution: Answer: A = 90^o, B = 30^o, C = 60^o.

Assume AB ≥ AC. Then BK/KC = AB/AC ≥ 1, so K lies between M and C. Put MH = x. Then since area AHM = 1/4 area ABC, we have BC = 4x, and hence BM = 2x, HC = x. Since area AKM = 1 - (√3)/2, we have MK = (4 - 2√3) x. Hence AB/AC = (BM + MK)/(MC - MK) = (6 - 2√3)/(2√3 - 2) = √3.

We have AB² - BH² = AH² = AC² - CH², so AB² - AC² = (3x)² - x². Hence AC² = 4x². So AC : AB : BC = 1 : √3 : 2. Hence A = 90^o, B = 30^o, C = 60^o.

5th Balkan 1988 Problem 2

Find all real polynomials p(x, y) such that p(x, y) p(u, v) = p(xu + yv, xv + yu) for all x, y, u, v.

Solution

Answer: The possible polynomials are p(x, y) = (x + y)^m(x - y)ⁿ for some positive integers m, n or p(x, y) = 0.

p(x, 0) p(x, 0) = p(x², 0) for all x (*). But p(x, 0) is a polynomial in x. Suppose it is a_nxⁿ + ... + a₀ with a_n non-zero. Then we must have a_n² = a_n and hence a_n = 1. Now suppose there is another non-zero term. Take a_m to be the highest such. Then the lhs of (*) has a non-zero x^m+n term and the rhs does not. Contradiction. So we must have p(x, 0) = xⁿ for some n, or p(x, 0) = 0 for all x.

If p(x, 0) = 0 for all x, then p(u, v) = p(1, 0) p(u, v) = 0, so p is identically zero. Assume then that p(x, 0) = xⁿ for some n. Then p(xu, xv) = p(x, 0) p(u, v) = xⁿ p(u, v). So p(x, y) is homogeneous of degree n. In other words, all its terms have the form a x^h y^k, where h and k are non-negative integers with sum n.

Now put g(x, y) = p( (x+y)/2, (x-y)/2). Then p(x, y) = g(x + y, x - y) and g(x, y) g(u, v) = g(xu, yv). The definition shows that g must also be homogeneous of degree n. So suppose g(x, y) = c₀yⁿ + c₁x y^n-1 + c₂x²y^n-2 + ... + c_nyⁿ. But now if we compare the expansions of g(x, y) g(u, v) and g(xu, yv) the former has terms with x and u to different powers unless all but one c_i is zero. So we must have g(x, y) = c x^hy^n-h for some h. We then require c² = c and hence c = 1. So g(x, y) = x^hy^k for some h, k. Hence p(x, y) = (x + y)^h(x - y)^k for some h, k. 

5th Balkan 1988 Problem 3

The sum of the squares of the edges of a tetrahedron is S. Prove that the tetrahedron can be fitted between two parallel planes a distance √(S/12) apart.

Solution

Let the tetrahedron be PQRS. Put PQ = a, PR = b, PS = c, QR = d, RS = e, SQ = f. We find the distance x between the midpoints of PS and QR. Let M be the midpoint of QR and let PM = m. Using the median result (see below), we have m² = a²/2 + b²/2 - d²/4. Similarly, if SM = n, then n² = e²/2 + f²/2 - d²/4. Let N be the midpoint of PS, then MN is the median of MPS, so x² = m²/2 + n²/2 - c²/4 = S/4 - (c² + d²)/2.

Similarly, if y is the distance between the midpoints of PR and QS we have y² = S/4 - (b² + f²)/2, and if z is the distance between the midpoints of PQ and RS, then z² = S/4 - (a² + e²)/2. Hence x² + y² + z² = S/4. Assume that x is the smallest of x, y, z. Then x² ≤ S/12. But if we take a plane through PS parallel to QR and a parallel plane through QR, then the tetrahedron fits between the planes, and the distance between them is at most x.

Notes

The median result is as follows. If ABC is a triangle with M the midpoint of BC, then 2 AM² + 2 BM² = AB² + AC². The proof is immediate by applying the cosine formula to triangles AMB and AMC and using the fact that cos AMB + cos AMC = 0.

Note that we cannot always fit a tetrahedron between two planes a distance √(S/12) apart if we put a face flat on one of the planes. For consider a regular tetrahedron side 1. It has height √(2/3), whereas √(S/12) = √(1/2), which is smaller.

5th Balkan 1988 Problem 4

x_n is the sequence 51, 53, 57, 65, ... , 2ⁿ + 49, ... Find all n such that x_n and x_n+1 are each the product of just two distinct primes with the same difference.

Solution: Answer: n = 7 (x₇ = 3·59, x₈ = 5·61).

Calculating the first few terms we find: x₁ = 3·17, x₂ = 53, x₃ = 3·19, x₄ = 5·13, x₅ = 3⁴, x₆ = 113, x₇ = 3·59, x₈ = 5·61.

We notice that for n odd, the term is divisible by 3. This is easily proved: 2ⁿ + 49 = (3 - 1)ⁿ + 3.16 + 1, which is a multiple of 3 for n odd. We also notice that for n divisible by 4, the term is divisible by 5. That is also easily proved: 2⁴ⁿ + 49 = (15 + 1)ⁿ + 50 - 1, which is obviously divisible by 5.

Hence there can be no solutions with n a multiple of 4, for then 5 divides x_n and 3 divides x_n+1 which would imply x_n = 5(p+2), x_n+1 = 3p < x_n. Contradiction.

If n is a multiple of 4 plus 3, then we have 3p = 2ⁿ + 49, 5(p+2) = 2ⁿ⁺¹ + 49. Subtracting, 2p + 10 = 2ⁿ. Subtracting again, p - 10 = 49, so p = 59. It is easily checked that this is the solution n = 7, x₇ = 3·59, x₈ = 5·61.

n cannot be a multiple of 4 plus 2, for then x_n+1 would be divisible by 3. There is no prime less than 3 except 2 and it is obvious that 2 cannot divide x_n. So if x_n+1 = 3p, then we would have x_n = (3+k)(p+k) > x_n+1. Contradiction.

Finally, we consider the case n a multiple of 4 plus 1. Then 3 divides x_n, so we have 2ⁿ + 49 = 3p, 2ⁿ⁺¹ + 49 = (3 + k)(p + k). Subtracting, 2ⁿ = 3p + kp + k². Subtracting again, 49 = -(k-3)p - 3k - k² (*). But we know that k is not 2, because we have just found all those solutions. Hence k ≥ 3. But that means the rhs of (*) is negative. Contradiction. So there are no solutions for n a multiple of 4 plus 2.