8th Balkan Mathematical Olympiad Problems 1991

8th Balkan Mathematical Olympiad Problems 1991

A1.  The circumcircle of the acute-angled triangle ABC has center O. M lies on the minor arc AB. The line through M perpendicular to OA cuts AB at K and AC at L. The line through M perpendicular to OB cuts AB at N and BC at P. MN = KL. Find angle MLP in terms of angles A, B and C.

A2.  Find an infinite set of incongruent triangles each of which has integral area and sides which are relatively prime integers, but none of whose altitudes are integral.

A3.  A regular hexagon area H has its vertices on the perimeter of a convex polygon of area A. Prove that 2A ≤ 3H. When do we have equality?

A4.  A is the set of positive integers and B is A ∪ {0}. Prove that no bijection f: A → B can satisfy f(mn) = f(m) + f(n) + 3 f(m) f(n) for all m, n. 

Solutions

8th Balkan 1991 Problem 1

The circumcircle of the acute-angled triangle ABC has center O. M lies on the minor arc AB. The line through M perpendicular to OA cuts AB at K and AC at L. The line through M perpendicular to OB cuts AB at N and BC at P. MN = KL. Find angle MLP in terms of angles A, B and C.

Solution : Answer: angle C.

∠MKN = ∠AKL (opposite angles) = 90^o - ∠OAB = 1/2 ∠AOB = ∠C. A similar argument shows that ∠MNK = ∠C, so MNK is isosceles and MN = MK.

Also triangles AKL and ACB are similar, so AK/KL = AC/CB (*). Extend ML to meet the circumcircle again at R. Then since OA is perpendicular to MR, we have AM = AR. Triangles AKR and MKB are similar, because ∠AKR = ∠MKB (opposite angles) and ∠ARK = ∠ARM (same angle) = ∠ABM (ARBM cyclic) = ∠MBK (same angle). So AK/AR = MK/MB. Hence MK/AK = MB/AR = MB/AM. Multiplying by (*) we get MK/KL = (AC/BC) (BM/AM) (**). But MK = MN = KL. Hence (AC/BC) (BM/AM) = 1.

But a similar argument to that used to show (**) gives that MN/NP = (BC/AC) (AM/BM). Hence MN = NP. So N is the midpoint of MP and K is the midpoint of ML. Hence ∠MLP = ∠MKN = ∠C.

8th Balkan 1991 Problem 2

Find an infinite set of incongruent triangles each of which has integral area and sides which are relatively prime integers, but none of whose altitudes are integral.

Solution

One possible answer is: r² + 4, r⁴ + 3r² + 1, r⁴ + 4r² + 3 with r any integer not ±2 mod 5.

We use Heron's formula. Take the sides as a + b, n, n + a - b. This involves no loss of generality and makes Heron's expression more convenient. Let the area be A. Heron gives A² = (n+a)(n-b)ab. Suppose we take n - b = ar², n + a = bs². Then A = abrs. We require a(r² + 1) = b(s² - 1). The simplest idea is not to take a = s² - 1, b = r² + 1, but then the area is rs(s² - 1)(r² + 1), n = s²r² + 1 and n + a - b = (s² - 1)(r² + 1) which divides the area, so one of the altitudes is integral.

The next simplest idea is to take b = 1, s - 1 = r² + 1, a = s + 1 = r² + 3. This gives sides r² + 4, r⁴ + 3r² + 1, r⁴ + 4r² + 3 and area A = r(r² + 2)(r² + 3). We need to check that each pair of sides is relatively prime and that each altitude is non-integral.

Any common factor of r² + 4 and r⁴ + 3r² + 1 must also divide r²(r² + 4) - (r⁴ + 3r² + 1) = r² - 1 and hence also (r² + 4) - (r² - 1) = 5. But 5 only divides r² + 4 if r = ±2 mod 5, so for other values of r, these two sides are relatively prime.

Any common factor of r² + 4 and r⁴ + 4r² + 3 must also divide r⁴ + 4r² + 3 - r²(r² + 4) = 3. But any square is 0 or 1 mod 3, so r² + 4 cannot be divisible by 3. Hence these two sides are relatively prime. Finally, any common factor of (r⁴ + 3r² + 1) and (r⁴ + 4r² + 3) must also divide their difference r² + 2. Hence also (r⁴ + 3r² + 1) - r²(r² + 2) = r² + 1 and hence 1. So these two sides are relatively prime.

r² + 4 is relatively prime to r² + 3, has at most a factor 2 in common with r² + 2 and at most a factor 4 in common with r. So r² + 4 can only divide A if r = 2. If (r⁴ + 3r² + 1) divides A then it also divides (r² + 2)(r⁴ + 3r² + 1) - rA= r² + 2. But (r⁴ + 3r² + 1) > r² + 2. So (r⁴ + 3r² + 1) does not divide A. Finally, if (r⁴ + 4r² + 3) = (r² + 3)(r² + 1) divides A, then r² + 1 divides r(r² + 2), but that is impossible since r² + 1 and r² + 2 are relatively prime and r² + 1 > r. 

8th Balkan 1991 Problem 3

A regular hexagon area H has its vertices on the perimeter of a convex polygon of area A. Prove that 2A ≤ 3H. When do we have equality?

Solution

We use the basic result that there is at least one support line through any point on the boundary of a convex set. The support line has the property that all points of the convex set lie on one side of the line (or on the line itself). Since each vertex of the hexagon lies on the boundary of the polygon, there is a support line through each vertex of the hexagon.

Let the hexagon be ABCDEF. Take points P, Q outside the hexagon so that ABP, BCQ are equilateral triangles. Let the support lines through A, B, C, D, E, F be L_A, L_B, L_C, L_D, L_E, L_F respectively. Then these lines bound a hexagon H' which contains the convex polygon, so it is sufficient to show that area H' ≤ 3H/2. Let L_A and L_B meet at X and let L_B and L_C meet at Y. The area of H' is H plus the area of the triangle ABX and the area of the five corresponding triangles on the other sides. We claim that area ABX + area BCY ≤ H/6. It follows that the area of all six triangles is ≤ H/2 and so the required result follows.

Extend XY to meet AP at R and CQ at S. Since the hexagon area H is regular, AP is parallel to CQ and hence angle ARB = angle QSB. Also angle ABR = angle QBS and AB = BQ. So triangles ABR and QBS are congruent and hence have equal area. So area ABX + area BCY <= area ABR + area BCS = area QBS + area BCS = area BCQ = H/6, which establishes the claim. If we have equality then area ARX = 0, so L_A must coincide with the line AB or the line FA. Without loss of generality, it coincides with FA. Then L_B must coincide with the line BC, so H' is an equilateral triangle whose sides contain three sides of the hexagon. It is easy to see that this does indeed give equality. 

8th Balkan 1991 Problem 4

A is the set of positive integers and B is A ∪ {0}. Prove that no bijection f: A → B can satisfy f(mn) = f(m) + f(n) + 3 f(m) f(n) for all m, n.

Solution

We must take f(1) = 0, since if f(k) = 0, then f(k²) = 0 also and f is (1, 1). So for any n > 1 we have f(n) ≥ 1. Now suppose n is composite, n = rs with r, s > 1. Then f(n) = f(r) + f(s) + 3 f(r) f(s) ≥ 5. Take p and q so that f(p) = 1 and f(q) = 3. Then we have just shown that p and q are prime. Now take r so that f(r) = 8.

But now f(q²) = f(q) + f(q) + 3 f(q) f(q) = 3 + 3 + 27 = 33, and f(pr) = f(p) + f(r) + 3 f(p)f(r) = 1 + 8 + 3·8 = 33. f is a bijection so we must have q² = pr, but that is impossible, since p and q are distinct primes.

Comment. How did I come by that? I started playing with small values, but could not reach any obvious contradiction - unfortunately, I did not go as far as 33. So I started to look harder. I noticed a pattern in f(pⁿ). It seemed to be (4ⁿ-1)/3. That suggested looking at other f(qⁿ). Sure enough it appeared that if f(q) = 2, then f(qⁿ) = (7ⁿ-1)/3. If f(s) = 3, then f(sⁿ) = (10ⁿ-1)/3. I then noticed that that implied that f(pⁿq^m) = (4ⁿ7^m-1)/3. At that point I knew I was almost there because it seemed likely that we could find two different factorisations. So I worked out a few more and found that if f(r) = 8, then f(rⁿ) = (25ⁿ-1)/3. Since 25¹.4¹=10² I was home, apart from some routine tidying up. At that point I had not proved any of the patterns. In fact, it is easy to show that if f(p) = k, then f(pⁿ) = ( (3k+1)ⁿ-1)/3, but it is not necessary for the proof.