3rd Iberoamerican Mathematical Olympiad Problems 1988

A1.  The sides of a triangle form an arithmetic progression. The altitudes also form an arithmetic progression. Show that the triangle must be equilateral.

A2.  The positive integers a, b, c, d, p, q satisfy ad - bc = 1 and a/b > p/q > c/d. Show that q ≥ b + d and that if q = b + d, then p = a + c.

A3.  P is a fixed point in the plane. Show that amongst triangles ABC such that PA = 3, PB = 5, PC = 7, those with the largest perimeter have P as incenter.

B1.  Points A₁, A₂, ... , A_n are equally spaced on the side BC of the triangle ABC (so that BA₁ = A₁A₂ = ... = A_n-1A_n = A_nC). Similarly, points B₁, B₂, ... , B_n are equally spaced on the side CA, and points C₁, C₂, ... , C_n are equally spaced on the side AB. Show that (AA₁² + AA₂² + ... + AA_n² + BB₁² + BB₂² + ... + BB_n² + C₁² + ... + CC_n²) is a rational multiple of (AB² + BC² + CA²).

B2.  Let k³ = 2 and let x, y, z be any rational numbers such that x + y k + z k² is non-zero. Show that there are rational numbers u, v, w such that (x + y k + z k²)(u + v k + w k²) = 1.

B3.  Let S be the collection of all sets of n distinct positive integers, with no three in arithmetic progression. Show that there is a member of S which has the largest sum of the inverses of its elements (you do not have to find it or to show that it is unique). 

Solutions

Problem A1

The sides of a triangle form an arithmetic progression. The altitudes also form an arithmetic progression. Show that the triangle must be equilateral.
Solution

Let the sides be a, a+d, a+2d with d >= 0. Then the altitudes are k/a ≥ k/(a+d) ≥ k/(a+2d), where k is twice the area. We claim that k/a + k/(a+2d) > 2k/(a + d) unless d = 0. This is equivalent to (a + d)(a + 2d) + a(a + d) > 2a(a + 2d) or 2d² > 0, which is obviously true. So the altitudes can only form an arithmetic progression if d = 0 and hence the triangle is equilateral.

Problem A2

The positive integers a, b, c, d, p, q satisfy ad - bc = 1 and a/b > p/q > c/d. Show that q >= b + d and that if q = b + d, then p = a + c.
Solution

p/q > c/d implies pd > cq and hence pd >= cq + 1, so p/q ≥ c/d + 1/(qd). Similarly, a/b > p/q implies a/b ≥ p/q + 1/(bq). So a/b - c/d ≥ 1/(qd) + 1/(qb) = (b + d)/(qbd). But a/b - c/d = 1/bd. Hence q ≥ b + d.
Now assume q = b + d. We have ad - bc = 1 ≤ d, so ad + cd - d ≤ bc + cd and hence (a+c-1)/(b+d) ≤ c/d. So p ≥ a + c. Similarly ad - bc ≤ b, so bc + b + ab ≥ ad + ab, so (a+c+1)/(b+d) ≥ a/b. So p ≤ a + c. Hence p = a + c.

Problem A3

P is a fixed point in the plane. Show that amongst triangles ABC such that PA = 3, PB = 5, PC = 7, those with the largest perimeter have P as incenter.
Solution

Given points P, B, C and a fixed circle center P, we show that the point A on the circle which maximises AB + AC is such that PA bisects angle BAC. Consider a point A' close to A. Then the change in AB + AC as we move A to A' is AA'(sin PAC - sin PAB) + O(AA'² ). So for a maximal configuration we must have sin PAC = sin PAB, otherwise we could get a larger sum by taking A' on one side or the other. This applies to each vertex of the triangle, so P must be the incenter.

Comment. This is arguably too advanced, because it uses calculus concepts like O(x). Can anyone supply a more elementary solution?

Problem B1

Points A₁, A₂, ... , A_n are equally spaced on the side BC of the triangle ABC (so that BA₁ = A₁A₂ = ... = A_n-1A_n = A_nC). Similarly, points B₁, B₂, ... , B_n are equally spaced on the side CA, and points C₁, C₂, ... , C_n are equally spaced on the side AB. Show that (AA₁² + AA₂² + ... + AA_n² + BB₁² + BB₂² + ... + BB_n² + C₁² + ... + CC_n²) is a rational multiple of (AB² + BC² + CA²).

Solution

Using the cosine formula, AA_k² = AB² + k²BC²/(n+1)² - 2k AB.BC/(n+1) cos B. So ∑ AA_k² = n AB² + BC²/(n+1)² (1² + 2² + ... + n²) - 2 AB.BC cos B (1 + 2 + ... + n)/(n+1). Similarly for the other two sides.

Thus the total sum is n (AB² + BC² + CA²) + n(2n+1)/(6(n+1) ) (AB² + BC² + CA²) - n (AB.BC cos B + BC.CA cos C + CA.AB cos A). But AB.BC cos B = (AB² + BC² - CA²)/2, so AB.BC cos B + BC.CA cos C + CA.AB cos A = (AB² + BC² + CA²)/2. Thus the sum is rational multiple of (AB² + BC² + CA²). 

Problem B2

Let k³ = 2 and let x, y, z be any rational numbers such that x + y k + z k² is non-zero. Show that there are rational numbers u, v, w such that (x + y k + z k²)(u + v k + w k²) = 1.

Solution

We need xu + 2zv + 2yw = 1, yu + xv + 2zw = 0, zu + yv + xw = 0. This is just a straightforward set of linear equations. Solving, we get: u = (x² - 2yz)/d, v = (2z² - xy)/d, w = (y² - xz)/d, were d = x³ + 2y³ + 4z³ - 6xyz.

This would fail if d = 0. But if d = 0, then multiplying through by a suitable integer we have 6mnr = m³ + 2n³ + 4r³ for some integers m, n, r. But we can divide by any common factor of m, n, r to get them without any common factor. But 6mnr, 2n³, 4r³ are all even, so m must be even. Put m = 2M. Then 12Mnr = 8M³ + 2n³ + 4r³, so 6Mnr = 4M³ + n³ + 2r³. But 6Mnr, 4M³ and 2r³ are all even, so n must be even. Put n = 2N. Then 12MNr = 4M³ + 8N³ + 2r³, so 6MNr = 2M³ + 4N³ + r³, so r must be even. So m, n, r had a common factor 2. Contradiction. So d cannot be zero. 

Problem B3

Let S be the collection of all sets of n distinct positive integers, with no three in arithmetic progression. Show that there is a member of S which has the largest sum of the inverses of its elements (you do not have to find it or to show that it is unique).

Solution

Induction on n. For n = 1, {1} is obviously maximal. Now suppose a₁ < a₂ < ... < a_n is a maximal set for n. Take a_n+1 to be the smallest integer > a_n such that {a₁, a₂, ... , a_n+1} has no three members in AP. Now consider the sequences b₁ < b₂ < ... < b_n which have no three in AP and b_n+1 ≤ a_n+1. There are only finitely many such sequences. So we can find one which is maximal. Suppose it is c₁ < c₂ < ... < c_n+1. Now take whichever of a_i, c_i has the larger sum of inverses. It is clearly maximal with respect to sequences whose largest member is ≤ a_n+1. Suppose we have a sequence x₁ < x₂ < ... < x_n+1 with no three in AP and x_n+1 > a_n+1. Then we have 1/x_n+1 < 1/a_n+1 and, by induction, 1/x₁ + ... + 1/x_n ≤ 1/a₁ + ... + 1/a_n, so 1/x₁ + ... + 1/x_n+1 < 1/a₁ + ... + 1/a_n+1, so it is worse than the sequence we have chosen.