5th International Mathematical Olympiad 1963 Problems & Solutions

A1.  For which real values of p does the equation         √(x² - p) + 2 √(x² - 1) = x have real roots? What are the roots?

A2.  Given a point A and a segment BC, determine the locus of all points P in space for which ∠APX = 90^o for some X on the segment BC.

A3.  An n-gon has all angles equal and the lengths of consecutive sides satisfy a₁ ≥ a₂ ≥ ... ≥ a_n. Prove that all the sides are equal.

B1.  Find all solutions x₁, ... , x₅ to the five equations x_i + x_i+2 = y x_i+1 for i = 1, ... , 5, where subscripts are reduced by 5 if necessary.

B2.  Prove that cos π/7 - cos 2π/7 + cos 3π/7 = 1/2.

B3.  Five students A, B, C, D, E were placed 1 to 5 in a contest with no ties. One prediction was that the result would be the order A, B, C, D, E. But no student finished in the position predicted and no two students predicted to finish consecutively did so. For example, the outcome for C and D was not 1, 2 (respectively), or 2, 3, or 3, 4 or 4, 5. Another prediction was the order D, A, E, C, B. Exactly two students finished in the places predicted and two disjoint pairs predicted to finish consecutively did so. Determine the outcome. 

Solutions

Problem A1

For which real values of p does the equation

        √(x² - p) + 2 √(x² - 1) = x have real roots? What are the roots?

Solution

I must admit to having formed rather a dislike for this type of question which came up in almost every one of the early IMOs. Its sole purpose seems to be to teach you to be careful with one-way implications: the fact that a² = b² does not imply a = b.

The lhs is non-negative, so x must be non-negative. Moreover 2√(x² - 1) ≤ x, so x ≤ 2/√3. Also √(x² - p) ≤ x, so p ≥ 0.

Squaring etc gives that any solution must satisfy x² = (p - 4)²/(16 - 8p). We require x ≤ 2/√3 and hence (3p - 4)(p + 4) ≤ 0, so p ≤ 4/3.

Substituting back in the original equality we get |3p-4| + 2|p| = |p - 4|, which is indeed true for any p satisfying 0 ≤ p ≤ 4/3.   

Problem A2

Given a point A and a segment BC, determine the locus of all points P in space for which ∠APX = 90^o for some X on the segment BC.

Solution

Take the solid sphere on diameter AB, and the solid sphere on diameter AC. Then the locus is the points in one sphere but not the other (or on the surface of either sphere). Given P, consider the plane through P perpendicular to AP and the parallel planes through the other two points of intersection of AP with the two spheres (apart from A) which pass through B and C. 

Problem A3

An n-gon has all angles equal and the lengths of consecutive sides satisfy a₁ ≥ a₂ ≥ ... ≥ a_n. Prove that all the sides are equal.

Solution

For n odd consider the perpendicular distance of the shortest side from the opposite vertex. This is a sum of terms a_i x cosine of some angle. We can go either way round. The angles are the same in both cases, so the inequalities give that a₁ = a_n-1, and hence a₁ = a_i for all i < n. We get a₁ = a_n by repeating the argument for the next shortest side. The case n even is easier, because we take a line through the vertex with sides a₁ and a_n making equal angles with them and look at the perpendicular distance to the opposite vertex. This gives immediately that a₁ = a_n. 

Problem B1

Find all solutions x₁, ... , x₅ to the five equations x_i + x_i+2 = y x_i+1 for i = 1, ... , 5, where subscripts are reduced by 5 if necessary.

Solution

Successively eliminate variables to get x₁(y - 2)(y² + y - 1)² = 0. We have the trivial solution x_i = 0 for any y. For y = 2, we find x_i = s for all i (where s is arbitrary). Care is needed for the case y² + y - 1 = 0, because after eliminating three variables the two remaining equations have a factor y² + y - 1, and so they are automatically satisfied. In this case, we can take any two x_i arbitrary and still get a solution. For example, x₁ = s, x₂ = t, x₃ = - s + yt, x₄ = - ys - yt, x₅ = ys - t. 

Problem B2

Prove that cos π/7 - cos 2π/7 + cos 3π/7 = 1/2.

Solution

Consider the roots of x⁷ + 1 = 0. They are e^iπ/7, e^i3π/7, ... , e^i13π/7 and must have sum zero since there is no x⁶ term. Hence, in particular, their real parts sum to zero. But cos7π/7 = - 1 and the others are equal in pairs, because cos(2π - x) = cos x. So we get cos π/7 + cos 3π/7 + cos 5π/7 = 1/2. Finally since cos(π - x) = - cos x, cos 5π/7 = - cos 2π/7. 

Problem B3

Five students A, B, C, D, E were placed 1 to 5 in a contest with no ties. One prediction was that the result would be the order A, B, C, D, E. But no student finished in the position predicted and to two students predicted to finish consecutively did so. For example, the outcome for C and D was not 1, 2 (respectively), or 2, 3, or 3, 4 or 4, 5. Another prediction was the order D, A, E, C, B. Exactly two students finished in the places predicted and two disjoint pairs predicted to finish consecutively did so. Determine the outcome.

Solution

Start from the second prediction. The disjoint pairs can only be: DA, EC; DC, CB; or AE, CB. The additional requirement of just two correct places means that the only possibilities (in the light of the information about the second prediction) are: DABEC, DACBE, EDACB, AEDCB. The first is ruled out because AB are consecutive. The second is ruled out because C is in the correct place. The fourth is ruled out because A is in the correct place. This leaves EDACB, which is indeed a solution.

Solutions are also available in:   Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in   István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.