9th International Mathematical Olympiad 1967 Problems & Solutions

A1.  The parallelogram ABCD has AB = a, AD = 1, angle BAD = A, and the triangle ABD has all angles acute. Prove that circles radius 1 and center A, B, C, D cover the parallelogram iff             a ≤ cos A + √3 sin A.

A2.  Prove that a tetrahedron with just one edge length greater than 1 has volume at most 1/8.

A3.  Let k, m, n be natural numbers such that m + k + 1 is a prime greater than n + 1. Let c_s = s(s+1). Prove that:         (c_m+1 - c_k)(c_m+2 - c_k) ... (c_m+n - c_k)

is divisible by the product c₁c₂ ... c_n.

B1.  A₀B₀C₀ and A₁B₁C₁ are acute-angled triangles. Construct the triangle ABC with the largest possible area which is circumscribed about A₀B₀C₀ (BC contains A₀, CA contains B₀, and AB contains C₀) and similar to A₁B₁C₁.

B2.  a₁, ... , a₈ are reals, not all zero. Let c_n = a₁ⁿ + a₂ⁿ + ... + a₈ⁿ for n = 1, 2, 3, ... . Given that an infinite number of c_n are zero, find all n for which c_n is zero.

B3.  In a sports contest a total of m medals were awarded over n days. On the first day one medal and 1/7 of the remaining medals were awarded. On the second day two medals and 1/7 of the remaining medals were awarded, and so on. On the last day, the remaining n medals were awarded. How many medals were awarded, and over how many days? 

Solutions

Problem A1

The parallelogram ABCD has AB = a, AD = 1, angle BAD = A, and the triangle ABD has all angles acute. Prove that circles radius 1 and center A, B, C, D cover the parallelogram iff

            a ≤ cos A + √3 sin A.

Solution

Evidently the parallelogram is a red herring, since the circles cover it iff and only if the three circles center A, B, D cover the triangle ABD.

The three circles radius x and centers the three vertices cover an acute-angled triangle ABD iff x is at least R, the circumradius. The circumcenter O is a distance R from each vertex, so the condition is clearly necessary. If the midpoints of BD, DA, AB are P, Q, R, then the circle center A, radius R covers the quadrilateral AQOR, the circle center B, radius R covers the quadrilateral BROP, and the circle center D radius R covers the quadrilateral DPOQ, so the condition is also sufficient.

We need an expression for R in terms of a and A. We can express BD two ways: 2R sin A, and √(a² + 1 - 2a cos A). So a necessary and sufficient condition for the covering is 4 sin²A ≥ (a² + 1 - 2a cos A), which reduces to a ≤ cos A + √3 sin A, since cos A ≤ a (the foot of the perpendicular from D onto AB must lie between A and B). 

Problem 2

Prove that a tetrahedron with just one edge length greater than 1 has volume at most 1/8.

Solution

Let the tetrahedron be ABCD and assume that all edges except AB have length at most 1. The volume is the 1/3 x area BCD x height of A above BCD. The height is at most the height of A above CD, so we maximise the volume by taking the planes ACD and BCD to be perpendicular. If AC or AD is less than 1, then we can increase the altitude from A to CD whilst keeping BCD fixed by taking AC = AD = 1. A similar argument shows that we must have BC = BD = 1.

But the volume is also the 1/3 x area ABC x height of D above ABC, so we must adjust CD to maximise this height. We want the angle between planes ABC and ABD to be as close as possible to 90^o. The angle increases with increasing CD until it becomes 90^o. CMD is then a right-angled triangle. Now the angle ACB must be less than the angle between the planes ACD and BCD and hence < 90^o, so angle ACM < 45^o, so CM > 1/√2. Similarly DM. Hence when CMD = 90^o we have CD > 1. Thus we maximise the height of D above ABC by taking CD = 1.

So BCD is equilateral with area (√3)/4. ACD is also equilateral with altitude (√3)/2. Since the planes ACD and BCD are perpendicular, that is also the height of A above BCD. So the volume is 1/3 x(√3)/4 x (√3)/2 = 1/8.

Problem A3

Let k, m, n be natural numbers such that m + k + 1 is a prime greater than n + 1. Let c_s = s(s+1). Prove that:

        (c_m+1 - c_k)(c_m+2 - c_k) ... (c_m+n - c_k)

is divisible by the product c₁c₂ ... c_n.

Solution

The key is that c_a - c_b = (a - b)(a + b + 1). Hence the product (c_m+1 - c_k)(c_m+2 - c_k) ... (c_m+n - c_k) is the product of the n consecutive numbers (m - k + 1), ... , (m - k + n), times the product of the n consecutive numbers (m + k + 2), ... , (m + k + n + 1). The first product is just the binomial coefficient (m-k+n)Cn times n!, so it is divisible by n!. The second product is 1/(m + k + 1) x (m + k + 1)(m + k + 2) ... (m + k + n + 1) = 1/(m + k + 1) x (m+k+n+1)C(n+1) x (n+1)!. But m + k + 1 is a prime greater than n + 1, so it has no factors in common with (n+1)!, hence the second product is divisible by (n+1)!. Finally note that c₁c₂ ... c_n= n! (n+1)!. 

Problem B1

A₀B₀C₀ and A₁B₁C₁ are acute-angled triangles. Construct the triangle ABC with the largest possible area which is circumscribed about A₀B₀C₀ (BC contains A₀, CA contains B₀, and AB contains C₀) and similar to A₁B₁C₁.

Solution

Take any triangle similar to A₁B₁C₁ and circumscribing A₀B₀C₀. For example, take an arbitrary line through A₀ and then lines through B₀ and C₀ at the appropriate angles to the first line. Label the triangle's vertices X, Y, Z so that A₀ lies on YZ, B₀ on ZX, and C₀ on XY. Now any circumscribed ABC (labeled with the same convention) must have C on the circle through A₀, B₀ and Z, because it has ∠C = ∠Z = ∠C₁. Similarly it must have B on the circle through C₀, A₀ and Y, and it must have A on the circle through B₀, C₀ and X.

Consider the side AB. It passes through C₀. Its length is twice the projection of the line joining the centers of the two circles onto AB (because each center projects onto the midpoint of the part of AB that is a chord of its circle). But this projection is maximum when it is parallel to the line joining the two centers. The area is maximised when AB is maximised (because all the triangles are similar), so we take AB parallel to the line joining the centers. [Note, in passing, that this proves that the other sides must also be parallel to the lines joining the respective centers and hence that the three centers form a triangle similar to A₁B₁C₁.] 

Problem B2

a₁, ... , a₈ are reals, not all zero. Let c_n = a₁ⁿ + a₂ⁿ + ... + a₈ⁿ for n = 1, 2, 3, ... . Given that an infinite number of c_n are zero, find all n for which c_n is zero.

Solution

Take |a₁| ≥ |a₂| ≥ ... ≥ |a₈|. Suppose that |a₁|, ... , |a_r| are all equal and greater than |a_r+1|. Then for sufficiently large n, we can ensure that |a_s|ⁿ < 1/8 |a₁|ⁿ for s > r, and hence the sum of |a_s|ⁿ for all s > r is less than |a₁|ⁿ. Hence r must be even with half of a₁, ... , a_r positive and half negative.

If that does not exhaust the a_i, then in a similar way there must be an even number of a_i with the next largest value of |a_i|, with half positive and half negative, and so on. Thus we find that c_n = 0 for all odd n. 

Problem B3

In a sports contest a total of m medals were awarded over n days. On the first day one medal and 1/7 of the remaining medals were awarded. On the second day two medals and 1/7 of the remaining medals were awarded, and so on. On the last day, the remaining n medals were awarded. How many medals were awarded, and over how many days?

Solution

Let the number of medals remaining at the start of day r be m_r. Then m₁ = m, and 6(m_k - k)/7 = m_k+1 for k < n with m_n = n.

After a little rearrangement, we find that m = 1 + 2(7/6) + 3(7/6)² + ... + n(7/6)^n-1. Summing, we get m = 36(1 - (n + 1)(7/6)ⁿ + n (7/6)ⁿ⁺¹) = 36 + (n - 6)7ⁿ/6^n-1. 6 and 7 are coprime, so 6^n-1 must divide n - 6. But 6^n-1 > n - 6, so n = 6 and m = 36.

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.