11th International Mathematical Olympiad 1969 Problems & Solutions

A1.  Prove that there are infinitely many positive integers m, such that n⁴ + m is not prime for any positive integer n.

A2.  Let f(x) = cos(a₁ + x) + 1/2 cos(a₂ + x) + 1/4 cos(a₃ + x) + ... + 1/2^n-1 cos(a_n + x), where a_i are real constants and x is a real variable. If f(x₁) = f(x₂) = 0, prove that x₁ - x₂ is a multiple of π.

A3.  For each of k = 1, 2, 3, 4, 5 find necessary and sufficient conditions on a > 0 such that there exists a tetrahedron with k edges length a and the remainder length 1.

B1.  C is a point on the semicircle diameter AB, between A and B. D is the foot of the perpendicular from C to AB. The circle K₁ is the in-circle of ABC, the circle K₂ touches CD, DA and the semicircle, the circle K₃ touches CD, DB and the semicircle. Prove that K₁, K₂ and K₃ have another common tangent apart from AB.

B2.  Given n > 4 points in the plane, no three collinear. Prove that there are at least (n-3)(n-4)/2 convex quadrilaterals with vertices amongst the n points.

B3.  Given real numbers x₁, x₂, y₁, y₂, z₁, z₂, satisfying x₁ > 0, x₂ > 0, x₁y₁ > z₁², and x₂y₂ > z₂², prove that:       8/((x₁ + x₂)(y₁ + y₂) - (z₁ + z₂)²) ≤ 1/(x₁y₁ - z₁²) + 1/(x₂y₂ - z₂²).

Give necessary and sufficient conditions for equality. 

Solutions

Problem A1

Prove that there are infinitely many positive integers m, such that n⁴ + m is not prime for any positive integer n.

Solution

n⁴ + 4 r⁴ = (n² + 2rn + 2r²)(n² - 2rn + 2r²). Clearly the first factor is greater than 1, the second factor is (n - r)² + r², which is also greater than 1 for r greater than 1. So we may take m = 4 r⁴ for any r greater than 1. 

Problem A2

Let f(x) = cos(a₁ + x) + 1/2 cos(a₂ + x) + 1/4 cos(a₃ + x) + ... + 1/2^n-1 cos(a_n + x), where a_i are real constants and x is a real variable. If f(x₁) = f(x₂) = 0, prove that x₁ - x₂ is a multiple of π.

Solution

f is not identically zero, because f(-a₁) = 1 + 1/2 cos(a₂ - a₁) + ... > 1 - 1/2 - 1/4 - ... - 1/2^n-1 > 0.

Using the expression for cos(x + y) we obtain f(x) = b cos x + c sin x, where b = cos a₁ + 1/2 cos a₂ + ... + 1/2^n-1 cos a_n, and c = - sin a₁ - 1/2 sin a₂ - ... - 1/2^n-1 sin a_n. b and c are not both zero, since f is not identically zero, so f(x) = √(b² + c²) cos(d + x), where cos d = b/√(b² + c²), and sin d = c/√(b² + c²). Hence the roots of f(x) = 0 are just mπ + π/2 - d. 

Problem A3

For each of k = 1, 2, 3, 4, 5 find necessary and sufficient conditions on a > 0 such that there exists a tetrahedron with k edges length a and the remainder length 1.

Solution

A plodding question. Take the tetrahedron to be ABCD.

Take k = 1 and AB to have length a, the other edges length 1. Then we can hinge triangles ACD and BCD about CD to vary AB. The extreme values evidently occur with A, B, C, D coplanar. The least value, 0, when A coincides with B, and the greatest value √3, when A and B are on opposite sides of CD. We rule out the extreme values on the grounds that the tetrahedron is degenerate, thus obtaining 0 < a < √3.

For k = 5, the same argument shows that 0 < 1 < √3 a, and hence a > 1/√3.

For k = 2, there are two possible configurations: the sides length a adjacent, or not. Consider first the adjacent case. Take the sides length a to be AC and AD. As before, the two extreme cases gave A, B, C, D coplanar. If A and B are on opposite sides of CD, then a = √(2 - √3). If they are on the same side, then a = √(2 + √3). So this configuration allows any a satisfying √(2 - √3) < a < √(2 + √3).

The other configuration has AB = CD = a. One extreme case has a = 0. We can increase a until we reach the other extreme case with ADBC a square side 1, giving a = √2. So this configuration allows any a satisfying 0 < a < √2. Together, the two configurations allow any a satisfying: 0 < a < √(2 + √3).

This also solves the case k = 4, and allows any a satisfying: a > 1/√(2 + √3) = √(2 - √3).

For k = 3, any value of a > 0 is allowed. For a <= 1, we may take the edges length a to form a triangle. For a ≥ 1 we may take a triangle with unit edges and the edges joining the vertices to the fourth vertex to have length a.  

Problem B1

C is a point on the semicircle diameter AB, between A and B. D is the foot of the perpendicular from C to AB. The circle K₁ is the in-circle of ABC, the circle K₂ touches CD, DA and the semicircle, the circle K₃ touches CD, DB and the semicircle. Prove that K₁, K₂ and K₃ have another common tangent apart from AB.

Solution

Let the three centers be O₁, O₂ and O₃. We show that O₁ is the midpoint of O₂O₃. In fact it is sufficient to show that O₁ lies on O₂O₃, because then we can reflect the known tangent AB in the line O₂O₃.

As usual, let AB = c, BC = a, CA = b. Let the in-circle touch AB at P, AC at Q and BC at R. Then since angle ACB = 90, O₁QCR is a square. Also AQ = AP and BP = BR, so r₁ = b - AP, and r₁ = a - BP = a - (c - AP). Adding: r₁ = (a + b - c)/2, and AP = (b + c - a)/2.

Let the circle center O₂ touch AB at X, and the circle center O₃ touch AB at Y. Let O be the midpoint of AB. Now consider the right-angled triangle OXO₂. Since the circle center O₂ touches the semicircle, OO₂ = c/2 - r₂. OX = OD + DX = (c/2 - AD) + r₂. Also, by similar triangles, AD = b²/c. So, using Pythagoras: (c/2 - r₂)² = r₂² + (c/2 - b²/c + r₂)². Multiplying out and rearranging: r₂² - 2r₂(c - b²/c) - (b² - b⁴/c²). But ABC is right-angled, so c² = a² + b², and hence c - b²/c = a²/c and b² - b⁴/c² = a²b²/c². So r₂² + 2r₂ a²/c - a²b²/c² = 0, which has roots r₂ = a - a²/c (positive) and - a + a²/c (negative). So r₂ = a - a²/c. Similarly, r₃ = b - b²/c. So O₂X + O₃Y = XY = r₂ + r₃ = a + b - c = 2 r₁.

XP = AP - AX = AP - (AD - DX) = (b + c - a)/2 - (b²/c - r₂) = (b + c - a)/2 - (c - a) = (a + b - c)/2 = r₁. We now have all we need: XP = PY = PO₁, and XO₂ + YO₃ = 2 PO₁.

 Problem B2

Given n > 4 points in the plane, no three collinear. Prove that there are at least (n-3)(n-4)/2 convex quadrilaterals with vertices amongst the n points.

Solution

(n-3)(n-4)/2 is a poor lower bound.

Observe first that any 5 points include 4 forming a convex quadrilateral. For take the convex hull. If it consists of more than 3 points, we are done. If not, it must consist of 3 points, A, B and C, with the other 2 points, D and E, inside the triangle ABC. Two vertices of the triangle must lie on the same side of the line DE and they form convex quadrilateral with D and E.

Given n points, we can choose 5 in n(n-1)(n-2)(n-3)(n-4)/120 different ways. Each choice gives us a convex quadrilateral, but any given convex quadrilateral may arise from n-4 different sets of 5 points, so we have at least n(n-1)(n-2)(n-3)/120 different convex quadrilaterals. We now show that n(n-1)(n-2)(n-3)/120 ≥ (n-3)(n-4)/2 for all n ≥ 5.

We wish to prove that n(n-1)(n-2) ≥ 60(n-4), or n(n-1)(n-2) - 60(n-4) ≥ 0. Trial shows equality for n = 5 and 6, so we can factorise and get (n-5)(n-6)(n+8), which is clearly at least 0 for n at least 5. 

Problem B3

Given real numbers x₁, x₂, y₁, y₂, z₁, z₂, satisfying x₁ > 0, x₂ > 0, x₁y₁ > z₁², and x₂y₂ > z₂², prove that:

      8/((x₁ + x₂)(y₁ + y₂) - (z₁ + z₂)²) ≤ 1/(x₁y₁ - z₁²) + 1/(x₂y₂ - z₂²).

Give necessary and sufficient conditions for equality.

Solution

Let a₁ = x₁y₁ - z₁² and a₂ = x₂y₂ - z₂². We apply the arithmetic/geometric mean result 3 times:

(1) to a₁², a₂², giving 2a₁a₂ ≤ a₁² + a₂²;

(2) to a₁, a₂, giving √(a₁a₂) ≤ (a₁ + a₂)/2;

(3) to a₁y₂/y₁, a₂y₁/y₂, giving √(a₁a₂) ≤ (a₁y₂/y₁ + a₂y₁/y₂)/2;

We also use (z₁/y₁ - z₂/y₂)² ≥ 0. Now x₁y₁ > z₁² ≥ 0, and x₁ > 0, so y₁ > 0. Similarly, y₂ > 0. So:

(4) y₁y₂(z₁/y₁ - z₂/y₂)² ≥ 0, and hence z₁²y₂/y₁ + z₂²y₁/y₂ ≥ 2z₁z₂.

Using (3) and (4) gives 2√(a₁a₂) ≤ (x₁y₂ + x₂y₁) - (z₁²y₂/y₁ + z₂²y₁/y₂) ≤ (x₁y₂ + x₂y₁ - 2z₁z₂).

Multiplying by (2) gives: 4a₁a₂ ≤ (a₁ + a₂)(x₁y₂ + x₂y₁ - 2z₁z₂).

Adding (1) and 2a₁a₂ gives: 8a₁a₂ ≤ (a₁ + a₂)² + (a₁ + a₂)(x₁y₂ + x₂y₁ - 2z₁z₂) = a(a₁ + a₂), where a = (x₁ + x₂)(y₁ + y₂) - (z₁ + z₂)². Dividing by a₁a₂a gives the required inequality.

Equality requires a₁ = a₂ from (1), y₁ = y₂ from (2), z₁ = z₂ from (3), and hence x₁ = x₂. Conversely, it is easy to see that these conditions are sufficient for equality. 

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.