10th International Mathematical Olympiad 1968 Problems & Solutions

A1.  Find all triangles whose side lengths are consecutive integers, and one of whose angles is twice another.

A2.  Find all natural numbers n the product of whose decimal digits is n² - 10n - 22.

A3.  a, b, c are real with a non-zero. x₁, x₂, ... , x_n satisfy the n equations:         ax_i² + bx_i + c = x_i+1, for 1 ≤ i < n

        ax_n² + bx_n + c = x₁

Prove that the system has zero, 1 or >1 real solutions according as (b - 1)² - 4ac is <0, =0 or >0.

B1.  Prove that every tetrahedron has a vertex whose three edges have the right lengths to form a triangle.

B2.  Let f be a real-valued function defined for all real numbers, such that for some a > 0 we have         f(x + a) = 1/2 + √(f(x) - f(x)²) for all x.

Prove that f is periodic, and give an example of such a non-constant f for a = 1.

B3.  For every natural number n evaluate the sum     [(n+1)/2] + [(n+2)/4] + [(n+4)/8] + ... + [(n+2^k)/2^k+1] + ... , where [x] denotes the greatest integer ≤ x. 

Solutions

Problem A1

Find all triangles whose side lengths are consecutive integers, and one of whose angles is twice another.

 

Solution

Let the sides be a, a+1, a+2, the angle oppose a be A, the angle opposite a+1 be B, and the angle opposite a+2 be C.

Using the cosine rule, we find cos A = (a+5)/(2a+4), cos B = (a+1)/2a, cos C = (a-3)/2a. Finally, using cos 2x = 2 cos²x - 1, we find solutions a = 4 for C = 2A, a = 1 for B = 2A, and no solutions for C = 2B.

a = 1 is a degenerate solution (the triangle has the three vertices collinear). The other solution is 4, 5, 6. 

Problem A2

Find all natural numbers n the product of whose decimal digits is n² - 10n - 22.

 

Solution

Suppose n has m > 1 digits. Let the first digit be d. Then the product of the digits is at most d.9^m-1 < d.10^m-1 <= n. But (n² - 10n - 22) - n = n(n - 11) - 22 > 0 for n >= 13. So there are no solutions for n ≥ 13. But n² - 10n - 22 < 0 for n ≤ 11, so the only possible solution is n = 12 and indeed that is a solution. 

Problem A3

a, b, c are real with a non-zero. x₁, x₂, ... , x_n satisfy the n equations:

        ax_i² + bx_i + c = x_i+1, for 1 ≤ i < n

        ax_n² + bx_n + c = x₁

Prove that the system has zero, 1 or >1 real solutions according as (b - 1)² - 4ac is <0, =0 or >0.

 

Solution

Let f(x) = ax² + bx + c - x. Then f(x)/a = (x + (b-1)/2a)² + (4ac - (b-1)²)/4a². Hence if 4ac - (b-1)² > 0, then f(x) has the same sign for all x. But f(x) > 0 means ax² + bx + c > x, so if {x_i} is a solution, then either x₁ < x₂ < ... < x_n < x₁, or x₁ > x₂ > ... > x_n > x₁. Either way we have a contradiction. So if 4ac - (b-1)² > 0 there cannot be any solutions.

If 4ac - (b-1)² = 0, then we can argue in the same way that either x₁ ≤ x₂ ≤ ... ≤ x_n ≤ x₁, or x₁ ≥ x₂ ≥ ... ≥ x_n ≥ x₁. So we must have all x_i = the single root of f(x) = 0 (which clearly is a solution).

If 4ac - (b-1)² < 0, then f(x) = 0 has two distinct real roots y and z and so we have at least two solutions to the equations: all x_i =y, and all x_i = z. We may, however, have additional solutions. For example, if a = 1, b = 0, c = -1 and n is even, then we have the additional solution x₁ = x₃ = x₅ = ... = 0, x₂ = x₄ = ... = -1. 

Problem B1

Prove that every tetrahedron has a vertex whose three edges have the right lengths to form a triangle.

 

Solution

The trick is to consider the longest side. That avoids getting into lots of different possible cases for which edge is longer than the sum of the other two.

So assume the result is false and let AB be the longest side. Then we have AB > AC + AD and BA > BC + BD. So 2AB > AC + AD + BC + BC. But by the triangle inequality, AB < AC + CB, AB < AD + DB, so 2AB < AC + CB + AD + DB. Contradiction. 

Problem B2

Let f be a real-valued function defined for all real numbers, such that for some a > 0 we have

        f(x + a) = 1/2 + √(f(x) - f(x)²) for all x.

Prove that f is periodic, and give an example of such a non-constant f for a = 1.

 

Solution

Directly from the equality given: f(x+a) ≥ 1/2 for all x, and hence f(x) ≥ 1/2 for all x.

So f(x+2a) = 1/2 + √( f(x+a) - f(x+a)² ) = 1/2 + √f(x+a) √(1 - f(x+a)) = 1/2 + √(1/4 - f(x) + f(x)²) = 1/2 + (f(x) - 1/2) = f(x). So f is periodic with period 2a.

We may take f(x) to be arbitrary in the interval [0,1). For example, let f(x) = 1 for 0 ≤ x < 1, f(x) = 1/2 for 1 ≤ x < 2. Then use f(x+2) = f(x) to define f(x) for all other values of x. 

Problem B3

For every natural number n evaluate the sum

    [(n+1)/2] + [(n+2)/4] + [(n+4)/8] + ... + [(n+2^k)/2^k+1] + ... , where [x] denotes the greatest integer ≤ x.

 

Solution

For any real x we have [x] = [x/2] + [(x+1]/2]. For if x = 2n + 1 + k, where n is an integer and 0 ≤ k < 1, then lhs = 2n + 1, and rhs = n + n + 1. Similarly, if x = 2n + k.

Hence for any integer n, we have: [n/2^k] - [n/2^k+1] = [(n/2^k + 1)/2] = [(n + 2^k)/2^k+1]. Hence summing over k, and using the fact that n < 2^k for sufficiently large k, so that [n/2^k ] = 0, we have: n = [(n + 1)/2] + [(n + 2)/4] + [(n + 4)/8] + ... .

 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.