18th Canadian Mathematical Olympiad Problems 1986

18th Canadian Mathematical Olympiad Problems 1986

1.  The triangle ABC has angle B = 90^o. The point D is taken on the ray AC, the other side of C from A, such that CD = AB. ∠CBD = 30^o. Find AC/CD.

2.  Three competitors A, B, C compete in a number of sporting events. In each event a points is awarded for a win, b points for second place and c points for third place. There are no ties. The final score was A 22, B 9, C 9. B won the 100 meters. How many events were there and who came second in the high jump?

3.  A chord AB of constant length slides around the curved part of a semicircle. M is the midpoint of AB, and C is the foot of the perpendicular from A onto the diameter. Show that angle ACM does not change.

4.  Show that (1 + 2 + ... + n) divides (1^k + 2^k + ... + n^k) for k odd.

5.  The integer sequence a₁, a₂, a₃, ... is defined by a₁ = 39, a₂ = 45, a_n+2 = a_n+1² - a_n. Show that infinitely many terms of the sequence are divisible by 1986.

Solutions

Problem 1

The triangle ABC has ∠B = 90^o. The point D is taken on the ray AC, the other side of C from A, such that CD = AB. The ∠CBD = 30^o. Find AC/CD.

Solution

Let angle ACB = x. Wlog we may take AC = 1. Then BC = cos x and AB = sin x. So CD = sin x. Now apply the sine rule to the triangle BCD: cos x/(sin(x-30^o) = sin x/sin 30^o. Hence cos x = √3 sin²x - sin x cos x. Hence (1 + sin x)²(1 - sin²x) = 3 sin⁴x. So 4 sin⁴x + 2 sin³x - 2 sin x - 1 = 0. Factorising, (2 sin³x - 1)(2 sin x + 1) = 0. But x must lie between 0 and 90^o, so we cannot have sin x = -1/2. Hence sin x = 1/2^1/3. Hence the ratio AC/CD = 2^1/3.

Problem 2

Three competitors A, B, C compete in a number of sporting events. In each event a points is awarded for a win, b points for second place and c points for third place. There are no ties. The final score was A 22, B 9, C 9. B won the 100 meters. How many events were there and who came second in the high jump?

Solution

A 5, 5, 5, 5, 2; B 5, 1, 1, 1, 1; C 2, 2, 2, 2, 1. A must have come second in the event in which B won. So C came second in every other event.

Problem 3

A chord AB of constant length slides around the curved part of a semicircle. M is the midpoint of AB, and C is the foot of the perpendicular from A onto the diameter. Show that angle ACM does not change.

Solution

Let O be the center of the semicircle. Angle OMA = angle OCA = 90^o, so OMAC is cyclic. Hence angle ACM = angle AOM, which is independent of the position of the chord since it has constant length.

Problem 4

Show that (1 + 2 + ... + n) divides (1^k + 2^k + ... + n^k) for k odd.

Solution

Assume k is odd and let N = (1^k + 2^k + ... + n^k). We have 2N = (n^k + 1^k) + ( (n-1)^k + 2^k) + ... + (1^k + n^k). But expanding (n+1 - m)^k by the binomial theorem we find that it is (-m)^k mod n+1 and hence m^k + (n+1 - m)^k is divisible by n+1. So n+1 divides 2N. Similarly, we may write 2N = 2n^k + ( (n-1)^k + 1^k) + ( (n-2)^k + 2^k) + ... + (1^k + (n-1)^k) and each term is divisible by n. But n and n+1 are relatively prime, hence n(n+1) divides 2N. Hence 1 + 2 + ... + n = n(n+1)/2 divides N.

Problem 5

The integer sequence a₁, a₂, a₃, ... is defined by a₁ = 39, a₂ = 45, a_n+2 = a_n+1² - a_n. Show that infinitely many terms of the sequence are divisible by 1986.

Solution

We find a₃ = 1986. Define b_n = residue of a_n mod 1986. Then b_n satisfies the relation b_n+2 = b_n+1² - b_n mod 1986. Now there are only 1986² possibile values for the pair (b_n, b_n+1), so some pair must repeat. But any pair of consecutive values of b_n determines the rest of the sequence and all preceding terms of the sequence (since b_n = b_n+1² - b_n+2). So the values of b_n must be periodic for some period N <= 1986². We know that the value 0 occurs for n = 3, so it must occur infinitely often.

Comment. But it takes a while, the values 39, 45 next occur at b₁₃₃₉ , b₁₃₄₀ .