5th Canadian Mathematical Olympiad Problems 1973

5th Canadian Mathematical Olympiad Problems 1973

1.  (1) For what x do we have x < 0 and x < 1/(4x) ? (2) What is the greatest integer n such that 4n + 13 < 0 and n(n+3) > 16? (3) Give an example of a rational number between 11/24 and 6/13. (4) Express 100000 as a product of two integers which are not divisible by 10. (5) Find 1/log₂36 + 1/log₃36.

2.  Find all real numbers x such that x + 1 = |x + 3| - |x - 1|.

3.  Show that if p and p+2 are primes then p = 3 or 6 divides p+1.

4.  Let P₀, P₁, ... , P₈ be a convex 9-gon. Draw the diagonals P₀P₃, P₀P₆, P₀P₇, P₁P₃, P₄P₆, thus dividing the 9-gon into seven triangles. How many ways can we label these triangles from 1 to 7, so that P_n belongs to triangle n for n = 1, 2, ... , 7.

5.  Let s(n) = 1 + 1/2 + 1/3 + ... + 1/n. Show that s(1) + s(2) + ... + s(n-1) = n s(n) - n.

6.  C is a circle with chord AB (not a diameter). XY is any diameter. Find the locus of the intersection of the lines AX and BY.

7.  Let a_n = 1/(n(n+1) ). (1) Show that 1/n = 1/(n+1) + a_n. (2) Show that for any integer n > 1 there are positive integers r < s such that 1/n = a_r + a_r+1 + ... + a_s.

Solutions

Problem 1

(1) For what x do we have x < 0 and x < 1/(4x) ? (2) What is the greatest integer n such that 4n + 13 < 0 and n(n+3) > 16? (3) Give an example of a rational number between 11/24 and 6/13. (4) Express 100000 as a product of two integers which are not divisible by 10. (5) Find 1/log₂36 + 1/log₃36.

Solution

(1) x < -2. (2) n = -6. (3) 17/37. (4) 32 x 3125. (5) Let k = log₂3, then 1/k = log₃2. We have log₂36 = log₂4 + log₂9 = 2 + 2k, log₃36 = log₃9 + log₃4 = 2 + 2/k = 2(1+k)/k. So 1/log₂36 + 1/log₃36 = 1/(2(1+k)) + k/(2(1+k)) = 1/2.

Problem 2

Find all real numbers x such that x + 1 = |x + 3| - |x - 1|.

Solution

If x ≥ 1, then x+1 = x+3 - (x-1) = 4, so x = 3. If -3 ≤ x < 1, then x+1 = (x+3) - (1-x) = 2x+2, so x = -1. If x < -3, then x+1 = -(x+3) - (1-x) = -4, so x = -5.

Problem 3

Show that if p and p+2 are primes then p = 3 or 6 divides p+1.

Solution

One of p, p+1, p+2 must be a multiple of 3, for p > 3, p and p+2 cannot be, so p+2 must be a multiple of 3. Similarly, p+1 or p+2 must be even, p+1 cannot be, so p+1 must be even.

Problem 4

Let P₀, P₁, ... , P₈ be a convex 9-gon. Draw the diagonals P₀P₃, P₀P₆, P₀P₇, P₁P₃, P₄P₆, thus dividing the 9-gon into seven triangles. How many ways can we label these triangles from 1 to 7, so that P_n belongs to triangle n for n = 1, 2, ... , 7.

Solution

P₀P₇P₈ must be 7. Hence P₀P₆P₇ must be 6. Hence P₀P₃P₆ must be 3. Hence P₀P₁P₃ must be 1. Hence P₁P₂P₃ must be 2. Also P₃P₄P₆ must be 4, so P₄P₅P₆ must be 5. So there is only one way.

Problem 5

Let s(n) = 1 + 1/2 + 1/3 + ... + 1/n. Show that s(1) + s(2) + ... + s(n-1) = n s(n) - n.

Solution

Induction. Obviously true for n = 1 (taking lhs to be empty) and n = 2. Suppose it is true for n. Then s(1) + s(2) + ... + s(n) = n s(n) - n + s(n) = (n+1) s(n) - n = (n+1)s(n+1) - (n+1).

Problem 6

C is a circle with chord AB (not a diameter). XY is any diameter. Find the locus of the intersection of the lines AX and BY.

Solution

Answer: the orthogonal circle through A and B.

Let AX and BY intersect at P. Suppose X and Y are both on the major arc AB. ∠BAY + ∠ABY = 180^o - ∠AYB. So ∠ABY + ∠BAX = 270^o - ∠AYB. Hence ∠PAB + ∠PBA = 90 + ∠AYB, so ∠APB = 90 - ∠AYB. But ∠AYB is constant, so ∠APB is constant and hence P lies on a circle through A and B. If we take the limit as X tends to A, then AY becomes a diameter, so ∠ABY becomes 90^o, so AP becomes a diameter. But ∠YAP = 180^o - ∠APB - ∠AYB = 90^o, so the circles are orthogonal.

A similar argument works if we take Y on the major arc, but X on the minor arc. So any point on the locus must belong to the orthogonal circle.

Let C have center O and the orthogonal circle through A and B have center O'. Let the lines PA and PB meet C again at X and Y respectively. We show that XY is a diameter (so that P belongs to the locus).

If P is any point on the minor arc AB, the angle between BY and the tangent OB equals ∠PAB. The angle betwen BX and the tangent O'B equals ∠BAX. But PAB and BAX are the same angle and OB and O'B are perpendicular, so BX and BY are perpendicular. Hence XY is a diameter.

If P is any point on the major arc AB, then the angle between BY and the tangent O'B equals angle BAY. The angle between BP and the tangent OB equals ∠BAP. But these two angles sum to 90^o, so ∠BAY + ∠BAP = 90^o. Hence AY is perpendicular to AX, so XY is a diameter.

Problem 7

Let a_n = 1/(n(n+1) ). (1) Show that 1/n = 1/(n+1) + a_n. (2) Show that for any integer n > 1 there are positive integers r < s such that 1/n = a_r + a_r+1 + ... + a_s.

Solution

(1) 1/(n+1) + a_n = (n + 1)/(n(n+1) ) = 1/n. (2) We have a_r + a_r+1 + ... + a_s = 1/r - 1/(r+1) + 1/(r+1) - 1/(r+2) + ... + 1/s - 1/(s+1) = 1/r - 1/(s+1). Take r = n-1 and s = n(n-1) - 1. Then 1/r - 1/(s+1) = 1/(n-1) - 1/(n(n-1) ) = 1/n.