7th Canadian Mathematical Olympiad Problems 1975

7th Canadian Mathematical Olympiad Problems 1975

1.  Evaluate (1·2·4 + 2·4·8 + 3·6·12 + 4·8·16 + ... + n·2n·4n)^1/3/(1·3·9 + 2·6·18 + 3·9·27 + 4·12·36 + ... + n·3n·9n)^1/3.

2.  Define the real sequence a₁, a₂, a₃, ... by a₁ = 1/2, n²a_n = a₁ + a₂ + ... + a_n. Evaluate a_n.

3.  Sketch the points in the x, y plane for which [x]² + [y]² = 4.

4.  Find all positive real x such that x - [x], [x], x form a geometric progression.

5.  Four points on a circle divide it into four arcs. The four midpoints form a quadrilateral. Show that its diagonals are perpendicular.

6.  15 guests with different names sit down at a circular table, not realizing that there is a name card at each place. Everyone is in the wrong place. Show that the table can be rotated so that at least two guests match their name cards. Give an example of an arrangement where just one guest is correct, but rotating the table does not improve the situation.

7.  Is sin(x²) periodic?

8.  Find all real polynomials p(x) such that p(p(x) ) = p(x)ⁿ for some positive integer n. 

Solutions

Problem

Evaluate (1·2·4 + 2·4·8 + 3·6·12 + 4·8·16 + ... + n·2n·4n)^1/3/(1·3·9 + 2·6·18 + 3·9·27 + 4·12·36 + ... + n·3n·9n)^1/3.

Solution

The numerator cubed is 8(1³ + 2³ + ... + n³), the denominator cubed is 27(1³ + 2³ + ... + n³). Hence the expression is 2/3. 

Problem 2

Define the real sequence a₁, a₂, a₃, ... by a₁ = 1/2, n²a_n = a₁ + a₂ + ... + a_n. Evaluate a_n.

Solution

We show by induction that a_n = 1/(n(n+1) ). Obviously true for n = 1. Suppose it is true for n. We need to show that (n+1)²/( (n+1)(n+2) ) - n²/( n(n+1) ) = 1/( (n+1)(n+2) ), or (n+1)/(n+2) - n/(n+1) = 1/( (n+1)(n+2) ), or (n+1)² - n(n+2) = 1, which is true. 

Problem 3

Sketch the points in the x, y plane for which [x]² + [y]² = 4.

Solution

There are four squares with vertices at lattice points and centered at (2 1/2, 1/2), (1/2, 2 1/2), (-1 1/2, 1/2), (1/2, -1 1/2). The first square includes all its interior points, the vertex (2, 0) (but none of the others), all other points on the bottom and left-hand boundaries, but no points on the top or right-hand boundaries. The others are similar. 

Problem 4

Find all positive real x such that x - [x], [x], x form a geometric progression.

Solution

Let [x] = n. We have x - n < 1, so if n ≥ 2, then n/(x - n) > 2, but x/n < 2. We must have n ≥ 1, otherwise n < x - n. Hence n = 1. Let x = 1 + y. Then 1/y = 1 + y, so y = (√5 - 1)/2 and x = (√5 + 1)/2. 

Problem 5

Four points on a circle divide it into four arcs. The four midpoints form a quadrilateral. Show that its diagonals are perpendicular.

Solution

Let the circle have center O and the four points be A, B, C, D going around the circle counter-clockwise. Measure angles at O counter-clockwise from A, so angle A = 0. Then the midpoint of the arc AB has angle B/2 and the midpoint of the arc CD has angle C/2 + D/2, so the midpoint of the chord joining them has angle (B + C + D)/4. The midpoint of the arc BC has angle B/2 + C/2 and the midpoint of the arc DA has angle 180 + D/2. Hence the midpoint of the chord joining them has angle 90 + (B + C + D)/4. So the two diagonals of the midpoint quadrilateral are perpendicular. 

Problem 6

15 guests with different names sit down at a circular table, not realizing that there is a name card at each place. Everyone is in the wrong place. Show that the table can be rotated so that at least two guests match their name cards. Give an example of an arrangement where just one guest is correct, but rotating the table does not improve the situation.

Solution

Pigeon-hole principle. There are 15 possible positions for the table. Everyone must be correct in one position. No one is correct in the initial position, so 15 people are correct in 14 positions, so at least two people must be correct in the same position.

Consider the arrangement:

card     1   2   3   4   5   6   7   8   9  10  11  12  13  14  15

person   1   9   2  10   3  11   4  12   5  13   6  14   7  15   8

A rotation of 1 brings 2 into line, a rotation of 3 brings 2 into line and so on. So no rotation lines up two or more guests. 

Problem 7

Is sin(x²) periodic?

Solution

No. The zeros are √(nπ), but these do not form an arithmetic progression. 

Problem 8

Find all real polynomials p(x) such that p(p(x) ) = p(x)ⁿ for some positive integer n.

Solution

Let p(x) have highest term a x^m. Then the highest term in p(p(x) ) is a^m+1x^M, where M = m², and the highest term in p(x)ⁿ is aⁿx^mn. Hence m = n and a = 1. But that means p(p(x) ) = p(x)ⁿ + the other terms, so the other terms must all be zero. Hence p(x) = xⁿ.