20th Canadian Mathematical Olympiad Problems 1988

20th Canadian Mathematical Olympiad Problems 1988

1.  For what real values of k do 1988x² + kx + 8891 and 8891x² + kx + 1988 have a common zero?

2.  Given a triangle area A and perimeter p, let S be the set of all points a distance 5 or less from a point of the triangle. Find the area of S.

3.  Given n > 4 points in the plane, some of which are colored red and the rest black. No three points of the same color are collinear. Show that we can find three points of the same color, such that two of the points do not have a point of the opposite color on the segment joining them.

4.  Define two integer sequences a₀, a₁, a₂, ... and b₀, b₁, b₂, ... as follows. a₀ = 0, a₁ = 1, a_n+2 = 4a_n+1 - a_n, b₀ = 1, b₁ = 2, b_n+2 = 4b_n+1 - b_n. Show that b_n² = 3a_n² + 1.

5.  If S is a sequence of positive integers let p(S) be the product of the members of S. Let m(S) be the arithmetic mean of p(T) for all non-empty subsets T of S. S' is formed from S by appending an additional positive integer. If m(S) = 13 and m(S') = 49, find S'.

Solutions

Problem 1

For what real values of k do 1988x² + kx + 8891 and 8891x² + kx + 1988 have a common zero?

Solution

If x is a common zero, then 1988x² + kx + 8891 = 0, and 8891x² + kx + 1988 = 0. Subtracting one equation from the other gives 6903(x² - 1) = 0, so x = 1 or -1. Substituting back into the first equation gives k = ±10879. Check: for k = 10879, the polynomials are (1998x + 8891)(x + 1) and (8891x + 1998)(x + 1); or k = -10879, the polynomials are (1998x - 8891)(x - 1) and (8891x - 1998)(x - 1). In both cases, there is indeed a common zero.

Problem 2

Given a triangle area A and perimeter p, let S be the set of all points a distance 5 or less from a point of the triangle. Find the area of S.

Solution

S is the triangle itself (and all its interior points), a rectangle width 5 on each side and the sector of a circle radius 5 at each vertex to join together the two rectangles. The three sectors together form a circle. Hence the area of S is A + 5p + 25 π.

Problem 3

Given n > 4 points in the plane, some of which are colored red and the rest black. No three points of the same color are collinear. Show that we can find a three points of the same color, such that two of the points do not have a point of the opposite color on the segment joining them.

Solution

There are at least 5 points, so there must be at least 3 of the same color. Label them A₁, B₁, C₁. If one pair does not have a point of the opposite color on the segment joining them, then we are done. If not, take the three points of the opposite color to be A₂, B₂, C₂. Now repeat. The process must terminate because at each step we get three new points (none of the previous points can lie on the sides of A_nB_nC_n, because all points of A_nB_nC_n except the vertices lie inside the triangle A_n-1B_n-1C_n-1). So for some n A_n, B_n, C_n are the required points.

Problem 4

Define two integer sequences a₀, a₁, a₂, ... and b₀, b₁, b₂, ... as follows. a₀ = 0, a₁ = 1, a_n+2 = 4a_n+1 - a_n, b₀ = 1, b₁ = 2, b_n+2 = 4b_n+1 - b_n. Show that b_n² = 3a_n² + 1.

Solution

We note that a₂ = 4 and b₂ = 7. Let S_n be the statement that b_n² = 3a_n² + 1 and b_nb_n-1 = 3a_na_n-1 + 2. So S₁ is true, because b₁² = 4 = 3a₁² + 1 and b₁b₀ = 2 = 3a₁a₀ + 2. S₂ is also true, because b₂² = 49 = 3a₂² + 1 and b₂b₁ = 14 = 3a₂a₁ + 2.

We show that S_n is true for all n by induction. Suppose it is true for n and n-1. We have b_n+1b_n = b_n(4b_n - b_n-1) = 4b_n² - b_nb_n-1 = 12a_n² + 4 - (3a_na_n-1 + 2) = 3a_n(4a_n - a_n-1) + 2 = 3a_na_n+1 + 2, as required. Similarly, b_n+1² - 3a_n+1² = (4b_n - b_n-1)² - 3(4a_n - a_n-1)² = 16(b_n² - 3_n² - 1) - 8(b_nb_n-1 - 3a_na_n-1 - 2) + b_n-1² - 3a_n-1² = 0 - 0 + 1, as required.

Problem 5

If S is a sequence of positive integers let p(S) be the product of the members of S. Let m(S) be the arithmetic mean of p(T) for all non-empty subsets T of S. S' is formed from S by appending an additional positive integer. If m(S) = 13 and m(S') = 49, find S'.

Solution

Suppose there are n elements in S, namely a₁, a₂, ... , a_n. Then m(S) = (K - 1)/(2ⁿ - 1), where K = (a₁ + 1)(a₂ + 1) ... (a_n + 1). If the additional integer is h, then m(S') = (K(h + 1) - 1)/(2ⁿ⁺¹ - 1). So we have K = 13(2ⁿ - 1) + 1, (h + 1)K = 49(2ⁿ⁺¹ - 1) + 1. Hence h + 1 = (49·2ⁿ⁺¹ - 48)/(13·2ⁿ - 12). But 7(13·2ⁿ - 1) = 91·2ⁿ - 84 < 98.2ⁿ - 48 = 49·2ⁿ⁺¹ - 48. For n > 3, 8(13·2ⁿ - 12) = (52·2ⁿ⁺¹ - 96) > (49·2ⁿ⁺¹ - 48). So we must have n = 1, 2, or 3. If n = 1, then h+1 = 148/14, which is not an integer. If n = 2, then h+1 = 344/40, which is not an integer. If n = 3, then h+1 = 736/92 = 8, which is ok.

If n = 3, then if the elements of S are a, b, c, we must have (a + 1)(b + 1)(c + 1) = 92 = 2.2.23. So S = {2, 2, 23} and S' = {2, 2, 7, 23}.