12th Canadian Mathematical Olympiad Problems 1980

12th Canadian Mathematical Olympiad Problems 1980

1.  If the 5-digit decimal number a679b is a multiple of 72 find a and b.

2.  The numbers 1 to 50 are arranged in an arbitrary manner into 5 rows of 10 numbers each. Then each row is rearranged so that it is in increasing order. Then each column is arranged so that it is in increasing order. Are the rows necessarily still in increasing order?

3.  Find the triangle with given angle A and given inradius r with the smallest perimeter.
4.  A fair coin is tossed repeatedly. At each toss 1 is scored for a head and 2 for a tail. Show that the probability that at some point the score is n is (2 + (-1/2)ⁿ)/3.
5.  Do any polyhedra other than parallelepipeds have the property that all cross sections parallel to any given face have the same perimeter?

Solutions

Problem 1

If the 5-digit decimal number a679b is a multiple of 72 find a and b.

Solution

100 = 28 mod 72, 1000 = -8 mod 72, 10000 = -8 mod 72, so we require -8a - 48 + 196 + 90 + b = 0 mod 72, or 22 - 8a + b = 0 mod 72. We have -72 < -50 = 22 - 8.9 + 0 ≤ 22 - 8a + b ≤ 22 + 0 + 9 = 31 < 72, so 22 - 8a + b = 0, or 8a = 22 + b. So b = 2, a = 3. Check: 36792 = 511·72.

Problem 2

The numbers 1 to 50 are arranged in an arbitrary manner into 5 rows of 10 numbers each. Then each row is rearranged so that it is in increasing order. Then each column is arranged so that it is in increasing order. Are the rows necessarily still in increasing order?

Solution

Answer: yes.

After the row rearrangement, let the numbers be a₁, a₂, ... , a₁₀ in the first row, b₁, b₂, ... , b₁₀ in the second row and so on.

Suppose that after the column rearrangement we have x_i in column i of row k (where x is one of a, b, c, d, e) and y_j in column j of row k (where j > i and y is one of a, b, c, d, e). We wish to show that x_i < y_j. There is one more numbers in column j   in row k or above than there are in column i   above row k. So there must be some z_j in column j and in row k or above with the corresponding z_i in column i in row k or below (z may equal x or y). So x_i ≤ z_i < z_j ≤ y_j.

Problem 3

Find the triangle with given angle A and given inradius r with the smallest perimeter.

Solution

It is fairly obvious that we take angle B = angle C. But we need to prove it.

The perimeter is 2r (cot A/2 + cot B/2 + cot C/2). Both 2r and cot A/2 are fixed, so we have to minimise cot B/2 + cot C/2. Write B/2 = x + y, C/2 = x - y, where x = (B+C)/4, y = (B-C)/4. So we have cot(x + y) + cot(x - y) = (1 - tan x tan y)/(tan x + tan y) + (1 + tan x tan y)/(tan x - tan y) = (2 tan x + 2 tan x tan²y)/(tan²x - tan²y). Now tan x is fixed (because A and hence B+C is fixed), so we both minimise the numerator and maximise the denominator by taking B = C.

Problem 4

A fair coin is tossed repeatedly. At each toss 1 is scored for a head and 2 for a tail. Show that the probability that at some point the score is n is (2 + (-1/2)ⁿ)/3.

Solution

Induction on n. True for n = 1 because a score of 1 can only be achieved after the frist toss - thereafter the score must exceed 1. Suppose the result is true for n.

If a score of n is achieved, then a score of n+1 can only be achieved by tossing a head on the next throw. If a score of n is not achieved, then a score of n+1 must be achieved, because the gap between scores can be at most 2. Hence the prob of achieving n+1 = (2 + (-1/2)ⁿ)/6 + (1 - (2 + (-1/2)ⁿ)/3 = 2/3 + (1/3 - 1/6) (-1/2)ⁿ⁺¹ = (2 + (-1/2)ⁿ⁺¹)/3.

Problem 5

Do any polyhedra other than parallelepipeds have the property that all cross sections parallel to any given face have the same perimeter?

Solution

Yes. The octahedron. Suppose it has edge length 1. You can cut along edges and fold the faces flat, so that the six faces with a vertex in common with the top face form a 1 x 3 parallelogram. A plane parallel to the top face cuts this parallelogram in a line of length 3 parallel to the long sides.