15th Canadian Mathematical Olympiad Problems 1983

15th Canadian Mathematical Olympiad Problems 1983

1.  Find all solutions to n! = a! + b! + c! .

2.  Find all real-valued functions f on the reals whose graphs remain unchanged under all transformations (x, y) → (2^kx, 2^k(kx + y) ), where k is real.

3.  Is the volume of a tetrahedron determined by the areas of its faces?

4.  Show that we can find infinitely many positive integers n such that 2ⁿ - n is a multiple of any given prime p.

5.  Show that the geometric mean of a set S of positive numbers equals the geometric mean of the geometric means of all non-empty subsets of S.

Solutions

Problem 1

Find all solutions to n! = a! + b! + c! .

Solution

We must have n > a, b, c or n! < a! + b! + c! . But if n > a, b, c and n > 3, then n! > 3 (n-1)! ≥ a! + b! + c! . So we must have n = 1, 2, or 3. It is easy to check that this gives 3! = 2! + 2! + 2! as the only solution.

Problem 2

Find all real-valued functions f on the reals whose graphs remain unchanged under all transformations (x, y) → (2^kx, 2^k(kx + y) ), where k is real.

Solution

Answer: for any fixed real a, b we can take f(x) = x ln₂|x| + ax for x positive and x ln₂|x| + bx for x negative.

Suppose f(0) = h. So (0, h) is a point on the graph. Hence so is (0, 2^kh) for all k. But f is single-valued, so h must be 0.

Suppose f(1) = h. Then (1, h) is a point on the graph. Hence so is (2^k, 2^kk + 2^kh) for all real k. Put k = ln₂x. Then (x, x ln₂x + xh) is a point on the graph. In other words, f(x) = x ln₂x + hx. We need to check that this works. So we have a point (x, x ln₂x + hx). Under the transformation this goes to the point P (2^kx, 2^k(kx + x ln₂x + hx) ). Put x' = 2^kx, so ln₂x' = k + ln₂x. So we can write P as (x', x'(ln₂x' - ln₂x) + x' ln₂x + hx') = (x', x' ln₂x' + hx'), which is indeed another point on the graph.

We now have to consider negative x. So suppose f(-1) = h. So (-1, h) is a point on the graph. Hence so is (-2^k, -2^kk - 2^kh) for all real k. Put k = ln₂x. Then (-x, -x ln₂x - hx) is a point on the graph. In other words, f(x) = x ln₂|x| + hx for x negative. Again we can easily check that this works. Note that h does not have to be the same for positive and negative x.

Problem 3

Is the volume of a tetrahedron determined by the areas of its faces?

Solution

No. For example, take a tetrahedron with each side a 3, 4, 5 triangle (one pair of opposite edges is 3, another pair is 4, and the third pair is 5). Then each face has area 6 and the volume of the tetrahedron is nil. But if we take a regular tetrahedron with side 2^3/23^1/2 then each side has area 6 but the volume is non-zero.

Problem 4

Show that we can find infinitely many positive integers n such that 2ⁿ - n is a multiple of any given prime p.

Solution

We have 2^p-1 = 1 mod p. So 2^k(p-1) - k(p-1) = k+1 mod p. Thus if we take k to be hp - 1 for any integer h, then we get 2^k(p-1) - k(p-1) = 0 mod p. In other words n = (hp -1)(p - 1) works for any h. For example, 2⁴ - 4 = 12 = 0 mod 3, 2¹⁰ - 10 = 1014 = 0 mod 3, 2¹⁶ - 16 = 65520 = 0 mod 3 and so on.

Problem 5

Show that the geometric mean of a set S of positive numbers equals the geometric mean of the geometric means of all non-empty subsets of S.

Solution

A geometric mean is a product x^ay^b ... , where the exponents a, b, are positive reals with sum 1. This remains true if some of the x, y, ... are equal and we group like terms together, so that not all the a, b, are equal. It also remains true if each of the x, y, ... is itself a geometric mean and we express the product in terms of the underlying numbers. Because if (a + b + ... ) = 1, (c + d + ... ) = 1, ... and we take x(a + b + ... ) + y(c + d + ... ) + ... ,where x + y + ... = 1 and we expand then we get a collection of terms with sum 1.

Thus if the elements of S are x, y, z, ... the geometric mean of the geometric mean of all non-empty subsets of S simplifies to an expression of the form x^ay^b ... where a + b + ... = 1. But by symmetry we must have a = b = ... and hence the expression is just the geometric mean of S.