31st Canadian Mathematical Olympiad Problems 1999

31st Canadian Mathematical Olympiad Problems 1999

1.  Find all real solutions to the equation 4x² - 40[x] + 51 = 0.

2.  ABC is equilateral. A circle with center on the line through A parallel to BC touches the segment BC. Show that the length of arc of the circle inside ABC is independent of the position of the circle.

3.  Find all positive integers which equal the square of their number of positive divisors.

4.  X is a subset of eight elements of {1, 2, 3, ... , 17}. Show that there are three pairs of (distinct) elements with the same difference.

5.  x, y, z are non-negative reals with sum 1, show that x²y + y²z + z²x ≤ 4/27. When do we have equality?

Solutions

Problem 1

Find all real solutions to the equation 4x² - 40[x] + 51 = 0.

Solution

Answer: 2x = √29, √189, √229 or √269.

To orient ourselves notice that 4x² - 40x + 51 = (2x - 17)(2x - 3), so we expect roots to lie in the range 1 to 9.

Let f(x) = 4x² - 40[x] + 51. For x <= 1 f(x) > 51, so there are no solutions with x < 1.

f(1) = 15, f(2^-) = 16 - 40 + 51 = 27, so there are no solutions in the interval [1, 2).

f(2) = -13, f(3^-) = 7, so we expect a solution in [2, 3). For all x in this interval we have f(x) = 4x² - 29, so the unique solution is 2x = √29.

f(3) = -33, f(4^-) = -5, so there are no solutions in [3, 4).

f(4) = -45, f(5^-) = -9, so there are no solutions in [4, 5).

f(5) = -49, f(6^-) = -5, so there are no solutions in [5, 6).

f(6) = -45, f(7^-) = 7, so we expect a solution in [6, 7). For all x in this interval we have f(x) = 4x² - 189, so the unique solution is 2x = √189.

f(7) = -33, f(8^-) = 27. For all x in the interval [7, 8) we have f(x) = 4x² - 229, so the unique solution is 2x = √229.

f(8) = -13, f(9^-) = 55. For all x in the interval [8, 9) we have f(x) = 4x² - 269, so the unique solution is 2x = √269.

f(9) = 15. f(10^-) = 91, so there are no solutions in [9, 10). For x >= 10, we have 4x² - 40[x] + 51 ≥ 4x² - 40x + 51 = 4x(x - 10) + 51 > 0, so there are no solutions.

Problem 2

ABC is equilateral. A circle with center on the line through A parallel to BC touches the segment BC. Show that the length of arc of the circle inside ABC is independent of the position of the circle.

Solution

Let the circle meet the segment AC at X and the line AB at Y with Y outside the segment AB. Let O be the center of the circle. Then ∠OAY = ∠OAC = 60^o. But the line OA is a diameter of the circle, so XY must be perpendicular to AO and hence ∠AYX = 30^o. So the arc inside the circle subtends the constant angle 30^o at the circumference, so it must have constant length.

Problem 3

Find all positive integers which equal the square of their number of positive divisors.

Solution

Let N = p^aq^b ... . Then the square of the number of positive divisors of N is (a + 1)²(b + 1)² ... . Any square has all its prime factors to an even power, so a, b, ... must all be even. So writing a = 2m, b = 2n etc, we have p^mqⁿ ... = (2m+1)(2n+1) ... . Since the rhs is odd, all the primes p, q, ... must be odd.

A trivial induction shows that 2n + 1 < 3ⁿ for n > 1. Hence 2n + 1 < pⁿ for n > 1 for any odd prime p. For n = 1, 2n + 1 = 3 < p unless p = 3. So 2n+1 < pⁿ, where p is an odd prime and n is a positive integer, except in the case n = 1, p = 3.

Thus we can have at most one prime p in the factorisation of N and N = 1 or 3².

Problem 4

X is a subset of eight elements of {1, 2, 3, ... , 17}. Show that there are three pairs of (distinct) elements with the same difference.

Solution

Let the elements of X be a₁ < a₂ < ... < a₈. The 7 differences a₂ - a₁, a₃ - a₂, ... a₈ - a₇ have sum a₈ - a₁. The 6 differences a₃ - a₁, a₄ - a₂, ... , a₈ - a₆ have sum a₈ + a₇ - a₁ - a₂. So the 13 differences together have sum 2a₈ + a₇ - a₂ - 2a₁ ≤ 34 + 16 - 2 - 2 = 46. But if no difference occurs more than twice then the sum must be at least 2(1 + 2 + 3 + 4 + 5 + 6) + 7 = 49.

Problem 5

x, y, z are non-negative reals with sum 1, show that x²y + y²z + z²x ≤ 4/27. When do we have equality?

Solution

Assume x ≥ y, z. We have (x + z/2)²(y + z/2) - x²y - y²z - z²x = (x - y)yz + xz(x - z)/2 + yz²/4 + z³/8 > 0, unless z = 0. So if z > 0, we get a larger sum for x' = x+z/2, y' = y+z/2, z' = 0. So we can assume z = 0.

By AM/GM applied to x, x, 2y, we have 2x²y ≤ ( (2x + 2y)/3)³ = 8/27 and hence x²y ≤ 4/27 with equality iff x = 2/3, y = 1/3. Thus the original inequality holds with equality iff (x, y, z) = (2/3, 1/3, 0), (0, 2/3, 1/3) or (1/3, 0, 2/3).