9th Swedish Mathematical Society Problems 1969

1.  Find all integers m, n such that m³ = n³ + n.

2.  Show that tan π/3n is irrational for all positive integers n.

3.  a₁ ≥ a₂ ≥ ... ≥ a_n is a sequence of reals. b₁, b₂, b₃, ... b_n is any rearrangement of the sequence B₁ ≥ B₂ ≥ ... ≥ B_n. Show that ∑ a_ib_i ≤ &sum a_iB_i.

4.  Define g(x) as the largest value of |y² - xy| for y in [0, 1]. Find the minimum value of g (for real x).

5.  Let N = a₁a₂ ... a_n in binary. Show that if a₁ - a₂ + a₃ - ... + (-1)^n-1a_n = 0 mod 3, then N = 0 mod 3.

6.  Given 3n points in the plane, no three collinear, is it always possible to form n triangles (with vertices at the points), so that no point in the plane lies in more than one triangle? 

Solutions

Problem 1

Find all integers m, n such that m³ = n³ + n.

Answer

m = n = 0.

Solution

n and n² + 1 are relatively prime, so both must be cubes. Hence n² is also a cube. But the only consecutive integers which are both cubes are 0 and 1. 

Problem 2

Show that tan π/3n is irrational for all positive integers n.

Solution

tan π/3 = √3, which is irrational by the usual argument. (If √3 = a/b in lowest terms, then 3b² = a², so 3 must divide a and hence b. Contradiction.).

Now cos nx + i sin nx = (cos x + i sin x)ⁿ. Put c = cos x, s = sin x, t = tan x. Expanding by the binomial theorem and equating real and imaginary parts, we get sin nx = nC1 c^n-1s - nC3 c^n-3s³ + ... , cos nx = cⁿ - nC2 c^n-2s² + ... . Hence tan nx = (nC1 t - nC3 t³ + ... )/(1 - nC2 t² + ... ). Hence if t is rational, then so is tan nx. But tan π/3 is irrational, so tan π/3n must also be irrational. 

Problem 3

a₁ ≥ a₂ ≥ ... ≥ a_n is a sequence of reals. b₁, b₂, b₃, ... b_n is any rearrangement of the sequence B₁ ≥ B₂ ≥ ... ≥ B_n. Show that ∑ a_ib_i ≤ &sum a_iB_i.

Solution

This is the well-known rearrangement inequality. Let S = ∑ a_ib_i. Let S' be the sum after swapping b_i and b_j. Then S' - S = (a_i - a_j)(b_j - b_i). So if i ≤ j, then a_i - a_j ≥ 0, so S' - S has the same sign as b_j - b_i. In other words, we get the largest sum if the bs are sorted the same way as the as. 

Problem 4

Define g(x) as the largest value of |y² - xy| for y in [0, 1]. Find the minimum value of g (for real x).

Answer

3 - √8

Solution

y² - xy = (y - x/2)² - x2/4, so y² - xy has a minimum at y = x/2. It is decreasing for y < x/2 and increasing for y > x/2. Thus the largest value of y² - xy must be at y = 0, x/2 or 1. The values there are 0, x²/4, |1-x|. So for x outside the interval (0,1), g(x) ≥ 1/4. For x ∈ (0,1) we have g(x) = max(x²/4, 1-x). The quadratic x² + 4x - 4 = 0 has roots -2 ± √8, which are approx and -4.83 and 0.83. Hence g(x) = 1-x for x ≤ √8 - 2, and x²/4 for x ≥ √8 - 2, and the minimum value of g(x) in the interval (0,1) is g(√8 - 2) = 3 - √8, which is approx 0.18 and < 1/4. Thus this is also the minimum value of g(x) for all x. 

Problem 5

Let N = a₁a₂ ... a_n in binary. Show that if a₁ - a₂ + a₃ - ... + (-1)^n-1a_n = 0 mod 3, then N = 0 mod 3.

Solution

We have N = a₁2^n-1 + a₂2^n-2 + ... + a_n = a₁(-1)^n-1 + a₂(-1)^n-2 + ... + a_n = (-1)^n-1(a₁ - a₂ + a₃ - ... + (-1)^n-1a_n) = 0 mod 3. 

Problem 6

Given 3n points in the plane, no three collinear, is it always possible to form n triangles (with vertices at the points), so that no point in the plane lies in more than one triangle?

Answer

Yes

Solution

Take a slope different from that of the line joining any two of the points. Now divide the points into groups of three by parallel lines of that slope. (Start with a line distant from all the points and move it gradually towards the points, without changing its slope. It crosses the points one at a time.) Now join the points in each group.