11th Swedish Mathematical Society Problems 1971

1.  Show that (1 + a + a²)² < 3(1 + a² + a⁴) for real a ≠ 1.

2.  An arbitrary number of lines divide the plane into regions. Show that the regions can be colored red and blue so that neighboring regions have different colors.

3.  A table is covered by 15 pieces of paper. Show that we can remove 7 pieces so that the remaining 8 cover at least 8/15 of the table.

4.  Find (65533³ + 65534³ + 65535³ + 65536³ + 65537³ + 65538³+ 65539³)/(32765·32766 + 32767·32768 + 32768·32769 + 32770·32771).

5.  Show that max_|x|≤t |1 - a cos x| ≥ tan²(t/2) for a positive and t ∈ (0, π/2).

6.  99 cards each have a label chosen from 1, 2, ... , 99, such that no (non-empty) subset of the cards has labels with total divisible by 100. Show that the labels must all be equal. 

Solutions

Problem 1

Show that (1 + a + a²)² < 3(1 + a² + a⁴) for real a ≠ 1.

Solution

We have a² + a + 1 = (a + 1/2)² + 3/4 > 0. So for a ≠ 1, we have 0 < (a - 1)²(a² + a + 1) = (a - 1)(a³ - 1) = a⁴ - a³ - a + 1. Hence 2(1 + a⁴) > 2a(1 + a²), so (1 + a + a²)² = 1 + 2a + 3a² + 2a³ + a⁴ < 3(1 + a² + a⁴). 

Problem 2

An arbitrary number of lines divide the plane into regions. Show that the regions can be colored red and blue so that neighboring regions have different colors.

Solution

Induction on the number of lines. Obvious for 1 line. Suppose true for n lines. Now add the n+1st line and change the color of every region on one side of the new line. 

Problem 4

Find (65533³ + 65534³ + 65535³ + 65536³ + 65537³ + 65538³+ 65539³)/(32765·32766 + 32767·32768 + 32768·32769 + 32770·32771).

Answer

7·2¹⁶

Solution

Put N = 65536 = 2¹⁶. Then numerator = (N-3)³ + (N-2)³ + (N-1)³ + N³ + (N+1)³ + (N+2)³ + (N+3)³ = ((N-3)³ + (N+3)³) + ((N-2)³ + (N+2)³) + ((N-1)³ + (N+1)³) + N³ = 2(N³ + 27N) + 2(N³ + 12N) + 2(N³ + 3N) + N³ = 7(N³ + 12N).

Put M = N/2 = 32768. Denominator = (M-3)(M-2) + (M-1)M + M(M+1) + (M+2)(M+3) = 4M² + 12 = N² + 12. Hence expr = 7N. 

Problem 6

99 cards each have a label chosen from 1, 2, ... , 99, such that no (non-empty) subset of the cards has labels with total divisible by 100. Show that the labels must all be equal.

Solution

Let the labels be a₁, a₂, ... , a₉₉. Use cyclic subscripts so a₁₀₀ means a₁, a₁₀₁ means a₂ etc. The 99 sums a_i, a_i + a_i+1, a_i + a_i+1 + a_i+2, ... , a_i + a_i+1 + ... + a_i+98 must all be non-zero mod 100 and all unequal (otherwise their difference would give a subset with zero sum mod 100). So one of them must equal -a_i-1 mod 100. If it is any but the last, then a_i-1 + a_i + ... + a_i+k gives a subset with sum 0 mod 100. So it must be the last. In other words a_i-1 = -(a_i + a_i+1 + ... + a_i+98). But this is independent of i because a_i + a_i+1 + ... + a_i+98 = a₁ + a₂ + ... + a₉₉. So a_i-1 are all equal mod 100 and hence are all equal.