1st Swedish Mathematical Society Problems 1961

1.  Let S be the system of equations (1) y(x⁴ - y² + x²) = x, (2) x(x⁴ - y² + x²) = 1. Take S' to be the system of equations (1) and x·(1) - y·(2) (or y = x²). Show that S and S' do not have the same set of solutions and explain why.

2.  Show that x₁/x_n + x₂/x_n-1 + x₃/x_n-2 + ... + x_n/x₁ ≥ n for any positive reals x₁, x₂, ... , x_n.

3.  For which n is it possible to put n identical candles in a candlestick and to light them as follows. For i = 1, 2, ... , n, exactly i candles are lit on day i and burn for exactly one hour. At the end of day n, all n candles must be burnt out. State a possible rule for deciding which candles to light on day i.

4.  288 points are placed inside a square ABCD of side 1. Show that one can draw a set S of lines length 1 parallel to AB joining AD and BC, and additional lines parallel to AD joining each of the 288 point to a line in S, so that the total length of all the lines is less than 24. Is there a stronger result?

5.  n is a positive integer. Show that x⁶/6 + x² - nx has exactly one minimum a_n. Show that for some k, lim_n→∞ a_n/n^k exists and is non-zero. Find k and the limit.

Solutions

Problem 1

Let S be the system of equations (1) y(x⁴ - y² + x²) = x, (2) x(x⁴ - y² + x²) = 1. Take S' to be the system of equations (1) and x·(1) - y·(2) (or y = x²). Show that S and S' do not have the same set of solutions and explain why.

Solution

S' has the solution x = 0, y = 0, but S does not. Replacing (2) by x·(1) - y·(2) introduces the extra solution. 

Problem 2

Show that x₁/x_n + x₂/x_n-1 + x₃/x_n-2 + ... + x_n/x₁ ≥ n for any positive reals x₁, x₂, ... , x_n.

Solution

AM/GM. 

Problem 3

For which n is it possible to put n identical candles in a candlestick and to light them as follows. For i = 1, 2, ... , n, exactly i candles are lit on day i and burn for exactly one hour. At the end of day n, all n candles must be burnt out. State a possible rule for deciding which candles to light on day i.

Answer

n odd.

If i < n/2, light i candles starting from 1. If i > n/2, light i candles starting from n. Eg 1; 1,2; 3,4,5; 2,3,4,5; 1,2,3,4,5.

Solution

Total candle hours = 1 + 2 + ... + n = n(n+1)/2. Hence (n+1)/2 per candle. But each candle is lit for a whole number of hours, so n must be odd.

Rule above gives pairs of days 1, 2, ... , i and i+1, i+2, ... , n, plus the final day when all candles are lit. Each candle is lit once for each pair of days (and is lit on the final day). 

Problem 4

288 points are placed inside a square ABCD of side 1. Show that one can draw a set S of lines length 1 parallel to AB joining AD and BC, and additional lines parallel to AD joining each of the 288 point to a line in S, so that the total length of all the lines is less than 24. Is there a stronger result?

Answer

≤ 18 6/23

Solution

Place 12 horizontal non-overlapping strips width 1/12 across the square. Take S to be the center-lines of the strips. Join each point to the center-line of its strip. So we have 12 lines length 1 and 288 lines length at most 1/24, total at most 24.

Take 24 equally spaced parallel lines, including two edges of the square. So the distance between two adjacent lines is 1/23. Color the lines alternately red and blue. Let R be the sum of the distances of the 288 points from the nearest red line and B the sum of the distances from the nearest blue line. For each point the sum of the distance to the nearest blue line and the distance to the nearest red line is 1/23, because each point lies in a strip width 1/23 bounded by the nearest red line and the nearest blue line. Hence B + R = 288/23. So at least one of B, R is ≤ 144/23. So if we take that set of 12 lines as the lines we get a total length for the lines and perpendicular segments of at most 12 + 144/23 = 18 6/23. 

Problem 5

n is a positive integer. Show that x⁶/6 + x² - nx has exactly one minimum an. Show that for some k, lim_n→∞ a_n/n^k exists and is non-zero. Find k and the limit.

Answer

k = 6/5, limit = -5/6.

Solution

Derivative is x⁵ + 2x - n, which is strictly monotonic increasing. It is -n at 0. So it has a single zero for some h > 0 and is negative for x < h, and positive for x > h. Hence the zero represents a minimum.

At n^1/5 derivative is n + 2n^1/5 - n = 2n^1/5 > 0, so minimum is before n^1/5. At n^1/5-1, derivative is n - 5n^4/5 + O(n^3/5) + 2n^1/5 - n which is less than 0 for sufficiently large n. So the minimum occurs (for sufficiently large n) for x = n^1/5-h for some 0 < h < 1. The value at n^1/5 - h is (1/6) n^6/5 + O(n) - n^6/5 = -5/6 n^6/5 + O(n).

Comment. Early Western olympiad questions often used basic calculus.