32nd Eötvös Competition Problems 1928

1.  Show that for any real x, at least one of x, 2x, 3x, ... , (n-1)x differs from an integer by no more than 1/n.

2.  The numbers 1, 2, ... , n are arranged around a circle so that the difference between any two adjacent numbers does not exceed 2. Show that this can be done in only one way (treating rotations and reflections of an arrangement as the same arrangement).

3.  Given two points A, B and a line L in the plane, find the point P on the line for which max(AP, BP) is as short as possible. 

Solutions

Problem 1

Show that for any real x, at least one of x, 2x, 3x, ... , (n-1)x differs from an integer by no more than 1/n.

Solution

For i = 1, 2, ... , n let S_i = {y | (i-1)/n ≤ [y] < i/n}. We wish to show that one of the given n-1 points belongs to S¹ or Sⁿ. Suppose not. Then they belong to the n-2 sets S², S³, ... , S^n-1. So two of them rx and sx must belong to the same Si. But now |r-s|x belongs to S₁ or S_n. Contradiction. 

Problem 2

The numbers 1, 2, ... , n are arranged around a circle so that the difference between any two adjacent numbers does not exceed 2. Show that this can be done in only one way (treating rotations and reflections of an arrangement as the same arrangement).

Solution

The unique solution is 1,3,5, ... ,largest odd, largest even, ... , 6, 4, 2.

Start with 1. Its neighbors must be 2 and 3. Now only 4 is available to go on the other side of 2, now only 5 is available to go next to 3, now only 6 is available to go next to 4, and so on. 

Problem 3

Given two points A, B and a line L in the plane, find the point P on the line for which max(AP, BP) is as short as possible.

Solution

Let B be the point further from the line (or one of them). Let B' be the foot of the perpendicular from B to the line. If BB' > AB', then we must obviously take P = B'. If not, then let X be the point where the perpendicular bisector of AB meets the line. We take P to be X. For X must lie between A' and B'. If we take P to the right of X then max(AP,BP) = AP and AP > AX. If we take P to the left of X, then max(AP, BP) = BP and BP > BX.