39th Eötvös Competition Problems 1935

1.  x₁, x₂, ... , x_n is any sequence of positive reals and y_i is any permutation of x_i. Show that ∑ x_i/y_i ≥ n.

2.  S is a finite set of points in the plane. Show that there is at most one point P in the plane such that if A is any point of S, then there is a point A' in S with P the midpoint of AA'.

3.  Each vertex of a triangular prism is labeled with a real number. If each number is the arithmetic mean of the three numbers on the adjacent vertices, show that the numbers are all equal.

Solutions

Problem 1

x₁, x₂, ... , x_n is any sequence of positive reals and y_i is any permutation of x_i. Show that ∑ x_i/y_i ≥ n.

Solution

By AM/GM (∑ x_i/y_i)/n ≥ geometric mean of x_i/y_i which is 1 since ∏ x_i = ∏ y_i.

Problem 2

S is a finite set of points in the plane. Show that there is at most one point P in the plane such that if A is any point of S, then there is a point A' in S with P the midpoint of AA'.

Solution

The points must form disjoint pairs (A,A') with P the midpoint of each pair. But P is the centroid of each pair and hence the centroid of all the points, but that is a uniquely defined point. 

Problem 3

Each vertex of a triangular prism is labeled with a real number. If each number is the arithmetic mean of the three numbers on the adjacent vertices, show that the numbers are all equal.

Solution

We have 3a₁ = a₂ + a₃ + b₁, 3a₂ = a₁ + a₃ + b₂, 3a₃ = a₁ + a₂ + ba2 ₃. Adding gives a₁ + a₂ + a₃ = b₁ + b₂ + b₃. Hence (a₁ - b₁) = -(a₂ - b₂) - (a₃ - b₃) (*).

We also have 3b₁ = b₂ + b₃ + a₁. Subtracting from the first equation above, 4(a₁ - b₁) = (a₂ - b₂) + (a₃ - b₃). Adding (*) gives 5(a₁ - b₁) = 0. Hence a₁ = b₁. Similarly a₂ = b₂ and a₃ = b₃. But now the first two equations give 2a₁ = a₂ + a₃ and 2a₂ = a₁ + a₃, subtracting, 3(a₁ - a₂) = 0, so a₁ = a₂. Similarly a₂ = a₃. Hence all the numbers are equal.