11th Eötvös Competition Problems 1904

1.  A pentagon inscribed in a circle has equal angles. Show that it has equal sides.
2.  Let a be an integer, and let p(x₁, x₂, ... , x_n) = ∑₁ⁿ k x_k. Show that the number of integral solutions (x₁, x₂, ... , x_n) to p(x₁, x₂, ... , x_n) = a, with all x_i > 0 equals the number of integral solutions (x₁, x₂, ... , x_n) to p(x₁, x₂, ... , x_n) = a - n(n + 1)/2, with all x_i ≥ 0.

3.  R is a rectangle. Find the set of all points which are closer to the center of the rectangle than to any vertex.

Solutions

Problem 1
A pentagon inscribed in a circle has equal angles. Show that it has equal sides.
Solution

Consider triangles ABC and DCB. We have ∠ABC = ∠DCB, BC = CB, ∠BAC = ∠CDB (because the pentagon is cyclic). Hence AB = DC. Similarly, AB = DE, BC = ED and BC = EA. So all sides are equal.

Problem 2
Let a be an integer, and let p(x₁, x₂, ... , x_n) = ∑₁ⁿ k x_k. Show that the number of integral solutions (x₁, x₂, ... , x_n) to p(x₁, x₂, ... , x_n) = a, with all x_i > 0 equals the number of integral solutions (x₁, x₂, ... , x_n) to p(x₁, x₂, ... , x_n) = a - n(n + 1)/2, with all x_i ≥ 0.
Solution
If x_i is a positive integral solution to p = a, then x_i-1 is a non-negative integral solution to p = a - n(n+1)/2. Conversely if x_i is a non-negative integral solution to p = a - n(n+1)/2, then x_i+1 is a positive integral solution to p = a. So we have a bijection between the two solution sets.

Problem 3

R is a rectangle. Find the set of all points which are closer to the center of the rectangle than to any vertex.
Solution

The area enclosed by the perpendicular bisectors of OA, OB, OC and OD.