45th Eötvös Competition Problems 1941

1.  Prove that (1+x)(1+x²)(1+x⁴) ... (1+x^2n-1) = 1 + x + x² + x³ + ... + x^2n-1.

2.  The a parallelogram has its vertices at lattice points and there is at least one other lattice point inside the parallelogram or on its sides. Show that its area is greater than 1.

3.  ABCDEF is a hexagon with vertices on a circle radius R (in that order). The three sides AB, CD, EF have length R. Show that the midpoints of BC, DE, FA form an equilateral triangle.

Solutions

Problem 1

Prove that (1+x)(1+x²)(1+x⁴) ... (1+x^2n-1) = 1 + x + x² + x³ + ... + x^2n-1.

Solution

Multiply the lhs by 1-x. We have (1-x)(1+x) = 1-x², (1-x²)(1+x²) = 1-x⁴, (1-x⁴)(1+x⁴) = 1-x⁸, and so on. So (1-x)lhs = 1-x²ⁿ. Similarly, (1-x)rhs = 1-x²ⁿ. 

Problem 2

The a parallelogram has its vertices at lattice points and there is at least one other lattice point inside the parallelogram or on its sides. Show that its area is greater than 1.

Solution

This follows immediately from Pick's theorem, which makes one worry that a proof of Pick's theorem might be expected. An alternative is to use the formula for the area of a triangle: the triangle with coordinates (x₁,y₁), (x₂,y₂), (x₃,y₃) has area ±(x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂))/2. So the area of a lattice triangle must be a multiple of ½ and hence ≥½. Now take the extra lattice point and join it to the vertices of the parallelogram. If the point is inside the parallelogram, we get 4 lattice triangles, each area ≥½. If the point is on a side of the parallelogram, then we get 3 lattice triangles, each area ≥½.

However, proving the area theorem is also fairly tedious. The idea is to enclose the triangle in a lattice rectangle, so that its area is a rectangle area less various right-angle triangle areas, all of which can be written straight down. The problem is that there are several different cases to consider. 

Problem 3

ABCDEF is a hexagon with vertices on a circle radius r (in that order). The three sides AB, CD, EF have length r. Show that the midpoints of BC, DE, FA form an equilateral triangle.

Solution

Use vectors. Take the origin at the center of the circle. Denote the vector OX by X. Let the midpoint of BC, DE, FA be P, Q, R respectively. Then P = (B+C)/2, Q = (D+E)/2. So the vector QP is (B+C-D-E)/2. If we rotate through 60^o, then B becomes A, C becomes C-D, D becomes C, and E becomes E-F, so the vector QP becomes (A+C-D-C-E+F)/2 = (A+F-D-E)/2, which is the vector QR. Hence PQR is equilateral.