18th Eötvös Competition Problems 1911

1.  Real numbers a, b, c, A, B, C satisfy b² < ac and aC - 2bB + cA = 0. Show that B² ≥ AC.

2.  L₁, L₂, L₃, L₄ are diameters of a circle C radius 1 and the angle between any two is either π/2 or π/4. If P is a point on the circle, show that the sum of the fourth powers of the distances from P to the four diameters is 3/2.

3.  Prove that 3ⁿ + 1 is not divisible by 2ⁿ for n > 1. 

Solutions

Problem 1

Real numbers a, b, c, A, B, C satisfy b² < ac and aC - 2bB + cA = 0. Show that B² ≥ AC.
Solution

2bB = aC + cA, so 4b²B² = (aC + cA)². But (aC - cA)² ≥ 0, so 4b²B² ≥ 4acAC. But ac > b² ≥ 0, so dividing by 4ac gives, b²/ac B² ≥ AC. But 0 ≤ b²/ac < 1, so B² ≥ AC. Note that we can have equality if A = B = C = 0.

Problem 2

L₁, L₂, L₃, L₄ are diameters of a circle C radius 1 and the angle between any two is either π/2 or π/4. If P is a point on the circle, show that the sum of the fourth powers of the distances from P to the four diameters is 3/2.
Solution

Take rectangular coordinates with origin at the center of the circle. Take two of the diameters as axes. Then the other two diameters are y = x and y = -x. A point (a, b) on the circle is a distance |a|, |b|, |a-b|/√2, |a+b|/√2 from the four diameters. Thus the sum of the fourth powers is a⁴ + b⁴ + (a-b)⁴/4 + (a+b)⁴/4 = 3a⁴/2 + 3b⁴/2 + 3a⁴b⁴ = (3/2)(a² + b²)² = 3/2.

Problem 3

Prove that 3ⁿ + 1 is not divisible by 2ⁿ for n > 1.
Solution

We have 3² = 1 mod 8, so for n even 3ⁿ + 1 = 2 mod 8, and 3ⁿ + 1 is not divisible by 2². So since n ≥ 2 for n even, 3ⁿ + 1 is not divisible by 2ⁿ.
For n odd we have 3ⁿ + 1 = 4 mod 8, which is not divisible by 2³. So since n ≥ 3 for n odd, 3ⁿ + 1 is not divisible by 2ⁿ.