10th Eötvös Competition Problems 1903

1.  Prove that 2^p-1(2^p - 1) is perfect when 2^p - 1 is prime. [A perfect number equals the sum of its (positive) divisors, excluding the number itself.]

2.  α and β are real and a = sin α, b = sin β, c = sin(α+β). Find a polynomial p(x, y, z) with integer coefficients, such that p(a, b, c) = 0. Find all values of (a, b) for which there are less than four distinct values of c.

3.  ABCD is a rhombus. C_A is the circle through B, C, D; C_B is the circle through A, C, D; C_C is the circle through A, B, D; and C_D is the circle through A, B, C. Show that the angle between the tangents to C_A and C_C at B equals the angle between the tangents to C_B and C_D at A. 

Solutions

Problem 1

Prove that 2^p-1(2^p - 1) is perfect when 2^p - 1 is prime. [A perfect number equals the twice sum of its (positive) divisors (excluding the number itself.]

Solution

In this case sum of the divisors is 1+2+2²+ ... +2^p-1 + (2^p-1)(1+2+2²+ ... +2^p-2) = (2^p-1) + (2^p-1)(2^p-1-1) = 2^p-1(2^p-1). 

Problem 2

α and β are real and a = sin α, b = sin β, c = sin(α+β). Find a polynomial p(x, y, z) with integer coefficients, such that p(a, b, c) = 0. Find all values of (a, b) for which there are less than four distinct values of c.

Solution

c = sin α cos β + cos α sin β = a√(1-b²) + b√(1-a²). Squaring, c² = a²(1-b²) + b²(1-a²) +2ab√(1-a²)√(1-b²). Leaving the remaining radical on one side and squaring again, we get c⁴ - 2(a²(1-b²)+b²(1-a²))c² + (a²(1-b²)+b²(1-a²))² - 4a²b²(1-a²)(1-b²) = 0, or c⁴ - 2(a²-2a²b²+b²)c² + (a²-b²)² = 0.

We know that one root of this quartic in c is c₁ = a√(1-b²) + b√(1-a²). Since a and b only appear in the quartic as squares, the other roots must be c₂ = -a√(1-b²) + b√(1-a²), c₃ = a√(1-b²) - b√(1-a²), and c₄ = -a√(1-b²) - b√(1-a²). Now c₁ = c₂ iff a√(1-b²) = 0 ie a = 0 or b = ±1. Similarly c₁ = c₃ iff a = ±1 or b = 0. We find c₁ = c₄ iff a√(1-b²) = -b√(1-a²) or a = -b. Similarly, we find c₂ = c₃ iff a = b, whilst c2 = c4 and c3 = c4 do not give any new conditions. Thus we have less than 4 distinct roots if a = 0, ±1, or b = 0, ±1, or a = ±b. 

Problem 3

ABCD is a rhombus. C_A is the circle through B, C, D; C_B is the circle through A, C, D; C_C is the circle through A, B, D; and C_D is the circle through A, B, C. Show that the angle between the tangents to C_A and C_C at B equals the angle between the tangents to C_B and C_D at A.

Solution

The fact that ABCD is a rhombus is irrelevant. The result holds for any quadrilateral.

We can mark the angles as shown. We wish to show that c = e. We have:
a + b + c = 180^o (looking at A) (1)
a + d + e = 180^o (looking at B) (2)
c + d + f = 180^o (looking at C) (3)
b + f + e = 180^o (looking at D) (4)
Taking (1) - (2), we get c + e + b - d = 0 and taking (3) - (4) we get c + e + d - b = 0. Adding gives c = e as required.

Note that the derivation still holds for different configurations, such as