6th Eötvös Competition Problems 1899

1.  ABCDE is a regular pentagon (with vertices in that order) inscribed in a circle of radius 1. Show that AB·AC = √5.

2.  The roots of the quadratic x² - (a + d) x + ad - bc = 0 are α and β. Show that α³ and β³ are the roots of x² - (a³ + d³ + 3abc + 3bcd) x + (ad - bc)³ = 0.

3.  Show that 2903ⁿ - 803ⁿ - 464ⁿ + 261ⁿ is divisible by 1897. 

Solutions

Problem

ABCDE is a regular pentagon (with vertices in that order) inscribed in a circle of radius 1. Show that AB·AC = √5.

Solution

AB subtends 72^o at the center and AC subtends 144^o at the center. So AB·AC = 4 sin 36^o sin 72^o.

Using e^iθ = cos θ + i sin θ or otherwise, we get sin 5k = sin k (16 sin⁴k - 20 sin²k + 5) = 0. The roots of sin 5k = 0 are k = 0, ±36^o, ±72^o, so the roots of 16s² - 20s + 5 = 0 are sin²36^o and sin²72^o. Hence sin²36^o sin²72^o = 5/16, so 4 sin 36^o sin 72^o = √5. 

Problem 2

The roots of the quadratic x² - (a + d) x + ad - bc = 0 are α and β. Show that α³ and β³ are the roots of x² - (a³ + d³ + 3abc + 3bcd) x + (ad - bc)³ = 0.

Solution

We have αβ = ad-bc and α+β = a+d. Hence, α³β³ = (αβ)³ = (ad-bc)³, and α³+β³ = (α+β)³ - 3αβ(α+β) = (a+d)³ - 3(ad-bc)(a+d) = a³+d³ + 3bc(a+d). 

Problem 3

Show that 2903ⁿ - 803ⁿ - 464ⁿ + 261ⁿ is divisible by 1897.

Solution

Note that 1897 = 7·271. Since 7 and 271 are relatively prime (indeed both prime), it is sufficient to show that the expression is always divisible by 7 and always divisible by 271. We use the fact that a-b divides aⁿ-bⁿ. Note that 2903 - 803 = 2100 which is a multiple of 7. It is eash to check that 464 - 261 is also a multiple of 7. The only other choice is 2903 - 464 and 803 - 261 and it is easy to check that they are multiples of 271.