17th Eötvös Competition Problems 1910

1.  α, β, γ are real and satisfy α² + β² + γ² = 1. Show that -1/2 ≤ αβ + βγ + γα ≤ 1.

2.  If ac, bc + ad, bd = 0 (mod n) show that bc, ad = 0 (mod n).

3.  ABC is a triangle with angle C = 120^o. Find the length of the angle bisector of angle C in terms of BC and CA. 

Solutions

Problem 1

α, β, γ are real and satisfy α² + β² + γ² = 1. Show that -1/2 ≤ αβ + βγ + γα ≤ 1.

Solution

We have (α + β + γ)2 ≥ 0, so α² + β² + γ² ≥ -2(αβ + βγ + γα), so -½ ≤ αβ + βγ + γα.

Also (α - β)² + (β - γ)² + (γ + α)² ≥ 0, so 2(α² + β² + γ²) ≥ 2(αβ + βγ + γα), so αβ + βγ + γα ≤ 1. 

Problem 2

If ac, bc + ad, bd = 0 (mod n) show that bc, ad = 0 (mod n).

Solution

We have (bc - ad)² = (bc + ad)² - 4ac·bd. So ( (bc-ad)/n)² = ( (bc+ad)/n)² - 4(ac/n)(bd/n), which is an integer. Hence (bc - ad)/n is an integer. Moreover ( (bc+ad)/n)² - ( (bc-ad)/n)² is an even integer, so (bc+ad)/n and (bc-ad)/n have the same parity. Hence bc/n = half their sum, and ad/n = half their difference, are both integers. 

Problem 3

ABC is a triangle with angle C = 120^o. Find the length of the angle bisector of angle C in terms of BC and CA.

Solution

Let the bisector be CX with length k. Take AC, BC to have lengths b, a as usual. Then area ACX = ½bk sin 60^o, area BCX = ½ak sin 60^o, and area ABC = ½ab sin 60^o. But area ABC = area ACX + area BCX, so k = ab/(a+b).