16th Eötvös 1910

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).

17th Eötvös 1910

© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03