(i)     Consider the line from the common centre of C₁, C₂ and C₃ through the centre of one of the unit circles. This gives r₁+2=r₃. (ii)
Consider the equilateral triangle with vertices at the centres of the unit circles and side length 2. The altitude of this triangle is therefore of length Ö3, which gives 1+r₁+r₂=Ö3.

(iii)
The intersection of the altitudes of the triangle in (ii) divides each altitude in the ratio 2:1, hence 2r₂=r₁+1.

These three equations give

r1=  2 Ö3 -1,    r2=  1 Ö3 ,    r3=  2 Ö3 +1.