This solution together with the diagram was sent in by Derek Wan, age 17 of Sha Tin College, Hong Kong

Solution to Circles in Circles

To find OA=r1, OB=r2 and OD=r3 we must examine the diagram and make appropriate considerations. By dropping a perpendicular from O to V, the midpoint of UC, as the radius is perpendicular to the tangent, we can find OA = r1. Since triangle TCU is equilateral ÐTCU=60o and OC bisects ÐOCV so ÐOCV = 30o.

Since OC=OA+AC
 CV
OA+AC
 =  Ö3
2
As AC=CV=1 and OA=r1,
 1
r1+1
=  Ö3
2
r1+1
=  2
Ö3
r1
=  2
Ö3
 -1.
With r1 we can find r2 and r3. Since OV=r2, CV=1 and ÐOCV=300,
r2=  1
Ö3
 .
Since OD=OA+AC+CD,
r3=  2
Ö3
 - 1 + 1 + 1 =  2
Ö3
 + 1.
Now
r1r2
= æ
è 		 2
Ö3
- 1 ö
ø 		æ
è 		 2
Ö3
 + 1 ö
ø
=  4
3
 +  2
Ö3
 -  2
Ö3
 - 1
=  1
3
= æ
è 		 1
Ö3
 ö
ø 		2

 
= r22.