Edward Wallace, age 18, Graveney School, Tooting, London sent this solution. Let us make a the radius of the largest circle centre A etc. Then the lengths of the sides of the triangle are: AB = a − b, AC = a − c and BC = b + c. The perimeter of the triangle is:

AB + BC + CA = (a − b) + (a − c) + (b + c) = 2a.

So the perimeter of the triangle is twice the radius of the large circle whatever the sizes of the small circles.