Edward from 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.