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.