As before, the largest possible pair of circles will be tangent to the longest side of the triangle: the 15 side.

Using Heron’s formula we can find that the area of the triangle is 84.

From this we can work out that the inradius of the entire triangle will be 4.

The version with two circles will effectively be a scaled down version of the full triangle with one incircle, split apart and a 2r wide strip placed between.

The ratio of the inradius to the base length is the same for this split-apart triangle as it is for the full triangle.

(15-2r)/r = 15/4

60-8r = 15r

60 = 23r

r = 60/23