If I counted well, then $r=\frac{abc}{(a+b+c)(a+b)}$ where $a,b,c$ are sides of triangle and $r$ is radius of circles Then $2r(a+b)=c(a+b-c)$ or $c^2=(a+b)(c-2r)$ or $a+b=\frac{c^2}{c-2r}$ So $c-2r|4r^2$ As $c<a+b \leq \sqrt{2}c<\frac{3}{2}c$ so $\frac{c^2}{c-2r}<\frac{3}{2}c \to 6r<c$ and so $c-2r=r^2,2r^2,4r^2$ Case 1: $a+b=\frac{(r^2+2r)^2}{r^2}=(r+2)^2$ $2ab=(r+2)^4-r^2(r+2)^2=4(r+2)^2(r+1)$ then $r^2-4r-4$ is square so $r=5 \to a=21,b=28,c=35$ Case 2: $a+b=\frac{(2r^2+2r)^2}{2r^2}=2(r+1)^2, 2ab=4(r+1)^4-4r^2(r+1)^2=4(r+1)^2(2r+1)$ then $r^2-2r-1$ is square, but it is impossible Case 3: $a+b=\frac{(4r^2+2r)^2}{4r^2}=(2r+1)^2, 2ab=(2r+1)^4-4r^2(2r+1)^2=(2r+1)^2(4r+1)$ then $4r^2-4r-1$ is square, but it is impossible Answer: $(a,b,c,r)=(21,28,35,5)$

indeed that is the result

official solution uses this equivalent result for r