Assume $AB=c , AC=b , BC=a$ $\angle A= \alpha , \angle B=\beta , \angle C= \gamma$ , easy to get the sum of perimeters of the 3 triangles can be expressed as $2\sum a\cos \alpha +(\sum a)(\sum \sin {\frac{\alpha}{2}})$ , Hence the ratio $k=\frac{2\sum a\cos \alpha +(\sum a)(\sum \sin {\frac{\alpha}{2}})}{R}=2\frac{2\sum a\cos \alpha +(\sum a)(\sum \sin {\frac{\alpha}{2}})}{2R}$ , By law of sine , $k=4\sum \sin \alpha \cos {\frac{\alpha}{2}}+2\sum \sin \alpha \sum \sin {\frac{\alpha}{2}}$ . By Jensen inequality , we have :$2\sum \sin \alpha \sum \sin {\frac{\alpha}{2}}<= 2*3*\frac{\sqrt 3}{2}*3*\frac{1}{2}=\frac{9\sqrt{3}}{2}$ And $4\sum \sin \alpha \cos {\frac{\alpha}{2}}=2\sum (\sin {\frac{3}{2}\alpha}+\sin \frac{\alpha}{2})$ $2\sum \sin \frac{\alpha}{2}<=3$ , obviously for any $x$ , $\sin x<=1$ , hence , $2\sum \sin {\frac{3}{2}\alpha}<=6$ Hence $4\sum \sin \alpha \cos {\frac{\alpha}{2}} <= 9$ Plus the two parts up we get $k=4\sum \sin \alpha \cos {\frac{\alpha}{2}}+2\sum \sin \alpha \sum \sin {\frac{\alpha}{2}}<= 9+\frac{9}{2}\sqrt{3}$ . The equality holds if and only if $\alpha =\beta=\gamma=\frac{\pi}{3}$ . Hence $k_{max}= 9+\frac{9}{2}\sqrt{3}$ .