Triangle Solving – Problem 14: Find the range of

In $\triangle ABC$, let the sides opposite $A,B,C$ be $a,b,c$. The circumradius is $R=\sqrt{3}$, and $$\frac{b\sin C}{\sin A+\sin B-\sin C}=a-b+c.$$ Find the range of $b+c$.

A. $(0,2]$ B. $(1,2]$ C. $(3,6]$ D. $(\sqrt{3},3]$

Step 1. By the Law of Sines and $\frac{b\sin C}{\sin A+\sin B-\sin C}=a-b+c$, we obtain $\frac{bc}{a+b-c}=a-b+c$.

Step 2. Hence $bc=(a+b-c)(a-b+c)=a^{2}-(b-c)^{2}$, so $b^{2}+c^{2}-a^{2}=bc$.

Step 3. Therefore $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{1}{2}$, so $A=\frac{\pi}{3}$.

Step 4. Since $R=\sqrt{3}$, $a=2R\sin A=2\sqrt{3}\cdot\sin\frac{\pi}{3}=3$.

Step 5. By the Law of Cosines, $a^{2}=b^{2}+c^{2}-bc=(b+c)^{2}-3bc\ge (b+c)^{2}-3\left(\frac{b+c}{2}\right)^{2}=\frac{1}{4}(b+c)^{2}$.

Step 6. Thus $(b+c)^{2}\le 4a^{2}=36\Rightarrow b+c\le 6$. Also $b+c>a=3$.

Step 7. Therefore $b+c\in(3,6]$, so the correct choice is C.

C