Triangle Solving – Problem 12: Find the range of

In $\triangle ABC$, let the sides opposite $A,B,C$ be $a,b,c$. Suppose $2b\cos A+a=2c$ and $b=2\sqrt{3}$. Find the range of $2a-c$.

Step 1. From $2b\cos A+a=2c$ and the Law of Sines $a=2R\sin A$, $b=2R\sin B$, $c=2R\sin C$, we obtain $$2\sin B\cos A+\sin A=2\sin C=2\sin(A+B).$$

Step 2. Expanding $2\sin(A+B)=2\sin A\cos B+2\cos A\sin B$ and comparing gives $\sin A=2\sin A\cos B$.

Step 3. Since $\sin A>0$, we have $\cos B=\frac12$, so $B=\frac{\pi}{3}$.

Step 4. Given $b=2\sqrt{3}=2R\sin B=2R\cdot\frac{\sqrt{3}}{2}$, we get $R=2$. Hence $a=4\sin A$ and $c=4\sin C$.

Step 5. Since $C=\pi-A-B=\frac{2\pi}{3}-A$, $$2a-c=8\sin A-4\sin\left(\frac{2\pi}{3}-A\right)=6\sin A-2\sqrt{3}\cos A=4\sqrt{3}\sin\left(A-\frac{\pi}{6}\right).$$

Step 6. Because $A\in(0,\frac{2\pi}{3})$, we have $A-\frac{\pi}{6}\in\left(-\frac{\pi}{6},\frac{\pi}{2}\right)$, so $\sin\left(A-\frac{\pi}{6}\right)\in\left(-\frac12,1\right)$.

Step 7. Therefore $2a-c\in(-2\sqrt{3},4\sqrt{3})$.

$$(-2\sqrt{3},4\sqrt{3})$$