Triangle Solving – Problem 11: Find the maximum possible value of the perimeter of

In $\triangle ABC$, let the sides opposite $A,B,C$ be $a,b,c$. Suppose $$c\cos A-\sqrt{3}\,c\sin A-b+2a=0,\qquad c=3.$$ Find the maximum possible value of the perimeter of $\triangle ABC$.

A. $\sqrt{3}$ B. $\frac{9\sqrt{3}}{4}$ C. 6 D. 9

Step 1. Using the Law of Sines $a=2R\sin A$, $b=2R\sin B$, $c=2R\sin C$, the condition becomes $$\sin C\cos A-\sqrt{3}\sin C\sin A-\sin B+2\sin A=0.$$

Step 2. Since $\sin B=\sin(A+C)=\sin A\cos C+\cos A\sin C$, substituting and simplifying yields $\sin\left(C+\frac{\pi}{6}\right)=1$.

Step 3. Because $0<C<\pi$, we get $C=\frac{\pi}{3}$.

Step 4. By the Law of Cosines, $c^{2}=a^{2}+b^{2}-2ab\cos C$. With $c=3$ and $C=\frac{\pi}{3}$, we have $9=a^{2}+b^{2}-ab$.

Step 5. Note that $9=(a+b)^{2}-3ab\ge (a+b)^{2}-3\left(\frac{a+b}{2}\right)^{2}=\frac14(a+b)^{2}$. Thus $a+b\le 6$, with equality when $a=b=3$.

Step 6. Hence $a+b+c\le 6+3=9$, so the maximum perimeter is 9. Therefore the correct choice is D.

D