Trigonometry – Problem 26: Find the maximum perimeter of

Maximum Perimeter Problem

In $\triangle ABC$, given $c\cos A + \sqrt{3}c\sin A - b + 2a = 0$ and $c = 3$. Find the maximum perimeter of $\triangle ABC$.

Using Law of Sines to convert sides to angles:

$$\sin C\cos A + \sqrt{3}\sin C\sin A - \sin B + 2\sin A = 0$$

Since $\sin B = \sin(A+C)$:

$$\sqrt{3}\sin C\sin A - \cos C\sin A = 2\sin A$$

For $\sin A \neq 0$:

$$\sqrt{3}\sin C - \cos C = 2$$

$$2\sin\left(C - \frac{\pi}{6}\right) = 2$$

$$\sin\left(C - \frac{\pi}{6}\right) = 1$$

Since $0 < C < \pi$: $C = \frac{\pi}{3}$.

Using Law of Cosines:

$$c^2 = a^2 + b^2 - 2ab\cos C$$

$$9 = a^2 + b^2 - ab = (a+b)^2 - 3ab$$

By AM-GM: $ab \leq \frac{(a+b)^2}{4}$, so:

$$9 \geq (a+b)^2 - \frac{3(a+b)^2}{4} = \frac{(a+b)^2}{4}$$

$$a + b \leq 6$$

Maximum perimeter: $a + b + c = 6 + 3 = 9$ (when $a = b = 3$).

$9$