Solving Triangles - Part 9

Given that:

$\begin{aligned} b + 2c \cos A & = 0 & \small \color{#3D99F6} \text{By sine rule: }\frac {\sin B}b = \frac {\sin C}c \\ {\color{#3D99F6}\left(\frac {\sin B}b\right)}b + 2{\color{#3D99F6}\left(\frac {\sin C}c\right)}c \cos A & = 0 \\ {\color{#3D99F6}\sin B} + 2\sin C \cos A & = 0 & \small \color{#3D99F6} \text{Note that }\sin B = \sin (\pi - A-C) = \sin (A+C) \\ {\color{#3D99F6}\sin (A+C)} + 2\sin C \cos A & = 0 \\ {\color{#3D99F6}\sin A \cos C + \sin C \cos A} + 2\sin C \cos A & = 0 \\ \sin A \cos C & = -3 \sin C \cos A \\ \implies \tan A & = - 3\tan C \end{aligned}$

For $\triangle ABC$ ,

$\begin{aligned} \tan A + \tan B + \tan C & = \tan A \tan B \tan C & \small \color{#3D99F6} \text{Since }\tan A = - 3\tan C \\ \tan B - 2\tan C & = - 3\tan B \tan^2 C \\ \implies \tan B & = \color{#3D99F6} \frac {2\tan C}{1+3\tan^2 C} & \small \color{#3D99F6} \text{Divide up and down by }\tan C \\ & = \color{#3D99F6} \frac 2{\color{#D61F06} \frac 1{\tan C} +3\tan C} & \small \color{#D61F06} \text{By AM-GM inequality: } \\ & \le \frac 2{\color{#D61F06} 2\sqrt 3} & \small \color{#D61F06} \text{Equality occurs when }\tan C = \frac 1{\sqrt 3} \\ & = \frac 1{\sqrt 3} \end{aligned}$

$\angle B$ is the maximum, when $\tan B$ is the maximum, $\implies \max (B) = \frac \pi 6$ , when $C = \frac \pi 6$ , and $A = \frac {2\pi}3$ . Since $B=C$ , by sine rule $b=c$ . And $bc = b^2 = 1\implies b = c =1$ and $a = 2 \cos \frac \pi 6 = \sqrt 3$ . Therefore $a+b+c = \sqrt 3 + 2$ $\implies p + q = 2 + 3 = \boxed 5$ .

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Reference:

• Sine rule
• AM-GM inequality