find the largest perimeter of this triangle

Question

Let $ABC$ be a triangle with angles $A=2B$ and

$C$ obtuse angle $(90≤C≤180)$

Then find the largest perimeter of this triangle

This is my homework , give a hint to solve :

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

And $P=a+b+c$

Also we find $C=3B-π$

Now :

$\frac{\sin (2B)}{a}=\frac{\sin B}{b}=\frac{\sin (3B-π}{c}$

Now how I complete this work ?

Answer 1

You can use $$a=\frac{b\sin(2\beta)}{\sin(\beta)}=2b\sin(\beta)$$ and $$c=\frac{\sin(\pi-2\beta)}{\sin(\beta)}$$ so you will get $$p=b\left(2\sin(\beta)+1+\frac{\sin(\pi-3\beta)}{\sin(\beta)}\right)$$ and with $b=const.$

Answer 2

Using diameter of circumcircle: $$\frac{\sin (2B)}{a}=\frac{\sin B}{b}=\frac{\sin (\pi-3B)}{c}=\frac{1}{D}$$ From here we get the function to maximize: $p(B)=D(\sin B + \sin 2B + \sin 3B)$

Since sine is concave on the interval of interest, we can use Jensen’s inequality to show that $\sin B + \sin 2B + \sin 3B \le 3\sin \frac{B+2B+3B}{3}=3\sin 2B$.

We also have restriction that $3B < \frac{\pi}{2}$ or $2B < \frac{\pi}{3}$. Again, since sine is increasing on this interval, the maximum value of $\sin 2B$ will be when $2B$ is as close to $\frac{\pi}{3}$ as possible so $B$ should be as close to $\frac{\pi}{6}$ as possible.

Thus, $p(B) < \frac{3D\sqrt{3}}{2}$.