Solving A Pink Triangle

Here is a sketch of another solution.

$1.$ Extend $AB$ to a point $D$ such that $AC=AD$ . Join $CD$ .

$2.$ Now notice that $\triangle ABC$ similar to $\triangle CBD$ .

$3.$ Set $|AB|=a$ and therefore $|BC|=a+2$ .

$4.$ $\frac{a}{a+2}=\frac{a+2}{a+5}$ because of the similarity.

This gives the solution $|AB|=a=\boxed{4}$ .

Using $sine$ rule: $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$ $\Rightarrow \dfrac{c+2}{\sin 2 C}=\dfrac{5}{\sin 3C}=\dfrac{c}{\sin C}$ $\small{\color{#3D99F6}{(\sin 2C=2\sin C \cos C, \sin 3C=3\sin C-4\sin^3 C)}}$ $\dfrac{c+2}{2\cos C}=\dfrac{5}{3-4\sin^2 C}=\dfrac{c}{1}$ From 1 and 3 find cos C while from 2 and 3 find $\sin^2 C$ in terms of c and use $\small{\color{#3D99F6}{(\sin^2 C=1-\cos^2 C)}}$ . $\Rightarrow \dfrac{3c-5}{4c}=1-(\dfrac{c+2}{2c})^2$ $\Rightarrow c=4$ Hence $AB=c=\huge\boxed{\color{#007fff}{4}}$ .

Here is another solution, this time using Euclidean methods only. Let bisector of $\angle A$ intersect $BC$ at $D$ . $AD=DC$ and $\triangle ABC$ similar to $\triangle DBA$ . To simplify notation let $AD=x, AB=y$ . $\frac{5}{x}=\frac{y+2}{y}=\frac{y}{y+2-x}$ This gives values of $x=10/3, y=\boxed{4}$

I did a straightforward solution.

Let $\angle C = \theta$ , then $\angle A = 2 \theta$ and $\angle B = 180^\circ - 3\theta$ . Let $AB = x$ , then $BC = x + 2$ .

Using sine rule, we have:

$\begin{cases} \dfrac{\sin (2\theta)}{x+2} = \dfrac{\sin \theta}{x} & ...(1) \\ \dfrac{\sin \left(180^\circ - 3\theta\right)}{5} = \dfrac{\sin \theta}{x} & ...(2) \end{cases}$

$\begin{aligned} (1): \quad x\sin (2\theta) & = (x+2)\sin \theta \\ 2x\sin \theta \cos \theta & = (x+2)\sin \theta \\ \color{#3D99F6}{2 x \cos \theta } & \color{#3D99F6}{= x + 2} \end{aligned}$

$\begin{aligned} (2): \quad x\sin \left(180^\circ - 3\theta\right) & = 5\sin \theta \\ x\sin (3\theta) & = 5\sin \theta \\ x (\sin \theta \cos (2\theta) + \cos \theta \sin (2\theta)) & = 5\sin \theta \\ x (\cos (2\theta) + 2 \cos^2 \theta) & = 5 \\ x (2\cos^2 \theta - 1 + 2 \cos^2 \theta) & = 5 \\ 4x \cos^2 \theta - x & = 5 \\ \frac{\color{#3D99F6}{(2x\cos \theta)}^2}{x} - x & = 5 \\ \frac{\color{#3D99F6}{(x+2)}^2}{x} - x & = 5 \\ x^2 + 4x + 4 - x^2 & = 5x \\ \Rightarrow x & = \boxed{4} \end{aligned}$

AN EASY OVER RATED PROBLEM AGAIN!!!!!!!!!!!!!