Saudi Arabia Mathematical Competitions

On the side $BC$ take point $M$ such that $CM=A{B}^{\prime }$. The angle bisector theorem implies $$\frac{A{B}^{\prime }}{B^{\prime }C}=\frac{AB}{BC}=\frac{AC}{BC}\text{, hence }\frac{MC}{B^{\prime }C}=\frac{AC}{BC}\text{.}$$ It follows $\frac{MC}{AC}=\frac{B^{\prime }C}{BC}$, that is $△MC{B}^{\prime }\sim △ACB$, and we get $MC=M{B}^{\prime }$. Moreover, $\hat{C}=\widehat{MC{B}^{\prime }}=\widehat{M{B}^{\prime }C}$ and $MC=M{B}^{\prime }$. From $B{B}^{\prime }+B^{\prime }A=BC$ it follows $B{B}^{\prime }=BC-B^{\prime }A=BC-MC=BM$, hence $△B^{\prime }BM$ is isosceles. In $△B{B}^{\prime }M$ we have $180^{\circ }=2\hat{C}+2\hat{C}+\frac{\hat{C}}{2}=\frac{9\hat{C}}{2}$, implying $\hat{C}=40^{\circ }$. It follows $$\hat{B}=\hat{C}=40^{\circ }\text{ and }\hat{A}=100^{\circ }.$$



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Alternative solution.

In $△AB{B}^{\prime }$, we have $$\frac{B{B}^{\prime }}{\sin 4x}=\frac{A{B}^{\prime }}{\sin x}=\frac{AB}{\sin 3x}.$$ In $△B{B}^{\prime }C$ we have $$\frac{B{B}^{\prime }}{\sin 2x}=\frac{B^{\prime }C}{\sin x}=\frac{BC}{\sin 3x}$$ and in $△ABC$ we can write $$\frac{BC}{\sin 4x}=\frac{AC}{\sin 2x}.$$

We get $$B{B}^{\prime }+B{B}^{\prime }\frac{\sin x}{\sin 4x}=B{B}^{\prime }\frac{\sin 3x}{\sin 2x}$$ hence $$\begin{array}{c}1+\frac{\sin x}{\sin 4x}=\frac{\sin 3x}{\sin 2x}\end{array}\text{\hspace{1em}(1)}$$ Relation (1) is equivalent to $$\frac{\sin x}{\sin 4x}=\frac{\sin 3x-\sin 2x}{\sin 2x},$$ that is $$\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{2\sin 2x\cos 2x}=\frac{2\sin \frac{x}{2}\cos \frac{5x}{2}}{\sin 2x}.$$ We get $$\cos \frac{x}{2}=2\cos 2x\cos \frac{5x}{2},$$ or $$\cos \frac{x}{2}=\cos \frac{9x}{2}+\cos \frac{x}{2}.$$ It follows $\cos \frac{9x}{2}=0$, that is $\frac{9x}{2}=\frac{\pi }{2}$, and we obtain $x=\frac{\pi }{9}$. Finally, $$\hat{B}=\hat{C}=\frac{2\pi }{9},\hat{A}=\frac{5\pi }{9}.$$