Problem 8.1. The bisectors of $\Varangle A, \Varangle B$ and $\Varangle C$ of $\triangle A B C$ meet its circumcircle at points $A_{1}, B_{1}$ and $C_{1}$, respectively. Set $A A_{1} \cap C C_{1}=I, A A_{1} \cap B C=N$ and $B B_{1} \cap A_{1} C_{1}=P$. Denote by $O$ the circumcenter of $\triangle I P C_{1}$ and let $O P \cap B C=M$. If $B M=M N$ and $\Varangle B A C=2 \Varangle A B C$, find the angles of $\triangle A B C$.

Chavdar Lozanov
8.1. Set $\Varangle C A B=\alpha, \Varangle A B C=\beta$ and $\Varangle B C A=\gamma$. Then

$$
\begin{aligned}
\Varangle I P C_{1} & =\frac{1}{2}\left(\overparen{B A_{1}}+\overparen{C_{1} B_{1}}\right) \\
& =\frac{1}{2}\left(\overparen{B A_{1}}+\overparen{A C_{1}}+\overparen{A B}_{1}\right) \\
& =\frac{1}{2}(\alpha+\beta+\gamma)=90^{\circ} .
\end{aligned}
$$

Hence $O$ is the midpoint of the segment $I C_{1}$ and it follows that

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-092.jpg?height=487&width=507&top_left_y=426&top_left_x=1111)

$$
\Varangle I O P=2 \Varangle I C_{1} P=\overparen{C A}_{1}=\alpha .
$$

We also have $\Varangle C C_{1} B=\alpha$ and therefore $O P \| C_{1} B$. Since $C_{1} O=O I$ and $B M=M N$, we conclude that $I N \| C_{1} B$, i.e. $\Varangle C I A_{1}=\alpha$. On the other hand,

$$
\Varangle C I A_{1}=\frac{1}{2}\left(\overparen{C A_{1}}+\overparen{A C_{1}}\right)=\frac{1}{2}(\alpha+\gamma)=90^{\circ}-\frac{\beta}{2} .
$$

This implies that $\alpha=90^{\circ}-\frac{\beta}{2}=\frac{\alpha+\gamma}{2}$, i.e. $\alpha=\gamma$. Since $\alpha=2 \beta$, we obtain that $\alpha=\gamma=72^{\circ}$ and $\beta=36^{\circ}$.