Problem 4. Around point $\mathrm{O}$, $\mathrm{n}$ angles are constructed such that $\mathrm{m}\left(\Varangle \mathrm{A}_{0} \mathrm{OA}_{1}\right)=1^{\circ}$, $\mathrm{m}\left(\Varangle \mathrm{A}_{1} \mathrm{OA}_{2}\right)=2^{\circ}, \mathrm{m}\left(\Varangle \mathrm{A}_{2} \mathrm{OA}_{3}\right)=3^{\circ}, \ldots, \mathrm{m}\left(\Varangle \mathrm{A}_{\mathrm{n}-1} \mathrm{OA}_{\mathrm{n}}\right)=\mathrm{n}^{\circ}$.

a) What is the maximum possible value of $\mathrm{n}$?

b) For the maximum possible $\mathrm{n}$, prove that $\Varangle \mathrm{A}_{5} \mathrm{OA}_{14}$ is a right angle.

c) Calculate, for the same $\mathrm{n}$, the measure of the angle formed by the bisector of $\Varangle \mathrm{A}_{2} \mathrm{OA}_{3}$ and the bisector of $\Varangle \mathrm{A}_{13} \mathrm{OA}_{14}$.
Solution:

a) $1^{\circ}+2^{\circ}+3^{\circ}+\ldots+\mathrm{n}^{\circ}<360^{\circ} \Rightarrow \frac{n(n+1)}{2}<360^{\circ} \Rightarrow \mathrm{n}=26^{\circ}$

$2 p$

b) $\mathrm{n}=26^{\circ} \Rightarrow \mathrm{m}\left(\angle \mathrm{A}_{5} \mathrm{OA}_{14}\right)=\mathrm{m}\left(\angle \mathrm{A}_{5} \mathrm{OA}_{6}\right)+\mathrm{m}\left(\angle \mathrm{A}_{6} \mathrm{OA}_{7}\right)+\ldots+\mathrm{m}\left(\angle \mathrm{A}_{13} \mathrm{OA}_{14}\right)=6^{\circ}+7^{\circ}+\ldots+14^{\circ}=90^{\circ}$

$\Rightarrow \angle \mathrm{A}_{5} \mathrm{OA}_{14}$ is a right angle.

$2 p$

c) $\left[\mathrm{OM}\right.$ bisector of $\angle \mathrm{A}_{2} \mathrm{OA}_{3} \Rightarrow \mathrm{m}\left(\angle \mathrm{A}_{2} \mathrm{OM}\right)=\mathrm{m}\left(\angle \mathrm{MOA}_{3}\right)=1^{\circ} 30'$

1p

[ON bisector of $\angle \mathrm{A}_{13} \mathrm{OA}_{14} \Rightarrow \mathrm{m}\left(\angle \mathrm{A}_{13} \mathrm{ON}\right)=\mathrm{m}\left(\angle \mathrm{NOA}_{14}\right)=7^{\circ}$

$1 p$

$\mathrm{m}(\angle \mathrm{MON})=\mathrm{m}\left(\angle \mathrm{MOA}_{3}\right)+\mathrm{m}\left(\angle \mathrm{A}_{3} \mathrm{OA}_{4}\right)+\ldots+\mathrm{m}\left(\angle \mathrm{A}_{12} \mathrm{OA}_{13}\right)+\mathrm{m}\left(\angle \mathrm{A}_{13} \mathrm{ON}\right)$

$=1^{\circ} 30'+4^{\circ}+\ldots+13^{\circ}+7^{\circ}=93^{\circ} 30'$.

$1 p$