2019 ROMANIAN MATHEMATICAL OLYMPIAD

a.

The sum of all angles around point $O$ is $360^{\circ }$.

We have: $$\widehat{A_{0}O{A}_1}+\widehat{A_{1}O{A}_2}+\widehat{A_{2}O{A}_3}+\cdots +\widehat{A_{25}O{A}_{26}}+\widehat{A_{26}O{A}_0}=360^{\circ }$$

The sum $\widehat{A_{0}O{A}_1}+\widehat{A_{1}O{A}_2}+\cdots +\widehat{A_{25}O{A}_{26}}$ is $1^{\circ }+2^{\circ }+3^{\circ }+\cdots +26^{\circ }$.

This is an arithmetic progression with first term $1^{\circ }$, last term $26^{\circ }$, and $26$ terms: $$S=\frac{(1^{\circ }+26^{\circ })\times 26}{2}=\frac{27^{\circ }\times 26}{2}=13.5^{\circ }\times 26=351^{\circ }$$

Therefore, $$\widehat{A_{26}O{A}_0}=360^{\circ }-351^{\circ }=9^{\circ }$$

b.

$\widehat{A_{0}O{A}_n}$ is the sum of the first $n$ angles: $$\widehat{A_{0}O{A}_n}=1^{\circ }+2^{\circ }+\cdots +n^{\circ }={\frac{n(n+1)}{2}}^{\circ }$$

$\widehat{A_{0}O{A}_{n+1}}=1^{\circ }+2^{\circ }+\cdots +(n+1{)}^{\circ }={\frac{(n+1)(n+2)}{2}}^{\circ }$

We want: $$\widehat{A_{0}O{A}_n}>\widehat{A_{0}O{A}_{n+1}}$$ But $\widehat{A_{0}O{A}_{n+1}}$ is always greater than $\widehat{A_{0}O{A}_n}$, unless we consider the angles modulo $360^{\circ }$ (i.e., the measure from $A_0$ to $A_n$ going one way, and from $A_0$ to $A_{n+1}$ going the other way).

But since the sum of all angles is $360^{\circ }$, after passing $180^{\circ }$, the measure from $A_0$ to $A_n$ is $360^{\circ }$ minus the sum $1^{\circ }+2^{\circ }+\cdots +n^{\circ }$.

So, for $n$ such that $\frac{n(n+1)}{2}>180$, the angle $\widehat{A_{0}O{A}_n}$ is measured as $360^{\circ }-\frac{n(n+1)}{2}$.

We need to find $n$ such that $\widehat{A_{0}O{A}_n}>\widehat{A_{0}O{A}_{n+1}}$.

Let $S_n=\frac{n(n+1)}{2}$ and $S_{n+1}=\frac{(n+1)(n+2)}{2}$.

If $S_n\le 180$, then $\widehat{A_{0}O{A}_n}=S_n$ and $\widehat{A_{0}O{A}_{n+1}}=S_{n+1}$, so $\widehat{A_{0}O{A}_n}<\widehat{A_{0}O{A}_{n+1}}$.

If $S_n>180$, then $\widehat{A_{0}O{A}_n}=360-S_n$ and $\widehat{A_{0}O{A}_{n+1}}=360-S_{n+1}$.

So $\widehat{A_{0}O{A}_n}>\widehat{A_{0}O{A}_{n+1}}$ if $360-S_n>360-S_{n+1}$, i.e., $S_n<S_{n+1}$, which is always true.

But as $S_n$ increases, $360-S_n$ decreases, so $\widehat{A_{0}O{A}_n}>\widehat{A_{0}O{A}_{n+1}}$ for those $n$ where $S_n>180$.

Find $n$ such that $S_n>180$: $$\frac{n(n+1)}{2}>180$$ $$n(n+1)>360$$ Try $n=19$: $19\times 20=380>360$ Try $n=18$: $18\times 19=342<360$

So for $n=19,20,21,22,23,24,25$.

Answer:

For $n=19,20,21,22,23,24,25$, one has $\widehat{A_{0}O{A}_n}>\widehat{A_{0}O{A}_{n+1}}$.