, 2019

a.

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

We have:
$$\widehat{A_0OA_1} + \widehat{A_1OA_2} + \widehat{A_2OA_3} + \cdots + \widehat{A_{25}OA_{26}} + \widehat{A_{26}OA_0} = 360^\circ$$

The sum $\widehat{A_0OA_1} + \widehat{A_1OA_2} + \cdots + \widehat{A_{25}OA_{26}}$ 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,
$$\widehat{A_{26}OA_0} = 360^\circ - 351^\circ = 9^\circ$$

b.

$\widehat{A_0OA_n}$ is the sum of the first $n$ angles:
$$\widehat{A_0OA_n} = 1^\circ + 2^\circ + \cdots + n^\circ = \frac{n(n+1)}{2}^\circ$$

$\widehat{A_0OA_{n+1}} = 1^\circ + 2^\circ + \cdots + (n+1)^\circ = \frac{(n+1)(n+2)}{2}^\circ$

We want:
$$\widehat{A_0OA_n} > \widehat{A_0OA_{n+1}}$$
But $\widehat{A_0OA_{n+1}}$ is always greater than $\widehat{A_0OA_n}$, 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 $\widehat{A_0OA_n}$ is measured as $360^\circ - \frac{n(n+1)}{2}$.

We need to find $n$ such that $\widehat{A_0OA_n} > \widehat{A_0OA_{n+1}}$.

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

If $S_n \leq 180$, then $\widehat{A_0OA_n} = S_n$ and $\widehat{A_0OA_{n+1}} = S_{n+1}$, so $\widehat{A_0OA_n} < \widehat{A_0OA_{n+1}}$.

If $S_n > 180$, then $\widehat{A_0OA_n} = 360 - S_n$ and $\widehat{A_0OA_{n+1}} = 360 - S_{n+1}$.

So $\widehat{A_0OA_n} > \widehat{A_0OA_{n+1}}$ 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 $\widehat{A_0OA_n} > \widehat{A_0OA_{n+1}}$ 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 $\widehat{A_0OA_n} > \widehat{A_0OA_{n+1}}$.