Estonian Mathematical Olympiad

Let $B_n$ for some $n > 1$ be definable (Fig. 30). By construction, $A$, $B_{n-2}$, and $B_n$ are collinear, whereby $A$ cannot lie between $B_{n-2}$ and $B_n$. Since $B_{n-1}B_{n-2} = B_{n-1}B_n$ and $B_{n-2} \neq B_n$, the triangle $B_{n-2}B_{n-1}B_n$ is isosceles, so that $\angle AB_{n-2}B_{n-1} + \angle AB_nB_{n-1} = 180^\circ$ and $\angle AB_{n-2}B_{n-1} \neq 90^\circ$. Consequently,

Fig. 30

$\angle AB_{n-1}B_n = 180^\circ - \alpha - \angle AB_nB_{n-1} = 180^\circ - \alpha - (180^\circ - \angle AB_{n-2}B_{n-1}) = \angle AB_{n-2}B_{n-1} - \alpha$. Since $\angle AB_0B_1 = \beta$, by induction we get $\angle AB_{n-1}B_n = \beta - (n-1)\alpha$. Hence defining of $B_n$ assumes that $\beta - (n-2)\alpha \neq 90^\circ$ and $\beta - (n-1)\alpha \ge 0^\circ$. These conditions are also sufficient, because if $\beta - (n-2)\alpha \neq 90^\circ$, meaning that $\angle AB_{n-2}B_{n-1} \neq 90^\circ$, then point $B_n$ can be chosen different from $B_{n-2}$ on the line $AB_{n-2}$, and if $\beta - (n-1)\alpha \ge 0^\circ$, meaning that $\angle AB_{n-2}B_{n-1} \ge \alpha$, then this point is located on the side $AB_{n-2}$.

Summing up, the largest $n$ for which $B_n$ is definable equals $\frac{\beta-90^\circ}{\alpha} + 1$ if $\beta - (n-2)\alpha = 90^\circ$ for some integer $n > 1$, or equivalently, $\frac{\beta-90^\circ}{\alpha}$ is a non-negative integer, and $\lfloor \frac{\beta}{\alpha} \rfloor + 1$ otherwise. Figures 31 and 32 depict the situation in the first and second case, respectively.

Fig. 31

Fig. 32