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 $B_{n-2}\ne B_n$, the triangle $B_{n-2}{B}_{n-1}{B}_n$ is isosceles, so that $\angle A{B}_{n-2}{B}_{n-1}+\angle A{B}_n{B}_{n-1}=180^{\circ }$ and $\angle A{B}_{n-2}{B}_{n-1}\ne 90^{\circ }$. Consequently,

Fig. 30

$\angle A{B}_{n-1}{B}_n=180^{\circ }-\alpha -\angle A{B}_n{B}_{n-1}=180^{\circ }-\alpha -(180^{\circ }-\angle A{B}_{n-2}{B}_{n-1})=\angle A{B}_{n-2}{B}_{n-1}-\alpha$. Since $\angle A{B}_0{B}_1=\beta$, by induction we get $\angle A{B}_{n-1}{B}_n=\beta -(n-1)\alpha$. Hence defining of $B_n$ assumes that $\beta -(n-2)\alpha \ne 90^{\circ }$ and $\beta -(n-1)\alpha \ge 0^{\circ }$. These conditions are also sufficient, because if $\beta -(n-2)\alpha \ne 90^{\circ }$, meaning that $\angle A{B}_{n-2}{B}_{n-1}\ne 90^{\circ }$, then point $B_n$ can be chosen different from $B_{n-2}$ on the line $A{B}_{n-2}$, and if $\beta -(n-1)\alpha \ge 0^{\circ }$, meaning that $\angle A{B}_{n-2}{B}_{n-1}\ge \alpha$, then this point is located on the side $A{B}_{n-2}$.

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