. Let $A_{0}, B, C$ be three non-collinear points in the plane. We denote $A_{1}$ as the center of the incircle of $A_{0} B C$, $A_{2}$ as the center of the incircle of $A_{1} B C$, and so on to construct the points $A_{2}, A_{3}, \ldots$. For an integer $n$, calculate the radius of the incircle of $A_{n} B C$.
Let $r_{n}$ be the sought radius and $X_{n}$ the point of tangency between the incircle of $A_{n} B C$ and $[B C]$. Note that by denoting $\beta=\widehat{B}$ and $\gamma=\widehat{C}$, we have $\widehat{A_{n} B C}=\beta / 2^{n}$ and $\widehat{A_{n} C B}=\gamma / 2^{n}$. We deduce that $\widehat{B A_{n} C}=180^{\circ}-(\beta+\gamma) / 2^{n}$. According to the law of sines applied in $A_{n} C B$:

$$
B A_{n}=B C \frac{\sin \left(\frac{\gamma}{2^{n}}\right)}{\sin \left(180^{\circ}-\frac{\beta+\gamma}{2^{n}}\right)}
$$

Thus, $B A_{n}=B C \sin \left(\gamma / 2^{n}\right) / \sin \left((\beta+\gamma) / 2^{n}\right)$. Since $B A_{n} X_{n}$ is a right triangle, we deduce that:

$$
r_{n}=B C \frac{\sin \left(\frac{\gamma}{2^{n}}\right) \sin \left(\frac{\beta}{2^{n}}\right)}{\sin \left(\frac{\beta+\gamma}{2^{n}}\right)}
$$

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