A unit radius circle has an inscribed triangle. What could be the distance between the centers of two excircles?
Solution. Let the vertices and angles of the triangle be denoted in the usual way by $A, B, C$, and $\alpha, \beta, \gamma$, respectively. Let the center of the inscribed circle be $O$, and the centers of the excircles be $O_{A}, O_{B}$, and $O_{C}$. We know that the center of the inscribed circle is the intersection of the three internal angle bisectors, while the centers of the excircles are the intersections of two external and one internal angle bisectors. Since the internal and external angle bisectors at any vertex of a triangle are perpendicular to each other, $A O_{A} \perp O_{B} O_{C}, B O_{B} \perp O_{A} O_{C}$, and $C O_{C} \perp O_{B} O_{A}$. Therefore, the segments $A O_{A}, B O_{B}$, and $C O_{C}$ are the altitudes of the triangle $O_{A} O_{B} O_{C}$, so the points $A, B$, and $C$ are the feet of the altitudes of this triangle, which means the circle passing through them (which is also the circumcircle of the triangle $A B C$) is the Feuerbach circle of the triangle $O_{A} O_{B} O_{C}$. It is known that the radius of the Feuerbach circle of any triangle is exactly half the radius of the circumcircle of the triangle, so the radius of the circumcircle of the triangle $O_{A} O_{B} O_{C}$ is $R=2$ units.

![](https://cdn.mathpix.com/cropped/2024_05_02_88518a9bf9d545bd7027g-1.jpg?height=591&width=616&top_left_y=553&top_left_x=749)

Due to the symmetric role of the centers of the excircles, it is sufficient to determine the possible values of the distance $O_{A} O_{B}$. According to the generalized sine theorem, $O_{A} O_{B}=2 R \sin \angle O_{B} O_{C} O_{A}$, so we need to determine the possible values of $\angle O_{B} O_{C} O_{A}$. Since

$$
\angle A B O_{C} = \angle O_{B} B O_{C} - \angle O_{B} B A = 90^{\circ} - \angle O B A = 90^{\circ} - \frac{\beta}{2},
$$

and similarly, we get that $\angle B A O_{C} = 90^{\circ} - \frac{\alpha}{2}$,

$$
\begin{aligned}
\angle O_{B} O_{C} O_{A} & = \angle A O_{C} B = 180^{\circ} - \left( \angle A B O_{C} + \angle B A O_{C} \right) = \\
& = 180^{\circ} - \left( \left( 90^{\circ} - \frac{\beta}{2} \right) + \left( 90^{\circ} - \frac{\alpha}{2} \right) \right) = \frac{\alpha + \beta}{2}.
\end{aligned}
$$

Thus,

$$
O_{A} O_{B} = 4 \sin \frac{\alpha + \beta}{2}
$$

In the triangle $A B C$, the angle $\gamma$ can take any value between $0^{\circ}$ and $180^{\circ}$, so $\frac{\alpha + \beta}{2}$ can be any angle between $0^{\circ}$ and $90^{\circ}$, meaning its sine can take any value between 0 and 1. Therefore, the distance between the centers of two excircles of a triangle inscribed in a unit circle is greater than 0 but less than 4, and with the appropriate choice of the triangle, it can be achieved that for any value $0 < d < 4$, there are two excircles of the triangle whose centers are exactly $d$ units apart.