Given the center of the inscribed circle and the centers of two excircles of a triangle, construct the triangle!

翻译结果如下：

Given the center of the inscribed circle and the centers of two excircles of a triangle, construct the triangle!
In triangle $ABC$, let the center of the inscribed circle be $O$, and the centers of the two given excircles be $O_{A}$ and $O_{B}$.

The points $O_{A}$ and $O_{B}$ lie on the external angle bisector passing through vertex $C$ of the triangle, while $O$ lies on the internal angle bisector passing through $C$. These two lines are perpendicular to each other, so point $C$ is the orthogonal projection of point $O$ onto the segment $O_{A} O_{B}$. Similarly, point $A$ is the orthogonal projection of $O_{B}$ onto $O O_{A}$, and point $B$ is the orthogonal projection of $O_{A}$ onto $O_{B} O$. Let $O^{*}$ be the orthocenter of triangle $O O_{A} O_{B}$.

Based on the above, the points $A, B, C$ are the vertices of a real triangle if $O, O_{A}$, and $O_{B}$ are not collinear and $\angle O_{A} O O_{B} \neq 90^{\circ}$. These three points form an orthocentric system, with $O, O_{A}, O_{B}$, and $O^{*}$ being the feet of the altitudes. The construction leads to either diagram 1/a or 1/b, depending on whether $\angle O_{A} O O_{B}$ is greater or less than $90^{\circ}$.

# $1987-12-455-1$. eps 

1.a diagram

$1987-12-455-2 . \mathrm{eps}$

1.b diagram

It is known that in an acute triangle, the altitudes are internal, and the sides are external angle bisectors in the pedal triangle. Accordingly, when $\angle O_{A} O O_{B} < 90^{\circ}$, the constructed triangle $ABC$ is not a solution to our problem, as in this case, $O$ becomes the center of the excircle opposite to $C$.

Therefore, the problem has a solution if and only if $90^{\circ} < \angle O_{A} O O_{B} < 180^{\circ}$, and in this case, there is exactly one solution.