Bulgarian Mathematical Olympiad

Solution:
Set $AB = c$, $BC = a$, $AC = b$, $p = \frac{a + b + c}{2}$ and let $O$, $P$ and $Q$ be the midpoints of $AB$, $AM$ and $BN$, respectively.
Since $OO_{1} \perp AC$ and $O_{1}P \perp AM$, we have $\triangle O_{1}OP \sim \triangle ABC$, implying that $\frac{OO_{1}}{c} = \frac{OP}{a}$. Further, it follows from $AM = p - c$ that
$$OP = OA + \frac{1}{2} AM = \frac{1}{2}c + \frac{1}{2}(p - c) = \frac{a + b + c}{4}$$
and therefore $OO_{1} = \frac{c(a + b + c)}{4a}$. Analogously $OO_{2} = \frac{c(a + b + c)}{4b}$.
Since $\angle ACB = 90^{\circ}$ we have that $\angle O_{1}OO_{2} = 90^{\circ}$.
Finally, using that $S_{O_{1}O_{2}C} = \left| S_{OO_{1}O_{2}} - S_{OO_{1}C} - S_{OO_{2}C} \right|$ we compute
$$\begin{aligned} S_{OO_{1}O_{2}} - S_{OO_{1}C} - S_{OO_{2}C} & = \frac{OO_{1} \cdot OO_{2}}{2} - \frac{OO_{1} \cdot b}{4} - \frac{OO_{2} \cdot a}{4} \\ & = \frac{c^{2}(a + b + c)^{2}}{32ab} - \frac{c(a + b + c)}{16} \left( \frac{b}{a} + \frac{a}{b} \right) \\ & = \frac{c^{2}(a + b + c)^{2}}{32ab} - \frac{c^{3}(a + b + c)}{16ab} = \frac{c^{2}}{16}, \end{aligned}$$
since $c^{2} = a^{2} + b^{2}$. This implies that $S_{O_{1}CO_{2}}$ does not depend on the choice of the point $C$.