1. In an acute triangle $ABC$, let $O$ be the center of the circumscribed circle, $S_{a}, S_{b}$ the midpoints of sides $BC, AC$ respectively, and $P$ the foot of the altitude from $A$ to side $AB$. Express the ratio $\left|O S_{a}\right| /\left|O S_{b}\right|$ in terms of $a=|BC|, b=|CA|$, and $k=|AP| /|BP|$.
1. Triangles $S_{b} O C$ and $P B C$ are similar because both are right-angled and the measure of the inscribed angle $P B C$ corresponding to the chord $A C$ is equal to half the measure of the central angle $A O C$ corresponding to the same chord (Fig. 1).

![](https://cdn.mathpix.com/cropped/2024_04_17_f31050e6eb0873694209g-2.jpg?height=645&width=642&top_left_y=631&top_left_x=707)

Fig. 1

For the same reason, triangles $S_{a} O C$ and $P A C$ are also similar. Therefore,

$$
\frac{\left|O S_{a}\right|}{|O C|}=\frac{|A P|}{|A C|}=\frac{|A P|}{b}, \quad \frac{\left|O S_{b}\right|}{|O C|}=\frac{|B P|}{|B C|}=\frac{|B P|}{a}
$$

from which

$$
\frac{\left|O S_{a}\right|}{\left|O S_{b}\right|}=\frac{|A P| \cdot|O C|}{b} \cdot \frac{a}{|B P| \cdot|O C|}=\frac{|A P|}{|B P|} \cdot \frac{a}{b} =k \cdot \frac{a}{b} .
$$

For a complete solution, award 6 points, with each of the following steps being worth one point: 1. one similarity (with proof), 2. second similarity (with proof), 3. correct ratio in the first similarity, 4. correct ratio in the second similarity, 5. combining into an equation (eliminating $|O C|$), 6. correct conclusion (expressing in terms of $a, b, k$).