9. (20 points) In an acute isosceles triangle $ABC (BC = CA)$, two medians $BB_1$ and $CC_1$ are drawn. The circumcircle of triangle $BB_1C$ intersects the median $CC_1$ at points $C$ and $Q$. Find the radius of the circumcircle of triangle $ABC$ if $CQ = m$.

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# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_e38b8a325740b6549434g-09.jpg?height=771&width=860&top_left_y=437&top_left_x=632)

Since triangle $ABC$ is isosceles, $CC_1$ is the height, median, and bisector. Therefore, point $Q$ is the midpoint of arc $BB_1$ that does not contain point $C$. Consequently, $B_1Q = QB$. Since $CC_1$ is the perpendicular bisector of segment $AB$, $AQ = QB$. Drop a perpendicular $QP$ to side $AC$. Since triangle $AQB_1$ is isosceles,

$$
AP = PB_1 \Longrightarrow \frac{CB_1}{CP} = \frac{2}{3}
$$

If $O$ is the center of the circumscribed circle of triangle $ABC$, then it lies on $CC_1$ and $OB_1 \perp AC$. Therefore, $OB_1 \parallel QP$, and by the theorem of proportional segments,

$$
R = CO = CQ \cdot \frac{CB_1}{CP} = \frac{2m}{3}
$$

## Answers:

| Option | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | 6 | 8 | 4 | 10 | 18 | 2 | 14 | 16 | 18 | 20 |