9.3. A circle of radius $R$ touches the base $A C$ of an isosceles triangle $A B C$ at its midpoint and intersects side $A B$ at points $P$ and $Q$, and side $C B$ at points $S$ and $T$. The circumcircles of triangles $S Q B$ and $P T B$ intersect at points $B$ and $X$. Find the distance from point $X$ to the base of triangle $A B C$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1ee9a86668a3e4f14940g-1.jpg?height=349&width=346&top_left_y=2347&top_left_x=912)
Answer: R.

Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_1ee9a86668a3e4f14940g-2.jpg?height=420&width=517&top_left_y=401&top_left_x=724)

Obviously, the diagram shows symmetry with respect to the line $B H$, which divides the angle $A B C$ into two equal angles. Note that the angles $Q B X$ and $Q S X$ are equal (as inscribed angles subtending the same arc). Similarly, the angles $X Q S$ and $X B S$ are equal, but the angle $Q B X$ is equal to the angle $X B S$ due to symmetry, so all four angles are equal. Therefore, triangle $Q S X$ is isosceles with base $Q S$. Consequently, point $X$ lies on the perpendicular bisector of the chord $Q S$, and it is also equidistant from points $Q$ and $T$, so $X$ is the center of the original circle. Therefore, the distance from $X$ to $A C$ is $R$.