4.2. $B_{1}$ is the midpoint of side $A C$ of triangle $A B C$, $C_{1}$ is the midpoint of side $A B$ of triangle $A B C$. The circumcircles of triangles $A B B_{1}$ and $A C C_{1}$ intersect at point $P$. Line $A P$ intersects the circumcircle of triangle $A B_{1} C_{1}$ at point $Q$. Find $\frac{A P}{A Q}$.
4.2. Answer: $\frac{3}{2}$.

Consider an inversion with center at point $A$ and an arbitrary radius. (see fig. 15).

We will denote the images of points after inversion with primes. Then $B^{\prime}-$ is the midpoint of segment $A C_{1}^{\prime}, C^{\prime}-$ is the midpoint of segment $A B_{1}^{\prime}$. The image of the circumcircle of $\triangle A B B_{1}$ will be the line $B^{\prime} B_{1}^{\prime}$, and the image of the circumcircle of $\triangle A C C_{1}$ will be the line $C^{\prime} C_{1}^{\prime}$. Since these lines are medians of triangle $A B_{1}^{\prime} C_{1}^{\prime}$, their intersection $P^{\prime}$ will be the centroid of this triangle. Since the image of the circumcircle of $\triangle A B_{1} C_{1}$ will be the line $B_{1}^{\prime} C_{1}^{\prime}$, then $Q^{\prime}$ is the intersection of $A P^{\prime}$ with segment $B_{1}^{\prime} C_{1}^{\prime}$, and, therefore, is the midpoint of this segment. The centroid divides the median in the ratio $1: 2$, so $\frac{A P}{A Q}=\frac{A Q^{\prime}}{A P^{\prime}}=\frac{3}{2}$.