Problem 1414

1. **Inversion around point $P$**:
- Let $P$ be one of the intersection points of the two circles with centers $O_1$ and $O_2$.
- Consider an inversion around point $P$ with an arbitrary radius. This inversion will map the circles to themselves because $P$ is a common point of both circles.

2. Rephrasing the problem in terms of the inverted image:
- Let $A'$ and $B'$ be the images of points $A$ and $B$ under the inversion.
- The common tangent at $A$ and $B$ will map to a line passing through $P$ in the inverted image.
- Let $P'$ be the antipode of $P$ in the circle passing through $A$ and $B$ after inversion.

3. **Considering the triangle $\Delta ABP'$**:
- In the inverted image, we have a triangle $\Delta ABP'$ where $P'$ is the antipode of $P$ in the circle $\odot (ABP)$.

4. Constructing the perpendiculars and rectangles:
- Let $X$ be a point on the circle $\odot (P, PA')$ such that $\angle BA'X = 90^\circ$.
- Let $R$ be a point such that $BA'XR$ forms a rectangle.

5. **Proving $R \in CA'$**:
- Since $P$ lies on the perpendicular bisectors of $BC$, $BR$, and $A'X$, it implies that $P$ is the center of the circle $\odot (BCR)$.
- Therefore, $\angle BCR = \frac{1}{2} \angle BPR = 90^\circ - \angle RBP = \angle BCA'$.

6. Conclusion:
- From the above steps, we have shown that $AP \perp PC$.

$\blacksquare$