5. Let $C$ and $C^{\prime}$ be two externally tangent circles with centers $O$ and $O^{\prime}$ and radii 1 and 2, respectively. From $O$, a tangent is drawn to $C^{\prime}$ with the point of tangency at $P^{\prime}$, and from $O^{\prime}$, a tangent is drawn to $C$ with the point of tangency at $P$ in the same half-plane as $P^{\prime}$ with respect to the line passing through $O$ and $O^{\prime}$. Find the area of the triangle $O X O^{\prime}$, where $X$ is the intersection of $O^{\prime} P$ and $O P^{\prime}$.
Solution. The triangles $O P O^{\prime}$ and $O P^{\prime} O^{\prime}$ are right-angled at $P$ and $P^{\prime}$, respectively, and $\angle P X O=\angle P^{\prime} X O^{\prime}$, so the triangles $P X O$ and $P^{\prime} X O^{\prime}$ are similar with a similarity ratio of $O^{\prime} P^{\prime} / O P=2$. The ratio between their areas $S^{\prime}$ and $S$ is then $S^{\prime} / S=4$. By the Pythagorean Theorem, $O P^{\prime}=\sqrt{5}$ and $O^{\prime} P=2 \sqrt{2}$, so if $A$ is the desired area, we have that

$$
A+S^{\prime}=\frac{1}{2} O^{\prime} P^{\prime} \cdot O P^{\prime}=\sqrt{5} ; \quad A+S=\frac{1}{2} O P \cdot O^{\prime} P=\sqrt{2}
$$

From the above relationships, it is easily obtained that $A=\frac{4 \sqrt{2}-\sqrt{5}}{3}$.

![](https://cdn.mathpix.com/cropped/2024_04_26_2ae3d060a9eccc7695c7g-5.jpg?height=606&width=882&top_left_y=425&top_left_x=587)