3. Given are two circles with centers $O$ and $O_{1}$ and radii $R$ and $\frac{R}{2}$. The circles touch each other internally at the point $T$. Construct a circle that touches the given circles and the line $O O_{1}$.
Solution. Analysis. Let $k$ and $k_{1}$ denote the given circles with centers $O$ and $O_{1}$. Suppose the circle $k_{2}$ with center $O_{2}$ and radius $x$ touches the circles $k$ and $k_{1}$ and the line $O O_{1}$, sequentially at points $A, B$, and $C$, (diagram on the right). Let $O C=c$. Since

$$
O_{1} O_{2}=\frac{R}{2}+x, O O_{2}=R-x, O_{2} C=x, C O_{1}=\frac{R}{2}+c
$$

and since triangles $\mathrm{O}_{2} \mathrm{CO}$ and $\mathrm{O}_{2} \mathrm{CO}_{1}$ are right-angled (with right angles at vertex $C$), from the Pythagorean theorem we get

$$
\begin{aligned}
& \left(\frac{R}{2}+c\right)^{2}+x^{2}=\left(\frac{R}{2}-c\right)^{2} \\
& c^{2}+x^{2}=(R-x)^{2}
\end{aligned}
$$

i.e.,

$$
R c+c^{2}=R x, c^{2}=R^{2}-2 R x
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_3d6c04de42baf650923eg-02.jpg?height=409&width=411&top_left_y=399&top_left_x=1068)

If we eliminate $c$ from these equations, we get $x=\frac{4 R}{9}$. Note that $O_{1} O_{2}=\frac{17 R}{18}, O O_{2}=\frac{5 R}{9}$.

Construction. We construct a circle $k_{3}$ with center $O_{1}$ and radius $\frac{17 R}{18}$ and a circle $k_{4}$ with center $O$ and radius $\frac{5 R}{9}$. Let $O_{2}^{\prime}$ be the intersection of circles $k_{3}$ and $k_{4}$. We construct a circle $k_{2}^{\prime}$ with center $O_{2}^{\prime}$ and radius $\frac{4 R}{9}$. Then $k_{2}^{\prime}$ is the desired circle.

Proof. The circles $k_{3}$ and $k_{4}$ intersect because the numbers $\frac{17 R}{18}, \frac{5 R}{9}$, and $\frac{R}{2}$ satisfy the necessary and sufficient condition to be the sides of a triangle. Since $\left(\frac{17 R}{18}\right)^{2}>\left(\frac{5 R}{9}\right)^{2}+\left(\frac{R}{2}\right)^{2}$, we conclude that $O_{1} \mathrm{O}_{2}^{\prime} O$ is an obtuse triangle with the obtuse angle at vertex $O$. Therefore, point $O$ lies between point $O_{1}$ and the foot $C^{\prime}$ of the altitude of triangle $\mathrm{O}_{1} \mathrm{O}_{2}^{\prime} \mathrm{O}$ from vertex $\mathrm{O}_{2}^{\prime}$. Let $C^{\prime} O=c^{\prime}$ and $C^{\prime} O_{2}^{\prime}=h$. Since triangles $O_{2}^{\prime} C^{\prime} O$ and $O_{2}^{\prime} C^{\prime} O_{1}$ are right-angled with right angles at vertex $C^{\prime}$, we have

$$
h^{2}=\left(\frac{5 R}{9}\right)^{2}-c^{\prime 2}=\left(\frac{17 R}{18}\right)^{2}-\left(\frac{R}{2}+c^{\prime}\right)^{2}
$$

Further, we easily get $c^{\prime}=\frac{R}{3}$ and $h=\frac{4 R}{9}$, which means that circle $k_{2}^{\prime}$ touches the line $O O_{1}$. From the equation

$$
O_{2}^{\prime} O_{1}=\frac{17 R}{18}=\frac{R}{2}+\frac{4 R}{9}
$$

it follows that circle $k_{2}^{\prime}$ touches circle $k_{1}$. Since

$$
O_{2}^{\prime} O=\sqrt{c^{\prime 2}+h^{2}}=\frac{5 R}{9}=R-\frac{4 R}{9}
$$

we conclude that circles $k_{2}^{\prime}$ and $k$ touch each other.

Discussion. Circles $k_{2}^{\prime}$ and $k_{1}$ have two intersection points, which means that the problem has two solutions.