$\begin{aligned} & {\left[\begin{array}{l}\text { Tangent Circles } \\ {[\text { Pythagorean Theorem (direct and inverse) }}\end{array}\right]}\end{aligned}$

On the segment $A B=2 R$ as a diameter, a circle is constructed. A second circle, whose radius is half the radius of the first circle, touches the first circle internally at point A. A third circle touches the first circle internally, the second circle externally, and also touches the segment $A B$. Find the radius of the third circle.
Let $O_{1}, O_{2}, O_{3}$ be the centers of the first, second, and third circles respectively, $x$ be the radius of the third circle, and $M$ be the point of tangency of the third circle with the segment $A B$. Then
$O_{1} M^{2}=O_{1} O_{3}^{2}-M O_{2}^{2}=(R-x)^{2}-x^{2}=R^{2}-2 R x, M O_{2}^{2}+M O_{3}^{2}=O_{2} O_{3}^{2}$, or

$\left(\frac{R}{2}+\sqrt{R^{2}-2 R x}\right)^{2}+x^{2}=\left(\frac{R}{2}+x\right)^{2}$. From this, $x=4 R / 9$

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-49.jpg?height=666&width=717&top_left_y=176&top_left_x=676)

Answer

$4 R / 9$.

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