A quarter circle with a radius of $12 \mathrm{~cm}$ has a semicircle cut out over one of its bounding radii, as if the radius were a diameter. What is the radius of the largest circle that can be inscribed in the resulting shape?
Solution. The inscribed circle will be the largest when it touches all three boundary lines of the shape. The line connecting the centers of the tangent circles always passes through the point of tangency.

Let the radius of the largest inscribed circle be denoted by $r$, and use the notation in the diagram: $O_{3} E_{1}=O_{3} E_{2}=O_{3} E_{3}=r, O_{1} O_{3}=$ $12-r, O_{2} O_{3}=6+r$, and $O_{1} T=r$, since the quadrilateral $E_{3} O_{3} T O_{1}$ is a rectangle. Thus, $O_{2} T=6-r$. Let $O_{3} T=y$, then for the right triangle $O_{1} T O_{3}$, by the Pythagorean theorem: $(12-r)^{2}-r^{2}=y^{2}$, and for the right triangle $O_{2} T O_{3}$: $(6+r)^{2}-(6-r)^{2}=y^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_76dbb7efa27791192ebag-1.jpg?height=512&width=507&top_left_y=430&top_left_x=798)

Based on the two equations,

$$
(12-r)^{2}-r^{2}=(6+r)^{2}-(6-r)^{2}
$$

which is a linear equation in $r$, and from this $r=3$.

Therefore, the radius of the desired circle is $3 \mathrm{~cm}$.