Let $A O B$ be a quarter circle, and the first third points from $O$ on its bounding radii be $A_{1}$ and $B_{1}$. Calculate the radius of the circle that touches the semicircles with diameters $A A_{1}$ and $B B_{1}$, and the arc $A B$. Use the segment $O A_{1}$ as the unit. Also, construct the center of this circle.

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Since the problem statement only specifies the diameters of the semicircles involved, let's consider the circles with diameters $A A_{1}$ and $B B_{1}$ (and of course the quarter circle $\mathrm{AB}$) as the shapes that the sought circle $k$ must touch. Let the circles with diameters $A A_{1}$ and $B B_{1}$ be denoted by $k_{A}$ and $k_{B}$, respectively, and the sought circle by $k$. If $k$ were to touch the arc $A B$ from the outside, or if the arc $A B$ were to touch $k$ from the inside, then $k$ would have to touch the arc $A B$ at both points $A$ and $B$, since $k_{A}$ and $k_{B}$ are inside the circle containing the arc $A B$. Under these conditions, $k$ could only be the circle containing the arc $A B$ itself. However, we do not consider the identity of two arcs as a touching condition.

Therefore, $k$ touches the arc $A B$ from the inside, and the possible configurations for the mutual positions of $k_{A}$, $k_{B}$, and $k$ are: 1. both $k_{A}$ and $k_{B}$ touch $k$ from the inside, 2. both $k_{A}$ and $k_{B}$ touch $k$ from the outside, 3. one touches from the inside and the other from the outside. Since the given circles are symmetric with respect to the angle bisector of $\angle A O B$, in the first case, the center of $k$ is on the angle bisector of $\angle A O B$, and in the third case, the center of $k$ is on the ray $O A$ or $O B$, depending on whether $k_{A}$ or $k_{B}$ touches $k$ from the inside.

Suppose $k$ touches $k_{A}$ from the inside (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_a1451649247b9ca7a3d7g-1.jpg?height=959&width=764&top_left_y=678&top_left_x=667)

Figure 1

Let the center of $k$ be $C$, its radius $x$, the point of tangency between the arc $A B$ and $k$ be $E_{1}$, the point of tangency between $k$ and $k_{A}$ be $E_{2}$, and the center of $k_{A}$ be $A_{2}$. Since $O C = 3 - x$, $C A_{2} = x - 1$, and $O A_{2} = 2$, we have $O C + C A_{2} = O A_{2}$. Therefore, the triangle $O C A$ is degenerate, so $C$ lies on the line $O A$.

The statements made for the third case can be applied twice to the first case, so in the first case, the center of $k$ can only be $O$. However, since we do not consider the circle containing the arc $A B$ as a solution, we discard the first possibility.

![](https://cdn.mathpix.com/cropped/2024_05_02_a1451649247b9ca7a3d7g-2.jpg?height=857&width=767&top_left_y=130&top_left_x=663)

In the second case (Figure 2), denoting the radius of $k$ by $x$, by the cosine rule,

$$
(1 + x)^2 = 4 + (3 - x)^2 - 4(3 - x) \cos 45^\circ
$$

from which

$$
x = \frac{6 - 3\sqrt{2}}{4 - \sqrt{2}} = \frac{9 - 3\sqrt{2}}{7} \approx 0.6796 \ldots
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_a1451649247b9ca7a3d7g-2.jpg?height=973&width=946&top_left_y=1340&top_left_x=579)

Figure 3

In the third case, in the right triangle $\mathrm{OB}_{2} C$ (Figure 3),

$$
(1 + x)^2 = 4 + (3 - x)^2
$$

from which

$$
x = 1.5
$$

Thus, in the third case, the center of the sought circle can be the midpoint of segments $O A$ or $O B$. It is a matter of agreement whether to consider these as solutions, since in this case, the points of tangency are the endpoints of the arcs mentioned in the problem.

We still need to address the construction of the center of $k$ in the second case. Increase the radius of $k$ by one unit. The resulting circle passes through the centers $A_{2}$ and $B_{2}$ of the circles $k_{A}$ and $k_{B}$, respectively, and through the point $D$ on the angle bisector of $\angle A O B$ that is 4 units away from $O$ (Figure 4).

![](https://cdn.mathpix.com/cropped/2024_05_02_a1451649247b9ca7a3d7g-3.jpg?height=892&width=897&top_left_y=424&top_left_x=598)

Therefore, we need to construct the center of the circumcircle of triangle $A_{2} B_{2} D$. The distance from the point $C$ thus obtained to the intersection point $E$ of the line $O D$ and the quarter circle $A B$ is the radius of the sought circle.

László Heckenast (Budapest, Fazekas M. Gyak. Gimn., IV. o. t.)