I. solution. The center $O$ of the sought circle $k$ lies on the segment $CD$, because $k$ externally touches the equal-radius semicircles $AC = k_1$ and $CB = k_2$, and internally touches the equal-radius arcs $BD = k_3$ and $AD = k_4$. Therefore, $O$ is equidistant from the midpoints $A_1$ and $B_1$ of the diameters $AC$ and $CB$, respectively, and from points $A$ and $B$. The perpendicular bisectors of segments $A_1B_1$ and $AB$ coincide: the line $CD$. This is also the axis of symmetry of the arc quadrilateral, so to more precisely determine $O$, it is sufficient to ensure that $k$ touches one of the semicircles and one of the arcs $k_3$ or $k_4$, from which it follows that it touches all four arcs.

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Let $k$ touch $k_1$ at $T_1$ and $k_4$ at $T_4$, denote its radius by $r$, and take the segment $AA_1$ as the unit length (Figure 1). Then $OA_1 = OT_1 + T_1A_1 = r + 1$, $OB = BT_4 - OT_4 = 4 - r$. Expressing $OC$ in two ways from the right triangles $OA_1C$ and $OBC$, we get a one-variable equation for $r$:

$$OC^2 = (r + 1)^2 - 1^2 = (4 - r)^2 - 2^2$$

from which $r = 6/5$.

This segment can be constructed in several ways using proportional division. It is convenient to use the fact that segment $AB$ is divided into 4 equal parts. Let the reflection of $C$ and $B_1$ over $B$ be $C'$ and $B_1'$, respectively, so $AC' = 6$ and $AB_1' = 5$; measure the segments $AA_1$ and $AB_1'$ from a point $A$ on an arbitrary ray, let the endpoints be $A_1''$ and $B_1''$, respectively, and finally cut the line $AB$ with the line parallel to $B_1''C'$ through $A_1''$ at point $F$. Then $AF = 6/5 = r$. Furthermore, $BF = 4 - r = BO$, so $O$ can be found by intersecting the circle with radius $BF$ centered at $B$ with $CD$.

$k$ indeed touches $k_4$ because $BO + r = BF + FA = BA$, the radius of $k_4$, and also $k_1$ because

$$OA_1^2 = OC^2 + CA_1^2 = BO^2 - BC^2 + CA_1^2 = (14/5)^2 - 4 + 1 = 121/25$$

$OA_1 = 11/5$ and this is equal to the sum of the radii of $k$ and $k_1$. - Only one suitable circle can be constructed.

György Pásztor (Szeged, Petőfi-telepi I. sz. ált. isk. VII. o. t.)

II. solution. Thinking of the method of circle inversion, consider the circle $k^*$ concentric with $k$ that passes through $A_1$ and $B_1$ (Figure 2). Its radius is $r + AA_1$, so it touches the circle $k'$ with radius $AB + AA_1$ centered at $A$. Therefore, $O$ can also be found as the center of the circle passing through $A_1$ and $B_1$ and touching $k'$. Let the point of tangency be $T$, and the intersection of the common tangent of $k^*$ and $k'$ with $A_1B_1$ be $G$.

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$G$ can be determined as follows. Consider a helper circle $k_s$ that passes through $A_1$ and $B_1$ and intersects $k'$ at one of its common points, $H$. Let the second intersection of line $GH$ with $k'$ be $L$, and with $k_s$ be $M$. We will show that $L$ and $M$ coincide, and thus are the same as the second common point of $k'$ and $k_s$, $J$. Apply the known theorem for a secant and a tangent from a point outside a circle first to $k^*$ and $G$

$$GT^2 = GA_1 \cdot GB_1$$

then to $k'$ and $G$

$$GL \cdot GH = GT^2$$

and finally to the two secants of $k_s$ from $G$, with the tangent $GN$ from $G$ inserted

$$GA_1 \cdot GB_1 = GN^2 = GM \cdot GH$$

From these, $GL \cdot GH = GM \cdot GH$, so $GL = GM$, and this proves our statement.

Therefore, $G$ is the intersection of the line $HJ$ connecting the intersection points of $k'$ and $k_s$ with $AB$, and $T$ can be found by intersecting the Thales circle with diameter $GA$ above $G$ with $k'$. $O$ can then be