Consider the semicircles $S_1, S_2$ as full circles. Let $R_1 = \dfrac{AB}{2}$ be the radius of the initial circle $S_1$ and $r_1, R_2, r_2$ radii of the remaining circles $C_1, S_2, C_2$. Obviously, the circle $C_1$ is centered on the perpendicular bisector of the segment AB and its radius is $r_1 = PO = \dfrac{R_1}{2}$. Let U be the tangency point of the circles $S_1, C_1$ and let D be the other intersection of the circle $S_2$ with the line AB different from the point B. Invert the circles $C_1, S_2$ in the circle $S_1$. The circle $C_1$ tangent to the inversion circle and passing through the inversion center O is carried into a line tangent to the inversion circle $S_1$ at the point U, which is parallel to the line AB. The tangency point B of the semicircles $S_1, S_2$ is on the inversion circle and it stays in place. The circle $S_2$ is carried into a circle $S_2'$ tangent to the inversion circle $S_1$ at the point B and centered on the line AB. The other intersection D of the circle $S_2$ with the line AB is carried into the other intersection D' of the inverted circle $S_2'$ with the line AB. Since the circle $S_2'$ is centered on the line AB and tangent to the line parallel to AB through the point U, i.e., at a distance $OU = R_1$, its radius is equal to $R_2' = R_1$. Thus the distance of the inverted point D' from the inversion center is equal to ${OD' = OB + BD' = R_1 + 2R_2' = 3R_1}$. By the basic property of inversion, $OD \cdot OD' = R_1^2$ and consequently, $OD = \dfrac{R_1^2}{OD'} = \dfrac{R_1^2}{3R_1} = \dfrac{R_1}{3}$ This allows to calculate the radius $R_2$ of the circle $S_2$: $R_2 = \dfrac{DB}{2} = \dfrac{OB - OD}{2} = \dfrac{R_1 - \dfrac{R_1}{3}}{2} = \dfrac{R_1}{3}$ The pairwise tangent circles $S_1, C_1, S_2, C_2$ form a quadruple of Soddy's circles and their radii $R_1, r_1, R_2, r_2$ are bound by the relation $\dfrac{1}{(-R_1)^2} + \dfrac{1}{r_1^2} + \dfrac{1}{R_2^2} + \dfrac{1}{r_2^2} = \frac 1 2 \left(\dfrac{1}{-R_1} + \dfrac{1}{r_1} + \dfrac{1}{R_2} + \dfrac{1}{r_2}\right)^2$ The radius of the circle $S_1$ has to be taken with the negative sign, because its tangencies with the remaining 3 circles are internal. (If the 4 circles had all tangencies external, all radii would have positive signs.) This yields an equation for the radius of the circle $C_2$: $\dfrac{1}{R_1^2} + \dfrac{4}{R_1^2} + \dfrac{9}{R_1^2} + \dfrac{1}{r_2^2} = \frac 1 2 \left(-\dfrac{1}{R_1} + \dfrac{2}{R_1} + \dfrac{3}{R_1} + \dfrac{1}{r_2}\right)^2$ $\dfrac{14}{R_1^2} + \dfrac{1}{r_2^2} = \dfrac{1}{2} \left(\dfrac{4}{R_1} + \dfrac{1}{r_2}\right)^2$ $14 \dfrac{r_2^2}{R_1^2} + 1 = \frac 1 2 \left(4 \dfrac{r_2}{R_1} + 1\right)^2$ Denoting $x = \dfrac{r_2}{R_1}$, we get the following quadratic equation for x: $2(14x^2 + 1) = (4x + 1)^2$ $28x^2 + 2 = 16x^2 + 8x + 1$ $12x^2 - 8x + 1 = 0$ $x = \dfrac{8 \pm \sqrt{64 - 48}}{24} = \dfrac{8 \pm \sqr{16}}{24} = \dfrac{8 \pm 4}{24} = \dfrac{2 \pm 1}{6} = \frac 1 2\ \text{or}\ \frac 1 6$ The first root obviously corresponds to a circle congruent to the circle $C_1$ symmetrical to it with respect to the line AB. By the problem conditions (the semicircles), this root is not acceptable. Thus $r_2 = \dfrac{R_1}{6}$. Now we calculate the sides of the quadrilateral OPRQ: $OP = r_1 = \dfrac{R_1}{2}$ $QR = R_2 + r_2 = \dfrac{R_1}{3} + \dfrac{R_1}{6} = \dfrac{R_1}{2}$ $PR = r_1 + r_2 = \dfrac{R_1}{2} + \dfrac{R_1}{6} = \frac 2 3\ R_1$ $OQ = OD + R_2 = \dfrac{R_1}{3} + \dfrac{R_1}{3} = \frac 2 3\ R_1$ Since the opposite sides of the quadrilateral OPRQ are equal, it is a parallelogram. Since one of its internal angles $\angle POQ = 90^\circ$ is right, it is a rectangle. Hi shobber, nice to meet you again!

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I think that this sangaku-like problem can be solved in a simple way: Let $OA=OB$ be $r$. We have $OP = \dfrac{r}{2}$. If $OQ=t$ we have \[ \sqrt{t^2 + \dfrac{{r^2 }} {4}} = PQ = \dfrac{r} {2} + (r - t) = \dfrac{{3r}} {2} - t. \] Hence $OQ = t = \dfrac{2r}{3}$ and $QB=\dfrac{r}{3}$. We draw parallel lines $PR$ and $QR$ to $OQ$ and $OP$, respectively, and we get the intersection point $R$ such that $RP=QO=\dfrac{2r}{3}$ and $RQ=PO=\dfrac{r}{2}$. Being $\dfrac{{2r}} {3} - \dfrac{r} {2} = \dfrac{r} {6} = \dfrac{r} {2} - \dfrac{r} {3}$, the circle $C_2$ with center $R$ and radius $\dfrac{r}{6}$ is tangent to $C_1$ y $S_2$. On the other way, the relations \[ r - OR = r - \sqrt{\dfrac{{4r^2 }} {9} + \dfrac{{r^2 }} {4}} = r - \dfrac{{5r}} {6} = \dfrac{r} {6}, \] tell us that the circle $C_2$ is tangent to $S_1$. So, by construction, $OPRQ$ is a rectangle.