Problem 1754

1. Define the Problem and Setup:
- Let circles $\mathcal{C}_1$ and $\mathcal{C}_2$ have centers $C_1$ and $C_2$ respectively, and intersect at points $A$ and $B$.
- Points $P$ and $Q$ are on $\mathcal{C}_1$ and $\mathcal{C}_2$ respectively, such that $P$, $Q$, and $B$ are collinear with $B$ between $P$ and $Q$.
- Let lines $PC_1$ and $QC_2$ intersect at $R$.
- Let $I$ be the incenter of $\Delta PQR$ and $S$ be the circumcenter of $\Delta PIQ$.

2. **Claim 1: $XZ = YZ = R_{\Gamma_1} + R_{\Gamma_2}$**
- Let $X$ and $Y$ be the intersections of the external angle bisector of $\angle C_1BC_2$ with $\mathcal{C}_1$ and $\mathcal{C}_2$ respectively, distinct from $B$.
- Let $Z = XC_1 \cap YC_2$.
- Since $XU$ and $YV$ are diameters, $\angle UBX = 90^\circ$ and $\angle VBY = 90^\circ$, making $U, V, B$ collinear.
- Using directed angles, $\measuredangle ZXY = \measuredangle XYZ$, thus $XZ = YZ$.
- Since $\measuredangle ZUV = \measuredangle UVZ$, $UZ = VZ$.
- Therefore, $XZ + YZ = XU + VZ + YZ = 2R_{\Gamma_1} + 2R_{\Gamma_2}$, implying $XZ = YZ = R_{\Gamma_1} + R_{\Gamma_2}$.

3. **Claim 2: Points $A, C_1, C_2, O, R, Z$ are cyclic**
- Since $\measuredangle OM_1Z = \measuredangle OM_2Z = 90^\circ$, $ZM_1OM_2$ is cyclic.
- Using the midpoints $M_1$ and $M_2$ of $XZ$ and $YZ$ respectively, and the fact that $C_1M_1 = C_2M_2$, we have $\triangle OM_1C_1 \cong \triangle OM_2C_2$.
- Thus, $\measuredangle ZC_1O = \measuredangle ZC_2O$, so $O$ lies on $(ZC_1C_2)$.
- Since $\measuredangle C_1AC_2 = \measuredangle C_1ZC_2$, $A$ lies on $(ZC_1C_2)$.
- Similarly, $\measuredangle PAQ = \measuredangle PRQ$, so $R$ lies on $(ZC_1C_2)$.

4. **Claim 3: Points $A, P, Q, R$ are cyclic**
- Since $\measuredangle C_1PA = \measuredangle C_2QA$, $\triangle AC_1P \sim \triangle AC_2Q$.
- Thus, $A$ is the center of the spiral similarity sending $PC_1$ to $QC_2$, implying $\triangle AC_1C_2 \sim \triangle APQ$.
- Therefore, $\measuredangle PAQ = \measuredangle PRQ$, making $A, P, Q, R$ cyclic.

5. **Claim 4: $S = PX \cap QY$**
- By the Incenter-Excenter Lemma, $S$ is the midpoint of arc $\text{PQ}$ of $(PQR)$.
- Let $S' = PX \cap QY$. It suffices to show $S'$ is also the midpoint of $\text{PQ}$ in $(PQR)$.
- Since $A$ is the Miquel point of the complete quadrilateral $BXS'Q$, $A$ is the center of the spiral similarity sending $PQ$ to $XY$.
- Thus, $\measuredangle PS'Q = \measuredangle PAQ = \measuredangle PZQ$, so $S'$ lies on $(PQR)$.
- Since $\measuredangle S'PQ = \measuredangle PQS'$, $S'P = S'Q$, making $S'$ the midpoint of $\text{PQ}$.

6. **Claim 5: Points $K, S, X, T, Z$ are cyclic**
- Since $K, X, Y, Z$ are cyclic by $\measuredangle KXZ = 90^\circ$, $\measuredangle XSY = \measuredangle PAQ = \measuredangle XZY$.
- Thus, $S$ is cyclic with $K, X, Y, Z$.

Since $S$ lies on $(XYZ)$, which is fixed, as $P$ and $Q$ vary, $S$ will trace an arc of the circle $(XYZ)$, and its center $O$ is indeed concyclic with $A, C_1, C_2$.

$\blacksquare$