Problem 1731

Given the points $A, V_1, V_2, B, U_2, U_1$ on a circle $\Gamma$ in that order, with $BU_2 > AU_1 > BV_2 > AV_1$, and a variable point $X$ on the arc $V_1 V_2$ of $\Gamma$ not containing $A$ or $B$, we need to prove the existence of a fixed point $K$ and a real number $c$ such that $OK^2 - \rho^2 = c$, where $O$ and $\rho$ denote the circumcenter and circumradius of $\triangle XCD$, respectively.

To solve this, we proceed as follows:

1. Define points $B'$ and $A'$ on $\Gamma$ such that $BB' \parallel U_2V_2$ and $AA' \parallel U_1V_1$.
2. Let $K$ be the intersection point of lines $AB'$ and $BA'$.

We claim that $K$ is the fixed point we are looking for.

3. Let $AB' \cap U_2V_2 = B_1$ and $BA' \cap U_1V_1 = A_1$. Note that $\angle AXB = 180^\circ - \angle AB'B = 180^\circ - \angle AB_1D$, implying that quadrilateral $XADB_1$ is cyclic. Similarly, $BXA_1C$ is cyclic.

4. Using the power of a point theorem, we have:
$$p_K((AXD)) = KA \cdot KB_1 \quad \text{and} \quad p_K((BXC)) = KB \cdot KA_1,$$
both of which are fixed values.

5. Since $p_K((AXB))$ is fixed because the circle $(AXB)$ does not change, it follows that $p_K((CXD))$ is also fixed. This is because for any point $Q$, the sum of the powers of $Q$ with respect to the circles $(AXB)$ and $(CXD)$ equals the sum of the powers of $Q$ with respect to the circles $(AXD)$ and $(BXC)$.

Thus, we have shown that there exists a fixed point $K$ and a constant $c$ such that $OK^2 - \rho^2 = c$ for any choice of $X$.

The answer is: $\boxed{K \text{}}$ is the intersection of } AB' $\text{}$ and } BA', $\text{}$ and } c $\text{}$ is a constant}}.