Let $\Omega$ and $\omega$ be circles with radii 123 and 61 , respectively, such that the center of $\Omega$ lies on $\omega$. A chord of $\Omega$ is cut by $\omega$ into three segments, whose lengths are in the ratio $1: 2: 3$ in that order. Given that this chord is not a diameter of $\Omega$, compute the length of this chord.
Solution: Denote the center of $\Omega$ as $O$. Let the chord intersect the circles at $W, X, Y, Z$ so that $W X=t, X Y=2 t$, and $Y Z=3 t$. Notice that $Y$ is the midpoint of $W Z$; hence $\overline{O Y} \perp \overline{W X Y Z}$. The fact that $\angle O Y X=90^{\circ}$ means $X$ is the antipode of $O$ on $\omega$, so $O X=122$. Now applying power of point to $X$ with respect to $\Omega$ gives \[ 245=123^{2}-O X^{2}=W X \cdot X Z=5 t^{2} \Longrightarrow t=7 \] Hence the answer is $6 t=42$. $\fbox{42}$.