, 2022

Solution:

Let $X = AC \cap BD$, $Q = AB \cap CD$ and $R = BC \cap AD$. Since $QA \cdot QB = QC \cdot QD$, $Q$ is on the radical axis of $(ABP)$ and $(CDP)$, so $Q$ lies on the common tangent at $P$. Thus, $QP^{2} = QA \cdot QB$. Similarly, $RA \cdot RC = RP^{2}$. Let $M$ be the Miquel point of quadrilateral $ABCD$: in particular, $M = OX \cap QR$ is the foot from $O$ to $QR$. By properties of the Miquel point, $ABMR$ and $ACMQ$ are cyclic. Thus,
$$\begin{aligned} QP^{2} & = QA \cdot QB \\ RP^{2} & = RA \cdot RC \\ QP^{2} + RP^{2} & = QM \cdot QR + RM \cdot RQ = (QR + RM) QR = QR^{2} \end{aligned}$$
As a result, $\angle QPR = 90^{\circ}$.

Now, let $P'$ be the inverse of $P$ with respect to $\omega$. Note that by properties of inversion, $(ABP')$ and $(CDP')$ are tangent, and $(ACP')$ and $(BDP')$ are also tangent.

But now,
$$\begin{aligned} QP^{2} = QP'^{2} & = QA \cdot QB \\ RP^{2} = RP'^{2} & = RA \cdot RC \\ QP^{2} + RP^{2} = QP'^{2} + RP'^{2} & = QR^{2}. \end{aligned}$$
Thus, $PQ P'R$ is a cyclic kite, so $P$ and $P'$ are reflections of each other across $QR$. In particular, since $O, P, P'$ are collinear, then $M$ lies on line $OPP'$.

We can now compute $OP$ by using the fact that $OP + \frac{r^{2}}{OP} = 2OM = \frac{2r^{2}}{OX}$, where $r = 8$. Since $OX$ can be computed to equal $2$ quite easily, then $OP + \frac{64}{OP} = 64$, or $OP^{2} - 64OP + 64 = 0$. Solving this yields $OP = 32 \pm 8\sqrt{15}$, and because $P$ is inside the circle, $OP = 32 - 8\sqrt{15} = \sqrt{1024} - \sqrt{960}$.