Convex quadrilateral geometry and circumradius

Question

ABCD is a convex quadrilateral in which AC and BD meet at P. Given PA =1, PB = 2 , PC = 6 and PD = 3 . Let O be the circumcenter of triangle PBC. If OA is perpendicular to AD, find the circumradius of triangle PBC.

I have tried using similarity of the four triangles formed by the two diagonals by utilizing the dimensions given. However I am not able to proceed further.

Answer 1

Let circumradius of $\triangle PBC$ be $R$. Since $OAD$ is right triangle, $$AD^2=OD^2-OA^2=(OD^2-R^2)-(OA^2-R^2)$$ $$=\text{Pow}_{(PBC)}(D)-\text{Pow}_{(PBC)}(A)$$ $$=DP\cdot DB-AP\cdot AC=15-7=8$$

But $PD^2=9$ and $PA^2=1$. Therefore $AD^2=PD^2-PA^2 $ implies $\triangle PAD$ is right with $\angle PAD=90^{\circ}$. This means $P$ lies on $OA$.

Hence $PC$ is the diameter and $$R=PC/2=3$$

Answer 2

Since $PC = 2 \, PA \,$ and $\, PB = 2\, PD$ and $\angle\, APD = \angle \, BPC$ the two triangles $\Delta \, PAD$ and $\Delta \, PBC$ are similar, so $$ \angle \, ADP = \angle \, BCP = \alpha$$ (On the side note, although we do not need this fact, since $\angle\, ADB = \angle \, ADP = \alpha = \angle \, BCP = \angle \, BCA $ the quadrilateral $ABCD$ is inscribed in a circle)

As $O$ is the circumcenter of triangle $\Delta \, PBC$, angle $\angle \, BOP = 2\, \angle \, BCP = 2\, \alpha$. Furthermore, $OB = OP$ so $\Delta\, BOP$ is isosceles and therefore $$\angle \, OPB = 90 - \angle \, BOP = 90 - \alpha$$ Let line $OP$ intersect line $AD$ at point $A^*$. Then triangle $\Delta \, A^*DP$ has angles $$\angle\, A^*DP = \angle \, ADP = \alpha \,\, \text{ and } \,\, \angle \, A^*PD = \angle \, OPB = 90 - \alpha$$ Therefore $$\angle\, DA^*P = 180 - \angle \, A^*DP - \angle \, A^*PD = 180 - \alpha - (90 - \alpha) = 90$$ i.e. the line $$OP \, \text{ is perpendicular to } \, AD$$ However, by assumption, the line $OA$ is also perpendicular to $AD$. Since there can be only one line passing through point $O$ and perpendicular to the line $AD$, the two lines $OP$ and $OA$ coinside, i.e. the points $$O, \, \, P, \,\, A \,\, \text{ are collinear } \,\,$$ This means that point $O$ lies on the line $AP$. However, point $C$ also lies on the line $AP$, which means that point $$O, \, \, P, \,\, A, \,\, C\,\, \text{ are collinear } \,\,$$ But since $O$ is the circumcenter of triangle $\Delta \, PBC$ and it lies on the triangle's edge $CP$, the edge $CP$ itself is the diameter of the circumcircle of $\Delta\, PBC$ so its radius is $$OC = OP = OB = \frac{1}{2} \, CP = \frac{1}{2} \, 6 = 3$$