Let $ABC$ be a triangle with $\angle A = 45^\circ$. Let $P$ be a point on side $\overline{BC}$ with $PB = 3$ and $PC = 5$. Let $O$ be the circumcenter of triangle $ABC$. Determine the length $OP$.
Using the extended Sine law, we find the circumradius of $ABC$ to be $R = \frac{BC}{2\sin A} = 4\sqrt 2$.

[asy]
unitsize(0.8 cm);

pair A, B, C, O, P;

A = (0,0);
B = (2,2);
C = (5,0);
P = interp(B,C,3/8);
O = circumcenter(A,B,C);

draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
draw(O--P);

label("$A$", A, W);
label("$B$", B, N);
label("$C$", C, E);
dot("$O$", O, S);
dot("$P$", P, NE);
[/asy]

By considering the power of point $P$, we find that $R^2 - OP^2 = PB \cdot PC = 15$. So $OP = \sqrt{R^2 - 15} = \sqrt{ 16 \cdot 2 - 15} = \boxed{\sqrt{17}}$.