HMMT February

We translate the problem into one about 2-D geometry. Consider the right triangle $ABO$, and $P$ is some point on $AO$. Then, the choice of $P$ minimizes $PA+6PB$. Construct the line $\ell$ through $A$ but outside the triangle $ABO$ so that $\sin \angle(AO, \ell)=\frac{1}{6}$. For whichever $P$ chosen, let $Q$ be the projection of $P$ onto $\ell$, then $PQ=\frac{1}{6}AP$. Then, since $PA+6PB=6(PQ+PB)$, it is equivalent to minimize $PQ+PB$. Observe that this sum is minimized when $B, P, Q$ are collinear and the line through them is perpendicular to $\ell$ (so that $PQ+PB$ is simply the distance from $B$ to $\ell$). Then, $\angle AQB=90^{\circ}$, and since $\angle AOB=90^{\circ}$ as well, we see that $A, Q, P, B$ are concyclic. Therefore, $\angle PBO=\angle OPA=\angle(AO, \ell)$, and the sine of this angle is therefore $\frac{1}{6}$.