19. [10] Let $A B C D$ be a regular tetrahedron, and let $O$ be the centroid of triangle $B C D$. Consider the point $P$ on $A O$ such that $P$ minimizes $P A+2(P B+P C+P D)$. Find $\sin \angle P B O$.
Answer: $\quad \frac{1}{6}$ We translate the problem into one about 2-D geometry. Consider the right triangle $A B O$, and $P$ is some point on $A O$. Then, the choice of $P$ minimizes $P A+6 P B$. Construct the line $\ell$ through $A$ but outside the triangle $A B O$ so that $\sin \angle(A O, \ell)=\frac{1}{6}$. For whichever $P$ chosen, let $Q$ be the projection of $P$ onto $\ell$, then $P Q=\frac{1}{6} A P$. Then, since $P A+6 P B=6(P Q+P B)$, it is equivalent to minimize $P Q+P B$. Observe that this sum is minimized when $B, P, Q$ are collinear and the line through them is perpendicular to $\ell$ (so that $P Q+P B$ is simply the distance from $B$ to $\ell$ ). Then, $\angle A Q B=90^{\circ}$, and since $\angle A O B=90^{\circ}$ as well, we see that $A, Q, P, B$ are concyclic. Therefore, $\angle P B O=\angle O P A=\angle(A O, \ell)$, and the sine of this angle is therefore $\frac{1}{6}$.