Example 14 (2004 National High School Competition Question) As shown in Figure 1-6, the axial section of a cone with vertex $P$ is an isosceles right triangle. $A$ is a point on the circumference of the base circle, $B$ is a point inside the base circle, $O$ is the center of the base circle, $A B \perp O B$, with the foot of the perpendicular at $B$, $O H \perp P B$, with the foot of the perpendicular at $H$, and $P A=4$, $C$ is the midpoint of $P A$. When the volume of the tetrahedron $O-H P C$ is maximized, the length of $O B$ is ( ).
A. $\frac{\sqrt{5}}{3}$
B. $\frac{2 \sqrt{5}}{3}$
C. $\frac{\sqrt{6}}{3}$
D. $\frac{2 \sqrt{6}}{3}$
Solution: Choose D. Reason: The goal difference in this problem is between the volume and the line segment, i.e., when the volume of the tetrahedron $O-HPC$ is maximized, find the length of the line segment $OB$. In the tetrahedron $O-HPC$, points $P, O, C$ are fixed points. At this point, the goal difference is reduced to finding the distance from point $B$ to point $O$ when point $H$ is in a special position.
Since $AB \perp OB, AB \perp OP$, then $AB \perp PB$.
Since $OH \perp PB$, then plane $PAB \perp$ plane $POB$. Therefore, $OH \perp HC, OH \perp PA$. Since $C$ is the midpoint of $PA$, then $OC \perp PA$. So, $PC$ is the height of the tetrahedron $P-HOC$, and $PC=2$.
In the right triangle $\triangle OHC$, $OC=2$, so when $HO=HC$, $S_{\triangle HXC}$ is maximized, which also means $V_{(OHP C}=V_{P-H()}$ is maximized.
At this point, $HO=\sqrt{2}$, then $HO=\frac{1}{2} OP$, hence $\angle HPO=30^{\circ}$.
Therefore, $OB=OP \tan 30^{\circ}=\frac{2 \sqrt{6}}{3}$.