China Mathematical Competition

As shown in Fig. 1, consider the situation where the ball is in a corner of the container. Draw the plane $A_1{B}_1{C}_1\parallel ABC$, tangent to the ball at point $D$. Then the ball center $O$ is also the center of the tetrahedron $P-A_1{B}_1{C}_1$, with $PO\perp A_1{B}_1{C}_1$ and the foot point $D$ being the center of $△A_1{B}_1{C}_1$.

Since

$$\begin{array}{rl}{V}_{P-A_1{B}_1{C}_1}& =\frac{1}{3}\times S_{△A_1{B}_1{C}_1}\times PD\\ & =4\times V_{O-A_1{B}_1{C}_1}\\ & =4\times \frac{1}{3}\times S_{△A_1{B}_1{C}_1}\times OD,\end{array}$$ we have $PD=4OD=4r$, where $r$ is the radius of the ball. It follows that $$PO=PD-OD=3r.$$ Suppose that the ball is tangent to the plane $PAB$ at point $P_1$. Then we have $$P{P}_1=\sqrt{P{O}^2-O{P}_1^2}=\sqrt{(3r{)}^2-r^2}=2\sqrt{2}r.$$ As shown in Fig. 2, it is easy to see that the locus of the ball on the plane $PAB$ is also a regular triangle, denoted by $P_{1}EF$. Through $P_1$ draw $P_{1}M\perp PA$ with point $M$ on $PA$. Then $\angle MP{P}_1=\frac{\pi }{6}$, and $$PM=P{P}_1\times \cos \angle MP{P}_1=2\sqrt{2}r\times \frac{\sqrt{3}}{2}=\sqrt{6}r.$$

It follows that $P_{1}E=PA-2PM=a-2\sqrt{6}r$, where $a=PA$. Now, the space on $PAB$ which the ball will never touch is the shaded part of Fig. 2, and its size is equal to $$\begin{array}{rl}{S}_{△PAB}-S_{△P_{1}EF}& =\frac{\sqrt{3}}{4}(a^2-(a-2\sqrt{6}r{)}^2)\\ & =3\sqrt{2}ar-6\sqrt{3}{r}^2\\ & =24\sqrt{3}-6\sqrt{3}\\ & =18\sqrt{3},\end{array}$$ since $r=1$ and $a=4\sqrt{6}$ under given conditions. Then the total untouched area is $$4\times 18\sqrt{3}=72\sqrt{3}.$$