[Geometry]
Problem: A rectangular box has dimensions $8 \times 10 \times 12$. Compute the fraction of the box's volume that is not within 1 unit of any of the box's faces.

Solution: Let the box be defined by the product of the intervals on the $x, y$, and $z$ axes as $[0,8] \times$ $[0,10] \times[0,12]$ with volume $8 \times 10 \times 12$. The set of points inside the box that are not within 1 unit of any face is defined by the product of the intervals $[1,7] \times[1,9] \times[1,11]$ with volume $6 \times 8 \times 10$. This volume is $\frac{6 \times 8 \times 10}{8 \times 10 \times 12}=\frac{1}{2}$ of the whole box.

Final answer: $\frac{1}{2}$