Problem (Geometry):
The taxicab distance between points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by

$$
d\left(\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)\right)=\left|x_{1}-x_{2}\right|+\left|y_{1}-y_{2}\right|+\left|z_{1}-z_{2}\right| .
$$

The region $\mathcal{R}$ is obtained by taking the cube $\{(x, y, z): 0 \leq x, y, z \leq 1\}$ and removing every point whose taxicab distance to any vertex of the cube is less than $\frac{3}{5}$. Compute the volume of $\mathcal{R}$.

Solution:
For a fixed vertex $V$ on the cube, the locus of points on or inside the cube that are at most $\frac{3}{5}$ away from $V$ form a corner at $V$ (that is, the right pyramid $V W_{1} W_{2} W_{3}$ in the figure shown at left below, with equilateral triangular base $W_{1} W_{2} W_{3}$ and three isosceles right triangular lateral faces $V W_{1} W_{2}, V W_{2} W_{3}, V W_{3} W_{1}$ ). Thus $\mathcal{R}$ is formed by removing eight such congruent corners from the cube. However, each two neighboring corners share a common region along their shared edge. This common region is the union of two smaller right pyramids, each similar to the original corners. (See the figure shown at right below.)


We compute the volume of $\mathcal{R}$ as

$$
1-8 \cdot \frac{1}{6}\left(\frac{3}{5}\right)^{3}+12 \cdot 2 \cdot \frac{1}{6}\left(\frac{1}{10}\right)^{3}=\frac{\mathbf{1 7 9}}{\mathbf{2 5 0}}
$$

Final answer: $\frac{179}{250}$