[Geometry]
Problem: 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.)
<img_4047>

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: $\frac{179}{250}$