A wooden cube with a side length of 10 cm is subject to 3 consecutive chamfers along three edges that meet in a single vertex. Each chamfer is angled at $45^{\circ}$ with the faces of cube that join at the chamfer edge. The chamfer indents a distance of 2.5 cm along each of these two faces.

Find the volume of wood (in
$\text{cm}^3$
) that was
**
cut out
**
by the three chamfers. If the volume is
$V$
then enter
$\lfloor 10^5 V \rfloor$
.

The answer is 8203125.

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Brute force:$\text{Volume}\left[\text{RegionUnion}\left[ \\ \text{RegionIntersection}\left[\text{Cuboid}[\{-5,-5,-5\},\{5,5,5\}],\text{HalfSpace}\left[\{0,-1,-1\},\left\{0,\frac{15}{4},\frac{15}{4}\right\}\right]\right], \\ \text{RegionIntersection}\left[\text{Cuboid}[\{-5,-5,-5\},\{5,5,5\}],\text{HalfSpace}\left[\{-1,0,-1\},\left\{\frac{15}{4},0,\frac{15}{4}\right\}\right]\right], \\ \text{RegionIntersection}\left[\text{Cuboid}[\{-5,-5,-5\},\{5,5,5\}],\text{HalfSpace}\left[\{-1,-1,0\},\left\{\frac{15}{4},\frac{15}{4},0\right\}\right]\right]\right]\right] \Rightarrow \frac{2625}{32}$Volume of a single chamfer is $\frac{125}{4}$ .

Volume of intersection of 2 chamfers is $\frac{125}{24}$ .

Volume of intersection of all 3 chamfers is $\frac{125}{32}$ .

$3\times\frac{125}{4}- 3\times\frac{125}{24} + \frac{125}{32} \Rightarrow \frac{2625}{32} \Rightarrow 82\frac{1}{32}$ as an improper fraction.

The first expression is in Wolfram Mathematica 12 language, which is

notthe same as Wolfram/Alpha.