3-Way Chamfer

Geometry Level 3

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 4 5 ∘ 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 cm 3 \text{cm}^3 ) that was cut out by the three chamfers. If the volume is V V then enter ⌊ 1 0 5 V ⌋ \lfloor 10^5 V \rfloor .


The answer is 8203125.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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1 solution

Brute force: Volume [ RegionUnion [ RegionIntersection [ Cuboid [ { − 5 , − 5 , − 5 } , { 5 , 5 , 5 } ] , HalfSpace [ { 0 , − 1 , − 1 } , { 0 , 15 4 , 15 4 } ] ] , RegionIntersection [ Cuboid [ { − 5 , − 5 , − 5 } , { 5 , 5 , 5 } ] , HalfSpace [ { − 1 , 0 , − 1 } , { 15 4 , 0 , 15 4 } ] ] , RegionIntersection [ Cuboid [ { − 5 , − 5 , − 5 } , { 5 , 5 , 5 } ] , HalfSpace [ { − 1 , − 1 , 0 } , { 15 4 , 15 4 , 0 } ] ] ] ] ⇒ 2625 32 \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 125 4 \frac{125}{4} .

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

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

3 × 125 4 − 3 × 125 24 + 125 32 ⇒ 2625 32 ⇒ 82 1 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 not the same as Wolfram/Alpha.

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