Problem 46746. Volume of a truncated cube

Created by Alex

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The surface of a truncated cube consists of 6 octagons, and 8 equilateral triangles. Our truncated cube is parametrised by (h1, h2), where
h1 refers to the distance between the center of the volume and the center of the octagons, and
h2 refers to the distance between the center of the volume and the center of the triangles.
For  we get the special case of a cube, and for  we get a cuboctahedron.
Your task: Write a function which returns the volume V of the truncated cube as function of the heights h1 and h2. If the ratio h2/h1 violates the constraints  return the volume of the related special case (cube or cuboctahedron) instead.

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Last Solution submitted on Aug 04, 2024

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William on 13 Oct 2020

Test Suite problems 1 and 2 seem to be incorrect. Did you mean to put parentheses around h1+abs(randn) and h1-abs(randn)?

Alex on 14 Oct 2020

I don't think that there is a problem.. For example test suite problem 1: h1 = abs(randn); h2 = sqrt(3)*h1; is a cube of random size. By adding + abs(randn), I violate the constraint h2 <= sqrt(3)*h1. If I put parentheses around h1+abs(randn), I have the same effect as before, just with another scaling of the random number that I added.. --> Result is the same: violation of the constraint for the cube. Same with test suit problem 2

William on 14 Oct 2020

Oh, sorry. I see now that these are the cases that are supposed to check that our code returns the correct responses when the inputs are "out of bounds". Thanks for pointing that out!

Rafael S.T. Vieira on 15 Oct 2020

Tip: the 8 tetrahedra at the cube corners are not regular, but they have equilateral triangles as base.

PS: Kudos for the author, nice figures. It's so rare to find a problem with graphics at Cody. And sometimes a picture conveys much more than a long text.

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Problem 46746. Volume of a truncated cube

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Problem 46746. Volume of a truncated cube

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