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Imagine the world (Posted on 2019-04-05) Difficulty: 3 of 5
There is a collection of points inside a unit cube that are closer to the center of the cube than to any of the cube’s vertices.

What is the volume of this 3D region?

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution delayed aha! moment after too much algebra Comment 1 of 1
I've kept my notes in an attempted solution, which I gave up upon getting my aha! moment:

Some of the faces of this shape will be portions of planes defined as midway between the center of the cube and each of the eight vertices of the cube.  If they were extended indefinitely (full planes) they'd form a regular octahedron.

However, the center of each of the cube's faces is only 0.5 units from the center of the cube but 0.5 * sqrt(2) from each of the nearest vertices of the cube, and so is well within the set of points described. Therefore the octahedron is truncated, having six extra faces--each replacing one of the vertices of what would have been the octahedron. Each of these faces is a square coincident with a portion of a face of the original cube: smaller and at a 45° angle to the edges of the cube's face on which it lies.

Each of the eight triangular faces of the octahedron is then reduced to a hexagon, in which each of the angles is the same (120°), but the sides alternate short and long, so I don't believe the shape is Archimedean (though if the sides actually turn out equal, it would be). Further, each of these eight hexagons is the base of a pyramid whose apex is the nearest corner of the cube. As the pyramids are congruent, eight times the volume of one of them can be subtracted from the unit volume of the cube to get the volume sought in the puzzle.

If we take a plane passing through two opposite edges of the cube they will intersect the midpoints of the edges of two of the the square faces of the truncated octahedron, and we can focus on one such square to see its size. This cross section of the cube is one unit high and sqrt(2) units wide.

That midpoint of the small square is of course equidistant from the center of the cube, which is the center of the rectangular cross section, and the nearest corner (of the cube and of the rectangle). It's the apex of an isosceles triangle whose base is sqrt(1/4 + 2/4) = sqrt(3/4), extending from the center of the cube to the corner in the rectangular cross section.

The base angle of this isosceles triangle is arctan(1/sqrt(2)). Rather than identify this angle, consider that it is an angle of a right triangle that's half the isosceles triangle. The adjacent leg is just half of that sqrt(3/4). The altitude of the isosceles triangle is therefore sqrt(3/4)/(2*sqrt(2)) = sqrt(3/32). We're actually interested in the hypotenuse of that right triangular half of the isosceles triangle, and that's sqrt(3/16 + 3/32) by the Pythagorean theorem, or sqrt(9/32). That's how distant the midpoint of the little surface square is from the nearest vertex of the containing cube. Twice this, or sqrt(9/8) is subtracted from the sqrt(2) length of the diagonal of the cubic face to get the length of the edge of the square that's a face of the truncated octahedron: sqrt(2) - sqrt(9/8).

.............  Oh, oh; aha moment ..................

I was going to continue in this vein. I realized that so much algebra and arithmetic would be error prone and needed to be checked against a Monte Carlo simulation.  To simplify that, I thought about using just one octant of the cube. But that simplifies even the analytic solution. Within any octant one corner is the center of the whole cube and the opposite corner is the nearest corner to any point in the octant. Half the points are nearer the center and the other half are nearer the corner.

The volume is 1/2. (and the shape is indeed Archimedean as the hexagonal faces are regular).

  Posted by Charlie on 2019-04-05 18:06:57
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