
Re: Hexagonal base prism with three faces at top
Posted:
Jun 8, 2017 10:17 PM


On 6/8/2017 8:57 PM, Mike Terry wrote: > On 09/06/2017 00:39, Jim Burns wrote: >> On 6/8/2017 5:52 PM, matt wrote:
>>> How can I calculate the dimensions of >>> the top of a hexagonal prism when the top has >>> three faces, each of equal dimensions >>> (as though it were the three faces of a cube)? >> >> If I understand you, if we say we're talking about a >> a cube with corners at >> >> (0,0,1)  (1,0,1) >> \ \ >>  \  \ >>  (0,1,1)  (1,1,1) >>     >>     >> (0,00)  (1,00) >> \  \  >> \ \ >> (0,1,0)  (1,1,0) >> >> then it's sliced through the middle to make your prism, >> where "through the middle" is perpendicular to >> an axis running through opposite corners such as >> (0,0,0) and (1,1,1). >> >> The hexagonal base would run from midpoint to >> midpoint of the edges in the middle, >> so each edge would be sqrt(1/2), and regular. >> >> From poletopole of the cube it's sqrt(3), >> so the height of the sliced cube is sqrt(3)/2. >> >> The faces touching that hexagon would be, alternately, >> 1/2,1/2,sqrt(1/2) triangles and 1,1/2,sqrt(1/2),1/2,1 >> pentagons. >> >> The volume of the whole shape would be half the cube, >> or 1/2. >> >> Is this what you mean by the dimensions? >> > > I interpreted the description a bit differently. > If we project a cube onto a plane perpendicular to > an axis through opposite corners, we see a hexagon > shape (i.e. as in the shadow of a cube held at the > appropriate angle), and this is what I took to be > the crosssection (base?) of the prism. It is a bit > larger than your hexagon, and rotated by Pi/6, > (assuming a fixed cube). Another way I visualise it is > to imagine a hexagonal pencil where three plane cuts > have been made at the top as though to sharpen the pencil. > > Anyway perhaps the OP will clarify the description, and > also what measurements are being sought.
It would be nice for the OP to clarify, but your interpretation looks more reasonable than mine.
Trickier to work out  which might be why the question.
The six corners of the cube that correspond to the vertices of the shadow hexagon form two groups of three, in two planes perpendicular to the (0,0,0)(1,1,1) axis. Two equilateral triangles rotated pi/3 relative to each other, with sqrt(2) sides.
The distance from the vertices of a sqrt(2),sqrt(2),sqrt(2) triangle to its center is (scribble,scribble) sqrt(2/3). So, the distance from the vertices of the shadow hexagon to its center is also sqrt(2/3). And, regular hexagon, so its sides are sqrt(2/3).
More could be said, but that's a start.

