inspired by How many faces of a solid can one "see"? at first I thought 3 faces of a cube and half a sphere seem opposite extremes both with 50% surface area visible. then I considered a regular triangular pyramid which can be seen with 75% area. Is this the maximal visible percentage for a regular platonic solid? what about a regular polyhedron?
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1visible not visable. – John Alexiou May 23 '16 at 01:19
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This is sort of trivial because you can just hand check all of the possibilities. For regular solids I believe that you are correct that a Triangular Pyramid has the largest visible surface area. For non regular solids it is unbounded. – Zaros May 23 '16 at 02:17
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@lordoftheshadows, nice name. I'm not sure it's trivial for non platonic yet regular polyhedron such as https://en.m.wikipedia.org/wiki/Kepler%E2%80%93Poinsot_polyhedron – shai horowitz May 23 '16 at 02:23
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1You're right that it wasn't as trivial as I thought. I'm not 100% on the great stellated dodecahedra as it doesn't actually follow the rules I though it did. I'll have to check but I'm pretty sure that it still doesn't even surpass 50%. – Zaros May 23 '16 at 02:40
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1For all polyhedron which is centrally symmetric (i.e. invariant under reflection wrt to its centroid), you can see at most 50% of it. In fact, if the polyhedron is convex, you are observing it from infinity and not on any direction parallel to one of its faces, exactly 50% of the faces are visible. – achille hui May 23 '16 at 04:20
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@achillehui so now we have an upper and lower bounds for symmetric polyhedron,are all polyhedron symmetric in the way you describe? I'm guessing no because I can see 75% of pryrimid – shai horowitz May 23 '16 at 04:23
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@shaihorowitz it is clear there are polyhedron that isn't symmetric. I just rule out the obvious. – achille hui May 23 '16 at 04:37
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@achillehui alright I just didn't understand how you used symmetry there. this is not my strong field. – shai horowitz May 23 '16 at 04:38