polyhedral we call a $12$-gon in plane good whenever: first, it should be regular, second, it's inner plane must be filled!!, third, it's center must be the origin of the coordinates, forth, it's vertices must have points $(0,1)$,$(1,0)$,$(-1,0)$ and $(0,-1)$. find the faces of the massivest polyhedral that it's image on every three plane $xy$,$yz$ and $zx$ is a good $12$-gon. (it's obvios that centers of these three $12$-gons are the origin of coordinates for three dimensions.) time allowed for this question is 1 hour.
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Tags: analytic geometry, geometry proposed, geometry
P-H-David-Clarence
14.06.2016 16:06
72 by imagination.
Reynan
14.06.2016 16:11
what do you mean by its inner plane must be filled?
P-H-David-Clarence
14.06.2016 18:27
I just think it doesn't make sense
abk2015
14.06.2016 21:15
I'm pretty sure that "inner plane being filled" means that the lane must be continuous. Please correct me if I'm wrong.
Pathological
11.11.2019 02:05
I think the answer is $\boxed{36}.$ Consider three "cylindrical" figures which are formed by extending the three good polygons perpendicular to the planes they are on. Then the polyhedron in question is simply the intersection of the three figures. Henceforth these three spaces will be referred to as cylinders even though they are not really cylinders. Notice that each cylinder is convex, and hence so is their intersection. This means that no plane can contain two distinct faces of the polyhedron. We must know show that every "side" of the cylinders actually contains a face of the polyhedron, which hence upper bounds the number of faces by $3 \cdot 12 = 36.$ Let's show that each of the $12$ "sides" of the cylinders indeed contain at least one face of the polyhedron. Simply note that the intersection of that face with the plane perpendicular to the cylinder is contained in the polyhedron's surface, and so we're done. $\square$
rchokler
12.11.2019 22:50
It is a Triacontahexahedron. I leave it as an exercise to compute the following. $38$ vertices: $(1,0,0)$, $\left(\frac{1+\sqrt{3}}{4},\frac{1+\sqrt{3}}{4}, \frac{1+\sqrt{3}}{4}\right)$, $\left(\frac{1}{2},\frac{1}{2},\frac{\sqrt{3}}{2}\right)$ and all their permutations and sign flip combinations. $72$ edges: $24$ of each with lengths $\frac{3-\sqrt{3}}{4}$, $\frac{\sqrt{9-4\sqrt{3}}}{2}$, and $1$. Volume: $\frac{7+\sqrt{3}}{2}$ Surface area: $9\sqrt{6}-6\sqrt{2}$