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Let a convex polyhedron $P$ be given, all of whose faces are congruent. Consider any pyramid formed by a face of $P$ as its base and the centroid of $P$ as its vertex. Allowing congruence to admit reflection, are all such pyramids comprising $P$ congruent?

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Look at this. –  dtldarek Feb 6 '13 at 23:29

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up vote 4 down vote accepted

Not all such pyramids are congruent. Start with an octahedron and split it in half down the bisecting axis between two opposite vertices; take one of the halves, twist it 45 degrees, and pull it apart from its partner until the distances are such that the intermediate band of triangles is equilateral. Clearly the triangles making up the central band are closer to the centroid than the triangles making up the end caps, so their pyramids can't be the same.

To make this explicit: imagine one end cap consisting of the coordinates $(0, 0, 1+c)$, $(\pm 1, 0, c)$ and $(0, \pm1, c)$ and the other endcap (rotated 45 degrees) having the vertex coordinates $(0, 0, -(1+c))$ and $(\pm\frac{\sqrt2}{2}, \pm\frac{\sqrt2}{2}, -c)$. Then the squared distance between e.g. $(1,0,c)$ and $(\frac{\sqrt2}{2}, \frac{\sqrt2}{2}, -c)$ is $\left(1-\frac{\sqrt2}{2}\right)^2+\frac{1}{2}+4c^2$; setting this equal to 2 (i.e. the squared length of the other edges of the shape) gives $4c^2 = 1\!\frac{1}{2}-\left(1-\frac{\sqrt2}{2}\right)^2 = \sqrt{2}$, or $c=\dfrac{\sqrt[4]{2}}{2}$. By symmetry, all of the other intermediate edges of the central band are of the same length, so all of the triangles in the central band are equilateral triangles.

This solid is apparently known as the gyroelongated square bipyramid, and it's one of the Johnson solids. Wikipedia claims that it's convex (which was the only question I had about the construction above), so this should be a clean counterexample.

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Are you sure you can construct such an object? what is it called? –  nbubis Feb 6 '13 at 23:54
    
@nbubis such a thing is easily constructed, but it doesn't really have a name. I'll update in a moment with coordinates. –  Steven Stadnicki Feb 6 '13 at 23:54
    
But all sides are regular polygons, meaning it should be a platonic solid, which it's not. –  nbubis Feb 6 '13 at 23:57
    
@nbubis There are many more polyhedra with regular faces for sides than just the platonic solids - consider for instance the bipyramid made of stacking together two tetrahedra. This is a polyhedron with 5 equilateral triangular faces, but it's not regular because it's not vertex-transitive. If you impose vertex-transitivity as well then I suspect the answer becomes yes. –  Steven Stadnicki Feb 7 '13 at 0:02
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Probably the snub disphenoid also has this property, but I can't tell for sure. I'm finding it really hard to get a handle on the shape of that thing. –  Rahul Feb 7 '13 at 1:17

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