The five platonic solids have all faces regular Polyhedra. They are also all convex. But there are other properties they satisfy. Like each vertex should have the same number of edges meeting there. Let's say we relax all other criterion and only require that the solid be convex and that it have all faces regular. How many such solids can exist? One example is pasting two tetrahedra together along a face to get a solid with 6 sides and triangular faces.
$\begingroup$
$\endgroup$
asked Jan 4, 2022 at 21:15
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
Convex polyhedra with regular faces which are not Platonic are called Johnson solids. There are 5 Johnson solids with the same faces. It happens they are all triangular faced (can you see why?)
- J12 Triangular bipyramid
- J13 Pentagonal bipyramid
- J17 Gyroelongated square bipyramid
- J51 Triaugmented triangular prism
- J84 Snub disphenoid
So together with 5 Platonic solids, you have 10.
answered Jan 4, 2022 at 22:23
-
$\begingroup$ Why are they all triangular faced? Sorry, I can't even see how to being trying to see why. Especially since we have the cube and Dodecahedron. $\endgroup$ Jan 4, 2022 at 22:51
-
1$\begingroup$ If we have identical regular faces and identical vertices, we would get platonic solids. So some vertices should be non-identical, meaning we should have different numbers of faces meet at vertices. However, you can only have 3 squares or 3 pentagons at vertex of a strictly-convex polyhedron. Thus, only deltahedra meet your criteria. $\endgroup$ Jan 4, 2022 at 23:46