# Why can't a pyramid stay balanced on a vertex? [duplicate]

A pyramid, such as a tetrahedron/3-simplex, or any other isohedron, falls from some height and lands on a vertex. It will eventually end up with a face to the ground.

Why can't it stay balanced on a vertex?

EDIT: this is not the same question as Can we theoretically balance a perfectly symmetrical pencil on its one-atom tip? because there symmetry is assumed. For this question I am not making any assumptions about symmetry. Also, in my situation the object is falling from a height. The issue there is assuming it starts on the ground.

## marked as duplicate by Kyle Kanos, AccidentalFourierTransform, sammy gerbil, ZeroTheHero, Emilio PisantyJun 11 '18 at 11:46

• Why a downboat? – GFauxPas Jun 8 '18 at 20:07
• When you compute the area of a triangle you takr one side as base and its perpendicular through the opposite vertex is the height. And the base is usually at the bottom. – rodrigo Jun 8 '18 at 21:32
• A pyramid (or any other polyhedron) absolutely can land on a vertex. The question that you really are asking is, why can a polyhedron not stay balanced on a vertex? – Solomon Slow Jun 8 '18 at 21:43
• @jameslarge good call – GFauxPas Jun 8 '18 at 22:06
• Practically, since the pyramid falls from some height and lands on a vertex the vertex could get embedded in the surface material, so that a torque prevents it from toppling. – sammy gerbil Jun 9 '18 at 11:18

In terms of Newtonian mechanics, the state of a rigid body, which uniquely determines the time evolution, is fully described by the position $\mathbb R^3$ and orientation, an element of $SO(3)$, linear momentum, an element of $\mathbb R^3$, and angular momentum, an element of $\mathfrak{so}(3)$, the Lie algebra of infinitesimal rotations. This is a 12-dimensional space.
• Ah, excellent, thank you :) I'm not familiar with $\mathfrak{so}(3)$ but 5 < 9 anyway – GFauxPas Jun 8 '18 at 22:09