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

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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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 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

## 1 Answer

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.

Let's consider a state at the moment the tetrahedron is released in which it ends up balanced on one of its vertices. It is clear that there is not a 12-dimensional neighborhood of states that also end up on a vertex: translations would not change it, small changes in initial orientation would most probably make it fall, a change in the direction of the linear momentum would keep it ending up on its vertex, but a change in it's magnitude probably not (you would have to specify the problem more precisely, like in the initial orientation), and finally a small change in angular momentum would most of the time, at least in two of the dimensions, make it fall as well.

What this shows is that the space of initial conditions is probably 5-dimensional or so. It might have some components of higher dimension depending on the problem's specifics, but always strictly lower dimensional than the full space of initial conditions. So whatever continuous probably distribution you put on that space, it will have measure (hence probability) 0.

• Ah, excellent, thank you :) I'm not familiar with $\mathfrak{so}(3)$ but 5 < 9 anyway – GFauxPas Jun 8 '18 at 22:09
• I just realized that this also gives you that a tetrahedron can't stay balanced on an edge, either. :) – GFauxPas Jun 8 '18 at 23:24
• -1. I think this answer is incorrect. All that you are saying is that it cannot be balanced in a position of stable equilibrium. But in the ideal world of Newtonian Mechanics positions of unstable equilibrium can exist which are infinitesimally narrow. The slightest disturbance would cause it to topple - even the impact of a single photon - within a finite time. But in theory it could be placed in that ideal position and would remain there for eternity. And it could be placed so close to the ideal position that the time it would take to topple would be longer than the age of the universe. – sammy gerbil Jun 9 '18 at 0:21
• @sammygerbil I am not saying that it cannot happen, only that the probability is 0. If you throw a dart at the plane, the probability of hitting any given point or line is 0, still it will hit a point and infinitely many lines. It is the same here, there are many initial positions that make it end up balanced on a vertex, but they make up a measure 0 subset. – doetoe Jun 9 '18 at 5:12
• Ok I am retracting my down-vote. – sammy gerbil Jun 9 '18 at 11:15