# Rigid Motion Woes (Geometry)

1. Apr 7, 2009

### dismo

Working in a Hilbert plane, show that any rigid motion that fixes at least three noncollinear points must be the identity.

I am certain that I can claim that:
(i) any translation of the plane will fix none of the points
(ii) any rotation will fix a single point
(iii) any reflection will fix only the points on the line about which the plane is reflected

The trouble is I don't know how to prove that no composition of these could fix only three points in the plane...

Where do I go next?

2. Apr 7, 2009

### VKint

I'm unfamiliar with the axioms of a Hilbert plane. If this question were posed in the context of an ordinary Euclidean plane, however, this is how I would approach it. Rigid motions preserve distances (i.e., the distance between points $$f(a)$$ and $$f(b)$$ is the same as that between $$a$$ and $$b$$, where $$f$$ is a rigid motion). Given a two points $$A$$ and $$B$$ and a third point $$X$$ not on $$\overline{AB}$$, there exists exactly one other point $$Y \neq X$$ such that $$d(A,Y) = d(A,X)$$ and $$d(B,Y) = d(B,X)$$ (where $$d(\; , \;)$$ denotes the distance function). Furthermore, $$Y = r_{AB}(X)$$, the reflection of $$X$$ about $$\overline{AB}$$. If $$X$$ is on $$\overline{AB}$$, then $$X$$ is the unique point in the plane satisfying these equations. (Neither of these assertions is hard to prove.)

Let $$f$$ be a rigid motion fixing the three noncollinear points $$A,B,C$$. From the last observation, we know that $$f$$ fixes all of $$\overline{AB}, \overline{AC}$$, and $$\overline{BC}$$. Let $$X$$ be a point not on any of these lines. Suppose $$f(X) \neq X$$; then, by distance conservation, we must have $$f(X) = r_{AB}(X) = r_{AC}(X) = r_{BC}(X)$$ simultaneously, a contradiction since (by assumption) $$\overline{AB} \neq \overline{AC} \neq \overline{BC}$$.

3. Apr 8, 2009

### Doom of Doom

A Hilbert Plane is just a Euclidean Plane, but without the Parallel Axiom and the Circle–Circle Intersection Property.