Product of Isometries

Date: 10/05/97 at 13:00:04
From: Marie D.
Subject: Product of translation and rotation

Greetings!

I am currently enrolled in an enriched Euclidean geometry class. In a 
recent assignment, I was asked to use three isometries (translation, 
rotation, and reflection) in composition with each other and deduce 
the net result of the two transformations. For example, I worked out 
that the product of a reflection followed by a reflection through a 
line parallel to the first line results in a net translation.  
However, if the lines  intersect, the net result is a rotation.  

The major problem occurs when I consider a translation followed by a 
rotation. My teacher claims that there is a fixed point around which 
the image may be rotated to obtain the net result. I'm having 
difficulty convincing myself of this, as I can't find a point which 
results in the same image. Any help would be appreciated, including 
referral to other sources of information on the Web, etc.  

Thanks!

Marie

Date: 10/06/97 at 20:00:15
From: Doctor Tom
Subject: Re: Product of translation and rotation

Hi Marie,

Here's how I think about it. It's hard to draw a picture using only
typing, so I suggest you draw a picture as you try to follow what
I'm saying.

The rotation occurs about some point, right? Call that point P. The 
translation moved some point that was previously at Q to that point P.  
On your paper, make two points and label them Q and P.

Draw an arrow from Q to P to represent the translation. Every other
point in the plane moved parallel to that arrow and by exactly the
same amount.

Now draw a circle around P that has a larger diameter than the arrow 
from Q to P. If you imagine sliding that arrow parallel to itself 
until it touches the circle at two points, then if the rotation is 
opposite the direction of the arrow and by an angle such that the 
arrow touches the circle at two points that are exactly that angle 
apart, you will have found a point that is moved back to itself.  
The translation takes it along the arrow that you moved to touch the 
circle in 2 places, and then the rotation undoes it.

If the rotation is in the opposite direction, slide the arrow to the 
opposite side of the circle and you'll get the same result.

But of course this won't work if the angle is wrong.

But the size of the circle can be anything. What you need to do is 
pick a radius for that circle that is exactly large enough that the 
length of the arrow connecting 2 points on the circumference of the 
circle describes a sector of the circle that has exactly the angle of 
rotation.

I've been a little sloppy here, but you can calculate the size of the 
circle exactly, and from there determine exactly which point is fixed.

I hope this is enough of a hint to get you going!

When you learn either some linear algebra or topology, this problem 
will seem trivial. When I saw the problem, I instantly found 2 proofs 
- one involving "eigenvalues" of the transformation matrix, and one 
invoking Brower's fixed-point theorem on a mapping of projective 
space, but that wasn't going to help you much. It actually took about 
5 minutes of work for me to convert those "instant" proofs into 
something that was purely geometric.

Tell your teacher it's a good problem.

-Doctor Tom,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/