Product of IsometriesDate: 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/ |
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