Cut the knot: learn to enjoy mathematics
A math books store at a unique math study site. Learn to enjoy mathematics.
Google
Web CTK
Best sites for teachers
Sites for teachers
Sites for parents
Terms of use
Awards

Interactive Activities
CTK Exchange
CTK Insights - a blog

Games & Puzzles
What Is What
Arithmetic/Algebra
Geometry
Probability
Outline Mathematics
Make an Identity
Book Reviews
Eye Opener
Analog Gadgets
Inventor's Paradox
Did you know?...
Proofs
Math as Language
Things Impossible
Visual Illusions
My Logo
Math Poll
Cut The Knot!
MSET99 Talk
Other Math sites
Front Page
Movie shortcuts
Personal info
Reciprocal links
Privacy Policy

Guest book
News sites

Recommend this site

Best sites for teachers
Sites for teachers
Sites for parents

Education & Parenting

Manifesto: what CTK is about Search CTK Buying a book is a commitment to learning Table of content Things you can find on CTK Chronology of updates Email to Cut The Knot Recommend this page

Product of Rotations


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


Buy this applet

The product, or the sum as its often called, of two rotations is either a rotation or a translation, both of which can be trivial. Depending on the context, one may treat the product of two rotations which is trivial as either a rotation or a translation.

The product of two rotations through angles a and b is a rotation through angle (a + b). Its center, which exists if the product is not trivial, can be found constructively.

Let there be given two rotations gP and gQ with centers P and Q. Assume gP is executed first, gQ second. gP leaves P in place. Denote gQ(P) = P'. gP carries some point Q' onto Q which is then left in place by gQ. The point O that lies at the intersection of perpendicular bisectors of QQ' and PP' is equidistant from P and P' and also from Q and Q'.

Now prove that POP' = Q'OQ = PQP' + Q'PQ. Finally, consider what happens with an arbitrary point (or two) under two successive rotations.

(The fact just established provides a powerful tool for solving geometric problems, see, for example, proofs of Bottem's theorem and an analogue.)

Copyright © 1996-2008 Alexander Bogomolny

28773212Page copy protected against web site content infringement by Copyscape


Search:
Keywords:


Latest on CTK Exchange
Math
Posted by Laura
2 messages
06:56 AM, Apr-15-08

Divisibility rules - Jargon buste ...
Posted by Carolyn
2 messages
08:35 AM, Apr-04-08

drawing puzzle
Posted by martin gran
31 messages
06:53 PM, May-09-08

conway's game of life
Posted by frequency
0 messages
11:52 PM, May-12-08

Mistake on the page (an aside, Be ...
Posted by Max
4 messages
10:28 AM, Feb-28-08

Need details on a part of Proof o ...
Posted by Manuel S.
2 messages
05:24 PM, May-16-08

Josephus Flavius (correction)
Posted by David Turner
1 messages
09:42 AM, May-14-08