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

Shepard's Parallelogram Illusion


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

A few words

Copyright © 1996-2008 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

Shepard's Parallelogram Illusion

The two parallelograms shown appear at times different. However, they are always congruent and can be obtained from one another by rotation. This illusion has been published by Roger Shepard in 1981 and later in 1990 [Shepard, p. 48] The parallelograms look especially different when one is rather vertical while the other more horizontal. According to Shepard, we are commonly fooled by our depth perception. Almost in all cases the effect is enhanced when the parallelograms grow legs to resemble the table tops.

Jacobs makes a delightful use of this illusion to discuss a property of rotations. A rotation f can be determined by two arbitrary points P and Q and by their images f(P) and f(Q), provided neither is fixed. In the applet we think of E as being the image of A and H the image of D. The center of rotation O is bound to lie on the perpendicular bisectors of AE and DH and hence at their intersection. (In principle, the two two lines may coincide. Then the lines AE and DH intersect at the center of rotation.)


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

So assume the parallelograms ABCD and EFGH are equal and that O has been found as just described. In triangles ADO and EHO, AO = EO and DO = HO by construction, and AD = EH by our assumption. The triangles are equal by SSS. In particular, their angles at O are equal. If we add to each the angle DOE, we arrive at

  AOE = DOH,

So that indeed when A is rotated around O to coincide with E, D rotates onto H. The rest of the parallelogram ABCD comes along as a rigid body.

References

  1. H. R. Jacobs, Geometry, 3rd edition, W. H. Freeman and Company, 2003
  2. R. N. Shepard, Mind Sights, W. H. Freeman and Company, 1990
  3. I.M. Yaglom, Geometric Transformations I, MAA, 1962

Copyright © 1996-2008 Alexander Bogomolny

28733092Page 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

Deriving functions based on diffe ...
Posted by ke_45
1 messages
12:47 PM, May-10-08

A typo in
Posted by alexwajn
1 messages
11:36 PM, Apr-19-08