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

Three equal circles

Here is the problem:

If three circles having the same radius pass through a point, the circle through their other three points of intersection also has the same radius.

The circles are denoted k, l, m. The points of their intersection are 0, A, B, C.

Solution

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution

The picture above is somewhat incomplete. We are dealing with circles after all. A circle is defined by its center and its radius. In the original picture the center points are missing. Let denote them K,L,M. Now it also might be important to indicate what distances are equal. For this is our data: the three circles all have the same radius.
The nine equal segments appear to form a well known shape. It's a projection of a parallelepiped on a plane.
Normally we would draw such a projection with three more edges that were hidden above. Now we have 12 equal segments. The newly drawn edges meet at a previously hidden vertex E which is equidistant from points A, B, and C.

(There is a very interesting circumstance where the above configuration emerges naturally but still unexpectedly.)

References

  1. R. Honsberger, Mathematical Gems II, MAA, 1976, pp. 20-21
  2. G. Polya, Mathematical Discovery, John Wiley and Sons, 1981, Ch. 10.

2D problems that benefit from a 3D outlook

  1. 4 Travellers
  2. Desargues' Theorem
  3. Soddy Circles and Eppstein's Points
  4. Symmetries in a Triangle
  5. Three Circles and Common Chords
  6. Three Circles and Common Tangents
  7. Three Equal Circles

Copyright © 1996-2008 Alexander Bogomolny

28675321Page 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

product of fractions
Posted by ke_45
3 messages
08:37 AM, May-06-08

Distance to the horizon
Posted by Monty
3 messages
04:38 PM, May-08-08

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

Nim Games - a query
Posted by Akash Kumar
1 messages
08:53 AM, Apr-15-08

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