Cut the knot: learn to enjoy mathematics
A math books store at a unique math study site. Shopping at the store helps maintain the site. Thank you.
Learning Math Online
Sites for teachers
Sites for parents
Terms of use
Awards
Interactive Activities

CTK Exchange
CTK Wiki Math
CTK Insights - a blog
Math Help

III Millennium Olympiad

Games & Puzzles
What Is What
Arithmetic/Algebra
Geometry
Probability
Outline Mathematics
Make an Identity
Book Reviews
Stories for Young
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
Privacy Policy

Guest book
News sites

Recommend this site

Games to relax

Sites for teachers
Sites for parents

Education & Parenting

Manifesto  |  Bookstore  |  Contents  |  Amazon store  |  Term index  |  What changed?  |  Contact  |  Recommend
RSS Feed: Recent changes at CTK

Touching Circles With Given Centers: What Is This About?
A Mathematical Droodle


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
What if applet does not run?

Explanation

Copyright © 1996-2009 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

Touching Circles With Given Centers

The applet illustrates the following theorem [Beardon]:

Given an n-gon. If n is odd, there always exists a unique sequence of circles with centers at the vertices of the polygon in which each circle touches its immediate neighbors. If n is even, no such sequence may exist; if it does it is not unique.


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
What if applet does not run?

Proof

Let a1, a2, ..., an denote the successive side lengths of the polygon. The circles that solve the problem exist iff there exist real numbers r1, r2, ..., rn that could serve as the radii of the pairwise touching circles:

(1) r1 + r2 = a1
r2 + r3 = a2
...
rn + r1 = an.

Assume r1 has been found. Then, using the first n-1 equations (1), we can successively determine the other radii:

(2) r2 = a1 - r1
r3 = a2 - r2 = a2 - a1 + r1
r4 = a3 - r3 = a3 - a2 + a1 - r1
...
rn = an-1 - rn-1 = an-1 - an-2 + an-3 - ... ±r1.

What about the last equation in (1)? Apparently, we should be able to exclude rn as we did with all other equations (2):

(3) r1 = an - rn = an - an-1 + an-2 - ... ±r1,

where ±r1 means (-1)n-1, i.e., for n even, r1 comes with the sign "+", whereas, for n odd, the sign is "-". In the latter case, we get an equation for r1 from:

(4) r1 = (an - an-1 + an-2 - ... + a1) / 2.

Which serves as the starting point for (2), such that all the radii are found in a unique way. For n even, (3) becomes

(5) an - an-1 + an-2 - ... - a1 = 0.

Unless (5) holds, the problem has no solution. When a solution exists, it is not unique. Indeed, assume that the sequence of radii r1, ..., rn, where n is even, solves (1). Pick any number s. By direct inspection, it is clear that the sequence

(6) r1 + s, r2 - s, r3 + s, ..., rn-1 + s, rn - s

also satisfies (1). There are no other solutions. (For n even, the applet displays a scroll bar that permits modification of a single parameter. By shifting any of the vertices, one can "find" a geometric solution to the problem. Once one is found, toying with the scroll bar shows that indeed there are many more.)

Note 1

The system (1) may have some solutions negative. I strip the sign minus so that all the circles are still defined. Those whose radii is a negative number turned positive touch their neighbors internally.

Note 2

In an unexpected way exactly same system of equations describes another problem: construct a polygon by the midpoints of its edges. For n odd, the problem has a unique solution. For n even, it either has no solution, or infinitely many solutions.

Note 3

For n = 4, the condition (5), which says that the sums of the opposite side lengths are equal, is equivalent to saying that the four points where the circles touch their neighbors are concyclic.

References

  1. T. Beardon, The Mid-Edges Theorem, in The Changing Shape of Geometry: Celebrating a Century of Geometry and Geometry Teaching, edited by Chris Pritchard, Cambridge University Press and MAA, 2002, pp. 179-183

Copyright © 1996-2009 Alexander Bogomolny

34221874Page copy protected against web site content infringement by Copyscape


Search:
Keywords:

Google
Web CTK