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.
Ask a tutor for free
Learning Math Online

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

Sites for parents

Education & Parenting

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

Fuss' Theorem

The applet below illustrates Poncelet's porism for quadrilaterals. It is based on a formula by Nicolaus Fuss (1755-1826), a student and friend of L. Euler. Fuss also found the corresponding formulas for the bicentric pentagon, hexagon, heptagon, and octagon [Dörrie, p. 192].

 

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 © 1996-2007 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The circumradius (R), the inradius (r) and the distance between the circumcenter and the incenter (d) of a bicentric quadrilateral stand in an elegant relationship

  1/(R - d)2 + 1/(R + d)2 = 1/r2,

It looks very much the same as Euler's formula for triangles, except for the exponent of 2.

Here's a very short proof of this fact by J. C. Salazar:

  Fuss' formula

Let the two circles be O(R) and I(r), K and L be the points where circle I(r) touches AB and BC.

Since quadrilateral ABCD is cyclic the angles at A and C are supplementary making angles BAI and ICB complimentary:

  BAI + ICB = 90°.

By construction, IK = IL = r. Two triangles AIK and CIL combined form a right triangle with legs AI and CI and the hypotenuse AK + CL. Its area can be found in two ways:

(1) r·(AK + CL) = AI · CI.

Also, the Pythagorean theorem applied to that triangle gives

(2) (AK + CL)² = AI² + CI².

From (1) and (2),

  r²·(AI² + CI²) = AI² · CI²,

or,

(3) 1/r² = 1/AI² + 1/CI².

Let AI and CI produced intersect O(R) in F and E. Then EF is a diameter of C(R) because

 
DOF + DOE= 2(DAF + DCE)
 = BAD + BCD
 = 180°.

We are then in a position to apply the formula for the length of a median (IO) in a triangle (EFI):

(4)
EI² + FI²= 2 IO² + EF² / 2
 = 2 (d² + R²).

Considering the diameter of O(R) through I, the intersecting chords theorem gives

(5) AI · FI = CI · EI = R² - d²

It follows (from (4) and (5)) that

(6)
1/AI² + 1/CI²= FI²/(R² - d²)² + EI²/(R² - d²)²
 = (EI² + FI²) / (R² - d²)²
 = 2 (R² + d²) / (R² - d²)².

Finally, from (3) and (6),

 
1/r²= 2 (R² + d²) / (R² - d²)²
 = [(R + d)² + (R - d)²)] / (R² - d²)²
 = 1/(R + d)² + 1/(R - d)².

References

  1. N. Altshiller-Court, College Geometry, Dover, 1980
  2. J. Casey, A Sequel to Euclid, Scholarly Publishing Office, University of Michigan Library (December 20, 2005), reprint of the 1888 edition, pp. 107-110
  3. J. L. Coolidge, A Treatise On the Circle and the Sphere, AMS - Chelsea Publishing, 1971, p. 45
  4. H. Dörrie, 100 Great Problems Of Elementary Mathematics, Dover Publications, NY,1965
  5. F. G.-M., Exercices de Géométrie, Éditions Jacques Gabay, sixiéme édition, 1991, pp. 837-839
  6. J. C. Salazar, Fuss' Theorem, The Mathematical Gazette, v 90, n 518 (July 2006), pp. 306-307.

Poncelet Porism

Copyright © 1996-2009 Alexander Bogomolny

34383879Page copy protected against web site content infringement by Copyscape

Search:
Keywords:

Google
Web CTK