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Pappus Chain

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PappusChain

Starting with the circle P_1 tangent to the three semicircles forming the arbelos, construct a chain of tangent circles P_i, all tangent to one of the two small interior circles and to the large exterior one. This chain is called the Pappus chain (left figure).

In a Pappus chain, the distance from the center of the first inscribed circle P_1 to the bottom line is twice the circle's radius, from the second circle P_2 is four times the radius, and for the nth circle P_n is 2n times the radius. Furthermore, the centers of the circles P_i lie on an ellipse (right figure).

If r=AB/AC, then the center and radius of the nth circle P_n in the Pappus chain are

x_n=(r(1+r))/(2[n^2(1-r)^2+r])
(1)
y_n=(nr(1-r))/(n^2(1-r)^2+r)
(2)
r_n=((1-r)r)/(2[n^2(1-r)^2+r]).
(3)

This general result simplifies to r_n=1/(6+n^2) for r=2/3 (Gardner 1979). Further special cases when AC=1+AB are considered by Gaba (1940).

PappusCircle

The positions of the points of tangency for the first circle are

x_A=r/((1-r)^2)
(4)
y_A=(r(1-r))/((1-r)^2)
(5)
x_B=(r(1+r))/(1+r^2)
(6)
y_B=(r(1-r))/(1+r^2)
(7)
x_C=(r^2)/(1-2r+2r^2)
(8)
y_C=(r(1-r))/(1-2r+2r^2).
(9)

The diameter of the nth circle P_n is given by (1/n)th the perpendicular distance to the base of the semicircle. This result was known to Pappus, who referred to it as an ancient theorem (Hood 1961, Cadwell 1966, Gardner 1979, Bankoff 1981). Note that this is also valid for the chain of tangent circles starting with P_1 and tangent to the two interior semicircles of the arbelos. The simplest proof is via inversive geometry.

Eliminating n from the equations for x_n and y_n, the center (x_n,y_n) of the circle P_n, gives

4rx^2-2r(1+r)x+(1+r)^2y^2=0.
(10)

Completing the square gives

4r[x-1/4(1+r)]^2+(1+r^2)y^2=1/4r(1+r)^2,
(11)

which can be rearranged as

[(x-1/4(1+r))/(1/4(1+r))]^2+(y/(1/2sqrt(r)))^2=1,
(12)

which is simply the equation of an ellipse having center ((1+r)/4,0) and semimajor and semiminor axes (1+r)/4 and sqrt(r)/2 respectively. Since

c=sqrt(a^2-b^2)=1/4(1-r),
(13)

(1+r)/4+/-c=r/2 and 1/2, so the ellipse has foci at the centers of the semicircles bounding the chain.

PappusTangentChain

The circles T_n tangent to the first arbelos semicircle and adjacent Pappus circles P_(n-1) and P_n have positions and sizes

x_n^'=(r(7+r))/(2[4+4n(n-1)(1-r)^2+r(r-1)])
(14)
y_n^'=(2(2n-1)r(1-r))/(4+4n(n-1)(1-r)^2+r(r-1))
(15)
r_n^'=(r(1-r))/(2[4+4n(n-1)(1-r)^2+r(r-1)]).
(16)

A special case of this problem with r=1/2 (giving equal circles forming the arbelos) was considered in a Japanese temple tablet (Sangaku problem) from 1788 in the Tokyo prefecture (Rothman 1998). In this case, the solution simplifies to

x_n^'=(15)/(2(15-4n+4n^2))
(17)
y_n^'=(2(2n-1))/(15-4n+4n^2)
(18)
r_n^'=1/(2(15-4n+4n^2)).
(19)
PappusTangentChain2

Furthermore, the positions and radii of the three tangent circles surrounding this circle can also be found analytically, and are given by

x_n^((1))=(r(17+r))/(2[12+3n(3n-4)(1-r)^2+r(4r-7)])
(20)
y_n^((1))=(3(3n-2)(1-r)r)/(12+3n(3n-4)(1-r)^2+r(4r-7))
(21)
r_n^((1))=(r(1-r))/(2[12+3n(3n-4)(1-r)^2+r(4r-7)])
(22)
x_n^((2))=(r(17+r))/(2[9+3n(3n-2)(1-r)^2-r(1-r)])
(23)
y_n^((2))=(3(3n-1)(1-r)r)/(9+3n(3n-2)(1-r)^2-r(1-r))
(24)
r_n^((2))=(r(1-r))/(2[9+3n(3n-2)(1-r)^2-r(1-r)])
(25)
x_n^((3))=(r(17+7r))/(2[9+12n(n-1)(1-r)^2+r(4r-1)])
(26)
y_n^((3))=(6(2n-1)(1-r)r)/(9+12n(n-1)(1-r)^2+r(4r-1))
(27)
r_n^((3))=(r(1-r))/(2[9+12n(n-1)(1-r)^2+r(4r-1)]).
(28)

If B divides AC in the golden ratio phi, then the circles in the chain satisfy a number of other special properties (Bankoff 1955).

In each arbelos, there are two Pappus chains P_i and P_i^', with P_1=P_1^'. For fixed n, the line connecting the centers of P_n and P_n^' passes through the external similitude center S of the two smaller semicircles of the arbelos. The line connecting the point of tangency of P_n and P_(n+1) and the point of tangency of P_n^' and P_(n-1)^' passes through S as well. Also the line connecting the point of tangency of P_n and the large exterior semicircle (the smaller interior semicircle) and the point of tangency of P_n^' and the large exterior semicircle (the smaller interior semicircle) passes through S. This can be proven with circle inversion. In particular, since P_1=P_1^', the common tangent of P_1 and the large exterior semicircle passes through S.

See also

Arbelos, Coxeter's Loxodromic Sequence of Tangent Circles, Pappus's Centroid Theorem, Pappus's Harmonic Theorem, Pappus's Hexagon Theorem, Six Circles Theorem, Soddy Circles, Steiner Chain

Portions of this entry contributed by Floor van Lamoen

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References

Bankoff, L. "The Golden Arbelos." Scripta Math. 21, 70-76, 1955.Bankoff, L. "Are the Twin Circles of Archimedes Really Twins?" Math. Mag. 47, 214-218, 1974.Bankoff, L. "How Did Pappus Do It?" In The Mathematical Gardner (Ed. D. Klarner). Boston, MA: Prindle, Weber, and Schmidt, pp. 112-118, 1981.Cadwell, J. H. Topics in Recreational Mathematics. Cambridge, England: Cambridge University Press, 1966.Casey, J. A Sequel to the First Six Books of the Elements of Euclid, Containing an Easy Introduction to Modern Geometry with Numerous Examples, 5th ed., rev. enl. Dublin: Hodges, Figgis, & Co., p. 103, 1888.Gaba, M. G. "On a Generalization of the Arbelos." Amer. Math. Monthly 47, 19-24, 1940.Gardner, M. "Mathematical Games: The Diverse Pleasures of Circles that Are Tangent to One Another." Sci. Amer. 240, 18-28, Jan. 1979.Hood, R. T. "A Chain of Circles." Math. Teacher 54, 134-137, 1961.Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, p. 117, 1929.Rothman, T. "Japanese Temple Geometry." Sci. Amer. 278, 85-91, May 1998.Steiner, J. Jacob Steiner's gesammelte Werke, Band I. Bronx, NY: Chelsea, p. 47, 1971.

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Pappus Chain

Cite this as:

van Lamoen, Floor and Weisstein, Eric W. "Pappus Chain." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/PappusChain.html

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