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Arbelos


Arbelos

The term "arbelos" means shoemaker's knife in Greek, and this term is applied to the shaded area in the above figure which resembles the blade of a knife used by ancient cobblers (Gardner 1979). Archimedes himself is believed to have been the first mathematician to study the mathematical properties of this figure. The position of the central notch is arbitrary and can be located anywhere along the diameter.

The arbelos satisfies a number of unexpected identities (Gardner 1979, Schoch).

1. Call the diameters of the left and right semicircles r<1 and 1-r, respectively, so the diameter of the enclosing semicircle is 1. Then the arc length along the bottom of the arbelos is

 L=1/2[pir+pi(1-r)]=1/2pi,
(1)

so the arc length along the enclosing semicircle is the same as the arc length along the two smaller semicircles.

2. Draw the perpendicular BD from the tangent of the two semicircles to the edge of the large circle. Then the area of the arbelos is the same as the area of the circle with diameter BD. Let AC=1 and r=AB, then simultaneously solve the equations

r^2+h^2=x^2
(2)
(1-r)^2+h^2=y^2
(3)
x^2+y^2=1^2
(4)

for the sides

x=AD=sqrt(r)
(5)
y=CD=sqrt(1-r)
(6)
h=BD=sqrt(r(1-r)).
(7)
ArbelosRightTriangles

3. The circles C_1 and C_1^' inscribed on each half of BD on the arbelos (called Archimedes' circles) each have diameter r(1-r), or radius r(1-r)/2.

ArbelosAnnotatedArbelosTriangles

The positions of the circles can be found using the triangles shown above. The lengths of the horizonal legs and hypotenuses are known as indicated, so the vertical legs can be found using the Pythagorean theorem. This then gives the centers of the circles as

x_1=r-R=1/2r(1+r)
(8)
y_1=sqrt(2rR)=rsqrt(1-r)
(9)

and

x_1^'=r+R=1/2r(3-r)
(10)
y_1^'=sqrt(2R(1-r))=(1-r)sqrt(r).
(11)

4. Let A^' be the point at which the circle centered at A and of radius r=AB intersects the enclosing semicircle, and let C^' be the point at which the circle centered at C of radius 1-r=BC intersects the enclosing semicircle. Then the smallest circle C_2 passing through A^' and tangent to BD is equal to the smallest circle C_2^' passing through C^' and tangent to BD (Schoch). Moreover, the radii R of these circles are the same as Archimedes' circles. Solving

 (x-1/2)^2+y^2=(1/2)^2
(12)

 x^2+y^2=r^2
(13)

gives (x,y)=(r^2,rsqrt(1-r^2)), so the center of C_2 is

x_2=r^2+1/2r(1-r)=1/2r(r+1)
(14)
y_2=rsqrt(1-r^2).
(15)

Similarly, solving

 (x-1/2)^2+y^2=(1/2)^2
(16)
 (x-1)^2+y^2=(1-r)^2
(17)

gives (x,y)=(r(2-r),(1-r)sqrt(r(2-r))), so the center of C_2^' is

x_2^'=r(2-r)-1/2r(1-r)=1/2r(r-3)
(18)
y_2^'=(1-r)sqrt(r(2-r)).
(19)
ArbelosArcs

5. The Apollonius circle C_3 of the circles with arcs BA^', BC^', and AA^'DC^'C is located at a position

x=1/2r(1+3r-2r^2)
(20)
y=r(1-r)sqrt((2-r)(1+r))
(21)

and has radius R equal to that of Archimedes' circles (Schoch), as does the smallest circle C_3^' passing through B and tangent to C_3.

ArbelosApolloniusCircle

Furthermore, letting B^'D^' be the line parallel to BD through the center of circle C_3, the circle C_3^('') with center on B^'D^' and tangent to the small semicircles of the arbelos also has radius R (Schoch). The position of the center of C_3^('') is given by

x_3^('')=x=1/2r(1+3r-2r^2)
(22)
y_3^('')=sqrt((1/2r+R)-(x-1/2r)^2)
(23)
=r(1-r)sqrt(1+r-r^2).
(24)
ArbelosApolloniusCircle3

The vertical h^' position of D^' is

h^'=sqrt(1/4-1/4(2r^3-3r^2-r+1)^2)
(25)
=1/2sqrt(r(1-r)(2r^2-3r-1)(2r^2-r-2)).
(26)

6. Let P be the midpoint of AB, and let Q be the midpoint of BC. Then draw the semicircle having PQ as a diameter with center M. This circle has radius

 R_(PQ)=1/2{1-1/2[r+(1-r)]}=1/4.
(27)

The smallest circle C_4 through D^' touching arc PQ then has radius R (Schoch). Using similar triangles, the center of this circle is at

x_4=(r(2r^4-5r^3+3r+1))/(1+4r-4r^2)
(28)
y_4=(2r^2-2r-1)/(2(4r^2-4r-1))sqrt(r(1-r)(2r^2-3r-1)(2r^2-r-2)).
(29)

Similarly, let U be the point of intersection of B^'D^' and the semicircle PQ, then the circle through B, B^', and U also has radius R (Schoch). The center of this circle is at

x_4^'=1/4r(3+3r-2r^2)
(30)
y_4^'=1/4r(1-r)sqrt((2r+1)(3-2r)).
(31)
ArbelosC4-12

Consider the circle X of radius r_X which is tangent to the two interior semicircles. Its position and radius are obtained by solving the simultaneous equations

 h^2+z^2=(1/2r+r_X)^2
(32)
 h^2+(1/2-z)^2=[1/2(1-r)+r_X]^2
(33)
 (1/2r+r_X)^2+[1/2(1-r)+r_X]^2=(1/4)^2,
(34)

giving

z=1/4+1/4(2r-1)sqrt(1+4r-4r^2)
(35)
h=r(1-r)
(36)
r_X=1/4(sqrt(1+4r-4r^2)-1).
(37)

Letting C_4^('') be the smallest circle through X and tangent to ABC, the radius of C_4^('') is therefore h/2=r(1-r)/2=R (Schoch), and its center is located at

x_4^('')=1/4+1/2r+1/4(2r-1)sqrt(1+4r-4r^2)
(38)
y_4^('')=1/2r(1-r).
(39)
ArbelosC4-3

7. Within each small semicircle of an arbelos, construct arbeloses similar to the original. Then the circles C_5 and C_5^' are congruent and have radius R (Schoch). Moreover, connect the midpoints of the arcs and their cusp points to form the rectangles EFGH and E^'F^'G^'H^'. Then these rectangles are similar with respect to the point C_5^('') (Schoch). This point lies on the line B^'D^', and the circle with center C_5^('') and radius C_5^('')B^' also has radius R, so C_5^('') has coordinates (1/2r(1+3r-2r^2),1/2r(1-r)). The following tables summarized the positions of the rectangle vertices.

XcoordinatesX^'coordinates
E(1/2r,1/2r)E^'(r(2-r),0)
F(1/2r(1+r),1/2r(1-r))F^'(1/2r(3-r),1/2r(1-r))
G(r^2,0)G^'(1/2(1+r),1/2(1-r))
H(1/2r^2,1/2r^2)H^'(1/2(1+2r-r^2),1/2(1-r)^2)
ArbelosC5

8. Let MM^' be the perpendicular bisector of AC, let B be the cusp of the arbelos and D lie above it, let E and G^' be the tops of the large and small semicircles, respectively. Let EG^' intersect the lines MM^' and BD in points I and J, respectively. Then the smallest circle C_6 passing through I and tangent to arc AC at M^', the smallest circle C_6^' through J and tangent to the outside semicircle at P_C, and the circle C_6^('') with diameter JB are all Archimedean circles (Schoch). The circle C_6^('') is called the Bankoff circle, and is also the circumcircle of the point B and tangent points P_A and P_C of the first Pappus circle. The centers of the circles C_6, C_6^', and C_6^('') are given by

x_6=1/2
(40)
y_6=1/2(1-r+r^2)
(41)
x_6^'=(r(1-r+2r^2))/(2(1-2r+2r^2))
(42)
y_6^'=(r(1-r)(1-r+r^2))/(1-2r+2r^2)
(43)
x_6^('')=r
(44)
y_6^('')=1/2r(1-r).
(45)

Rather amazingly, the points E, M, B, G^', P_C, D, and M^' are concyclic (Schoch) in a circle with center ((1+2r)/4,1/4) and radius

 R_(EMBG^'P_CDM^')=1/4sqrt(2(1-2r+2r^2)).
(46)
ArbelosC6

9. The smallest circumcircle of Archimedes' circles has an area equal to that of the arbelos.

ArbelosCircumcircle

10. The line tangent to the semicircles AB and BC contains the point E and F which lie on the lines AD and CD, respectively. Furthermore, BD and EF bisect each other, and the points B, D, E, and F are concyclic.

ArbelosAnnotated2

11. Construct a chain of tangent circles starting with the circle tangent to the two small ones and large one. This chain is called a Pappus chain, and the centers of its circles lie on an ellipse having foci at the centers of the semicircles bounding it. Furthermore, the diameter of the nth circle C_n is (1/n)th the perpendicular distance to the base of the semicircle. This result is most easily proven using inversion, but was known to Pappus, who referred to it as an ancient theorem (Hood 1961, Cadwell 1966, Gardner 1979, Bankoff 1981).

12. The common tangent EF (see 10) and the common tangent of the great semicircle and the first Pappus circle meet on line AB.

PappusChain

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


See also

Archimedean Circle, Archimedes' Circles, Bankoff Circle, Coxeter's Loxodromic Sequence of Tangent Circles, Golden Ratio, Inversion, Pappus Chain, Schoch Line, Steiner Chain, Tangent Circles, Tomahawk, Woo Circle

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References

Allanson, B. "Pappus's Arbelos" Java applet. http://members.ozemail.com.au/~llan/arbelos.html.Bankoff, L. "The Fibonacci Arbelos." Scripta Math. 20, 218, 1954.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.Bankoff, L. "The Marvelous Arbelos." In The Lighter Side of Mathematics (Ed. R. K. Guy and R. E. Woodrow). Washington, DC: Math. Assoc. Amer., 1994.Boas, H. P. "Reflections on the Arbelos." Preprint based on the Nineteenth Annual Rose-Hulman Undergraduate Mathematics Conference, March 15, 2002. Oct. 26, 2004. http://www.math.tamu.edu/~harold.boas/preprints/arbelos.pdf.Bogomolny, A. "Arbelos--The Shoemaker's Knife." http://www.cut-the-knot.org/proofs/arbelos.shtml.Cadwell, J. H. Topics in Recreational Mathematics. Cambridge, England: Cambridge University Press, 1966.Coolidge, J. L. A Treatise on the Geometry of the Circle and Sphere. New York: Chelsea, pp. 35-36, 1971.Dodge, C. W. Euclidean Geometry and Transformations. New York: Dover, 2004.Dodge, C. W.; Schoch, T.; Woo, P. Y.; and Yiu, P. "Those Ubiquitous Archimedean Circles." Math. Mag. 72, 202-213, 1999.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.Heath, T. L. The Works of Archimedes with the Method of Archimedes. New York: Dover, p. 307, 1953.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, pp. 116-117, 1929.Ogilvy, C. S. Excursions in Geometry. New York: Dover, pp. 54-55, 1990.Okumura, H. and Watanabe, M. "The Archimedean Circles of Schoch and Woo." Forum Geom. 4, 27-34, 2004. http://forumgeom.fau.edu/FG2004volume4/FG200404index.html.Okumura, H. and Watanabe, M. "The Twin Circles of Archimedes in a Skewed Arbelos." Forum Geom. 4, 229-251, 2004. http://forumgeom.fau.edu/FG2004volume4/FG200428index.html.Okumura, H. and Watanabe, M. "The Arbelos in n-Aliquot Parts." Forum Geom. 5, 37-45, 2005. http://forumgeom.fau.edu/FG2005volume5/FG200506index.html.Power, F. "Some More Archimedean Circles in the Arbelos." Forum Geom. 5, 133-134, 2005. http://forumgeom.fau.edu/FG2005volume5/FG200517index.html.Schoch, T. "Arbelos: Amazing Properties." http://www.retas.de/thomas/arbelos/arbelos.html.Schoch, T. "A Dozen More Arbelos Twins." http://www.retas.de/thomas/arbelos/biola/.Soddy, F. "The Bowl of Integers and the Hexlet." Nature 139, 77-79, 1937.Wells, D. The Penguin Dictionary of Curious and Interesting Geometry. London: Penguin, pp. 5-6, 1991.Update a linkWoo, P. "The Arbelos." http://www-students.biola.edu/~woopy/math/arbelos.htmWoo, P. Y. "Simple Construction for the Incircle of an Arbelos." Forum Geom. 1, 133-136, 2001. http://forumgeom.fau.edu/FG2001volume1/FG200119index.html.Yiu, P. "The Archimedean Circles in the Shoemaker's Knife." Lecture at the 31st Annual Meeting of the Florida Section of the Math. Assoc. Amer., Boca Raton, FL, March 6-7, 1998.

Cite this as:

Weisstein, Eric W. "Arbelos." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/Arbelos.html

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