In geometry, an arbelos is a plane region bounded by three semicircles with three apexes such that each corner of each semicircle is shared with one of the others (connected), all on the same side of a straight line (the baseline) that contains their diameters.[1]

An arbelos (grey region)
Arbelos sculpture in Kaatsheuvel, Netherlands

The earliest known reference to this figure is in Archimedes's Book of Lemmas, where some of its mathematical properties are stated as Propositions 4 through 8.[2] The word arbelos is Greek for 'shoemaker's knife'. The figure is closely related to the Pappus Chain.


Two of the semicircles are necessarily concave, with arbitrary diameters a and b; the third semicircle is convex, with diameter a+b.[1]

Some special points on the arbelos.


The area of the arbelos is equal to the area of a circle with diameter  .

Proof: For the proof, reflect the arbelos over the line through the points   and  , and observe that twice the area of the arbelos is what remains when the areas of the two smaller circles (with diameters    ) are subtracted from the area of the large circle (with diameter  ). Since the area of a circle is proportional to the square of the diameter (Euclid's Elements, Book XII, Proposition 2; we do not need to know that the constant of proportionality is  ), the problem reduces to showing that  . The length   equals the sum of the lengths   and  , so this equation simplifies algebraically to the statement that  . Thus the claim is that the length of the segment   is the geometric mean of the lengths of the segments   and  . Now (see Figure) the triangle  , being inscribed in the semicircle, has a right angle at the point   (Euclid, Book III, Proposition 31), and consequently   is indeed a "mean proportional" between   and   (Euclid, Book VI, Proposition 8, Porism). This proof approximates the ancient Greek argument; Harold P. Boas cites a paper of Roger B. Nelsen[3] who implemented the idea as the following proof without words.[4]


Let   and   be the points where the segments   and   intersect the semicircles   and  , respectively. The quadrilateral   is actually a rectangle.

Proof: The angles  ,  , and   are right angles because they are inscribed in semicircles (by Thales' theorem). The quadrilateral   therefore has three right angles, so it is a rectangle. Q.E.D.


The line   is tangent to semicircle   at   and semicircle   at  .

Proof: Since angle BDA is a right angle, angle DBA equals π/2 minus angle DAB. However, angle DAH also equals π/2 minus angle DAB (since angle HAB is a right angle). Therefore triangles DBA and DAH are similar. Therefore angle DIA equals angle DOH, where I is the midpoint of BA and O is the midpoint of AH. But AOH is a straight line, so angle DOH and DOA are supplementary angles. Therefore the sum of angles DIA and DOA is π. Angle IAO is a right angle. The sum of the angles in any quadrilateral is 2π, so in quadrilateral IDOA, angle IDO must be a right angle. But ADHE is a rectangle, so the midpoint O of AH (the rectangle's diagonal) is also the midpoint of DE (the rectangle's other diagonal). As I (defined as the midpoint of BA) is the center of semicircle BA, and angle IDE is a right angle, then DE is tangent to semicircle BA at D. By analogous reasoning DE is tangent to semicircle AC at E. Q.E.D.

Archimedes' circlesEdit

The altitude   divides the arbelos into two regions, each bounded by a semicircle, a straight line segment, and an arc of the outer semicircle. The circles inscribed in each of these regions, known as the Archimedes' circles of the arbelos, have the same size.

Variations and generalisationsEdit

example of an f-belos

The parbelos is a figure similar to the arbelos, that uses parabola segments instead of half circles. A generalisation comprising both arbelos and parbelos is the f-belos, which uses a certain type of similar differentiable functions.[5]

In the Poincaré half-plane model of the hyperbolic plane, an arbelos models an ideal triangle.


The type of shoemaker's knife that gave its name to the figure

The name arbelos comes from Greek ἡ ἄρβηλος he árbēlos or ἄρβυλος árbylos, meaning "shoemaker's knife", a knife used by cobblers from antiquity to the current day, whose blade is said to resemble the geometric figure.

See alsoEdit


  1. ^ a b Weisstein, Eric W. "Arbelos". MathWorld.
  2. ^ Thomas Little Heath (1897), The Works of Archimedes. Cambridge University Press. Proposition 4 in the Book of Lemmas. Quote: If AB be the diameter of a semicircle and N any point on AB, and if semicircles be described within the first semicircle and having AN, BN as diameters respectively, the figure included between the circumferences of the three semicircles is "what Archimedes called arbelos"; and its area is equal to the circle on PN as diameter, where PN is perpendicular to AB and meets the original semicircle in P. ("Arbelos - the Shoemaker's Knife")
  3. ^ Nelsen, R B (2002). "Proof without words: The area of an arbelos". Math. Mag. 75 (2): 144. doi:10.2307/3219152.
  4. ^ Boas, Harold P. (2006). "Reflections on the Arbelos". The American Mathematical Monthly. 113 (3): 236–249. doi:10.2307/27641891. JSTOR 27641891.
  5. ^ Antonio M. Oller-Marcen: "The f-belos". In: Forum Geometricorum, Volume 13 (2013), pp. 103–111.


External linksEdit

  •   Media related to Arbelos at Wikimedia Commons
  •   The dictionary definition of arbelos at Wiktionary