# Arbelos

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.

## Properties

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.

### Area

The area of the arbelos is equal to the area of a circle with diameter ${\displaystyle HA}$ .

Proof: For the proof, reflect the arbelos over the line through the points ${\displaystyle B}$  and ${\displaystyle C}$ , and observe that twice the area of the arbelos is what remains when the areas of the two smaller circles (with diameters ${\displaystyle BA}$  ${\displaystyle AC}$ ) are subtracted from the area of the large circle (with diameter ${\displaystyle BC}$ ). 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 ${\displaystyle {\frac {\pi }{4}}}$ ), the problem reduces to showing that ${\displaystyle 2(AH)^{2}=(BC)^{2}-(AC)^{2}-(BA)^{2}}$ . The length ${\displaystyle (BC)}$  equals the sum of the lengths ${\displaystyle (BA)}$  and ${\displaystyle (AC)}$ , so this equation simplifies algebraically to the statement that ${\displaystyle (AH)^{2}=(BA)(AC)}$ . Thus the claim is that the length of the segment ${\displaystyle AH}$  is the geometric mean of the lengths of the segments ${\displaystyle BA}$  and ${\displaystyle AC}$ . Now (see Figure) the triangle ${\displaystyle BHC}$ , being inscribed in the semicircle, has a right angle at the point ${\displaystyle H}$  (Euclid, Book III, Proposition 31), and consequently ${\displaystyle (HA)}$  is indeed a "mean proportional" between ${\displaystyle (BA)}$  and ${\displaystyle (AC)}$  (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]

### Rectangle

Let ${\displaystyle D}$  and ${\displaystyle E}$  be the points where the segments ${\displaystyle BH}$  and ${\displaystyle CH}$  intersect the semicircles ${\displaystyle AB}$  and ${\displaystyle AC}$ , respectively. The quadrilateral ${\displaystyle ADHE}$  is actually a rectangle.

Proof: The angles ${\displaystyle BDA}$ , ${\displaystyle BHC}$ , and ${\displaystyle AEC}$  are right angles because they are inscribed in semicircles (by Thales' theorem). The quadrilateral ${\displaystyle ADHE}$  therefore has three right angles, so it is a rectangle. Q.E.D.

### Tangents

The line ${\displaystyle DE}$  is tangent to semicircle ${\displaystyle BA}$  at ${\displaystyle D}$  and semicircle ${\displaystyle AC}$  at ${\displaystyle E}$ .

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' circles

The altitude ${\displaystyle AH}$  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 generalisations

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.

## Etymology

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.