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.

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. 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 $a+b.$



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

Proof: For the proof, reflect the arbelos over the line through the points $B$ and $C$, and observe that twice the area of the arbelos is what remains when the areas of the two smaller circles (with diameters $BC$, $AB$) are subtracted from the area of the large circle (with diameter $AC$). 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 $\pi⁄4$), the problem reduces to showing that $$2|AH|^2 = |BC|^2 - |AC|^2 - |BA|^2$$. The length $H$ equals the sum of the lengths $BC$ and $A$, so this equation simplifies algebraically to the statement that $$|AH|^2 = |BA||AC|$$. Thus the claim is that the length of the segment $\overline{HA}$ is the geometric mean of the lengths of the segments $B$ and $C$. Now (see Figure) the triangle $\overline{BA}$, being inscribed in the semicircle, has a right angle at the point $\overline{AC}$ (Euclid, Book III, Proposition 31), and consequently $\overline{BC}$ is indeed a "mean proportional" between $|BC|$ and $|BA|$ (Euclid, Book VI, Proposition 8, Porism). This proof approximates the ancient Greek argument; Harold P. Boas cites a paper of Roger B. Nelsen who implemented the idea as the following proof without words.

Rectangle
Let $|AC|$ and $\overline{AH}$ be the points where the segments $\overline{BA}$ and $\overline{AC}$ intersect the semicircles $BHC$ and $H$, respectively. The quadrilateral $|HA|$ is actually a rectangle.
 * Proof: $|BA|$, $|AC|$, and $D$ are right angles because they are inscribed in semicircles (by Thales's theorem). The quadrilateral $E$ therefore has three right angles, so it is a rectangle. Q.E.D.

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

Archimedes' circles
The altitude $\overline{BA}$ 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
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.

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

Etymology
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.