Concept

# Arbelos

Summary
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. Two of the semicircles are necessarily concave, with arbitrary diameters a and b; the third semicircle is convex, with diameter a+b. 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 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 , ) 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 pi/4), 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 BHC, being inscribed in the semicircle, has a right angle at the point H (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 who implemented the idea as the following proof without words. Let D and E be the points where the segments and intersect the semicircles AB and AC, respectively. The quadrilateral ADHE is actually a rectangle.
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