Arbelos

{{Short description|Plane region bounded by three semicircles}} right|thumb|320px|An arbelos (grey region) [[File:Arbelos sculpture Netherlands 1.jpg|thumb|Arbelos sculpture in Kaatsheuvel, Netherlands]]

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.<ref name=wolfram/>

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.<ref name=archBLprop4/> 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 {{mvar|a}} and {{mvar|b}}; the third semicircle is convex, with diameter {{mvar|''a''+''b''.}}<ref name=wolfram/> Let the diameters of the smaller semicircles be {{mvar|{{overline|BA}}}} and {{mvar|{{overline|AC}}}}; then the diameter of the larger semircle is {{mvar|{{overline|BC}}}}.

right|thumb|320px|Some special points on the arbelos.

===Area=== Let {{mvar|H}} be the intersection of the larger semicircle with the line perpendicular to {{mvar|BC}} at {{mvar|A}}. Then the area of the arbelos is equal to the area of a circle with diameter {{mvar|{{overline|HA}}}}.

'''Proof''': For the proof, reflect the arbelos over the line through the points {{mvar|B}} and {{mvar|C}}, and observe that twice the area of the arbelos is what remains when the areas of the two smaller circles (with diameters {{mvar|{{overline|BA}}}}, {{mvar|{{overline|AC}}}}) are subtracted from the area of the large circle (with diameter {{mvar|{{overline|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 {{math|{{sfrac|{{pi}}|4}}}}), the problem reduces to showing that <math>2|AH|^2 = |BC|^2 - |AC|^2 - |BA|^2</math>. The length {{mvar|{{abs|BC}}}} equals the sum of the lengths {{mvar|{{abs|BA}}}} and {{mvar|{{abs|AC}}}}, so this equation simplifies algebraically to the statement that <math>|AH|^2 = |BA||AC|</math>. Thus the claim is that the length of the segment {{mvar|{{overline|AH}}}} is the geometric mean of the lengths of the segments {{mvar|{{overline|BA}}}} and {{mvar|{{overline|AC}}}}. Now (see Figure) the triangle {{mvar|BHC}}, being inscribed in the semicircle, has a right angle at the point {{mvar|H}} (Euclid, Book III, Proposition 31), and consequently {{mvar|{{abs|HA}}}} is indeed a "mean proportional" between {{mvar|{{abs|BA}}}} and {{mvar|{{abs|AC}}}} (Euclid, Book VI, Proposition 8, Porism). This proof approximates the ancient Greek argument; Harold P. Boas cites a paper of Roger B. Nelsen<ref name="RBNelsen_2002">{{cite journal |last1=Nelsen |first1=R B |title=Proof without words: The area of an arbelos |journal=Math. Mag. |date=2002 |volume=75 |issue=2 |page=144|doi= 10.2307/3219152|jstor=3219152 }}</ref> who implemented the idea as the following proof without words.<ref>{{cite journal| last=Boas | first=Harold P.| author-link1=Harold P. Boas | title=Reflections on the Arbelos | journal= The American Mathematical Monthly| year=2006| volume=113| issue=3| url=http://www.maa.org/programs/maa-awards/writing-awards/reflections-on-the-arbelos| pages=236–249 | doi=10.2307/27641891| jstor=27641891}}</ref> center

===Rectangle=== Let {{mvar|D}} and {{mvar|E}} be the points where the segments {{mvar|{{overline|BH}}}} and {{mvar|{{overline|CH}}}} intersect the semicircles {{mvar|AB}} and {{mvar|AC}}, respectively. The quadrilateral {{mvar|ADHE}} is actually a rectangle. :''Proof'': {{mvar|∠BDA}}, {{mvar|∠BHC}}, and {{mvar|∠AEC}} are right angles because they are inscribed in semicircles (by Thales's theorem). The quadrilateral {{mvar|ADHE}} therefore has three right angles, so it is a rectangle. ''Q.E.D.''

===Tangents=== The line {{mvar|DE}} is tangent to semicircle {{mvar|BA}} at {{mvar|D}} and semicircle {{mvar|AC}} at {{mvar|E}}. :''Proof'': Since {{mvar|ADHE}} is a rectangle, the diagonals {{mvar|AH}} and {{mvar|DE}} have equal length and bisect each other at their intersection {{mvar|O}}. Therefore, <math>|OD| = |OA| = |OE|</math>. Also, since {{mvar|{{overline|OA}}}} is perpendicular to the diameters {{mvar|{{overline|BA}}}} and {{mvar|{{overline|AC}}}}, {{mvar|{{overline|OA}}}} is tangent to both semicircles at the point {{mvar|A}}. Finally, because the two tangents to a circle from any given exterior point have equal length, it follows that the other tangents from {{mvar|O}} to semicircles {{mvar|BA}} and {{mvar|AC}} are {{mvar|{{overline|OD}}}} and {{mvar|{{overline|OE}}}} respectively.

===Archimedes' circles=== The altitude {{mvar|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 == thumb|right|upright=1.0|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.<ref>Antonio M. Oller-Marcen: [http://forumgeom.fau.edu/FG2013volume13/FG201310.pdf "The f-belos"]. In: ''Forum Geometricorum'', Volume 13 (2013), pp. 103–111.</ref>

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

==Etymology== right|thumb|upright=1.0|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.

==References== <references> <ref name=wolfram>{{Mathworld| title=Arbelos |urlname=Arbelos}}</ref> <ref name=archBLprop4>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.'' ([https://web.archive.org/web/20080509094946/http://www.cut-the-knot.org/proofs/arbelos.shtml "Arbelos - the Shoemaker's Knife"])</ref> </references>

==Bibliography== * {{cite book | author = Johnson, R. A. | year = 1929 | title = Modern Geometry: An elementary treatise on the geometry of the triangle and the circle | publisher = Houghton Mifflin | url = https://babel.hathitrust.org/cgi/pt?id=wu.89043163211 | pages = 116–117 | contribution = The arbelos | contribution-url = https://babel.hathitrust.org/cgi/pt?id=wu.89043163211&seq=134}} * {{Cite book | authorlink = C. Stanley Ogilvy | last = Ogilvy | first = C. S. | year = 1969 | title = Excursions in Geometry | publisher = Oxford University Press | pages = 51–54 }} * {{cite journal |last=Sondow | first = J. |arxiv=1210.2279 |title=The parbelos, a parabolic analog of the arbelos| journal = American Mathematical Monthly |year=2013| volume = 120 | issue = 10 | pages = 929–935 | doi = 10.4169/amer.math.monthly.120.10.929 | jstor = 10.4169/amer.math.monthly.120.10.929 }} * {{cite book | last = Wells | first = D. | year = 1991 | title = The Penguin Dictionary of Curious and Interesting Geometry | publisher = Penguin Books | location = New York | isbn = 0-14-011813-6 | pages = 5–6}}

==External links== * {{commons category-inline}} * {{wiktionary-inline}}

Category:Arbelos Category:Ancient Greek mathematics Category:Archimedes