arbelos

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.}} 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}}}}.

File:Arbelos diagram with points marked.svg

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 $2|AH|^2\; =\; |BC|^2\; -\; |AC|^2\; -\; |BA|^2$. 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 $|AH|^2\; =\; |BA||AC|$. 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{{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 }} who implemented the idea as the following proof without words.{{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}}

center

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.

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, $|OD|\; =\; |OA|\; =\; |OE|$. 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.

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.

File:F-belos.svg

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.Antonio M. Oller-Marcen: [http://forumgeom.fau.edu/FG2013volume13/FG201310.pdf "The f-belos"]. In: Forum Geometricorum, Volume 13 (2013), pp. 103–111.

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

File:Arbelos Shoemakers Knife.jpg

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.

• Archimedes' quadruplets
• Bankoff circle
• Schoch circles
• Schoch line
• Woo circles
• Pappus chain
• Salinon

{{Mathworld| title=Arbelos |urlname=Arbelos}}

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"])

• {{cite book | author = Johnson, R. A. | year = 1960 | title = Advanced Euclidean Geometry: An elementary treatise on the geometry of the triangle and the circle | edition = reprint of 1929 edition by Houghton Mifflin | publisher = Dover Publications | location = New York | isbn = 978-0-486-46237-0 | pages = 116–117}}
• {{Cite book | authorlink = C. Stanley Ogilvy | last = Ogilvy | first = C. S. | year = 1990 | title = Excursions in Geometry | publisher = Dover | isbn = 0-486-26530-7 | pages = [https://archive.org/details/excursionsingeom0000ogil/page/51 51–54] | url = https://archive.org/details/excursionsingeom0000ogil/page/51 }}
• {{cite journal |last=Sondow | first = J. |arxiv=1210.2279 |title=The parbelos, a parabolic analog of the arbelos| journal = Amer. Math. Monthly |year=2013| volume = 120 | issue = 10 | pages = 929–935 | doi = 10.4169/amer.math.monthly.120.10.929 | s2cid = 33402874 }} American Mathematical Monthly, 120 (2013), 929–935.
• {{cite book | last = Wells | first = D. | year = 1991 | title = The Penguin Dictionary of Curious and Interesting Geometry | url = https://archive.org/details/penguindictionar0000well/page/5 | url-access = registration | publisher = Penguin Books | location = New York | isbn = 0-14-011813-6 | pages = [https://archive.org/details/penguindictionar0000well/page/5 5–6] }}

• {{commons category-inline}}
• {{wiktionary-inline}}