Английская Википедия: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]
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 Шаблон:Mvar and Шаблон:Mvar; the third semicircle is convex, with diameter Шаблон:Mvar[1]
Area
The area of the arbelos is equal to the area of a circle with diameter Шаблон:Mvar.
Proof: For the proof, reflect the arbelos over the line through the points Шаблон:Mvar and Шаблон:Mvar, and observe that twice the area of the arbelos is what remains when the areas of the two smaller circles (with diameters Шаблон:Mvar, Шаблон:Mvar) are subtracted from the area of the large circle (with diameter Шаблон:Mvar). 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 Шаблон:Math), the problem reduces to showing that <math>2|AH|^2 = |BC|^2 - |AC|^2 - |BA|^2</math>. The length Шаблон:Mvar equals the sum of the lengths Шаблон:Mvar and Шаблон:Mvar, 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 Шаблон:Mvar is the geometric mean of the lengths of the segments Шаблон:Mvar and Шаблон:Mvar. Now (see Figure) the triangle Шаблон:Mvar, being inscribed in the semicircle, has a right angle at the point Шаблон:Mvar (Euclid, Book III, Proposition 31), and consequently Шаблон:Mvar is indeed a "mean proportional" between Шаблон:Mvar and Шаблон:Mvar (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 Шаблон:Mvar and Шаблон:Mvar be the points where the segments Шаблон:Mvar and Шаблон:Mvar intersect the semicircles Шаблон:Mvar and Шаблон:Mvar, respectively. The quadrilateral Шаблон:Mvar is actually a rectangle.
- Proof: Шаблон:Mvar, Шаблон:Mvar, and Шаблон:Mvar are right angles because they are inscribed in semicircles (by Thales's theorem). The quadrilateral Шаблон:Mvar therefore has three right angles, so it is a rectangle. Q.E.D.
Tangents
The line Шаблон:Mvar is tangent to semicircle Шаблон:Mvar at Шаблон:Mvar and semicircle Шаблон:Mvar at Шаблон:Mvar.
- Proof: Since Шаблон:Mvar is a right angle, Шаблон:Mvar equals Шаблон:Math minus Шаблон:Mvar. However, Шаблон:Mvar also equals Шаблон:Math minus Шаблон:Mvar (since Шаблон:Mvar is a right angle). Therefore triangles Шаблон:Mvar and Шаблон:Mvar are similar. Therefore Шаблон:Mvar equals Шаблон:Mvar, where Шаблон:Mvar is the midpoint of Шаблон:Mvar and Шаблон:Mvar is the midpoint of Шаблон:Mvar. But Шаблон:Mvar is a straight line, so Шаблон:Mvar and Шаблон:Mvar are supplementary angles. Therefore the sum of Шаблон:Mvar and Шаблон:Mvar is π. Шаблон:Mvar is a right angle. The sum of the angles in any quadrilateral is 2π, so in quadrilateral Шаблон:Mvar, Шаблон:Mvar must be a right angle. But Шаблон:Mvar is a rectangle, so the midpoint Шаблон:Mvar of Шаблон:Mvar (the rectangle's diagonal) is also the midpoint of Шаблон:Mvar (the rectangle's other diagonal). As Шаблон:Mvar (defined as the midpoint of Шаблон:Mvar) is the center of semicircle Шаблон:Mvar, and angle Шаблон:Mvar is a right angle, then Шаблон:Mvar is tangent to semicircle Шаблон:Mvar at Шаблон:Mvar. By analogous reasoning Шаблон:Mvar is tangent to semicircle Шаблон:Mvar at Шаблон:Mvar. Q.E.D.
Archimedes' circles
The altitude Шаблон:Mvar 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.[5]
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.
See also
References
- ↑ 1,0 1,1 Шаблон:Mathworld
- ↑ 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. ("Arbelos - the Shoemaker's Knife")
- ↑ Шаблон:Cite journal
- ↑ Шаблон:Cite journal
- ↑ Antonio M. Oller-Marcen: "The f-belos". In: Forum Geometricorum, Volume 13 (2013), pp. 103–111.
Bibliography
- Шаблон:Cite book
- Шаблон:Cite book
- Шаблон:Cite journal American Mathematical Monthly, 120 (2013), 929–935.
- Шаблон:Cite book
External links
