Arbelos

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An arbelos (grey region)

In geometry, an arbelos is a plane region bounded by three semicircles connected at the corners, 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 the Book of Lemmas by Archimedes, where some of its mathematical properties are stated as Propositions 4 through 8.[2]

Etymology[edit]

The type of shoemaker's knife that gave its name to the figure.

The name "arbelos", used by Archimedes, comes from Greek ἡ ἄρβηλος he arbelos or ἄρβυλος arbylos, meaning "shoemaker's knife", a knife used by cobblers from antiquity to the current day, whose blade is said to resemble the geometric figure.

Properties[edit]

Two of the semicircles are necessarily concave, with arbitrary diameters a and b; the third semicircle is convex, with diameter a+b.[1]

Some special points on the arbelos.

In the following sections, the corners of the arbelos are labeled Failed to parse (Cannot store math image on filesystem.): A , Failed to parse (Cannot store math image on filesystem.): B , and Failed to parse (Cannot store math image on filesystem.): C , such that the diameter of the outer semicircle is Failed to parse (Cannot store math image on filesystem.): BC , assumed to have unit length; and the diameters of the inner semicircles are Failed to parse (Cannot store math image on filesystem.): AB

and Failed to parse (Cannot store math image on filesystem.): AC

, assumed to have lengths r and 1−r, respectively. The letter Failed to parse (Cannot store math image on filesystem.): H

denotes the point where the outer semicircle intercepts the line that is perpendicular to the diameter Failed to parse (Cannot store math image on filesystem.): BC
through the point Failed to parse (Cannot store math image on filesystem.): A

.

Area[edit]

The area of the arbelos is equal to the area of a circle with diameter Failed to parse (Cannot store math image on filesystem.): HA .

Proof[edit]

If BC = 1 and BA = r, then

  • In triangle BHA: Failed to parse (Cannot store math image on filesystem.): r^2+h^2=x^2
  • In triangle CHA: Failed to parse (Cannot store math image on filesystem.): (1-r)^2+h^2=y^2
  • In triangle BHC: Failed to parse (Cannot store math image on filesystem.): x^2+y^2=1

By substitution: Failed to parse (Cannot store math image on filesystem.): y^2=(1-r)^2+x^2-r^2 . By expansion: Failed to parse (Cannot store math image on filesystem.): y^2=1-2r+x^2 . By substituting for y2 into the equation for triangle BHC and solving for x:

Failed to parse (Cannot store math image on filesystem.): x=\sqrt{r}

By substituting this, solve for y and h

Failed to parse (Cannot store math image on filesystem.): y=\sqrt{1-r}
Failed to parse (Cannot store math image on filesystem.): h=\sqrt{r-r^2}

The radius of the circle with center O is:

Failed to parse (Cannot store math image on filesystem.): \frac{1}{2}\sqrt{r-r^2}.

Therefore, the area is:

Failed to parse (Cannot store math image on filesystem.): A_{circle}=\pi\left(\frac{1}{2}\sqrt{r-r^2}\right)^2
Failed to parse (Cannot store math image on filesystem.): A_{circle}=\frac{\pi r}{4}-\frac{\pi r^2}{4}

The area of the arbelos is the area of the large semicircle minus the area of the two smaller semicircles. Therefore the area of the arbelos is:

Failed to parse (Cannot store math image on filesystem.): A_{arbelos}=\frac{\pi}{8}-\left(\frac{\pi}{2}\left(\frac{r}{2}\right)^2+\frac{\pi}{2}\left(\frac{1-r}{2}\right)^2\right)
Failed to parse (Cannot store math image on filesystem.): A_{arbelos}=\frac{\pi-\pi r^2-\pi+2\pi r-\pi r^2}{8}
Failed to parse (Cannot store math image on filesystem.): A_{arbelos}=\frac{\pi r}{4}-\frac{\pi r^2}{4}=A_{circle}

Q.E.D.[3]

Rectangle[edit]

Let Failed to parse (Cannot store math image on filesystem.): D

and Failed to parse (Cannot store math image on filesystem.): E
be the points where the segments Failed to parse (Cannot store math image on filesystem.): BH
and Failed to parse (Cannot store math image on filesystem.): CH
intersect the semicircles Failed to parse (Cannot store math image on filesystem.): AB
and Failed to parse (Cannot store math image on filesystem.): AC

, respectively. The quadrilateral Failed to parse (Cannot store math image on filesystem.): ADHE

is actually a rectangle.
Proof: The angles Failed to parse (Cannot store math image on filesystem.): BDA

, Failed to parse (Cannot store math image on filesystem.): BHC , and Failed to parse (Cannot store math image on filesystem.): AEC

are right angles because they are inscribed in semicircles (by Thales' theorem). The quadrilateral Failed to parse (Cannot store math image on filesystem.): ADHE
therefore has three right angles, so it is a rectangle. Q.E.D.

Tangents[edit]

The line Failed to parse (Cannot store math image on filesystem.): DE

is tangent to semicircle Failed to parse (Cannot store math image on filesystem.): BA
at Failed to parse (Cannot store math image on filesystem.): D
and semicircle Failed to parse (Cannot store math image on filesystem.): AC
at Failed to parse (Cannot store math image on filesystem.): E

.

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

Archimedes' circles[edit]

The altitude Failed to parse (Cannot store math image on filesystem.): 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.

See also[edit]

Arbelos skulpture in Kaatsheuvel, Netherlands

References[edit]

  1. ^ a b Weisstein, Eric W., "Arbelos", MathWorld.
  2. ^ 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")
  3. ^ Behnaz Rouhani. "The Arbelos". Retrieved 2008-06-13. 

Bibliography[edit]

  • Johnson, R. A. (1960). Advanced Euclidean Geometry: An elementary treatise on the geometry of the triangle and the circle (reprint of 1929 edition by Houghton Miflin ed.). New York: Dover Publications. pp. 116–117. ISBN 978-0-486-46237-0. 
  • Ogilvy, C. S. (1990). Excursions in Geometry. Dover. pp. 51–54. ISBN 0-486-26530-7. 
  • Sondow, J. (2012). "The parbelos, a parabolic analog of the arbelos". arXiv:1210.2279 [math.HO].  American Mathematical Monthly, 120 (2013), 929-935.
  • Wells, D. (1991). The Penguin Dictionary of Curious and Interesting Geometry. New York: Penguin Books. pp. 5–6. ISBN 0-14-011813-6. 

External links[edit]