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]

A 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 A, B, and C, such that the diameter of the outer semicircle is BC, assumed to have unit length; and the diameters of the inner semicircles are AB and AC, assumed to have lengths r and 1−r, respectively. The letter H denotes the point where the outer semicircle intercepts the line that is perpendicular to the diameter BC through the point A.

Area[edit]

The area of the arbelos is equal to the area of a circle with diameter HA.

Proof[edit]

If BC = 1 and BA = r, then

  • In triangle BHA: r^2+h^2=x^2
  • In triangle CHA: (1-r)^2+h^2=y^2
  • In triangle BHC: x^2+y^2=1

By substitution: y^2=(1-r)^2+x^2-r^2. By expansion: y^2=1-2r+x^2. By substituting for y2 into the equation for triangle BHC and solving for x:

x=\sqrt{r}

By substituting this, solve for y and h

y=\sqrt{1-r}
h=\sqrt{r-r^2}

The radius of the circle with center O is:

\frac{1}{2}\sqrt{r-r^2}.

Therefore, the area is:

A_{circle}=\pi\left(\frac{1}{2}\sqrt{r-r^2}\right)^2
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:

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)
A_{arbelos}=\frac{\pi-\pi r^2-\pi+2\pi r-\pi r^2}{8}
A_{arbelos}=\frac{\pi r}{4}-\frac{\pi r^2}{4}=A_{circle}

Q.E.D.[3]

Rectangle[edit]

Let D and E be the points where the segments BH and CH intersect the semicircles AB and AC, respectively. The quadrilateral ADHE is actually a rectangle.

Proof: The angles BDA, BHC, and AEC are right angles because they are inscribed in semicircles (by Thales' theorem). The quadrilateral ADHE therefore has three right angles, so it is a rectangle. Q.E.D.

Tangents[edit]

The line DE is tangent to semicircle BA at D and semicircle AC at 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 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]