Arbelos

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

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 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.

Properties

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

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

Proof

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

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

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

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

Arbelos skulpture in Kaatsheuvel, Netherlands

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

References

1. 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

• 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

• Media related to Arbelos at Wikimedia Commons
• The dictionary definition of arbelos at Wiktionary
• Arbelos – Amazing Properties (an interactive diagram illustrating many Arbelos properties) at www.retas.de
• Arbelos at cut-the-knot
• Arbelos and Parbelos at PlanetMath
• L’arbelos. Partie I and Partie II by Hamza Khelif at www.images.math.cnrs.fr of CNRS
• Review of Jonathan Sondow's "The parbelos, a parabolic analog of the arbelos" by Mowaffaq Hajja in Zentralblatt MATH