Lune of Hippocrates

Hippocrates wanted to solve the classic problem of squaring the circle, i.e. constructing a square by means of straightedge and compass, having the same area as a given circle. He proved that the lune bounded by the arcs labeled E and F in the figure has the same area as triangle ABO. This afforded some hope of solving the circle-squaring problem, since the lune is bounded only by arcs of circles. Heath concludes that, in proving his result, Hippocrates was also the first to prove that the area of a circle is proportional to the square of its diameter.

Hippocrates' book on geometry in which this result appears, Elements, has been lost, but may have formed the model for Euclid's Elements.{{citation|contribution=Hippocrates of Chios|year=2012|title=Encyclopædia Britannica|contribution-url=http://www.britannica.com/EBchecked/topic/266646/Hippocrates-of-Chios|access-date=2012-01-12}}. Hippocrates' proof was preserved through the History of Geometry compiled by Eudemus of Rhodes, which has also not survived, but which was excerpted by Simplicius of Cilicia in his commentary on Aristotle's Physics.{{citation|first=Thomas L.|last=Heath|author-link=Thomas Little Heath|title=A Manual of Greek Mathematics|publisher=Courier Dover Publications|year=2003|isbn=0-486-43231-9|pages=121–132|url=https://books.google.com/books?id=_HZNr_mGFzQC&pg=PA121}}.{{MacTutor|id=Hippocrates|title=Hippocrates of Chios}}

Not until 1882, with Ferdinand von Lindemann's proof of the transcendence of π, was squaring the circle proved to be impossible.{{citation|title=Invitation to Mathematics|first=Konrad|last=Jacobs|publisher=Princeton University Press|year=1992|isbn=978-0-691-02528-5|contribution=2.1 Squaring the Circle|pages=11–13|url=https://books.google.com/books?id=Mgajuf62voQC&pg=PA11}}.

Hippocrates' result can be proved as follows: The center of the circle on which the arc {{mvar|AEB}} lies is the point {{mvar|D}}, which is the midpoint of the hypotenuse of the isosceles right triangle {{mvar|ABO}}. Therefore, the diameter {{mvar|AC}} of the larger circle {{mvar|ABC}} is {{tmath|\sqrt{2} }} times the diameter of the smaller circle on which the arc {{mvar|AEB}} lies. Consequently, the smaller circle has half the area of the larger circle, and therefore the quarter circle {{mvar|AFBOA}} is equal in area to the semicircle {{mvar|AEBDA}}. Subtracting the crescent-shaped area {{mvar|AFBDA}} from the quarter circle gives triangle {{mvar|ABO}} and subtracting the same crescent from the semicircle gives the lune. Since the triangle and lune are both formed by subtracting equal areas from equal area, they are themselves equal in area.{{citation|first1=Lucas Nicolaas Hendrik|last1=Bunt|first2=Phillip S.|last2=Jones|first3=Jack D.|last3=Bedient|title=The Historical Roots of Elementary Mathematics|publisher=Courier Dover Publications|year=1988|isbn=0-486-25563-8|pages=90–91|url=https://books.google.com/books?id=7xArILpcndYC&pg=PA90|contribution=4-2 Hippocrates of Chios and the quadrature of lunes}}.

File:LunesOfAlhazen.svg

Using a similar proof to the one above, the Arab mathematician Hasan Ibn al-Haytham (Latinized name Alhazen, c. 965 – c. 1040) showed that where two lunes are formed, on the two sides of a right triangle, whose outer boundaries are semicircles and whose inner boundaries are formed by the circumcircle of the triangle, then the areas of these two lunes added together are equal to the area of the triangle. The lunes formed in this way from a right triangle are known as the lunes of Alhazen.[http://www.cut-the-knot.org/Curriculum/Geometry/PythagoreanLune.shtml Hippocrates' Squaring of the Lune] at cut-the-knot, accessed 2012-01-12.{{citation|title=Charming Proofs: A Journey into Elegant Mathematics|volume=42|series=Dolciani mathematical expositions|first1=Claudi|last1=Alsina|first2=Roger B.|last2=Nelsen|publisher=Mathematical Association of America|year=2010|isbn=978-0-88385-348-1|contribution=9.1 Squarable lunes|pages=137–144|url=https://books.google.com/books?id=mIT5-BN_L0oC&pg=PA137}}. The quadrature of the lune of Hippocrates is the special case of this result for an isosceles right triangle.{{citation|first=W. S.|last=Anglin|title=Mathematics, a Concise History and Philosophy|publisher=Springer|year=1994|isbn=0-387-94280-7|pages=51–53|url=https://books.google.com/books?id=13dKav77vGsC&pg=PA51|contribution=Hippocrates and the Lunes}}.

All lunes constructable by compass and straight-edge can be specified by the two angles formed by the inner and outer arcs on their respective circles; in this notation, for instance, the lune of Hippocrates would have the inner and outer angles (90°, 180°) with ratio 1:2. Hippocrates found two other squarable concave lunes, with angles approximately (107.2°, 160.9°) with ratio 2:3 and (68.5°, 205.6°) with ratio 1:3. Two more squarable concave lunes, with angles approximately (46.9°, 234.4°) with ratio 1:5 and (100.8°, 168.0°) with ratio 3:5 were found in 1766 by {{Interlanguage link multi|Martin Johan Wallenius|ru|3=Валлениус, Мартин Йохан}} and again in 1840 by Thomas Clausen. In the mid-20th century, two Russian mathematicians, Nikolai Chebotaryov and his student Anatoly Dorodnov, completely classified the lunes that are constructible by compass and straightedge and that have equal area to a given square. As Chebotaryov and Dorodnov showed, these five pairs of angles give the only constructible squarable lunes; in particular, there are no other constructible squarable lunes.

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{{Ancient Greek mathematics}}

Category:Circles

Category:Squaring the circle

Category:Greek mathematics