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Lebesgue Minimal Problem

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LebesguesMinimal

Find the plane lamina of least area A which is capable of covering any plane figure of unit generalized diameter. A unit circle is too small, but a hexagon circumscribed on the unit circle is larger than necessary. Pál (1920) showed that the hexagon can be reduced by cutting off two isosceles triangles on the corners of the hexagon which are tangent to the hexagon's incircle (Wells 1991; left figure above). Sprague subsequently demonstrated that an additional small curvilinear region could be removed (Wells 1991; right figure above). These constructions give upper bounds.

The hexagon having inradius r=1/2 (giving a diameter of 1) has side length

a=2rtan(pi/n)=1/3sqrt(3),
(1)

and the area of this hexagon is

A_1=nr^2tan(pi/n)=1/2sqrt(3)=0.866025...
(2)

(OEIS A010527).

LebesguesMinimalTriangle

In the above figure, the sagitta is given by

s=rtan(pi/n)tan(pi/(2n))
(3)
=1/6(2sqrt(3)-3),
(4)

and the other distances by

b=stan(pi/3)=sqrt(3)s
(5)
h=sqrt(s^2+b^2)=2s,
(6)

so the area of one of the equilateral triangles removed in Pál's reduction is

A_Delta=bs
(7)
=sqrt(3)s^2
(8)
=1/(12)(7sqrt(3)-12)
(9)
approx 0.010363,
(10)

so the area left after removing two of these triangles is

A_2=A_1-2A_Delta
(11)
=2/3(3-sqrt(3))
(12)
=0.845299...
(13)

(OEIS A093821).

Computing the area of the region removed in Sprague's construction is more involved. First, use similar triangles

(a-h)/h=(r_2)/(r_1)
(14)

together with r_1+r_2=r to obtain

r_2=(2r(a-h))/a=sqrt(3)-1.
(15)

Then

x=r_2cos(pi/3)=1/2(sqrt(3)-1),
(16)

and the angle theta is given by

theta=cos^(-1)(x/(2r))=cos^(-1)[1/2(sqrt(3)-1)],
(17)

and the angle phi is just

phi=theta-1/3pi.
(18)

The distance h^' is

h^'=2rtanphi
(19)
l=2rsecphi,
(20)

and the area between the triangle and sector is

dA_3^((1))=rh-1/2(2r)^2phi
(21)
=2r^2(tanphi-phi)
(22)
=1/2(tanphi-phi)
(23)
approx 0.000554738.
(24)

The area of the small triangle is

dA_3^((2))=1/2(l-2r)(h-h^')
(25)
=1/6(secphi-1)(2sqrt(3)-3-3tanphi)
(26)
approx 0.0000264307,
(27)

so the total area remaining is

A_3=A_2-2(dA_3^((1))-dA_3^((2)))
(28)
=-(109)/(121)-(82)/(121sqrt(3))+2/(121)sqrt(28634sqrt(3)-35139)-1/3pi+cos^(-1)[1/2(sqrt(3)-1)]
(29)
=0.844137...
(30)

(OEIS A093822).

It is also known that a lower bound for the area is given by

A>1/8pi+1/4sqrt(3) approx 0.825712
(31)

(Ogilvy 1990).

See also

Area, Borsuk's Conjecture, Generalized Diameter, Kakeya Needle Problem

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References

Ball, W. W. R. and Coxeter, H. S. M. Mathematical Recreations and Essays, 13th ed. New York: Dover, p. 99, 1987.Coxeter, H. S. M. "Lebesgue's Minimal Problem." Eureka 21, 13, 1958.Grünbaum, B. "Borsuk's Problem and Related Questions." Proc. Sympos. Pure Math, Vol. 7. Providence, RI: Amer. Math. Soc., pp. 271-284, 1963.Kakeya, S. "Some Problems on Maxima and Minima Regarding Ovals." Sci. Reports Tôhoku Imperial Univ., Ser. 1 (Math., Phys., Chem.) 6, 71-88, 1917.Ogilvy, C. S. Tomorrow's Math: Unsolved Problems for the Amateur, 2nd ed. New York: Oxford University Press, 1972.Ogilvy, C. S. Excursions in Geometry. New York: Dover, pp. 142-144, 1990.Pál, J. "Ueber ein elementares Variationsproblem." Det Kgl. Danske videnkabernes selskab, Math.-fys. meddelelser 3, Nr. 2, 1-35, 1920.Sloane, N. J. A. Sequences A010527, A093821, and A093822 in "The On-Line Encyclopedia of Integer Sequences."Wells, D. The Penguin Dictionary of Curious and Interesting Geometry. London: Penguin, p. 138, 1991.Yaglom, I. M. and Boltyanskii, V. G. Convex Figures. New York: Holt, Rinehart, & Winston, pp. 18 and 100, 1961.

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Lebesgue Minimal Problem

Cite this as:

Weisstein, Eric W. "Lebesgue Minimal Problem." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/LebesgueMinimalProblem.html

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