Find the plane lamina of least area 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 (giving a diameter of 1) has side length
(1)
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and the area of this hexagon is
(2)
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(OEIS A010527).
In the above figure, the sagitta is given by
(3)
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(4)
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and the other distances by
(5)
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(6)
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so the area of one of the equilateral triangles removed in Pál's reduction is
(7)
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(8)
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(9)
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(10)
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so the area left after removing two of these triangles is
(11)
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(12)
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(13)
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(OEIS A093821).
Computing the area of the region removed in Sprague's construction is more involved. First, use similar triangles
(14)
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together with to obtain
(15)
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Then
(16)
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and the angle is given by
(17)
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and the angle is just
(18)
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The distance is
(19)
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(20)
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and the area between the triangle and sector is
(21)
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(22)
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(23)
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(24)
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The area of the small triangle is
(25)
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(26)
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(27)
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so the total area remaining is
(28)
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(29)
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(30)
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(OEIS A093822).
It is also known that a lower bound for the area is given by
(31)
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(Ogilvy 1990).