Lebesgue Minimal Problem

The Lebesgue minimal problem asks one to find the convex plane lamina of least area which contains an isometric copy of every plane set of unit generalized diameter (Baez et al. 2015). A disk of diameter 1 is too small, but a regular hexagon circumscribed about such a disk 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 (1936, cited in Baez et al. 2015) demonstrated that an additional small curvilinear region could be removed (Wells 1991; right figure above). These constructions give upper bounds.

The regular hexagon with sides and inradius has an incircle of diameter 1 and side length

(1)

and the area of this hexagon is

(2)

(OEIS A010527).

In the above figure, the sagitta is given by

			(3)
			(4)

and the other distances by

			(5)
			(6)

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

			(7)
			(8)
			(9)
			(10)

so the area left after removing two of these triangles is

			(11)
			(12)
			(13)

(OEIS A093821).

Sprague's construction removes the portions near a third corner that lie outside two unit-radius disks centered at the appropriate vertices of Pál's reduced hexagon (Baez et al. 2015, §2). Place the original hexagon's center at the origin and this corner at . The two disk centers are , where . The removed area is symmetric about the vertical axis. On its right half, the upper boundary is the side and the lower boundary is the circular arc . They are tangent at , giving

(14)

The remaining area is therefore

			(15)
			(16)
			(17)

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

(18)

(Ogilvy 1990).