VII] OF THE BEE'S CELL 533 



trihedral angle at F, such that the surface of the whole figure may 

 be a minimum. 



Let the angle NVX, which is the inclination of the plane of the 

 rhonlbus to the axis of the prism, = d\ the side of the hexagon, as 

 AB, = s\ and the height, as Aa, = h. 



Then AC = 25 cos 30°, = s VS. And, from inspection of the 

 triangle LXB, 



VX 



sin 6 

 Therefore the area of the rhombus 



VAXC 



V3 



2sin^* 



And the area of AabX = -{2h— ^VX cos 6), 



2i " 



= ' (2A - \s cot Q). 



z 



Therefore the total area of the figure 



= the hexagonal base ahcdef + 3s (2h — |s cot 6) + 3 —-. — -, 



(^(area) Ss^ / 1 a/3 cos ^ 



Therefore — j^ — = -^ ~^~t^ . o n 



ad 2 Vsm^ d sm^ d. 



\ 



^ , . . . , d (area) ^ , ^1 



But this expression vanishes, or = 0, when cos U = -— =, 



^ dd V3 



that is to say, when 6 = 54° 44' 8" = } (109° 28' 16"). 



Such then are the conditions under which the total area of the 

 figure has its minimal value. 



The following is, in substance, Maclaurin's elementary but somewhat lengthy 

 proof of the minimal properties of the bee's cell, using "no higher Geometry 

 than was known to the Antients." 



Let ABCD, abed, represent one-half of a right prism on a regular hexagonal 

 base; and let AabE, EbcC be the trapezial portions of two adjacent sides, to 

 which one of the three rhombs, AECe, is fitted. 



Let be the centre of the hexagon, of which AB, EC are adjacent sides; 

 join ^C and OB, intersecting in P. Then, because AOC = ABC, and BE = Oe, 



