Four villages (hamlets, airport terminals, warehouses, or whatever), each being a vertex of a square, should be connected by a road network so that the total length of the road is minimal. If each side of the square is 2 units long, what is the total length of the road?

Source: Miodrag S. Petkovic, Famous Puzzles of Great Mathematicians, American Mathematical Society, 2009, Problem 4.25.

Level: Advanced

By symmetry, we can design the road network to be composed of line segments AE, BE, EF, CF, and DF, where AE = BE = CF = DF = x and OE = OF = y as in the figure, with the length L = AE+BE+CF+DF+EF = 4x+2y is as small as possible. Since x = 1/cos(theta) and z = tan(theta), so
L(theta)
= 4x+2*(1-z)
= 4*sec(theta)+2*[1-tan(theta)].
To find the minimum value of L, we solve for L'(theta) = 0:
4*sec(theta)*tan(theta)-2*sec^2(theta) = 0,
2*sec(theta)*[2*tan(theta)-sec(theta)] = 0.
sec(theta) = 0 or 1/2 = tan(theta)/sec(theta) = sin(theta),
theta = pi/6.
Next L''(theta) = 4*[sec(theta)*tan^2(theta)+sec^3(theta)]-4*sec^(theta)*tan(theta). Therefore L''(pi/6) = 4*[2/sqrt(3)*(1/3)+(8/3)/sqrt(3)]-4*(4/3)/sqrt(3)] = 8/sqrt(3) > 0, and so L is minimum when theta = pi/6. It follows that the total length of the road has minimum value L(pi/6) = 4*sec(pi/6)-2*[1-tan(pi/6)] = 2+2*sqrt(3) = 5.4641 units.