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.
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
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.
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