On Growth and Form

by D'Arcy Wentworth Thompson

And VX = a ⁄ sin θ (from inspection of the triangle LXB)

Therefore the area of the rhombus VAXC = (a² √3) ⁄ (2 sin θ).

And the area of AabX = (a ⁄ 2)(2h − ½VX cos θ)

= (a ⁄ 2)(2h − a ⁄ 2 · cot θ).

Therefore the total area of the figure

= hexagon abcdef + 3a(2h − (a ⁄ 2) cot θ) + (3a² √3) ⁄ (2 sin θ).

Therefore

d(Area) ⁄ dθ = (3a² ⁄ 2)((1 ⁄ sin² θ) − (√3 cos θ) ⁄ (sin² θ)).

But this expression vanishes, that is to say, d(Area) ⁄ dθ = 0, when cos θ = 1 ⁄ √3, that is when θ = 54° 44′ 8″

= ½(109° 28′ 16″).