Consider the hexagon of vertices $ A = (L,0)$, $ B = (\ell,h)$, $ C = ( - \ell,h)$, $ D = ( - L,0)$, $ E = ( - \ell, - h)$ and $ F = (\ell, - h)$, where $ 1 > L > \ell > 0$, $ 1 > h > 0$ and $ (L + \ell)^2 + h^2 = 1$. The hexagon is convex, and its diagonals $ AC = AE = BD = DF = 1$. The hexagon may be divided into two trapezii, with parallel sides of lengths $ 2L$ and $ 2\ell$ and altitude of length $ h$, or the area of the hexagon is $ 2(L + \ell)h$. Clearly, when $ L + \ell = h = \frac {1}{\sqrt {2}}$ (for example for $ L = \frac {2}{3\sqrt {2}}$ and $ \ell = \frac {1}{3\sqrt {2}}$), the area of the hexagon is exactly $ 1$, and this cannot be improved in this configuration because of the AM-GM inequality. Let now $ L = 2\ell$, $ \ell = \frac {\sqrt {1 - h^2}}{3}$. The area of the hexagon is then $ \sqrt {h^2(1 - h^2)}$, which for $ 0 < h < 1$ has a maximum at $ h = \frac {1}{\sqrt {2}}$, and varies continuously, with limits $ 0$ when $ h\to0$ and $ h\to1$. Therefore, all values of the area between $ 0$ (exclusive) and $ 1$ (inclusive) are obtained in this configuration. This solves part a), and seems to indicate that part b) should be "Show that the area of any $ \it{pretty}$ hexagon is at most $ 1$" instead of "less than $ 1$"...

In Part (b), we must prove that the area of any pretty hexagon is less than 3/2 instead of 1.