Seven in Four

Suppose that $p = \sin \tfrac{1}{14}\pi = \cos\tfrac{3}{7}\pi$ . If we put $z = e^{\frac{3\pi i}{7}}$ , then $z + z^{-1} = 2p$ and $\begin{aligned} 0 \; = \; z^7 + 1 & = \; (z + 1)(z^6 - z^5 + z^4 - z^3 + z^2 - z + 1) \; = \; (z+1)z^3\big[z^3 + z^{-3} - z^2 - z^{-2} + z + z^{-1} - 1\big] \\ & = \; (z + 1)z^3\big[(z + z^{-1})^3 - (z + z^{-1})^2 - 2(z + z^{-1}) - 1\big] \; =\; (z + 1)z^3(8p^3 - 4p^2 - 4p + 1) \end{aligned}$ so that $8p^3 - 4p^2 - 4p + 1 = 0$ . If we now put $t = \tfrac{1}{28}\pi$ , so that $p = \sin2t$ , then $\begin{aligned} 4\sin t(\cos7t + \cos t + \sin 5t) & = \; 4\sin t \cos7t + 4\sin t \sin5t + \sin2t \; =\; 2(\sin8t - \sin6t) + 2(\cos4t - \cos6t) + 2\sin2t \\ & = \; 2(-\sin6t + \cos4t + \sin2t) \; =\; 2(\sin10t - \sin6t + \sin2t) \\ & = \; 2\big[(16p^5 - 20p^3 + 5p) - (3p - 4p^3) + p\big] \; =\; 32p^5 - 32p^3 + 6p \\ & = \; (8p^3 - 4p^2 - 4p + 1)(4p^2 + 2p - 1) + 1 \; = \; 1 \end{aligned}$ using the fact that $14t = \tfrac12\pi$ , so that $\sin8t = \cos6t$ and $\cos4t = \sin10t$ , and also using the identities $\sin3u \equiv 3\sin u - 4\sin^3 u$ and $\sin5u \equiv 16\sin^5 u - 20\sin^3 u + 5\sin u$ .

If we consider the desired heptagon, it is clear that it is situated symmetrically with respect to the top-left/bottom-right diagonal, and so key angles are of size $5t$ , $t$ and $7t = \tfrac14\pi$ . Thus the side $x$ of the heptagon is such that $1 \; = \; x(\cos7t + \cos t + \sin5t)$ and hence we deduce that $x = 4\sin t$ . Thus the area of the heptagon is $7 \times \tfrac12\times x \times \tfrac12x\cot\tfrac17\pi \; =\; \tfrac74x^2\cot\tfrac17\pi \; = \;28\sin^2t\cot\tfrac{1}{7}\pi \; = \; 28\sin^2\tfrac{1}{28}\pi \cot\tfrac17\pi$ so that $a=28$ , $b=7$ , making the answer $\boxed{35}$ .