Chebyshev Magic

The fourth Chebyshev polynomial $T_4(x) = 8x^4 - 8x^2 + 1$ has the property that $T_4(\cos\theta) = \cos4\theta$ for any $\theta$ . But then $T_4\big(\sin\tfrac{2j\pi}{5}\big) \; = \; T_4\big(\cos(\tfrac12\pi - \tfrac{2j\pi}{5})\big) \; = \; \cos(2\pi - \tfrac{8j\pi}{5}) \; = \; \cos(\tfrac{2j\pi}{5})$ for $j=0,1,2,3,4$ , and so the curve $y = T_4(x)$ passes through $A,B,C,D,E$ . Also $T_4\big(\pm\cos\tfrac16\pi\big) \; =\; \cos\tfrac23\pi \; = \; -\sin\tfrac16\pi$ and hence the curve $y = T_4(x)$ also passes through $F,G$ where $\angle FOA = \angle GOA = \tfrac23\pi$ . If we let $T = \tfrac12\sin\tfrac25\pi$ , then $\begin{aligned} |ABCDE| & = \; 5T \\ |ABFCDGE| & = \; 3T + 2\left(\tfrac12\sin\tfrac{4}{15}\pi + \tfrac12\sin\tfrac{2}{15}\pi\right) \; =\; 3T + 2\sin\tfrac15\pi \cos\tfrac{1}{15}\pi \\ & =\; 3T + 2T\frac{\cos\frac{1}{15}\pi}{\cos\frac15\pi} \\ \frac{|ABFCDGE|}{|ABCDE|} & = \; \frac{3\cos\frac15\pi + 2\cos\frac{1}{15}\pi}{5\cos\frac15\pi} \end{aligned}$

Now $\begin{aligned} \cos\tfrac15\pi & = \; \tfrac14\big(\sqrt{5} + 1\big) \\ \cos\big(\tfrac{1}{15}\pi\big) & = \; \cos\big(\tfrac25\pi - \tfrac13\pi\big) \; =\; \tfrac12\cos\tfrac25\pi + \tfrac12\sqrt{3}\sin\tfrac25\pi \\ & = \; \tfrac18\left(\sqrt{5}-1 + \sqrt{3}\sqrt{10 + 2\sqrt{5}}\right) \; =\; \tfrac18\left(\sqrt{5}-1 + \sqrt{30 + 6\sqrt{5}}\right) \\ 3\cos\tfrac15\pi + 2\cos\tfrac{1}{15}\pi & = \; \tfrac14\left(4\sqrt{5} + 2 + \sqrt{30 + 6\sqrt{5}}\right) \\ \frac{|ABFCDGE|}{|ABCDE|} & = \; \frac{4\sqrt{5} + 2 + \sqrt{30 + 6\sqrt{5}}}{5(\sqrt{5}+1)} \; = \; \frac{9-\sqrt{5} + \sqrt{30-6\sqrt{5}}}{10} \end{aligned}$ which is a much more elegant expression than its inverse, which is $\frac{|ABCDE|}{|ABFCDGE|} \; = \; \frac{5}{302}\left(111 + 13\sqrt{5} - \sqrt{6(785 + 179\sqrt{5})}\right)$ making the answer $5 + 302 + 111 + 13 + 5 + 6 + 785 + 179 + 5 = \boxed{1411}$ .

By the properties of a regular pentagon inscribed in a unit circle, the coordinates of some of its vertices are $A(0, 1)$ , $B(\cos \frac{\pi}{10}, \sin \frac{\pi}{10}) = B(\sqrt{\frac{5}{8} + \frac{\sqrt{5}}{8}}, -\frac{1}{4} + \frac{\sqrt{5}}{4})$ , and $C(\cos \frac{3\pi}{10}, -\sin \frac{3\pi}{10}) = C(\sqrt{\frac{5}{8} - \frac{\sqrt{5}}{8}}, -\frac{1}{4} - \frac{\sqrt{5}}{4})$ , and its area is $A_{ABCDE} = 5 \cdot \frac{1}{2} \cdot 1^2 \cdot \sin \frac{2\pi}{5} = \frac{5}{2}\sqrt{\frac{5}{8} + \frac{\sqrt{5}}{8}}$ .

Let $\pm a$ and $\pm b$ be the roots of the quartic polynomial, so that $y = c(x^2 - a^2)(x^2 - b^2)$ . Since $A(0, 1)$ , $B(\sqrt{\frac{5}{8} + \frac{\sqrt{5}}{8}}, -\frac{1}{4} + \frac{\sqrt{5}}{4})$ , and $C(\sqrt{\frac{5}{8} - \frac{\sqrt{5}}{8}}, -\frac{1}{4} - \frac{\sqrt{5}}{4})$ are on this polynomial:

$1 = c(0^2 - a^2)(0^2 - b^2)$

$-\frac{1}{4} + \frac{\sqrt{5}}{4} = c\bigg(\bigg(\sqrt{\frac{5}{8} + \frac{\sqrt{5}}{8}}\bigg)^2 - a^2\bigg)\bigg(\bigg(\sqrt{\frac{5}{8} + \frac{\sqrt{5}}{8}}\bigg)^2 - b^2\bigg)$

$-\frac{1}{4} - \frac{\sqrt{5}}{4} = c\bigg(\bigg(\sqrt{\frac{5}{8} - \frac{\sqrt{5}}{8}}\bigg)^2 - a^2\bigg)\bigg(\bigg(\sqrt{\frac{5}{8} - \frac{\sqrt{5}}{8}}\bigg)^2 - b^2\bigg)$

And these equations solve to $a = \frac{\sqrt{2 - \sqrt{2}}}{2}$ , $b = \frac{\sqrt{2 + \sqrt{2}}}{2}$ , and $c = 8$ , so $y = 8(x^2 - (\frac{\sqrt{2 - \sqrt{2}}}{2})^2)(x^2 - (\frac{\sqrt{2 + \sqrt{2}}}{2})^2)$ , which simplifies to $y = 8x^4 - 8x^2 + 1$ .

Since $F$ is on $y = 8x^4 - 8x^2 + 1$ and the unit circle $x^2 + y^2 = 1$ , it has coordinates $F(\frac{\sqrt{3}}{2}, -\frac{1}{2})$ .

The area of $\triangle BCF$ is then $A_{\triangle BCF} = \frac{1}{2}|(B_x - F_x)(C_y - F_y) - (C_x - F_x)(B_y - F_y)| = \frac{1}{8}(\sqrt{15} - \sqrt{5 + 2\sqrt{5}})$

That means the area ratio of $ABCDE$ to $ABFCDGE$ is $\cfrac{A_{ABCDE}}{A_{ABCDE} + 2A_{\triangle BCF}} = \cfrac{\frac{5}{2}\sqrt{\frac{5}{8} + \frac{\sqrt{5}}{8}}}{\frac{5}{2}\sqrt{\frac{5}{8} + \frac{\sqrt{5}}{8}} + 2(\frac{1}{8}(\sqrt{15} - \sqrt{5 + 2\sqrt{5}}))}$ , which simplifies to:

$\frac{5}{302}\bigg(111 + 13\sqrt{5} - \sqrt{6(785 + 179\sqrt{5})}\bigg)$

Therefore, $a = 5$ , $b = 302$ , $c = 111$ , $d = 13$ , $e = 5$ , $f = 6$ , $g = 785$ , $h = 179$ , $i = 5$ , and $a + b + c + d + e + f + g + h + i = \boxed{1411}$ .