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Problem 841

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Problem 841

Regular Star Polygons

The regular star polygon $\{p/q\}$, for coprime integers $p,q$ with $p > 2q > 0$, is a polygon formed from $p$ edges of equal length and equal internal angles, such that tracing the complete polygon wraps $q$ times around the centre. For example, $\{8/3\}$ is illustrated below:

{8/3}

The edges of a regular star polygon intersect one another, dividing the interior into several regions. Define the alternating shading of a regular star polygon to be a selection of such regions to shade, such that every piece of every edge has a shaded region on one side and an unshaded region on the other, with the exterior of the polygon unshaded. For example, the above image shows the alternating shading (in green) of $\{8/3\}$.

Let $A(p, q)$ be the area of the alternating shading of $\{p/q\}$, assuming that its inradius is $1$. (The inradius of a regular polygon, star or otherwise, is the distance from its centre to the midpoint of any of its edges.) For example, in the diagram above, it can be shown that central shaded octagon has area $8(\sqrt{2}-1)$ and each point’s shaded kite has area $2(\sqrt{2}-1)$, giving $A(8,3) = 24(\sqrt{2}-1) \approx 9.9411254970$.

You are also given that $A(130021, 50008)\approx 10.9210371479$, rounded to $10$ digits after the decimal point.

Find $\sum_{n=3}^{34} A(F_{n+1},F_{n-1})$, where $F_j$ is the Fibonacci sequence with $F_1=F_2=1$ (so $A(F_{5+1},F_{5-1}) = A(8,3)$). Give your answer rounded to $10$ digits after the decimal point.

星形正多边形

对于互质且满足$p>2q>0$的整数对$p,q$，可以绘制星形正多边形$\{p/q\}$。这一多边形有$p$条等长的边和$p$个相等的内角，且沿多边形移动时环绕其中心恰好$q$圈。例如，星形正多边形$\{8/3\}$如下图所示：

{8/3}

星形正多边形的边可以彼此相交，将其内部分成多个区域。对星形正多边形进行交替上色，是指对其中一部分区域进行上色，使得每一条边的每一截都恰好只有一侧被上色而另一侧未上色（多边形外部不上色）。例如，上图展示的是对星形正多边形$\{8/3\}$进行交替上色（绿色部分）的结果。

对内半径为$1$的星形正多边形$\{p/q\}$进行交替上色后，记上色区域的面积为$A(p, q)$。（任意正多边形或星形正多边形的内半径是指其中心和任意一条边的中点的距离。）例如，在上图中，正中位置被上色的八边形面积为$8(\sqrt{2}-1)$，每个顶点处被上色的筝形面积为$2(\sqrt{2}-1)$，因此$A(8,3) = 24(\sqrt{2}-1) \approx 9.9411254970$。

已知$A(130021, 50008)\approx 10.9210371479$，保留$10$位小数。

求$\sum_{n=3}^{34} A(F_{n+1},F_{n-1})$，其中$F_j$是由$F_1=F_2=1$开始的斐波那契数列的第$j$项（例如，$A(F_{5+1},F_{5-1}) = A(8,3)$），并将你的答案保留$10$位小数。