Ratio Carnival

Let $n \geq 4$ be an even integer.

$\dfrac{x}{2} = r\sin(\dfrac{\pi}{n})$ and $h = r\cos(\dfrac{\pi}{n})$

$\implies A_{n} = n\dfrac{1}{2}xh = \dfrac{n}{2}\sin(\dfrac{2\pi}{n})r^2$

$M_{1}$ and $M_{2}$ are midpoints of $PQ$ and $RS \implies M_{1}Q = RM_{2} = \dfrac{x}{2}$ and the height of the parallelogram is $M_{1}M_{2} = 2h \implies A_{p} = \dfrac{x}{2}(2h) = xh = \sin(\dfrac{2\pi}{n})r^2$

$\implies A_{n} - A_{p} = \dfrac{n - 2}{2}\sin(\dfrac{2\pi}{n})r^2 \implies$ $\dfrac{A_{n}}{A_{n} - A_{p}} = \dfrac{n}{n - 2} = \dfrac{50}{49} \implies 50n - 100 = 49n \implies n = \boxed{100}$ .

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Note: $A_{\triangle{AOB}} + A_{\triangle{BOC}} = \sin(\dfrac{2\pi}{n})r^2 = A_{p}$

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Note: You obtain the same result $\dfrac{A_{n}}{A_{n} - A_{p}} = \dfrac{n}{n - 2}$ for $n$ odd using the diagram below:

I didn't use $n$ odd since the problem would have been a giveaway.

We note that a regular $n$ -gon is formed by $n$ congruent central isosceles triangle (the two purple triangle on the right figure). Let the area of one of these central isosceles triangle be $A_\triangle$ . Then $A_n = n A_\triangle$ . Now we note that the central rectangle $ABCD$ has an area of $4A_\triangle$ and the area of the purple parallelogram (right figure) is $A_p = 2A_\triangle$ .

Then we have:

$\begin{aligned} \frac {A_n}{A_n-A_p} & = \frac {nA_\triangle}{nA_\triangle - 2A_\triangle} & \small \color{#3D99F6} \text{For an even }n = 2m \\ & = \frac {2mA_\triangle}{2mA_\triangle - 2A_\triangle} \\ & = \frac m{m-1} = \frac {50}{49} \end{aligned}$

Implying that $m=50$ and $n = \boxed{100}$ .