Petite Piece of Pie

Firstly, we need to find a formula for the length of the sector. We can observe that the length from the centre of the pie to the outer boundary of Area $k$ , $r_k\ = 2 - \dfrac{1}{2^{k-1}}$ since the total radius has length 2.

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Now, Area $n$ is equal to the total slice up to the outer boundary of Area $n$ minus the total slice up to the outer boundary of Area $n - 1$ . Therefore, $\text{Area } n = \frac18 \times \pi \times r_n^2 - \frac18 \times \pi \times r_{n-1}^2$ $\text{Area } n = \frac{\pi}{8} \times (r_n^2 - r_{n-1}^2)$ $\text{Area } n = \frac{\pi}{8} \times \left[ \left( 2 - \dfrac{1}{2^{(n)-1}} \right)^2 - \left( 2 - \dfrac{1}{2^{(n-1)-1}} \right)^2 \right]$ $\text{Area } n = \frac{\pi}{8} \times \left[ \left( 2 - 2^{1-n} \right)^2 - \left( 2 - 2^{2-n} \right)^2 \right]$ Expanding and simplifying, we get that, $\text{Area } n = \frac{\pi}{8} \times \left[ 2^{2-2n} - 2^{4-2n} + 2^{4-n} - 2^{3-n} \right]$ $\text{Area } n = \frac{\pi}{8} \times \left[ 2^{2-2n}(1 - 2^2) + 2^{3-n}(2-1) \right]$ $\text{Area } n = \frac{\pi}{8} \times \left[ \dfrac{-3}{2^{2n-2}} + \dfrac{1}{2^{n-3}} \right]$ $\text{Area } n = \frac{\pi}{8} \times \left[ \dfrac{-3}{2^{2n-2}} + \dfrac{2^{n+1}}{2^{2n-2}} \right]$ $\text{Area } n = \frac{\pi}{8} \times \left[ \dfrac{2^{n+1} - 3}{2^{2n-2}} \right]$ $\text{Area } n = \left[ \dfrac{2^{n+1} - 3}{2^{2n+1}} \right] \times \pi$ Hence, $\text{Area } 2018 = \left[ \dfrac{2^{2019} - 3}{2^{4037}} \right] \times \pi$ This makes $a + b + c = 2019 - 3 + 4037 = \boxed{6053}$

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Note: This value for $a$ , $b$ and $c$ is unique. Since $b$ is an odd integer, this means that the value of $c$ must be $4037$ . Also, if $a$ is either $2018$ or $2020$ (or any other numbers except $2019$ ), $|b| >> 2018$

Let the " distance " from the center of the pie eaten by the first $n$ th persons be $R_n$ . Then:

$\begin{aligned} R_n & = \sum_{k=1}^n \frac 1{2^{k-1}} = \sum_k^{n-1} \frac 1{2^k} = \frac {1-\frac 1{2^n}}{1-\frac 12} = \frac {2^n-1}{2^{n-1}} \end{aligned}$

Then the area of pie eaten by the first $n$ th persons is $A_n = \dfrac {\pi R_n^2}8$ and the area eaten by $n$ th person is:

$\begin{aligned} a_n & = A_n - A_{n-1} \\ & = \frac \pi 8 \left(R_n^2 - R_{n-1}^2\right) \\ & = \frac \pi 8 \left(R_n + R_{n-1}\right) \left(R_n - R_{n-1}\right) \\ & = \frac \pi 8 \left(\frac {2^n-1}{2^{n-1}}+\frac {2^{n-1}-1}{2^{n-2}}\right) \left(\frac {2^n-1}{2^{n-1}}-\frac {2^{n-1}-1}{2^{n-2}}\right) \\ & = \frac \pi 8 \left(\frac {2^{n+1}-3}{2^{n-1}}\right) \left(\frac 1{2^{n-1}}\right) \\ & = \left(\frac {2^{n+1}-3}{2^{2n+1}} \right)\pi \end{aligned}$

Therefore, for $n=2018$ , $a+b+c = (2018+1) + (-3)+(2\times 2018+1) = \boxed{6053}$ .

Nice problem