2018 again?

Geometry Level 3

Let $C$ be a circumference of radius $2$ , and $B$ a regular polygon of 2018 sides inscribed in $C$ . Consider $A$ be a fixed vertex of $B$ , then the product of the distances from $A$ to the remaining vertices of $B$ can be written as $a \cdot 2^b$ , with $a, b$ positive integers and $\text{gcd (a, 2) = 1}$ .

Enter $a + b$

Inspiration

The answer is 3027.

1 solution

Guillermo Templado
Feb 11, 2018

Take $\displaystyle p(x) = x^{2018} - 2^{2018} = (x - 2)(x^{2017} + 2x^{2016} + 2^2 x^{2015} + ... + 2^{2016} x + 2^{2017}) = (x - 2) \cdot \prod_{k = 1}^{2017} (x - 2e^{i \frac{2k \pi}{2018}})$ . Let's call $A = 2$ then the product is $\displaystyle \prod_{k = 1}^{2017} |(2 - 2e^{i \frac{2k \pi}{2018}})| = |\prod_{k = 1}^{2017} (2 - 2e^{i \frac{2k \pi}{2018}})| =$ . $= |2^{2017} + 2 \cdot 2^{2016} + 2^2 \cdot 2^{2015} + ... + 2^{2016} \cdot 2 + 2^{2017}| = 2018 \cdot 2^{2017} = 1009 \cdot 2^{2018}$

Therefore, $a + b = 1009 + 2018 = 3027$ .

2018 again?

The answer is 3027.

1 solution

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