Crazy New Year Firework

Notice the relationship between distance traveled and change in angle. The point will end up landing on the circumference from initial point $P$ to final point $P'$ .

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Step I. Analyzing Pattern

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Before approaching the problem, consider the symmetry equation $y = \tan\left(\left(2^n\right)^{\circ}\right) x$ . Let $P_n$ denote the initial angle with respect to the horizontal axis and $D_n$ denote the reflection by the line of the equation. For each time the point is reflected, it arrives toward the point on the circumference, which does not change the displacement between the origin and the final point. Since the firework travels twice the shortest distance $d$ between the initial point and the line of symmetry, the change in angle is $2D_{n+1}$ . In that case, the resulting angle is the difference between $P_n$ and $2D_{n + 1}$ , which is presented as the following harmonic sequence $P_{n + 1} = -P_n + 2D_{n + 1}$ for $n \geq 0$ . Because the line of symmetry depends on the slope $\tan\left(\left(2^n\right)^{\circ}\right)$ , $D_n = 2^{n}$ , which implies that $\begin{array}{rl} P_{n + 1} &= -P_n + 2 \cdot 2^{n + 1}\\ &= -P_n + 2^{n + 2} \end{array}$ Since the firework exploded after odd number of steps, the second step is to determine $P_{n + (2k + 1)}$ , where we treat $n = 0$ and $2k + 1$ to be the odd number for positive integer $k$ . Evaluating $P_{n + (2k + 1)}$ , we have $\begin{array}{rl} P_{n + (2k + 1)} &= -P_{n + 2k} + 2D_{n + (2k + 1)}\\ &= -\left(P_n - 2^{n + 2} + 2^{n + 3} - \dots + 2^{n + 2k + 1}\right) + 2^{n + (2k + 1) + 1}\\ &= -P_n + 2^{n + 2} - 2^{n + 3} + \dots - 2^{n + 2k + 1} + 2^{n + 2k + 2} \end{array}$ Setting $n = 0$ and $2k + 1 = 365$ (which implies that $k = 182$ ), $\begin{array}{rl} P_{365} &= -P_0 + 2^2 - 2^{3} + 2^4 - 2^5 + \dots - 2^{365} + 2^{366}\\ &= -P_0 + \left(2^2 + 2^4 + \dots + 2^{364} + 2^{366}\right) - \left(2^3 + 2^5 + \dots + 2^{363} + 2^{365}\right)\\ &= -P_0 + 4\left(1 + 2^2 + 2^4 + \cdots + 2^{362}\right) - 8\left(1 + 2^2 + 2^4 + \cdots + 2^{360} + 2^{362}\right) + 2^{366}\\ &= -P_0 - 4\left(1 + 2^2 + 2^4 + \cdots + 2^{360} + 2^{362}\right) + 2^{366} \end{array}$

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Step II. Modular Arithmetic

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Considering the bounds of angle measurement (between $0^{\circ}$ and $360^{\circ}$ ), we evaluate the numerical expression in $\bmod \, 360$ . That is, we want to simplify $P = \left(-4\left(1 + 2^2 + 2^4 + \cdots + 2^{360} + 2^{362}\right) + 2^{366}\right) \bmod 360$ Since $360 = 2^3 \cdot 3^2 \cdot 5$ , evaluate $P$ in $\bmod\,2^3$ and $\bmod\,3^2\cdot 5$ separately. Clearly, $\begin{array}{rl} P &\equiv -4\left(1 + 4 + 0 + 0 + \cdots + 0\right) + 0 \bmod\,2^3\\ &\equiv -20 \equiv 4 \bmod\,2^3 \end{array}$ For $\bmod\,3^2 \cdot 5$ , since $a^{\text{lcd}(\varphi(9),\varphi(5))} = a^{12} \equiv 1 \bmod\,3^2\cdot 5$ where $a$ is a prime factor relatively prime to $3^2 \cdot 5$ , notice that for nonnegative integer $k$ $\begin{array}{rl} 2^{12k} + 2^{12k + 2} + 2^{12k + 4} + 2^{12k + 6} + 2^{12k + 8} + 2^{12k + 10} &\equiv 2^{12k}\left(1 + 2^2 + 2^4 + \cdots + 2^{10}\right) \bmod\,3^2\cdot 5\\ &\equiv 1 + 2^2 + 2^4 + \cdots + 2^{10} \bmod\,3^2\cdot 5\\ &\equiv 1 + 4 + 4^2 + 4^3 + 4^4 + 4^5 \bmod\,3^2\cdot 5\\ &\equiv \left(1 + 4 + 4^2\right) + 4^3\left(1 + 4 + 4^2\right) \bmod\,3^2\cdot 5\\ &\equiv \left(1 + 4^3\right)\left(1 + 4 + 4^2\right)\bmod\,3^2\cdot 5\\ &\equiv 65 \cdot 21 \bmod\,3^2\cdot 5\\ &\equiv 15 \cdot 91 \bmod\,3^2\cdot 5\\ &\equiv 15 \bmod\,3^2\cdot 5 \end{array}$ Since there are $6$ consecutive terms of even exponents in the sum, there are $\left\lfloor \dfrac{362 + 2}{12}\right\rfloor = 30$ sums of the same form and $\left\{ \dfrac{362 + 2}{12}\right\} \cdot 6 = 2$ extra terms left, namely $2^{360} \equiv 1 \bmod\,3^2\cdot 5$ and $2^{362} \equiv 2^2 \bmod\,3^2\cdot 5$ , $\begin{array}{rl} -4\left(1 + 2^2 + \cdots + 2^{360} + 2^{362}\right) + 2^{366} &\equiv -4\left(30\left(1 + 2^2 + 2^4 + 2^6 + 2^8 + 2^{10}\right) + 1 + 2^2\right) + 2^{6}\bmod\,3^2\cdot 5\\ &\equiv -4\left(30(15) + 5\right) + 19\bmod\,3^2\cdot 5\\ &\equiv 25 + 19 \bmod\,3^2\cdot 5\\ &\equiv 44 \bmod\,3^2\cdot 5 \end{array}$ So we have $\begin{array}{rl} P &\equiv 4 \bmod\,2^3\\ P &\equiv 44 \bmod\,3^2\cdot 5 \end{array}$ which is easy to see that $P \equiv 44 \bmod 360$ . This shows that we evaluate $P_{365} = -P_0 + 44$

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Step III: Answer the Question

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Because the given point is shifted by $2016$ units down, we have $(2017,-2015)$ . In polar form, $r = \sqrt{(2017)^2 + (-2015)^2} \quad \text{and}\quad \theta = \arctan\left( -\dfrac{2015}{2017}\right)^{\circ}$ Then, $P_{365} = \left(-\arctan\left( -\dfrac{2015}{2017}\right) + 44\right)^{\circ} = \left(\arctan\left(\dfrac{2015}{2017}\right) + 44\right)^{\circ}$ Thus, if $\begin{array}{rl} x_a &= r\cos\left(P_{365}\right)\\ y_a &= r\sin\left(P_{365}\right) + 2016 \end{array}$ the answer is $\left\lfloor x_a + y_a \right\rfloor = \boxed{4917}$

Note: You can simply substitute the whole angle displacement into the expression without having to apply modular arithmetic. This still gives the same result.

Happy New Years

Edit: The whole thing can be reduced to

$x_{365}=rCos(\theta_{365})$
$y_{365}=rSin(\theta_{365})+2016$

where $r=\sqrt{{(2017)}^{2}+{(2016-1)}^{2}}$

and $n=365$ for $\theta_{365}$ , where

$\theta_n=\dfrac{2\pi}{360} \left( \dfrac{4}{3}({2}^{n} - {(-1)}^{n})+ {(-1)}^{n} \dfrac {360}{2 \pi} ArcTan( \dfrac {2016-1}{0-2017} ) \right)$