Math in Pokemon: Gotta Catch 'Em All!

The amount of distance walked can be represented as the absolute value of the alternating harmonic series $\displaystyle \big| \sum_{n=1}^{\infty}\frac{(-1)^n2\pi}{n}\big| = 2\pi\ln{2}$

We find the angle $\theta$ by $\displaystyle\frac{\text{arc length}}{\text{circumference}}\times 2\pi = \frac{2\pi\ln{2}}{3\pi}$

Now using the Law of Cosines the green distance $\displaystyle D = \sqrt{2*3^2-2*3^2\cos{(\frac{2\pi\ln{2}}{3})}}$

$\Huge\boxed{25}$

Calculus

In polar coordinates, $x^2 + y^2 = 9 \quad \Longleftrightarrow \quad r = 3$ Define a "step" as one counterclockwise-clockwise travel successively. It is clear that

• Counter-clockwise steps increase the arc length between the given and arriving points.
• Clockwise steps decrease the arc length between the given and arriving points.

Let $\theta_d$ denote the final point (relative to the origin) after infinite steps. Since Ash starts at the point $(3,0)$ , it is clear that $\theta = 0$ (also relative to the origin). By the polar-coordinate formula of the arc length , the net arc distance traveled is $\begin{array}{rl} \int\limits_0^{\theta_d} \sqrt{r^2 + \left(\dfrac{dr}{d\theta}\right)^2}\,d\theta &= \int\limits_0^{\theta_d} \sqrt{3^2 + 0^2}\,d\theta\\ &= \int\limits_0^{\theta_d} 3\,d\theta\\ &= 3\theta_d \end{array}$ Since the net arc distance traveled alternates between two different directions, the summation is $\sum \dfrac{2\pi}{n} = 2\pi\left(\sum\limits_{n\text{ odd}}^{\infty} \dfrac{1}{n} - \sum\limits_{n\text{ even}}^{\infty} \dfrac{1}{n}\right)$ which is equivalent to $2\pi\sum\limits_{n = 1}^{\infty} \dfrac{(-1)^{n + 1}}{n}$ By the following Maclaurin series $\ln(1 + x) = \sum\limits_{n = 1}^{\infty} \dfrac{(-1)^{n + 1}x^n}{n}$ where $x = 1$ , $2\pi\sum\limits_{n = 1}^{\infty} \dfrac{(-1)^{n + 1}}{n} = 2\pi\ln(2)$ So $3\theta_d = 2\pi\ln(2) \quad \Longrightarrow \quad \theta_d = \dfrac{2\pi}{3}\ln(2)$

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Determining the Final Displacement

For $\theta_d$ , let $(x_d,y_d)$ be the final point on $x^2 + y^2 = 9$ . In Cartesian coordinates, $\begin{array}{rl} \theta_d = \arctan\left(\dfrac{y_d}{x_d}\right) &= \dfrac{2\pi}{3}\ln(2)\\ \dfrac{y_d}{x_d} &= \tan\left(\dfrac{2\pi}{3}\ln(2)\right)\\ y_d &= \tan\left(\dfrac{2\pi}{3}\ln(2)\right) x_d \end{array}$ Solving for $x_d$ , $\begin{array}{rlccl} \left(x_d\right)^2 + \left(\tan\left(\dfrac{2\pi}{3}\ln(2)\right)x_d\right)^2 &= 9\\ \left(x_d\right)^2\left(1 + \tan^2\left(\dfrac{2\pi}{3}\ln(2)\right)\right) &= 9 & & & {\color{#3D99F6}\text{Factor }(x_d)^2}\\ \left(x_d\right)^2\sec^2\left(\dfrac{2\pi}{3}\ln(2)\right) &= 9 & & & {\color{#3D99F6}\text{Identity: }1 + \tan^2(\alpha) = \sec^2(\alpha)}\\ \left(x_d\right)^2 &= 9\cos^2\left(\dfrac{2\pi}{3}\ln(2)\right) & & & {\color{#3D99F6}\text{since }\dfrac{1}{\sec^2(\alpha)} = \cos^2(\alpha)}\\ x_d &= 3\cos\left(\dfrac{2\pi}{3}\ln(2)\right) & & & {\color{#3D99F6}\text{since }\sum_{\text{odd}} > \sum_{\text{even}}, x_d > 0}\\ \end{array}$ So $\begin{array}{rl} y_d &= \tan\left(\dfrac{2\pi}{3}\ln(2)\right) \cdot 3\cos\left(\dfrac{2\pi}{3}\ln(2)\right)\\ &= \dfrac{\sin\left(\dfrac{2\pi}{3}\ln(2)\right)}{\cos\left(\dfrac{2\pi}{3}\ln(2)\right)} \cdot 3\cos\left(\dfrac{2\pi}{3}\ln(2)\right)\\ &= 3\sin\left(\dfrac{2\pi}{3}\ln(2)\right) \end{array}$ Therefore, the final displacement between Ash and the starting point is $\begin{array}{rlccl} \text{Displacement} &= \sqrt{\left(x_d - 3\right)^2 + \left(y_d\right)^2}\\ &= \sqrt{\left(3\cos\left(\dfrac{2\pi}{3}\ln(2)\right) - 3\right)^2 + \left(3\sin\left(\dfrac{2\pi}{3}\ln(2)\right)\right)^2}\\ &= \sqrt{9\cos^2\left(\dfrac{2\pi}{3}\ln(2)\right) - 18\cos\left(\dfrac{2\pi}{3}\ln(2)\right) + 9 + 9\sin^2\left(\dfrac{2\pi}{3}\ln(2)\right)}\\ &= \sqrt{9\left(\cos^2\left(\dfrac{2\pi}{3}\ln(2)\right) + \sin^2\left(\dfrac{2\pi}{3}\ln(2)\right)\right) + 9 - 18\cos\left(\dfrac{2\pi}{3}\ln(2)\right)}\\ &= \sqrt{9 + 9 - 18\cos\left(\dfrac{2\pi}{3}\ln(2)\right)} & & & {\color{#3D99F6}\text{since} \cos^2(\alpha) + \sin^2(\alpha) = 1}\\ &= \sqrt{18 - 18\cos\left(\dfrac{2\pi}{3}\ln(2)\right)} \end{array}$ Since $b = 18, c = 2, d = 3, e = 2$ , $b + c + d + e = \boxed{25}$ .