The odd theoretical ant

The distances travelled by the ant form a geometric progression. Let $(0,0)$ be the starting point of the ant. Let $r$ be the ratio of this geometric progression, which, in this case, is $\frac{2}{3}$ .

The first turning point of the ant has coordinates $(0, -x)$ . The second turning point has coordinates $(rx, -x)$ . The third turning point has coordinates $(rx, -x+r^{2}x)$ . The fourth turning point has coordinates $(rx-r^{3}x, -x + r^{2}x)$

and the pattern will go on such that P will have the coordinates of

$( rx-r^{3}x+r^{5}x-r^{7}x+...,\: -x + r^{2}x - r^{4}x + r^{6}x + ... )$ .

So we can now simplify the coordinates of P, by determining the infinite geometric sums above, that is,

$\large (\frac {(r-r^{3})x}{1-r^{4}}, \frac {(-1 +r^{2})x }{1-r^4} )\: \equiv \: (x_P, y_P)$ .

So the distance of P from the starting point is $\sqrt { (x_P)^{2} + (y_P)^{2} }$ . With some algebraic manipulation, we will find the distance to be equal to

$d =\frac{x}{\sqrt {1+r^{2} }}$

Substituting for the values of d and r, thereby eliminating x, we find out that $k = \frac {3\sqrt{13} }{13}$ . Thus the answer is $3 + 13 + 13 = \boxed {29}$ .

We embed the above picture in the coordinate system with the origin $O$ , and assign the starting point the letter $A$ . Its coordinates are $0$ and $x$ . To reach the apscise coordinate of $P$ , we start from the point $B(\frac23x,0)$ , subtracting or adding distances alternatively. First, we subtract the next parallel distance to the segment $OB$ , i.e. $\frac23x-(\frac23)^3x$ . In the next step, to this apscise we add $(\frac23)^5x$ . Proceeding in the same way, we obtain $\frac23x-\Big(\frac23\Big)^3x+\Big(\frac23\Big)^5x-\cdots+(-1)^{n-1}\Big(\frac23\Big)^{2n-1}x+\cdots=\frac6{13}x,$ and that is the apscise coordinate of $P$ . Similarly, we obtain the ordinate coordinate of $P$ , i.e. $\Big(\frac23\Big)^2x-\Big(\frac23\Big)^4x+\Big(\frac23\Big)^6x-\cdots+(-1)^{n-1}\Big(\frac23\Big)^{2n}x+\cdots=\frac4{13}x.$ So, the distance $AP$ equals $\sqrt{\Big(\frac6{13}x\Big)^2+\Big(x-\frac4{13}x\Big)^2}=\frac{3x\sqrt{13}}{13},\quad k=\frac{3\sqrt{13}}{13}.$ Here $a=3,b=c=13$ , so $a+b+c=29$ .

Let's assume that the ant starts from $(0,0)$ .

Now let's say that the ant first moves distance $x$ on $x-axis$ .

The displacement of the ant on $x-axis$ from $(0,0)$ can be written as :

$=x-\frac{4}{9}x+\frac{16}{81}x-\ldots \infty$

$=\frac{1}{1+\frac{4}{9}}x=\frac{9}{13}x$

Now let's say that the ant first moves distance $\frac{2}{3}x$ on $y-axis$ .

The displacement of the ant on $y-axis$ from $(0,0)$ can be written as :

$=\frac{2}{3}x-\frac{8}{27}x+\frac{32}{243}-\ldots \infty$

$=\frac{\frac{2}{3}}{1+\frac{4}{9}}x=\frac{6}{13}x$

Now we just have to find the distance from $(0,0)$ to $(\frac{9}{13}x,\frac{6}{13}x)$ .

$=\sqrt{\left(\frac{9}{13}x-0\right)^2+\left(\frac{6}{13}x-0\right)^2}$

$=\frac{\sqrt{117}}{13}x=\frac{3\sqrt{13}}{13}$

$\Rightarrow k=3+13×2=\boxed{29}$

Turn the figure by $90^\circ$ anticlockwise as shown above. Let the starting point be the origin of a complex plane and define the $n$ th straight path as a complex number $u_n$ . We note that $u_n$ has a modulus of $\left(\frac 23\right)^nx$ and an argument of $i^n$ for non-negative integer $n$ or $u_n = \left(\frac 23i\right)^nx$ . Then the coordinates of the end of the $n$ th straight path is given by $\left(\Re(z_n), \Im(z_n)\right)$ , where $z_n = \sum_{k=0}^n u_k$ and the distance from the starting point or origin is given by $\left|z_n \right|$ . Then, for $n \to \infty$ , we have:

$\begin{aligned} z_\infty & = \sum_{k=0}^\infty \left(\frac 23i\right)^nx = \frac x{1-\frac 23i} = \frac {3x}{3-2i} = \frac {3(3+2i)x}{13} \\ \implies \left|z_\infty \right| & = \frac {3\sqrt{13}}{13} \end{aligned}$

Therefore $a+b+c = 3+13+13 = \boxed{29}$ .

The displacement on $x$ axis is $\displaystyle \sum_{n=0}^{\infty}x(\frac{2}{3})^{1+4n} - \displaystyle \sum_{n=0}^{\infty} x(\frac{2}{3})^{3+4n}$ and on $y$ axis is $\displaystyle \sum_{n=0}^{\infty} x(\frac{2}{3})^{2+4n} - \displaystyle \sum_{n=0}^{\infty} x(\frac{2}{3})^{4n}$ , so if the inital point is $(a.b)$ , the final point, using geometric series, is $(a+\frac{6}{13} \cdot x, b- \frac{9}{13} \cdot x)$ . Therefore, the distance between the inital point and the final point is $\frac{3\sqrt{13}}{13}$