Is this a Maze?

If we place the spiral on an xy-plane with point E at the origin, the coordinates of the key points reveal a pattern.

From the table we can see that after finishing a side having length that is a multiple of $4$ , say $4k$ , we are at the point $(-2k;-2k)$

Therefore, after completing the side of length $20$ , we are at the point $(-10;-10)$ . All sides of length 4k travel toward the left.

We must now move vertically upward $21$ units from this point $(-10;-10).$ Moving upward, this last side of length $21$ will end at the point say $X$ with coordinates $(-10; 11)$ .

This point is left $10$ units and up $11$ units from $E (0; 0)$ . Using the Pythagorean Theorem, $( EX)^2 = 10^2+11^2 = 100+121 = 221\ \text{or} \ EX =\sqrt{221} = 14.866$

In the $y$ direction, we have moved a total of $1 - 3 + 5 - 7 + 9 - 11 + 13 - 15 + 17 - 19 + 21 = 11$ units. In the $x$ direction we have moved a total of $2-4+6-8+10-12+14-16+18-20 = -10$ units. Thus the distance between the initial and final points is $\sqrt{10^2 + 11^2} = \sqrt{221}$ which lies between $14 = \sqrt{196}$ and $15 = \sqrt{225}$ .

In the complex plane, when we multiply a complex number (represented by a line vector) by the imaginary unit $i = \sqrt{-1}$ , it is equivalent to turning the line vector $90^\circ$ anticlockwise. For ease of calculation, let us consider the spiral to turn anticlockwise instead of clockwise. Then the first line vector is $i$ , the second $2i^2$ , the third $3i^3$ , ... the $n$ th line is $ni^n$ . Notice that the vector from the origin $E$ and the end-point of $n$ th line is a complex number $z(n)$ . And $z(n)$ is given by the sum of all the lines from the first to $n$ th as follows:

$\begin{aligned} z(n) & = \sum_{k=1}^n ki^k = i \sum_{k=1}^n k{\color{#3D99F6}i}^{k-1} & \small \color{#3D99F6} \text{Replace }i \text{ with }x. \\ & = i \sum_{k=1}^n k{\color{#3D99F6}x}^{k-1} = i \frac {\partial}{\partial x} \sum_{k=1}^n x^k \\ & = i \frac {\partial}{\partial x} \left(\frac{ x(x^n-1)}{x-1} \right) \\ & = i\left(\frac {nx^{n+1}-(n+1)x^n+1}{(x-1)^2}\right) & \small \color{#3D99F6} \text{Put back }x \text{ as }i.\\ & = \frac {-ni^{n+1}+(n+1) i^n -1}2 \\ \implies z(21) & = \frac {-21i^{22}+22 i^{21} -1}2 \\ & = 10+11i \approx 14.866 e^{0.833i} \end{aligned}$

The final Sam's distance from $E$ is about 14.866, which is between 14 and 15 .

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Inspiration: @Hana Nakkache 's solution to this problem.