Starting at point $P$, Breenah constructs a straight sided spiral so that:

- all angles are $90^{\circ}$
- after starting with a line segment of length 1 , each side is 1 longer than the previous side.

After completing the side with length 21, Breenah's distance from her original starting point $P$ will be between
(A) 13 and 14
(B) 14 and 15
(C) 15 and 16
(D) 16 and 17
(E) 17 and 18

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-061.jpg?height=368&width=393&top_left_y=1643&top_left_x=1321)
Solution 1

Breenah travels along each of the sides in a direction that is either up, down, right or left.

The "up" sides occur every fourth segment, thus they have lengths $1,5,9,13,17,21, \ldots$ or lengths that are one more than a multiple of 4 . As we see, the segment of length 21 is an up side.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-282.jpg?height=420&width=444&top_left_y=2186&top_left_x=900)

We see that the upper endpoint of each up segment is 2 units to the left and 2 units above the upper endpoint of the previous up segment. Thus, when Breenah is standing at the upper endpoint of the up segment of length 21 , she is $2+2+2+2+2=10$ units to the left and $1+2+2+2+2+2=11$ units above $P$. The following paragraphs prove this in a more formal way.

We now determine the horizontal distance from point $P$ to the up side of length 21. Following the spiral outward from $P$, the first horizontal line segment moves right 2 , or +2 , where the positive sign indicates movement to the right.

The second horizontal segment moves left 4 , or -4 , where the negative sign indicates movement to the left.

After these two horizontal movements, we are at the line segment of length 5 , an up side. To get there, we moved a horizontal distance of $(+2)+(-4)=-2$ or 2 units to the left.

The horizontal distance from $P$ to the next up side (length 9 ), can be found similarly. Beginning on the segment of length 5 , we are already at -2 or 2 units left of $P$ and we move right 6 (or +6 ), then left 8 (or -8 ).

Thus, to reach the up side with length 9 , we have moved horizontally

$(-2)+(+6)+(-8)=-4$ or 4 units left of $P$.

This pattern of determining the horizontal distances from $P$ to each of the up sides is continued in the table below.

We now determine the vertical distance from point $P$ to the upper endpoint of the up side with length 21.

From $P$, the first vertical segment moves up 1 or +1 , the second moves down 3 or -3 and the third moves up 5 or +5 .

Therefore the vertical position of the endpoint of the up side with length 5 is $(+1)+(-3)+(+5)=+3$ or 3 units above $P$.

Similarly, we can calculate the vertical position of each of the up side endpoints relative to $P$ and have summarized this in the table below.

| Side Length | Horizontal Distance | Vertical Distance |
| :---: | ---: | ---: |
| 5 | $(+2)+(-4)=-2$ | $(+1)+(-3)+(+5)=+3$ |
| 9 | $(-2)+(+6)+(-8)=-4$ | $(+3)+(-7)+(+9)=+5$ |
| 13 | $(-4)+(+10)+(-12)=-6$ | $(+5)+(-11)+(+13)=+7$ |
| 17 | $(-6)+(+14)+(-16)=-8$ | $(+7)+(-15)+(+17)=+9$ |
| 21 | $(-8)+(+18)+(-20)=-10$ | $(+9)+(-19)+(+21)=+11$ |

We now calculate the distance, $d$, from $P$ to the upper endpoint, $F$, of the up side of length 21.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-283.jpg?height=377&width=336&top_left_y=2029&top_left_x=949)

Since the calculated distances are horizontal and vertical, we have created a right angle and may find the required distance using the Pythagorean Theorem.

Then $d^{2}=10^{2}+11^{2}=100+121=221$ or $d=\sqrt{221} \approx 14.866$.

## Solution 2

If we place the spiral on an $x y$-plane with point $P$ at the origin, the coordinates of the key points reveal a pattern.

| Side Length | Endpoint Coordinates |
| :---: | :---: |
| 1 | $(0,1)$ |
| 2 | $(2,1)$ |
| 3 | $(2,-2)$ |
| 4 | $(-2,-2)$ |
| 5 | $(-2,3)$ |
| 6 | $(4,3)$ |
| 7 | $(4,-4)$ |
| 8 | $(-4,-4)$ |
| $\vdots$ | $\vdots$ |

From the table we can see that after finishing a side having length that is a multiple of 4, say $4 k$, we are at the point $(-2 k,-2 k)$ (the basis for this argument is shown in solution 1 ).

Therefore, after completing the side of length 20 , we are at the point $(-10,-10)$.

All sides of length $4 k$ 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 $F(-10,11)$.

This point is left 10 units and up 11 units from $P(0,0)$.

Using the Pythagorean Theorem, $P F^{2}=10^{2}+11^{2}=100+121=221$ or $P F=\sqrt{221} \approx 14.866$.

ANSWER: (B)