2015 AMC 10B Problem 24

Trace the spiral and record where the ant is at perfect-square indices:
$$p_1 = (1,0),\quad p_4 = (-1,1),\quad p_9 = (2,-1),\quad p_{16} = (-2,2),\quad p_{25} = (3,-2).$$
The pattern: from $p_{k^2}$ the ant walks a leg of length $k$ (turning left first), then a leg of length $k+1$, using $2k + 1$ steps and arriving at $p_{(k+1)^2}$. Each pair of legs moves the ant to the opposite diagonal corner one ring further out. In particular, for even indices
$$p_{(2k)^2} = (-k, k),$$
reached while walking west, so the next leg goes south (length $2k$) and the one after goes east (length $2k+1$).

Since $44^2 = 1936 \le 2015 < 2025 = 45^2$, start from $p_{1936} = (-22, 22)$. The ant walks $44$ steps south to $p_{1980} = (-22, -22)$, then heads east. The remaining $2015 - 1980 = 35$ steps go east:
$$p_{2015} = (-22 + 35, -22) = (13, -22).$$

The answer is $\boxed{\textbf{(D)}\ (13,-22)}$.

Spirals are governed by squares: the $k$-th "corner turn" happens after $k^2$ steps because $1 + 1 + 2 + 2 + 3 + 3 + \cdots$ accumulates to consecutive squares. Once the anchor points $p_{k^2}$ are identified with their direction of travel, any index is a short walk from the nearest square below it. Compute a handful of small cases, locate the square pattern, then jump.

Use the odd squares instead: $p_{(2k+1)^2} = (k+1, -k)$, reached walking east. So $p_{2025} = (23, -22)$, and ten steps earlier along that eastward leg is $p_{2015} = (13, -22)$. Consistency between the two anchors is a good check.

• Off-by-one in the leg lengths (each new pair of legs has lengths k and k+1) or walking the last leg in the wrong direction, landing on (-13, -22) or (22, -13).
• Anchoring at $p_{1936}$ but forgetting that the southward leg has length $44$ (not $43$ or $45$), which shifts the final $x$-coordinate.