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2020 AMC 10B Problem 13

2020 AMC 10B Problem 13AMC 10 Step by Step

Name: Date: October 28, 2026

Coordinate Geometry · difficulty 3 of 5 · about 4 min · position P11-15

Andy the Ant lives on a coordinate plane and is currently at $(-20, 20)$ facing east (that is, in the positive $x$ -direction). Andy moves $1$ unit and then turns $90^{\circ}$ left. From there, Andy moves $2$ units (north) and then turns $90^{\circ}$ left. He then moves $3$ units (west) and again turns $90^{\circ}$ left. Andy continues his progress, increasing his distance each time by $1$ unit and always turning left. What is the location of the point which Andy makes the $2020$ left turn?

A) $(-1030, -994)$
B) $(-1030, -990)$
C) $(-1026, -994)$
D) $(-1026, -990)$
E) $(-1022, -994)$

Answer

Key insight

Group the moves in fours (east, north, west, south). Each group moves Andy 2 left and 2 down, and 2020 moves make exactly 505 groups.

Solution

When does each turn happen? Andy turns left right after each move. So the $2020$ th left turn happens at the point where his $2020$ th move ends.
Look at the first few moves. Move lengths go $1, 2, 3, 4, \ldots$ and the directions repeat east, north, west, south.

Move	Length	Direction	Position after
1	1	east	$(-19, 20)$
2	2	north	$(-19, 22)$
3	3	west	$(-22, 22)$
4	4	south	$(-22, 18)$
5	5	east	$(-17, 18)$
6	6	north	$(-17, 24)$
7	7	west	$(-24, 24)$
8	8	south	$(-24, 16)$

3. After moves 1 to 4 Andy is at $(-22, 18)$ , which is $2$ left and $2$ down from the start. After moves 5 to 8 he has again gone $2$ left and $2$ down.
4. Why every group of four does this. In a group, the west move is $2$ longer than the east move, so he ends up $2$ units further west. The south move is $2$ longer than the north move, so he ends up $2$ units further south.
5. Count the groups. $2020 = 4 \times 505$ , so the first $2020$ moves are exactly $505$ complete groups.
6. Total shift. $505 \times 2 = 1010$ units left and $1010$ units down:
$(-20 - 1010,\ 20 - 1010) = (-1030,\ -990).$

The answer is $\boxed{\textbf{(B)}\ (-1030, -990)}$ .

Why this works

The path looks like a growing spiral, but the four directions repeat, and in each group of four the opposite moves differ by the same amount. Grouping the moves turns the whole walk into "the same small shift, repeated". When a large number like $2020$ divides evenly into groups, the answer is the start plus that many copies of one group's shift.

The trap

Getting the timing wrong, for example using the position after 2019 or 2021 moves instead of 2020.

Concise solution

Andy's $k$ th move has length $k$ , and the directions cycle east, north, west, south. The $k$ th left turn happens right after the $k$ th move, so we need his position after $2020$ moves.

Look at four consecutive moves of lengths $m, m+1, m+2, m+3$ in the directions east, north, west, south. The net displacement is
$(m - (m+2),\ (m+1) - (m+3)) = (-2, -2),$
independent of $m$ . For example the first four moves take Andy from $(-20, 20)$ to $(-22, 18)$ .

Since $2020 = 4 \cdot 505$ , the first $2020$ moves form exactly $505$ such blocks (and the direction pattern restarts cleanly after each block). Total displacement: $505 \cdot (-2, -2) = (-1010, -1010)$ .

Position: $(-20 - 1010,\ 20 - 1010) = (-1030, -990)$ .

The answer is $\boxed{\textbf{(B)}\ (-1030, -990)}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)
Introduction to Algebra, Chapter 8: Graphing Lines

Techniques

Compute small cases, spot the pattern, generalize