AMC 10 · 2022 · #5

Grade 6 counting

Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Figure-N Growth Sequence

absolute-valuecoordinate-geometryconsecutive-integer-countperfect-squares identify-subproblemssystematic-enumerationpattern-recognition ↑ Prerequisites: absolute-value

📏 Medium solution 💡 2 insights

Problem

The taxicab distance between two points is the horizontal gap plus the vertical gap, because you may only travel along grid lines. Count how many points with both coordinates integers sit at taxicab distance at most 20 from the origin.

Pick an answer.

(A)

441

(B)

761

(C)

841

(D)

921

(E)

924

How to solve

Strategy Identify Subproblems

Counting all the points at once is hopeless, but the condition |x| + |y| ≤ 20 splits cleanly the moment you freeze one coordinate. Tool #7 (Identify Subproblems) says: sweep x across its possible values and, for each fixed x, count the allowed y separately — that turns one big count into 41 tiny counts. Tool #1 (Draw a Diagram) makes the shape visible: the allowed points fill a diamond with corners at (± 20, 0) and (0, ± 20), so each vertical column of the diamond is one subproblem. Tool #2 (Make a Systematic List) counts the integers inside one column, and Tool #5 (Look for a Pattern) spots that the column heights run 41, 39, 37, …, 1 — consecutive odd numbers, which add up in a way we can shortcut.

1STEP 1

Turn it into one inequality

Turn the words into one inequality.

|x - 0| + |y - 0| = |x| + |y| ≤ 20

2STEP 2

See the shape

The region is a diamond, not a circle.

corners: (20, 0), (0, 20), (-20, 0), (0, -20)

3STEP 3

Freeze one coordinate and count

Freezing one coordinate gives an interval.

|x| + |y| ≤ 20 ⟹ |y| ≤ 20 - |x|, -20 ≤ x ≤ 20

4STEP 4

Count one column

Each column holds an odd number of points.

#{y ∈ Z : |y| ≤ m} = 2m + 1, m = 20 - |a| ⟹ 41 - 2|a|

5STEP 5

The heights are the odd numbers

The heights run through the odd numbers.

41, 39, 37, …, 3, 1

6STEP 6

Add with the odd-number shortcut

Odd numbers sum to a square, giving 841.

41 + 2(1 + 3 + … + 39) = 41 + 2 · 20² = 41 + 800 = 841 → (C)

Answer

841

Two independent checks land on the same number. First, split the diamond by the parity of x + y: the points with x + y even form a straight 21 × 21 grid once you rotate the picture 45°, and the points with x + y odd form a 20 × 20 grid, giving 21² + 20² = 441 + 400 = 841. Second, a size check — the diamond has diagonals of length 40 each, so its area is (40 · 40)/2 = 800, and a lattice count should sit a little above the area once the boundary is included. 841 fits; the smaller choices do not. Both wrong-looking choices are traps built from real pieces of this count: (A) 441 is only the even-parity half 21², and (B) 761 is the count of strictly interior points, since the diamond has 80 boundary lattice points and 841 - 80 = 761.

💡Key takeaway

This AMC 12 problem needs nothing past Grade 6 absolute value and the coordinate plane: freeze x, count the y that still fit in the leftover budget, and the column heights turn out to be the odd numbers 41, 39, …, 1, which add to 841.

Turn the words into one inequality
See the region as a diamond
Freeze x and count the y
One column holds 2m + 1 points
Column heights are the odd numbers
Add the columns with the odd-number shortcut