AMC 10 · 2016 · #14

Grade 8 arithmetic

Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Lattice Constraint Analysis

coordinate-geometrysystematic-enumerationslope-intercept casework ↑ Prerequisites: coordinate-geometry

📏 Long solution 💡 3 insights

Problem

Count the axis-aligned squares whose four corners all have integer coordinates and that fit entirely inside the region bounded on the left/top by the line y=π x, below by the line y=-0.1, and on the right by the line x=5.1.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

First sketch the region (tool #1) to see it is a right triangle and to read off what "inside" means: an integer corner (x,y) is inside when y ≤ π x, y ≥ -0.1, and x ≤ 5.1. A square has four corners, but only one of them is the troublemaker against each boundary — that is the Extreme Principle (tool #14): the top-left corner is the one most likely to poke above the slant line, so it is the only corner I have to test. Then the count is a clean Systematic List (tool #2): group the squares by side length, and within each side length count column by column. Tool #7 (Identify Subproblems) just sums the three side-length cases at the end.

1STEP 1

Sketch the region

The three lines bound a right triangle, so an integer corner is inside exactly when y ≤ π x, y ≥ 0, and x ≤ 5.

inside ⇔ y ≤ π x, y ≥ -0.1, x ≤ 5.1

2STEP 2

Only the top-left corner matters

Only the top-left corner (a, b+s) can poke above the slant line, so the whole slant test reduces to b+s ≤ π a.

top-left corner (a, b+s): b+s ≤ π a

3STEP 3

Pi never ties an integer corner

Since π is irrational, π a is never an integer, so no corner sits on the line; the highest allowed height in column a is 3a for a=1..4.

⌊ π a ⌋ = 3a (a=1,2,3,4)

4STEP 4

Count the 1x1 squares

For s=1, columns a=1,2,3,4 give 3a spots each, so the total is 3+6+9+12 = 30.

Σ_a=1⁴ 3a = 3+6+9+12 = 30

5STEP 5

Count the 2x2 squares

For s=2, columns a=1,2,3 give 3a-1 spots each, so 2+5+8 = 15.

Σ_a=1³ (3a-1) = 2+5+8 = 15

6STEP 6

Count 3x3, rule out 4x4, then add

For s=3, columns a=1,2 give 1+4 = 5; s=4 fits nowhere. Adding the sizes: 30 + 15 + 5 = 50, choice (D).

30 + 15 + 5 = 50 = (D)

Answer

The 1×1 count alone is 30, which is choice (A) — a trap for anyone who forgets the bigger squares. Adding the 15 two-by-twos and 5 three-by-threes pushes the total to 50, comfortably inside the largest choice 57. The column counts 3,6,9,12 rising by a steady 3 match the slant line gaining height π ≈ 3 per step to the right, which is a good sanity check on the shape of the triangle.

💡Key takeaway

Check only the corner most likely to break each boundary, then count the squares size by size and add them up.

Sketch the region
Only the top-left corner matters
Pi never ties an integer corner
Count the 1x1 squares
Count the 2x2 squares
Count 3x3, rule out 4x4, then add