AMC 8 · 2011 · #19

Grade 4 counting
Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type
systematic-enumerationcombinations-basicarea-rectangles caseworksystematic-enumerationidentify-subproblems ↑ Prerequisites: systematic-enumeration
📏 Medium solution 💡 3 insights 📊 Diagram
📘 View easy version →
Problem
The figure is made of three overlapping rectangles: a central square, a wider horizontal rectangle that crosses through the left side, and a taller vertical rectangle that crosses through the right side. Count every rectangle whose four sides all lie along the drawn line segments — small pieces, the three big ones, and any in between.

Pick an answer.

(A)
8
(B)
9
(C)
10
(D)
11
(E)
12

AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

Counting rectangles in a tangled picture is risky if done by eye, so Tool #2 (Make a Systematic List) is the safest plan: sort the rectangles by size or by which big rectangle they live inside, and tally each group. Tool #1 (Draw a Diagram) helps to label the three original rectangles S (central square), H (horizontal), V (vertical) and mark where their edges cross. Tool #7 (Identify Subproblems) splits the count into clean subproblems: (a) the three big rectangles, (b) extra rectangles made when H crosses the square, (c) extra rectangles made when V crosses the square, (d) the small rectangle where H and V overlap each other.

1STEP 1

The three original rectangles — S (square), H (horizontal), V (vertical) — are themselves rectangles, giving 3 right away.

big rectangles = {S, H, V} → 3
2STEP 2

Inside S the two cuts make a 2×2 grid: 4 small quarter-pieces plus 4 half-square rectangles give 8 new rectangles.

4 small pieces + 4 half-square pieces = 8 new rectangles inside S
3STEP 3

The grid formula C(3,2)×C(3,2) = 9 confirms the count; subtracting the whole square S, already counted, leaves 8 new.

C(3, 2) × C(3, 2) = 3 × 3 = 9; 9 - 1 = 8 new
4STEP 4

The parts of H and V that stick outside S add no new rectangle: their outer edges are just H's and V's own sides, already counted.

3 big + 8 inside S + 0 from stickouts + 0 extra from H∩ V that wasn't in S-grid= ?
5STEP 5

Add it up: 3 big rectangles + 8 new ones inside the square = 11, matching choice (D).

3 + 4 + 4 = 11 → (D)
Answer
11
The answer choices run from 8 to 12, and a quick sanity count gives at least 3 big rectangles, plus the obvious 4 tiny pieces from the 2× 2 split of the square — that is already 7, so the answer must be larger. Counting the 4 "half-square" rectangles (top, bottom, left, right halves of S) bumps the total to 11. Adding any more (e.g., the whole square plus a stickout) would double-count edges that are not actually drawn, so 11 is the ceiling — exactly choice (D).
💡Key takeaway

This AMC 8 problem only needs Grade 4 careful counting — split the picture into clean pieces and add them up — that you already know!

More like this

Same archetype — closest grade level first.