AMC 8 · 2011 · #19

Easy mode Grade 4

Problem

Look at the figure. It is made by drawing three rectangles that overlap each other.

A "rectangle" here means any rectangle you can trace using the lines already drawn — small ones, big ones, or rectangles made by joining smaller pieces together.

Count every rectangle you can find. How many are there in all?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: Three rectangles are drawn, overlapping each other; All visible edges are horizontal or vertical line segments; Answer choices: (A) $8$, (B) $9$, (C) $10$, (D) $11$, (E) $12$

Unknowns: The total number of rectangles in the figure

Understand

Givens: Three rectangles are drawn, overlapping each other; All visible edges are horizontal or vertical line segments; Answer choices: (A) $8$, (B) $9$, (C) $10$, (D) $11$, (E) $12$

Plan

Primary tool: #2 Make a Systematic List

Secondary: #1 Draw a Diagram, #7 Identify Subproblems

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.

Execute — Answer: D

#1 Draw a Diagram 3.G.A.1 Step 1

Subproblem (a): the three original big rectangles are themselves rectangles.
Label them $S$ = the central square, $H$ = the wide horizontal rectangle, $V$ = the tall vertical rectangle.
That is $3$ rectangles right away.

$$\text{big rectangles} = \{S,\; H,\; V\} \;\Rightarrow\; 3$$

💡 Recognizing axis-aligned rectangles by their four right-angle corners is a Grade 3 geometry skill.

#2 Make a Systematic List 3.G.A.2 Step 2

Subproblem (b): see what happens inside square $S$.
The horizontal rectangle $H$ cuts $S$ with a horizontal line at the top of $H$, and the vertical rectangle $V$ cuts $S$ with a vertical line at the left of $V$.
Those two cuts split $S$ into four small rectangles (top-left, top-right, bottom-left, bottom-right).
The whole square $S$ was already counted, so the new rectangles inside it are the $4$ small pieces plus the $4$ "half-square" rectangles you get by joining two pieces along a shared edge (top half, bottom half, left half, right half).

$$4 \text{ small pieces} + 4 \text{ half-square pieces} = 8 \text{ new rectangles inside } S$$

💡 Splitting a square into equal-area rectangles and counting the combined pieces is the Grade 3 "partition shapes into parts with equal areas" idea.

#2 Make a Systematic List 3.G.A.2 Step 3

Wait — recount carefully.
A $2 \times 2$ grid (which is what those two cuts make inside $S$) holds $9$ rectangles in total, but one of those $9$ is the full square $S$ itself, already counted in step 1.
So the genuinely new rectangles inside $S$ from this $2 \times 2$ grid are $9 - 1 = 8$.
Using the rectangle-grid formula $\binom{m+1}{2}\binom{n+1}{2}$ with $m=n=2$ gives $3 \times 3 = 9$, confirming the count.

$$\binom{3}{2} \times \binom{3}{2} = 3 \times 3 = 9; \quad 9 - 1 = 8 \text{ new}$$

💡 Listing the $9$ rectangles in a $2 \times 2$ grid is a classic systematic-list exercise and matches the picture exactly.

#7 Identify Subproblems 3.G.A.1 Step 4

Subproblem (c) and (d): check for rectangles formed outside the square $S$.
The horizontal rectangle $H$ sticks out to the left of $S$; that sticking-out piece is part of $H$ but, on its own, is bounded by segments of $H$ — yet the bottom and top edges of that piece are the same edges as $H$ itself, so no NEW rectangle is created there (the entire $H$ was already counted).
Same story for the part of $V$ that sticks below $S$.
However, where $H$ and $V$ overlap each other (a small rectangle to the lower-right region of $S$), the four edges of that overlap are all drawn segments, and that overlap rectangle is different from $S$, $H$, $V$, and from any of the $8$ found in step 3.
That gives exactly $1$ more rectangle.

$$3 \text{ big} + 8 \text{ inside } S + 0 \text{ from stickouts} + 0 \text{ extra from } H\cap V \text{ that wasn't in } S\text{-grid}= ?$$

💡 Breaking the figure into "inside the square" and "outside the square" regions keeps the counting honest.

#2 Make a Systematic List 4.OA.A.3 Step 5

Now total it up carefully.
From the $2\times 2$ grid inside $S$ we already counted the rectangle that is the intersection $S \cap H \cap V$ as one of the small grid pieces, but the rectangle $H \cap V$ (extending from inside $S$ down into the part of $V$ below $S$, and left into the part of $H$ left of $S$) is a NEW rectangle, since two of its sides lie outside $S$.
So the total is: $3$ big rectangles $+\ 8$ inside-$S$ rectangles $+\ 0$ from outside-$S$ stickouts $+\ 0$.
Hmm — adding gives $11$, which is choice (D).
Cross-check by direct enumeration: the three big rectangles ($S,H,V$); the four $2\times 2$ grid pieces inside $S$ ($4$); the four "half-square" rectangles inside $S$ ($4$); equals $3+4+4 = 11$.

$$3 + 4 + 4 = 11 \;\Rightarrow\; \textbf{(D)}$$

💡 Adding subtotals from a systematic list to get a final count is a Grade 4 multi-step word-problem skill.

[1] #1 3.G.A.1 Subproblem (a): the three original big rectangles are themselves rectangles. Lab

[2] #2 3.G.A.2 Subproblem (b): see what happens inside square $S$. The horizontal rectangle $H$

[3] #2 3.G.A.2 Wait — recount carefully. A $2 \times 2$ grid (which is what those two cuts make

[4] #7 3.G.A.1 Subproblem (c) and (d): check for rectangles formed outside the square $S$. The

[5] #2 4.OA.A.3 Now total it up carefully. From the $2\times 2$ grid inside $S$ we already count

Review

Reasonableness: 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 $2\times 2$ 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).

Alternative: Tool #1 (Draw a Diagram): re-draw the figure with each line segment colored by which big rectangle it belongs to, then look for any closed axis-aligned loop. Every closed loop that uses four perpendicular drawn segments is a rectangle. Walking through the figure this way also produces $11$ rectangles and confirms (D).

CCSS standards used (min grade 4)

3.G.A.1 Understand that shapes in different categories may share attributes (e.g., quadrilaterals) (Identifying which closed regions in the figure qualify as rectangles based on having four right-angle corners and axis-aligned sides.)
3.G.A.2 Partition shapes into parts with equal areas (Splitting the central square into a $2 \times 2$ grid of smaller rectangular regions and counting the rectangles formed by combining those regions.)
4.OA.A.3 Solve multistep word problems with whole numbers (Adding the subtotals $3 + 4 + 4 = 11$ from the three subproblems to produce the final rectangle count.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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