AMC 8 · 2002 · #22

Grade 6 geometry-3d

Archetype Compound Area by Decomposition / Difference

surface-areaspatial-visualizationface-adjacency identify-subproblemscasework ↑ Prerequisites: area-rectanglesmulti-digit-arithmetic

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom, and sides.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: Six unit cubes (each $1$ inch on every edge) are fastened together into the composite solid shown. Find its total surface area, in square inches, counting the top, the bottom, and every visible side.

Givens: $6$ unit cubes, each with edge length $1$ inch; Each face of a single cube has area $1 \times 1 = 1$ square inch; The cubes are fastened together along whole faces (face-to-face), so wherever two cubes touch, exactly two faces (one from each cube) disappear from the outside; The arrangement, read from the figure, has $5$ such face-to-face contacts (a central vertical stack of $2$; one cube on each side of the bottom of that stack; one cube on the right of the top of that stack; and one cube on top of that right cube); Answer choices: (A) $18$, (B) $24$, (C) $26$, (D) $30$, (E) $36$

Unknowns: The total surface area of the composite solid, in square inches

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

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

Counting visible faces one by one on a lumpy stack is error-prone. Tool #16 (Count the Complement) flips the job: start with all the faces the $6$ cubes would have if they were separate ($6 \times 6 = 36$), then subtract only the faces that got hidden by being pressed against another cube. Tool #1 (Draw a Diagram) makes the hidden faces easy to find — each place where two cubes touch is one contact, and one contact hides two unit faces. Tool #7 (Identify Subproblems) splits the count into two clean pieces: (a) total faces of separate cubes, (b) hidden faces from contacts.

Execute — Answer: C

#7 Identify Subproblems 6.G.A.4 Step 1

Subproblem 1 — pretend the $6$ cubes are not joined and add up all their faces.
A single cube has $6$ faces, each of area $1$ square inch, so its surface area is $6$.
Six separate cubes have $6 \times 6 = 36$ unit faces in all.

$$\text{Separate total} = 6 \text{ cubes} \times 6 \text{ faces} \times 1 \text{ in}^2 = 36 \text{ in}^2$$

💡 Grade 6 surface-area-from-nets: a cube's net is $6$ unit squares, so $6$ cubes contribute $36$ unit squares before any are glued.

#1 Draw a Diagram 6.G.A.4 Step 2

Subproblem 2 — use the diagram to count contacts.
Walk through the figure and tally each pair of cubes that share a whole face: (1) the two cubes in the central vertical stack, (2) the bottom of that stack with the cube on its left, (3) the bottom of that stack with the cube on its right, (4) the top of the stack with the cube attached to its right, (5) that right cube with the cube placed on top of it.

$$\text{Contacts} = 5$$

💡 Drawing or tracing the figure cube-by-cube makes each contact visible exactly once, so nothing is double-counted.

#16 Change Focus / Count the Complement 6.G.A.4 Step 3

Convert contacts to hidden faces.
Every contact glues one face of one cube to one face of another cube — so each contact removes $2$ unit faces from the outside surface.
With $5$ contacts, that is $5 \times 2 = 10$ hidden faces, total hidden area $10$ in$^2$.

$$\text{Hidden faces} = 5 \times 2 = 10,\quad \text{Hidden area} = 10 \times 1 = 10 \text{ in}^2$$

💡 Each glued seam takes two unit squares off the outside — one from each cube — exactly like closing the flaps on a cardboard net.

#16 Change Focus / Count the Complement 6.G.A.4 Step 4

Subtract hidden area from the separate total.
The composite's surface area is the $36$ from the separate cubes minus the $10$ that got buried by the $5$ contacts.

$$\text{Surface area} = 36 - 10 = 26 \text{ in}^2 \;\Rightarrow\; \textbf{(C)}$$

💡 "All faces minus the hidden ones" is the complement move applied directly to surface area.

[1] #7 6.G.A.4 Subproblem 1 — pretend the $6$ cubes are not joined and add up all their faces.

[2] #1 6.G.A.4 Subproblem 2 — use the diagram to count contacts. Walk through the figure and ta

[3] #16 6.G.A.4 Convert contacts to hidden faces. Every contact glues one face of one cube to on

[4] #16 6.G.A.4 Subtract hidden area from the separate total. The composite's surface area is th

Review

Reasonableness: Two quick checks. First, bounds: every contact hides $2$ faces, so the surface area must drop by an even number from $36$. The choices $18, 24, 26, 30, 36$ are all even drops ($18, 12, 10, 6, 0$), and only $36 - 26 = 10$ matches $5$ contacts $\times 2 = 10$ — so (C) is the only contact-count that is consistent with the figure. Second, sanity: a single cube has surface area $6$; six cubes glued in a straight line would have $5$ contacts too and give $36 - 10 = 26$ — the same as this arrangement, which is a useful reminder that surface area depends on the number of contacts, not the exact shape.

Alternative: Tool #10 (Make a Physical Model): hold up $6$ paper or sugar cubes and tape them together exactly as the figure shows. Count visible unit squares face by face — top, bottom, front, back, left, right of the whole solid. Adding the six view-counts gives the same $26$ square inches, confirming (C) without relying on the contact-counting trick.

CCSS standards used (min grade 6)

6.G.A.4 Represent three-dimensional figures using nets and find surface area (Treating each of the $6$ cubes as a net of $6$ unit squares to get $36$ in$^2$, then subtracting the $10$ unit faces hidden by the $5$ face-to-face contacts to land on $26$ in$^2$.)

⭐ Imagine the cubes apart ($36$ faces), then erase $2$ faces for every spot where two cubes touch ($5$ spots = $10$ faces). What's left is the surface area: $36 - 10 = 26$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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