AMC 8 · 2009 · #25

Grade 6 geometry-3d

Archetype Pair-Sum Invariant └ Sub-archetype Rectilinear Perimeter Closure

surface-areafraction-arithmeticarea-rectanglesset-partition identify-subproblemspattern-recognition ↑ Prerequisites: surface-areafraction-arithmetic

📏 Long solution 💡 4 insights 📊 Diagram

Problem

A 1 × 1 × 1 ft cube is sliced by three horizontal cuts into four flat slabs A, B, C, D with heights

\frac{1}{2}

\frac{1}{3}

\frac{1}{17}

, and the leftover. The four slabs are then placed side by side on the ground in the order D, A, B, C. Find the total surface area, in square feet, of this new solid.

Pick an answer.

(A)

(B)

(C)

$:\frac{419}{51}$

(D)

$:\frac{158}{17}$

(E)

:11

AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The new solid is a row of four short rectangular blocks of different heights. Instead of counting faces one by one, Tool #7 (Identify Subproblems) lets us split the total surface area into five tidy groups — top, bottom, front, back, and the staircase of side faces — and add. Tool #1 (Draw a Diagram) is the picture that makes those groups obvious. The big payoff comes from Tool #11 (Symmetry / Invariants): the front and back are each one rectangle whose total width is 4 ft and whose heights sum to exactly 1 ft, so each has area 1 ft² without ever needing the messy value of h_D = $\frac{11}{102}$ .

1STEP 1

The four slab heights must add to 1 ft, so h_D = 1 - ( $\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{17}$ ) = $\frac{11}{102}$ ft — needed later for the side faces.

h_D = 1 - (

\frac{1}{2}

\frac{1}{3}

\frac{1}{17}

) = 1 -

\frac{51 + 34 + 6}{102}

\frac{11}{102}

2STEP 2

Split the new solid's surface into five groups — top, bottom, front, back, and the side steps — then compute each and add.

Total SA = Top + Bottom + Front + Back + Sides

3STEP 3

Each slab keeps its 1×1 ft top and bottom, so with four slabs the top is 4 ft² and the bottom another 4 ft².

Top = 4 × (1 × 1) = 4 ft², Bottom = 4 ft²

4STEP 4

From the front you see four 1-ft-wide rectangles whose heights sum to 1 ft, so the front is 1 ft²; the back matches.

Front = 1 × (h_A + h_B + h_C + h_D) = 1 × 1 = 1 ft², Back = 1 ft²

5STEP 5

In order D, A, B, C the exposed vertical faces are the two ends (h_D, h_C) plus three steps between neighbors, each 1 ft deep.

Sides = 1 · (h_D + |h_A - h_D| + |h_A - h_B| + |h_B - h_C| + h_C)

6STEP 6

Over the common denominator 102 the five side pieces are 11, 40, 17, 28, 6 — summing to $\frac{102}{102}$ = 1 ft².

Sides =

\frac{11}{102}

\frac{40}{102}

\frac{17}{102}

\frac{28}{102}

\frac{6}{102}

\frac{102}{102}

= 1 ft²

7STEP 7

Add the five subtotals: 4 + 4 + 1 + 1 + 1 = 11 ft².

Total SA = 4 + 4 + 1 + 1 + 1 = 11 ft² → (E)

Answer

:11

The original cube has surface area 6 ft². Cutting and rearranging never removes material, only exposes hidden faces, so the new surface area must be at least 6 — answer (A) is therefore suspicious. The rearrangement exposes three internal horizontal cuts (3 extra top faces + 3 extra bottom faces = 6 extra ft²) and hides three side strips of total area 1 where neighbors touch, while exposing roughly the same staircase area on the side. The net change works out to +5, giving 6 + 5 = 11 ft². This matches (E) and rules out the other choices.

💡Key takeaway

This AMC 8 problem only needs Grade 6 surface-area-from-nets reasoning you already know!

More like this