Competition · AMC preparation · step 4 of 4

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 12\frac{1}{2}, 13\frac{1}{3}, 117\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)
:6
(B)
:7
(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 = 11/102.

1STEP 1

Find the fourth slab height

The four slab heights must add to 1 ft, so h_D = 1 - (12\frac{1}{2} + 13\frac{1}{3} + 117\frac{1}{17}) = 11102\frac{11}{102} ft — needed later for the side faces.

h_D = 1 - (1/2 + 1/3 + 1/17) = 1 - (51 + 34 + 6)/102 = 11/102 ft
2STEP 2

Split the surface into parts

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

Find the top and bottom

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

Find the front and back

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

Measure the stepped side faces

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

Add the side rectangles

Over the common denominator 102 the five side pieces are 11, 40, 17, 28, 6 — summing to 102102\frac{102}{102} = 1 ft².

Sides = 11/102 + 40/102 + 17/102 + 28/102 + 6/102 = 102/102 = 1 ft²
7STEP 7

Add the five subtotals

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!

  • Find the fourth slab height
  • Split the surface into parts
  • Find the top and bottom
  • Find the front and back
  • Measure the stepped side faces
  • Add the side rectangles
  • Add the five subtotals

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