AMC 8 · 2009 · #4

Grade 3 geometry-2d

Archetype Grid Tiling / Packing with Divisibility └ Sub-archetype Fixed Piece Set — Feasibility by Elimination

area-rectanglesspatial-visualizationsystematic-enumeration physical-representationcasework ↑ Prerequisites: area-rectanglesmulti-digit-arithmetic

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Problem

Five vertical strip tiles — of heights 1, 2, 3, 4, and 5 unit squares — are given. Four of the five answer-choice figures (A)–(E) can be assembled from these strips with no overlaps and no leftover pieces; one figure cannot. Find the figure that cannot be formed.

Pick an answer.

(A)

(figure A)

(B)

(figure B)

(C)

(figure C)

(D)

(figure D)

(E)

(figure E)

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

How to solve

Strategy Create a Physical Representation

This is a hands-on tiling puzzle, so Tool #10 (Physical Representation) is the natural primary move — cut five paper strips of heights 1, 2, 3, 4, 5 and try to place them on each figure. Tool #1 (Draw a Diagram) lets us shortcut the physical step: for each answer figure, just list the column heights and check whether they can be partitioned into a column-by-column arrangement of the five strip heights. Tool #3 (Eliminate Possibilities) closes the deal: in a multiple-choice "which one cannot" problem, we systematically check (A)–(E) and discard the four that can be tiled. The key shortcut: the height-5 strip needs a column with at least 5 stacked squares; any figure whose tallest column is shorter than 5 is impossible.

1STEP 1

The strip heights 1, 2, 3, 4, 5 cover 1+2+3+4+5=15 squares — every figure also has 15, so area alone rules nothing out.

1 + 2 + 3 + 4 + 5 = 15

2STEP 2

Each strip fills one column, so list the column heights: (A) 5,3,2,5; (B) 2,3,4,3,3; (C) 5,4,3,2,1; (D) 5,5,5; (E) 1,4,5,4,1.

(A) 5,3,2,5 | (B) 2,3,4,3,3 | (C) 5,4,3,2,1 | (D) 5,5,5 | (E) 1,4,5,4,1

3STEP 3

The vertical 5-strip needs a column of height 5. (A),(C),(D),(E) each have one, but (B)'s tallest column is only 4 — nowhere for it to go.

max(B) = 4 ＜ 5

4STEP 4

The other four do tile: (C) 5+4+3+2+1, (D) 5+(4+1)+(3+2), (A) 5+(4+1)+3+2, (E) 1+4+5+4+1 — each strip slots into a column.

(A) 5+(4+1)+3+2, (C) 5+4+3+2+1, (D) 5+(4+1)+(3+2), (E) 1+4+5+4+1

5STEP 5

Only one figure survives the elimination, and by Step 3 it has no height-5 column — so (B) is the figure that cannot be formed.

Answer: (B)

Answer

(figure B)

Each tile is 1 wide and stays vertical, so the 5-strip must occupy 5 stacked squares in one column. Among the five figures, only (B) has every column shorter than 5 (heights 2, 3, 4, 3, 3). For the other four, an explicit tiling has been exhibited in Step 4, so the answer (B) is consistent.

💡Key takeaway

This AMC 8 problem is really a Grade 3 "tile the shape" puzzle — just check where the longest strip can fit!

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