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AMC 8 · 2009 · #4

Grade 3 geometry-2d
Archetype Grid Tiling / Packing with Divisibility
area-rectanglesspatial-visualizationsystematic-enumeration physical-representationcasework ↑ Prerequisites: area-rectanglesmulti-digit-arithmetic
📏 Medium solution 💡 3 insights 📊 Diagram
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Problem

The five pieces shown below can be arranged to form four of the five figures shown in the choices. Which figure cannot be formed?

Pick an answer.

(A)
(figure A)
(B)
(figure B)
(C)
(figure C)
(D)
(figure D)
(E)
(figure E)
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Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: Five tile pieces, all $1$-square wide vertical strips, with heights $1$, $2$, $3$, $4$, $5$; Total tile area $= 1 + 2 + 3 + 4 + 5 = 15$ unit squares; Each answer-choice figure is a shape made of exactly $15$ unit squares; Tiles must be placed without overlap and cover the figure exactly; From the figures shown, the tiles are used as vertical strips (no $90^\circ$ rotation)

Unknowns: Which one of the figures (A), (B), (C), (D), (E) cannot be tiled by the five strips

Understand

Restated: 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.

Givens: Five tile pieces, all $1$-square wide vertical strips, with heights $1$, $2$, $3$, $4$, $5$; Total tile area $= 1 + 2 + 3 + 4 + 5 = 15$ unit squares; Each answer-choice figure is a shape made of exactly $15$ unit squares; Tiles must be placed without overlap and cover the figure exactly; From the figures shown, the tiles are used as vertical strips (no $90^\circ$ rotation)

Plan

Primary tool: #10 Create a Physical Representation

Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities

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.

Execute — Answer: B

#1 Draw a Diagram 2.OA.B.2 Step 1
  • Add up the tile squares to confirm they fit a $15$-square figure.
  • The strip heights are $1, 2, 3, 4, 5$, so the total area is $1+2+3+4+5=15$ unit squares.
  • Every answer figure also has $15$ shaded squares, so area alone does not rule any choice out — the constraint must be about how the tiles fit into columns.
$$1 + 2 + 3 + 4 + 5 = 15$$

💡 Adding $1$ through $5$ is a Grade $2$ within-$20$ addition fact and matches the figure's area.

#1 Draw a Diagram 3.MD.C.7 Step 2
  • Record the column heights of each answer figure (reading left to right).
  • Each tile is a $1$-square-wide vertical strip, so each tile must occupy one whole column or a contiguous part of one column.
  • (A) heights $5, 3, 2, 5$ (four columns) (B) heights $2, 3, 4, 3, 3$ (five columns) (C) heights $5, 4, 3, 2, 1$ (five columns — staircase) (D) heights $5, 5, 5$ (three columns — a $3\times 5$ rectangle) (E) heights $1, 4, 5, 4, 1$ (five columns).
$$\text{(A)}\;5,3,2,5\;|\;\text{(B)}\;2,3,4,3,3\;|\;\text{(C)}\;5,4,3,2,1\;|\;\text{(D)}\;5,5,5\;|\;\text{(E)}\;1,4,5,4,1$$

💡 Decomposing a rectilinear figure into vertical rectangles is a Grade $3$ area move.

#3 Eliminate Possibilities 3.MD.C.5 Step 3
  • Look for the home of the $5$-strip in each figure.
  • Since the strip stays vertical and is $1$ square wide, it must sit inside a single column whose contiguous height is at least $5$.
  • Scan the heights from Step 2: (A) has columns of height $5$, (C) has a column of height $5$, (D) has three columns of height $5$, (E) has a column of height $5$ — all four contain a column tall enough.
  • (B)'s tallest column is only $4$, so the $5$-strip has nowhere to go.
$$\max\text{(B)} = 4 < 5$$

💡 Comparing heights of unit-square columns is a Grade $3$ "area by counting unit squares" check.

#10 Create a Physical Representation 3.MD.C.7 Step 4
  • Confirm that (A), (C), (D), (E) really can be tiled — Tool #10 in action with paper strips: (C) staircase has columns $5,4,3,2,1$ — drop the matching strip into each column.
  • (D) $3\times 5$ rectangle has columns $5,5,5$ — fill two columns with the $5$-strip and the $4{+}1$ or $3{+}2$ stack, the third with the remaining pair.
  • (A) columns $5,3,2,5$ — place the $5$-strips in the two height-$5$ columns and the $3$- and $2$-strips in the middle two columns; the height-$4$ strip splits as $4 = ?$ — wait, that uses six pieces.
  • Re-check (A): heights are $5,3,2,5$ total $= 15$ but we have five tiles $1,2,3,4,5$.
  • We can partition $5{=}5$, $3{=}3$, $2{=}2$, $5{=}4{+}1$.
  • Done with all five tiles.
  • (E) columns $1,4,5,4,1$ — use $1, 4, 5, 4, 1$ directly.
  • Every choice except (B) admits a valid tiling.
$$(A)\,5{+}(4{+}1){+}3{+}2,\;(C)\,5{+}4{+}3{+}2{+}1,\;(D)\,5{+}(4{+}1){+}(3{+}2),\;(E)\,1{+}4{+}5{+}4{+}1$$

💡 Building each figure column-by-column is exactly the Grade $3$ "decompose into rectangles" idea.

#3 Eliminate Possibilities 3.MD.C.5 Step 5
  • Only (B) is left after elimination.
  • By Step 3 it has no column of height $5$, so the $5$-strip cannot be placed without overlap or leaving the figure.
  • Therefore (B) is the figure that cannot be formed.
$$\textbf{Answer: (B)}$$

💡 Eliminating four valid choices on a five-choice problem forces the fifth — classic Tool #3 finish.

[1] #1 2.OA.B.2 Add up the tile squares to confirm they fit a $15$-square figure. The strip heig
[2] #1 3.MD.C.7 Record the column heights of each answer figure (reading left to right). Each ti
[3] #3 3.MD.C.5 Look for the home of the $5$-strip in each figure. Since the strip stays vertica
[4] #10 3.MD.C.7 Confirm that (A), (C), (D), (E) really can be tiled — Tool #10 in action with pa
[5] #3 3.MD.C.5 Only (B) is left after elimination. By Step 3 it has no column of height $5$, so

Review

Reasonableness: 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.

Alternative: Tool #10 (Physical) end-to-end: cut five paper strips $1, 2, 3, 4, 5$ squares tall. Try to lay them on each printed figure. (A), (C), (D), (E) all close up after a few tries; (B) keeps leaving holes because the $5$-strip never fits. That hands-on failure is the same conclusion: (B).

CCSS standards used (min grade 3)

  • 2.OA.B.2 Fluently add and subtract within $20$ using mental strategies (Adding $1+2+3+4+5 = 15$ to confirm the total tile area equals the area of each answer-choice figure.)
  • 3.MD.C.5 Recognize area as an attribute of plane figures and measure it by counting unit squares (Reading each answer figure as a collection of unit squares and counting the height of each column to spot which figures contain a vertical run of $5$ squares.)
  • 3.MD.C.7 Relate area to the operations of multiplication and addition (Decomposing each answer figure into vertical rectangles (columns) and matching them against the strip heights $1, 2, 3, 4, 5$ to test if a tiling exists.)

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

⭐ 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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