← Sensim Lab Sensim Math

AMC 8 · 2002 · #23

Grade 4 geometry-2dcounting
Archetype Lattice / Grid Pattern from Small Cases
pattern-recognitionspatial-visualizationfraction-arithmeticratio-proportion pattern-recognitionidentify-subproblems ↑ Prerequisites: fraction-arithmeticmulti-digit-arithmetic
📏 Long solution 💡 3 insights 📊 Diagram
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Problem

A corner of a tiled floor is shown. If the entire floor is tiled in this way and each of the four corners looks like this one, then what fraction of the tiled floor is made of darker tiles?

Pick an answer.

(A)
$\frac{1}3$
(B)
$\frac{4}9$
(C)
$\frac{1}2$
(D)
$\frac{5}9$
(E)
$\frac{5}8$
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Toolkit + CCSS Solution

Understand

Restated: A corner of a tiled floor is shown. The same pattern continues across the whole floor, and each of the four corners of the floor looks like the one in the picture. Find the fraction of the floor that is covered by the darker tiles.

Givens: The shown corner is a square block of unit tiles, some dark and some light; The same tiling pattern repeats across the entire floor; All four corners of the floor look like the corner shown; Answer choices: (A) $\tfrac{1}{3}$, (B) $\tfrac{4}{9}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{5}{9}$, (E) $\tfrac{5}{8}$

Unknowns: The fraction of the whole floor covered by darker tiles

Understand

Restated: A corner of a tiled floor is shown. The same pattern continues across the whole floor, and each of the four corners of the floor looks like the one in the picture. Find the fraction of the floor that is covered by the darker tiles.

Givens: The shown corner is a square block of unit tiles, some dark and some light; The same tiling pattern repeats across the entire floor; All four corners of the floor look like the corner shown; Answer choices: (A) $\tfrac{1}{3}$, (B) $\tfrac{4}{9}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{5}{9}$, (E) $\tfrac{5}{8}$

Plan

Primary tool: #5 Look for a Pattern

Secondary: #9 Solve an Easier Related Problem, #1 Draw a Diagram

The floor is infinite in the imagination but its tiling is periodic, so Tool #5 (Look for a Pattern) tells us the dark fraction is determined by one repeating block — we never have to count the whole floor. Tool #9 (Solve an Easier Related Problem) then shrinks the work even further: instead of working with the full corner shown, find the smallest square block that already captures the proportion. A small $3 \times 3$ block at the corner does the job. Tool #1 (Draw a Diagram) — really, reading the diagram already given — is what lets us mark each cell of that $3 \times 3$ block as dark or light and count.

Execute — Answer: B

#5 Look for a Pattern 4.OA.C.5 Step 1
  • Use the pattern to reduce the problem to a small block.
  • Because the same pattern repeats across the floor, the dark fraction of any whole repeating block equals the dark fraction of the whole floor.
  • So we only need to study one small block.
$$\dfrac{\text{dark on floor}}{\text{total on floor}} = \dfrac{\text{dark in one block}}{\text{total in one block}}$$

💡 Grade 4 "generate and analyze a repeating pattern" — once the unit repeats, every copy looks the same, so any one copy answers the whole-floor question.

#9 Solve an Easier Related Problem 4.OA.C.5 Step 2
  • Pick the smallest convenient block.
  • Look at the $3 \times 3$ square of tiles in the very corner of the picture.
  • Because the four corners of the floor all look like this one, the dark fraction inside this $3 \times 3$ corner block is the same dark fraction as the whole floor.
$$\text{block size} = 3 \times 3 = 9 \text{ tiles}$$

💡 Solving an easier related problem: instead of counting hundreds of tiles, work with $9$. The block size is small enough to count by eye.

#1 Draw a Diagram 3.G.A.2 Step 3
  • Read off each cell of the $3 \times 3$ corner block from the diagram and mark it dark (D) or light (L).
  • Going row by row from the top of the corner block:
$$\begin{array}{|c|c|c|}\hline \text{L} & \text{L} & \text{D} \\\hline \text{L} & \text{D} & \text{D} \\\hline \text{D} & \text{L} & \text{L} \\\hline \end{array}$$

💡 Grade 3 "partition a shape into equal parts" — the $3 \times 3$ grid is already partitioned into $9$ unit squares; we just label each one.

#5 Look for a Pattern 3.OA.D.8 Step 4
  • Count the dark cells in the block.
  • Row 1 has $1$ dark, row 2 has $2$, row 3 has $1$.
  • Add them.
$$\text{dark count} = 1 + 2 + 1 = 4$$

💡 Grade 3 multi-step addition: tally the dark tiles in each row, then add the row totals.

#5 Look for a Pattern 3.NF.A.1 Step 5
  • Form the fraction.
  • Dark tiles in the block over total tiles in the block gives the dark fraction of the whole floor.
$$\dfrac{\text{dark}}{\text{total}} = \dfrac{4}{9} \;\Rightarrow\; \textbf{(B)}$$

💡 Grade 3 fractions: $4$ shaded parts out of $9$ equal parts is the fraction $\tfrac{4}{9}$.

[1] #5 4.OA.C.5 Use the pattern to reduce the problem to a small block. Because the same pattern
[2] #9 4.OA.C.5 Pick the smallest convenient block. Look at the $3 \times 3$ square of tiles in
[3] #1 3.G.A.2 Read off each cell of the $3 \times 3$ corner block from the diagram and mark it
[4] #5 3.OA.D.8 Count the dark cells in the block. Row 1 has $1$ dark, row 2 has $2$, row 3 has
[5] #5 3.NF.A.1 Form the fraction. Dark tiles in the block over total tiles in the block gives t

Review

Reasonableness: Check that the choice $\tfrac{4}{9}$ fits the picture. A little less than half of the corner block is dark, which matches what the eye sees: the dark pinwheel covers most of one row, just a corner of another, and almost none of a third. The next-larger choice $\tfrac{1}{2}$ would mean exactly half dark — clearly too much; the next-smaller choice $\tfrac{1}{3}$ would mean only $3$ dark out of $9$ — clearly too few. The four-corner symmetry condition also rules out anything other than a single repeating block, so the small-block count is enough.

Alternative: Tool #16 (Change Focus / Count the Complement): count the light tiles in the $3 \times 3$ corner block instead. Row 1 has $2$ light, row 2 has $1$, row 3 has $2$, giving $2 + 1 + 2 = 5$ light tiles. The dark fraction is then $1 - \tfrac{5}{9} = \tfrac{4}{9}$, the same answer. This is a useful cross-check when the dark and light counts are close — recounting the other color is the easiest way to catch a miscount.

CCSS standards used (min grade 4)

  • 4.OA.C.5 Generate a number or shape pattern that follows a given rule (Using the fact that the tiling repeats to reduce the whole-floor question to a single small block.)
  • 3.G.A.2 Partition shapes into parts with equal areas (Reading the $3 \times 3$ corner block as $9$ equal-area unit squares, each labeled dark or light.)
  • 3.OA.D.8 Solve two-step word problems using the four operations (Adding the per-row dark counts $1 + 2 + 1$ to get the total of $4$ dark tiles in the block.)
  • 3.NF.A.1 Understand a fraction $a/b$ as the quantity formed by $a$ parts of size $1/b$ (Writing the answer as the fraction $\tfrac{4}{9}$ — $4$ shaded parts out of $9$ equal parts.)

⭐ When a pattern repeats, you don't need to count the whole floor — count one small block. A $3 \times 3$ corner has $4$ dark tiles out of $9$, so the answer is $\tfrac{4}{9}$.

⭐ When a pattern repeats, you don't need to count the whole floor — count one small block. A $3 \times 3$ corner has $4$ dark tiles out of $9$, so the answer is $\tfrac{4}{9}$.

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