AMC 8 · 2002 · #23
Easy mode Grade 4Problem
Imagine a floor covered with small square tiles. Some tiles are dark and some are light. The dark tiles form a repeating pattern that covers the whole floor.
The figure below shows one corner of the floor. The same pattern repeats across the entire floor, and each of the four corners looks just like this one.
If you could see the whole floor, what fraction of all the tiles would be dark?
Diagram
One 3x3 corner block — the same block repeats across the whole floor.
Pick an answer.
AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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.
💡 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.
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.
💡 Solving an easier related problem: instead of counting hundreds of tiles, work with $9$. The block size is small enough to count by eye.
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:
💡 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.
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.
💡 Grade 3 multi-step addition: tally the dark tiles in each row, then add the row totals.
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.
💡 Grade 3 fractions: $4$ shaded parts out of $9$ equal parts is the fraction $\tfrac{4}{9}$.
4.OA.C.5 Use the pattern to reduce the problem to a small block. Because the same pattern 4.OA.C.5 Pick the smallest convenient block. Look at the $3 \times 3$ square of tiles in 3.G.A.2 Read off each cell of the $3 \times 3$ corner block from the diagram and mark it 3.OA.D.8 Count the dark cells in the block. Row 1 has $1$ dark, row 2 has $2$, row 3 has 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.5Generate 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.2Partition 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.8Solve 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.1Understand 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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