Patterns in Mathematics · Lesson 6 of 7
Counting Within Shapes
“Count the pieces and boundary segments of growing shapes to uncover square and multiplication patterns.”
• Count the smallest squares in a growing square grid. • Explain why stacked triangles also give square-number totals. • Distinguish a shape’s outline from the objects being counted. • Describe the replacement rule that builds a Koch snowflake. • Predict segment counts through repeated multiplication by 4.
Choose exactly what to count
A shape may contain several kinds of objects. A square grid has small square cells, grid lines, corner points, and larger squares made from several cells. Different choices lead to different counts. Before counting a growing shape, state which objects count as one item.
Here, stacked squares means the smallest square cells filling each square figure. Stacked triangles means the smallest triangular pieces filling each triangle. Later, in the Koch snowflake, we count the straight boundary segments. This careful choice lets us link the pictures to the correct number sequence.
Square grids and triangular stacks
A square grid with n cells along each side contains n rows of n cells. For side lengths 1, 2, 3, 4, and 5, the cell counts are 1, 4, 9, 16, and 25. We recognise the square-number sequence because the construction directly uses equal row and column counts.
Problem
A stacked-square figure has 5 cells along each side. How many smallest cells fill it, and how many fill the next stage?
- 1.The current figure has 5 rows of 5 cells: 5 × 5 = 25.
- 2.The next square stage has 6 rows of 6 cells: 6 × 6 = 36.
- 3.The increase is 36 − 25 = 11 cells, the next odd-number layer around the old square.
The triangular stack has a different outer shape, but its smallest-piece count follows the same sequence. Count both upward-pointing and downward-pointing little triangles. The top row contains 1 piece, the next 3, the next 5, and the next 7. Each row has one more upward triangle than downward triangles, giving an odd total.
| Rows in the triangular stack | Smallest triangles by row | Total |
|---|---|---|
| 1 | 1 | 1 |
| 2 | 1 + 3 | 4 |
| 3 | 1 + 3 + 5 | 9 |
| 4 | 1 + 3 + 5 + 7 | 16 |
| 5 | 1 + 3 + 5 + 7 + 9 | 25 |
A row containing r upward triangles has r − 1 downward triangles between them. Here r is the row number counted from the top. The combined row count is r + (r − 1), or 2r − 1. Thus the rows are successive odd numbers, and their sum is a square number. The outline is triangular; the total of its smallest pieces is square.
Problem
How many smallest triangular pieces are in a five-row triangular stack?
- 1.The rows contain 1, 3, 5, 7, and 9 pieces, including both orientations.
- 2.These are the first five odd numbers, which build a square-number total.
- 3.The count is 5 × 5 = 25. Direct addition, 1 + 3 + 5 + 7 + 9 = 25, confirms it.
Counting only the upward-pointing triangles gives 1 + 2 + 3 + …, a triangular number. The task asks for all smallest triangular pieces, so include the downward-pointing ones too. Also, do not include larger triangles made by combining several small pieces unless a task explicitly asks for them.
This distinction also separates triangular dots from stacked triangular cells. In a triangular dot arrangement, rows of 1, 2, 3 dots give 6. In a three-row triangular stack, rows of 1, 3, 5 smallest triangles give 9. The outline alone cannot determine the answer; the counted objects and their row structure do.
Replace a segment with a speed bump
The Koch snowflake begins with an equilateral triangle. To make the next stage, change every boundary segment in the same way. Divide it into three equal lengths, keep the first and last parts, and replace the middle part with the two sides of a small outward triangle. The straight route across the middle is removed.
There are four new boundary segments where there used to be one. Apply the rule to all three sides of the original triangle to get 3 × 4 = 12 segments. At the following stage, each of those 12 produces four, giving 12 × 4 = 48. The same multiplier applies at every stage.
| Stage | How the count is obtained | Boundary segments |
|---|---|---|
| Starting triangle | 3 sides | 3 |
| First replacement | 3 × 4 | 12 |
| Second replacement | 12 × 4 | 48 |
| Third replacement | 48 × 4 | 192 |
| Fourth replacement | 192 × 4 | 768 |
The counts are three times the powers of 4: 3 × 1, 3 × 4, 3 × 16, 3 × 64, and 3 × 256. Powers of 4 follow repeated multiplication by 4, just as powers of 2 and 3 follow their own multipliers. If k is the number of replacement rounds, the initial triangle has k = 0 and the count after k rounds is 3 × 4ᵏ.
Problem
How many boundary segments appear after the third and fourth replacement rounds?
- 1.Begin with 3 segments. After one round there are 12, and after two there are 48.
- 2.For the third round, multiply 48 by 4: 48 × 4 = 192.
- 3.For the fourth, multiply 192 by 4: 192 × 4 = 768. The construction rule supplies the count even when a full drawing is difficult.
More pieces can mean smaller pieces
At each Koch replacement, every new segment has one third the length of its parent segment. The outline gains more segments, but those segments become smaller. Later changes are therefore increasingly fine. A simple sketch soon cannot show every detail clearly, even though the rule tells us exactly how the count grows.
This provides an important comparison. Stacked-square and stacked-triangle counts increase through larger odd additions. Koch segment counts increase through repeated multiplication by 4. Both are shape patterns, but their construction rules create different kinds of numerical growth.
Redraw each stacked-square and stacked-triangle stage, then add the next one and label its piece count. Sketch one Koch replacement carefully. For later Koch stages, explain why you may use a partial sketch and a numerical prediction instead of trying to fit every tiny segment on the page.
Quiz
A 7-by-7 grid contains how many smallest square cells?
How many smallest triangular pieces are in a four-row triangular stack?
Why do stacked triangular pieces give square-number totals?
What happens to one boundary segment in a Koch replacement?
What follows 3, 12, 48 in the Koch segment sequence?
Why is a later Koch stage difficult to draw completely?
Practice Problems
- Draw square grids with side lengths 2, 3, and 4. Count the smallest cells and show the newly added odd layer at each growth step.
- Draw a four-row triangular stack. Label the upward and downward pieces in each row and verify the total.
- Find the smallest-triangle total in a six-row stack. Compare it with the number of dots in a six-row triangular dot arrangement.
- A student counts ten upward triangles in a four-row stack and gives 10 as the total. Find the omitted downward triangles and correct the answer.
- Draw the replacement of a single Koch segment, showing the removed middle third and the four new boundary pieces.
- Write the first six Koch boundary-segment counts, including the initial triangle. Explain the multiplier.
- How many new boundary segments replace the 48 old ones in the next Koch stage? What is the net increase in the total count?
- Explain why counting a figure’s corners, boundary segments, and smallest cells can lead to different answers.
Key Takeaways
• Decide whether you are counting cells, points, or boundary segments before calculating. • An n-by-n square grid contains n × n smallest square cells. • Triangular stacks also have square-number totals because their rows contain 1, 3, 5, … smallest triangles. • A Koch replacement turns each segment into four shorter boundary segments. • Koch counts are 3, 12, 48, 192, …; a clear construction rule predicts counts beyond what we can easily draw.