Expressions using Letter-Numbers · Lesson 10 of 11
Matchstick Patterns and Growing Figures
“Derive growth formulas by counting new contributions, shared sides, squares, and cut points.”
• Build a formula from a first-step count and the number of later additions. • Explain connected-triangle and connected-square formulas using shared sides. • Show that two counting methods produce equivalent expressions. • Count squares in a growing X and state how vertices are counted. • Derive the rope-piece rule from distinct cuts in the illustrated bending pattern.
First Observe, Then Explain the Growth
Growing patterns become easier when you ask what is already present and what is added next. A chain of connected triangles uses 3 sticks at step 1, then 5, 7, and 9 at later steps. The increases are all 2 because a new triangle shares one side with the preceding arrangement.
| Step y | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Sticks | 3 | 5 | 7 | 9 | 11 | 13 |
| Growth description | 3 | 3 + 2 | 3 + 2 + 2 | 3 + 2 + 2 + 2 | 3 + four 2s | 3 + five 2s |
Step y contains the first triangle plus y − 1 further triangles. The first needs 3 sticks, and each further triangle adds 2. There are y − 1 additions because step 1 has already supplied the starting figure. Multiplying by y instead would count one extra growth operation.
Problem
Find the stick totals for steps 5, 33, 84, and 108.
- 1.Use M = 2y + 1, derived from the shared-side growth.
- 2.For steps 5 and 33: 2 × 5 + 1 = 11; 2 × 33 + 1 = 67.
- 3.For steps 84 and 108: 2 × 84 + 1 = 169; 2 × 108 + 1 = 217. No large drawing is required.
A Second Way to See the Triangle Count
The sticks in the triangle chain have two orientations. At step y there are y horizontal sticks, along the alternating top and bottom edges, and y + 1 diagonal sticks along the middle zigzag. Counting these two groups gives y + (y + 1) = 2y + 1.
Problem
Compare the growth count 3 + 2(y − 1) with the orientation count y + (y + 1).
- 1.Expand the growth count: 3 + 2y − 2.
- 2.Combine constants: 3 − 2 = 1, giving 2y + 1.
- 3.The orientation count also gives y + y + 1 = 2y + 1. At step 4 there are 4 horizontal and 5 diagonal sticks, totalling 9.
A numerical sequence suggests a rule, but the geometry explains it. If the triangles were separate, each would need all three sticks and the rule would be 3y. The connected arrangement is essential: shared sticks change the contribution of every triangle after the first.
Connected Squares Share One Side
A row of squares has a related growth pattern. The first square uses 4 sticks. A new square shares its joining side with the existing row, so it contributes just 3 new sticks. The first-step count changes from the triangle pattern, and so does the growth amount.
Problem
How many sticks make 10 squares in a row? Show a second counting route.
- 1.Growth count: 4 + 3(10 − 1) = 4 + 27 = 31.
- 2.There are 10 top horizontal sticks and 10 bottom horizontal sticks, giving 20.
- 3.There are 11 vertical boundary sticks, giving 20 + 11 = 31. In general, 2w + (w + 1) = 3w + 1.
A Growing X of Squares
Another pattern has one centre square and four diagonal arms. At step 1 each arm has one square, giving 5 squares altogether. At step 2 each arm has two, giving 9; at step 3 each arm has three, giving 13. Every new step adds one square on each of four arms.
Problem
Find the square totals at steps 4, 10, and 50.
- 1.The fixed part is one centre square. The repeated part is four arms, each containing n squares.
- 2.Step 4: 4 × 4 + 1 = 17. Step 10: 4 × 10 + 1 = 41.
- 3.Step 50: 4 × 50 + 1 = 201. These totals count squares, not corners or sticks.
Each square has four vertices, meaning its four corners. If you count the vertices of every square separately, including a shared point once for each square that uses it, the vertex total is 4S = 4(4n + 1) = 16n + 4. For step 4 this gives 68 vertex occurrences.
The vertex total 16n + 4 counts corners with multiplicity: a point shared by two squares is counted twice. Counting distinct geometric corner points is a different question. In this illustrated X, diagonal neighbours touch at corners, so state your counting convention before using a vertex formula.
Problem
At step 10 of the X, how many vertices are counted if every square contributes all four of its corners?
- 1.First count squares: 4 × 10 + 1 = 41.
- 2.Each square contributes four corner occurrences, so multiply 41 by 4.
- 3.The total is 164 vertex occurrences. This calculation does not remove repeated counts of a shared geometric point.
Bending and Cutting a Rope
For the illustrated rope pattern, a straight open rope has one strand across the cut line. Bending it once creates two parallel strands, and bending it twice creates three. Each new bend adds one further strand across the same cut line. The cutting line passes through each strand once, away from the bend itself.
One cut at an interior point of a straight rope gives two pieces. Two distinct cut points along the original rope give three pieces. In general, k distinct interior cuts produce k + 1 pieces: the cuts divide the rope into a segment before the first cut, segments between cuts, and a segment after the final cut.
Problem
How many pieces result from 10 bends followed by the illustrated single cutting line?
- 1.Ten bends create 10 + 1 = 11 strands across the cut line in this pattern.
- 2.Cutting those strands creates 11 distinct cut points along the original open rope.
- 3.The number of pieces is 11 + 1 = 12. The formula r + 2 gives the same result at r = 10.
The rule follows from this particular arrangement of bends and cuts. Repeatedly folding an entire bundle in half would create a different strand pattern. Look at how the figure grows before choosing a formula; the word “fold” alone does not determine a count.
Use drawn segments or craft sticks to build the first three connected-square steps. Record total sticks and new sticks added. Explain which side is shared, derive the general expression, then check step 4. Use the same observation–reasoning–generalisation method for a new arrangement that stays within these counting ideas.
Check Your Understanding
Use the ideas from this lesson to choose an answer. Explain your choice to yourself before opening the explanations below.
Quiz
Why does each additional triangle in the shown chain require two new sticks?
How many sticks are needed at step 33 of the connected-triangle pattern?
How many sticks make 10 connected squares in one row?
How many squares are in step 50 of the X pattern?
For the illustrated rope pattern, r bends create r + 1 cut points. How many pieces result?
The triangle still has three sides, but one side is already present.
Practice Problems
- Write both triangle-count formulas and prove they are equivalent.
- Find the connected-triangle stick totals for steps 84 and 108.
- Count horizontal and diagonal sticks at steps 3 and 4 of the triangle chain.
- Write the connected-square formula and find its value for w = 25.
- How would the formula change for w separate squares sharing no sticks?
- Find square totals for X-pattern steps 4, 10, and 50.
- Write the vertex formula when every square contributes four vertices even at shared points. Find the total at step 4.
- For the illustrated rope pattern, find pieces after 0, 1, 2, and 10 bends.
- Explain why the growth formula uses y − 1 additions after the first triangle.
- Why should the phrase “number of vertices” be clarified in the X pattern?
3 + 2(y − 1) = 3 + 2y − 2 = 2y + 1. Both count the same shared-side triangle chain.
Key Takeaways
• Observe how the figure changes before writing its formula. • Separate the first-step count from later repeated additions. • Shared sides reduce the number of new sticks contributed. • Different counting routes can justify equivalent expressions. • State whether shared vertices are counted once or with multiplicity. • The rope-piece formula depends on the illustrated strand and cutting arrangement.