AMC 10 · 2003 · #23

Grade 6 counting

Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Figure-N Growth Sequence

triangular-numberspattern-recognitionsystematic-enumeration easier-related-problempattern-recognition ↑ Prerequisites: triangular-numbers

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A large equilateral triangle is built from toothpicks laid out as rows of small congruent equilateral triangles, each toothpick forming one side. The sample figure has 3 rows, with 5 small triangles in the bottom row. How many toothpicks are needed when the bottom row consists of 2003 small triangles?

Pick an answer.

(A)

1,004,004

(B)

1,005,006

(C)

1,507,509

(D)

3,015,018

(E)

6,021,018

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

Counting toothpicks directly is a trap, because most toothpicks are shared between an up-pointing and a down-pointing small triangle, so naive counting double-counts. Tool #16 (Change Focus) fixes this with one clean observation: look only at the up-pointing triangles. Every toothpick is a side of exactly one up-pointing triangle, and no toothpick is shared between two up-pointing triangles. So the total number of toothpicks is exactly 3 times the number of up-pointing triangles — no overcounting to correct. Tool #5 (Look for a Pattern) then counts those up-pointing triangles row by row (1, 2, 3, … per row), and Tool #9 (Solve an Easier Related Problem) lets us test the whole idea on the tiny given figure before trusting it on 2003.

1STEP 1

Count the rows from the bottom row

Rows hold 1, 3, 5, … small triangles, so row $r$ holds $2r-1$ . Solving $2r-1=2003$ gives 1002 rows.

2r-1=2003 → r=1002 rows

2STEP 2

Look only at up-pointing triangles

Key move: each toothpick is a side of exactly one up-pointing triangle, and no two of those share one. So toothpicks = 3 × that count.

toothpicks = 3 × (number of up-pointing triangles)

3STEP 3

Test the idea on the sample figure

Test it on the 3-row figure: its up-pointing triangles number 1+2+3=6, so the rule predicts 3 × 6 = 18 toothpicks — matching the picture.

3×(1+2+3)=18 = 9_boundary + 9_inside

4STEP 4

Count all up-pointing triangles

The $k$ -th row holds $k$ up-pointing triangles, so add $1+2+\cdots+1002 = \frac{1002 \times 1003}{2}$ = 502,503.

Σ_k=1¹⁰⁰² k=(1002 · 1003)/2=502,503

5STEP 5

Multiply by three for the toothpick total

Each up-pointing triangle owns 3 toothpicks, so 3 × 502,503 = 1,507,509 toothpicks — choice (C).

3 × 502,503 = 1,507,509 (C)

Answer

1,507,509

The answer 1,507,509 is choice (C). A sanity check on size: choice (A) 1,004,004=1002² is exactly the number of small triangles, which is fewer than the toothpicks (every triangle carries 3 sides, most shared, so toothpicks should be roughly 1.5 times the triangle count) — and indeed 1,507,509 ≈ 1.5 × 1,004,004, which fits. A units-digit check also isolates (C): 1002 · 1003 ends in 6, halving gives a number ending in 3, and tripling ends in 9; only (C) ends in 9.

💡Key takeaway

When pieces share their edges, count the up-pointing pieces only — each edge belongs to exactly one of them, so the total edges are just three times that count.

Count the rows from the bottom row
Look only at up-pointing triangles
Test the idea on the sample figure
Count all up-pointing triangles
Multiply by three for the toothpick total

Count the rows from the bottom row

Look only at up-pointing triangles

Test the idea on the sample figure

Count all up-pointing triangles

Multiply by three for the toothpick total

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