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2003 AMC 10A Problem 23

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2003 AMC 10A Problem 23AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Counting · difficulty 4 of 5 · about 5 min · position P21-25

A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have 33 rows of small congruent equilateral triangles, with 55 small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of 20032003 small equilateral triangles?

figure

  1. A)

    1,004,0041,004,004

  2. B)

    1,005,0061,005,006

  3. C)

    1,507,5091,507,509

  4. D)

    3,015,0183,015,018

  5. E)

    6,021,0186,021,018

Answer

C

Key insight

Every toothpick is a side of exactly one upward-pointing small triangle, and there are 1 + 2 + ... + 1002 of those, so multiply that sum by 3.

Solution

Row kk from the top contains 2k−12k - 1 small triangles: kk of them point up and k−1k - 1 point down. A base row of 2003=2⋅1002−12003 = 2 \cdot 1002 - 1 triangles means there are 10021002 rows.

Look only at the upward-pointing triangles. Each toothpick in the figure is a side of exactly one of them: the horizontal toothpicks are the bottoms of upward triangles, and the slanted toothpicks are their left or right sides. (A downward triangle borrows all three of its sides from neighboring upward triangles.) So no toothpick is counted twice if we count three per upward triangle.

Number of upward triangles: 1+2+⋯+1002=1002⋅10032=501⋅1003=5025031 + 2 + \cdots + 1002 = \dfrac{1002 \cdot 1003}{2} = 501 \cdot 1003 = 502503.

Toothpicks: 3⋅502503=15075093 \cdot 502503 = 1507509.

Check on the given figure (33 rows): 3⋅(1+2+3)=183 \cdot (1 + 2 + 3) = 18 toothpicks, which matches a direct count.

The answer is (C) 1,507,509\boxed{\textbf{(C)}\ 1,507,509}.

Why this works

When objects are shared among several shapes, find a family of shapes that partitions them: here each edge belongs to exactly one upward triangle, so counting is a clean multiplication. Verifying the formula on the small pictured case guards against the two classic slips (miscounting rows, or double counting interior edges).

Alternative approach

Small cases: base rows of 1,3,5,71, 3, 5, 7 triangles need 3,9,18,303, 9, 18, 30 toothpicks, i.e. 3⋅(m+12)3 \cdot \binom{m+1}{2} for m=1,2,3,4m = 1, 2, 3, 4 rows. With m=1002m = 1002 this is 3⋅5025033 \cdot 502503. Even without finishing the multiplication, the result ends in 99, and only choice (C) does.

The trap

Using 2003 as the number of rows instead of the number of triangles in the base row, which yields 6,021,018, choice (E).

Common mistakes

  • Using 2003 as the number of rows instead of the number of triangles in the base row, which yields 6,021,018, choice (E).
  • Multiplying the total number of small triangles 100221002^2 by 33, which counts every interior toothpick twice.

Techniques

Map the objects to something easier to count · Compute small cases, spot the pattern, generalize