# Difference between revisions of "2003 AMC 10A Problems/Problem 23"

## Problem

A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure we have $3$ rows of small congruent equilateral triangles, with $5$ 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 $2003$ small equilateral triangles?

$\mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018$

## Solution

### Solution 1

There are $1+3+5+...+2003=1002^{2}=1004004$ small equilateral triangles.

Each small equilateral triangle needs $3$ toothpicks to make it.

But, each toothpick that isn't one of the $1002\cdot3=3006$ toothpicks on the outside of the large equilateral triangle is a side for $2$ small equilateral triangles.

So, the number of toothpicks on the inside of the large equilateral triangle is $\frac{10040004\cdot3-3006}{2}=1504503$

Therefore the total number of toothpicks is $1504503+3006=\boxed{\mathrm{(C)}\ 1,507,509}$

### Solution 2

We see that the bottom row of $2003$ small triangles is formed from $1002$ downward-facing triangles and $1001$ upward-facing triangles. Since each downward-facing triangle uses three distinct toothpicks, and since the total number of downward-facing triangles is $1002+1001+...+1=\frac{1003\cdot1002}{2}=502503$, we have that the total number of toothpicks is $3\cdot 502503=\boxed{\mathrm{(C)}\ 1,507,509}$

### Solution 3

Experimenting a bit we find that the number of toothpicks needs for a triangle with $1$, $2$ and $3$ rows is $1\cdot{3}$, $3\cdot{3}$ and $6\cdot{3}$ respectively. Since $1$, $3$ and $6$ are triangular numbers we know that depending on how many rows there are in the triangle, the number we multiply by $3$ to find total no.toothpicks is the corresponding triangular number. Since the triangle in question has $2n-1=2003\implies{n=1002}$ rows, we can use $\frac{n(n+1)}{2}$ to find the triangular number for that row and multiply by $3$, hence finding the total no.toothpicks; this is just $\frac{3\cdot{1002}\cdot{1003}}{2}=3\cdot{512}\cdot{1003}=\boxed{\mathrm{(C)}\ 1,507,509}$.