# 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?

$[asy] unitsize(15mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair Ap=(0,0), Bp=(1,0), Cp=(2,0), Dp=(3,0), Gp=dir(60); pair Fp=shift(Gp)*Bp, Ep=shift(Gp)*Cp; pair Hp=shift(Gp)*Gp, Ip=shift(Gp)*Fp; pair Jp=shift(Gp)*Hp; pair[] points={Ap,Bp,Cp,Dp,Ep,Fp,Gp,Hp,Ip,Jp}; draw(Ap--Dp--Jp--cycle); draw(Gp--Bp--Ip--Hp--Cp--Ep--cycle); for(pair p : points) { fill(circle(p, 0.07),white); } pair[] Cn=new pair[5]; Cn[0]=centroid(Ap,Bp,Gp); Cn[1]=centroid(Gp,Bp,Fp); Cn[2]=centroid(Bp,Fp,Cp); Cn[3]=centroid(Cp,Fp,Ep); Cn[4]=centroid(Cp,Ep,Dp); label("1",Cn[0]); label("2",Cn[1]); label("3",Cn[2]); label("4",Cn[3]); label("5",Cn[4]); for (pair p : Cn) { draw(circle(p,0.1)); }[/asy]$ $\mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018$

## 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 are formed from $1002$ upward-facing triangles and $1001$ downward-facing triangles. Since each upward-facing triangle uses three distinct toothpicks, and since the total number of upward-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 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 $\frac{3\cdot{1002}\cdot{1003}}{2}=3\cdot{501}\cdot{1003}=\boxed{\mathrm{(C)}\ 1,507,509}$.

### Note

In the final step of the problem, we know that the units digit of the answer must $9$, so the only answer choice applicable must be $\boxed{\mathrm{(C)}\ 1,507,509}$. This saves you the time it takes to compute $3\cdot{501}\cdot{1003}$.

Alternatively, we can note that $\frac{3}{2}\cdot{1002}\cdot{1003}\approx 1.5 \cdot 1,000,000 = 1,500,000 \approx \boxed{\mathrm{(C)}\ 1,507,509}$.~ dolphin7

### Solution 4

We first see how many small equilateral triangles share a side with the bigger triangle, here we see there are $3$ such triangles in case of the perimeter being $3$, after some observation, we find that if the perimeter is $n$ the no. of triangles is $\frac{n(n-1)}{2}$, we know that the perimeter when there are $2003$ base triangles is, $\frac{2003+1}{2}$ which is $1002$, so the no. of small triangles would be $\frac{1001\cdot{1002}}{2}$, hence the total no. of toothpicks is, $3({501\cdot1001}+1002)$ which is $\boxed{1,507,509}$. ~RMOAspirantFaraz

 2003 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 24 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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