# 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?

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

## Solution 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 $1504503+3006=\boxed{\mathrm{(C)}\ 1,507,509}$ ~dolphin7

## Solution 2

We just need to count upward facing triangles because if we exclude the downward-facing triangles, we won't be overcounting any toothpicks. The first row of triangles has $1$ upward-facing triangle, the second row has $2$ upward-facing triangles, the third row has $3$ upward-facing triangles, and so on having $n$ upward-facing triangles in the $n^\text{th}$ row. The last row with $2003$ small triangles has $1002$ upward-facing triangles. By Gauss's formula, the number of the upward-facing triangles in the entire triangle are now $\frac{1002\times1003}{2}$, meaning that the number of toothpicks are $\frac{1002\times1003}{2}\times3$, or $\boxed{\text{C}}$.

~mathpro12345

### Note

You don't have to calculate the value of $\frac{1002\times1003}{2}\times3$, and you can use units digits to find the answer easily. The units digit of $1002\times1003$ is $6$, and has a unit digit of $3$ after being divided by $2$. Then this is multiplied by $3$, now the final number ending with a $9$. This leaves only one answer choice possible, which is $\boxed{\text{C}}$

### Solution 3

Test out some fewer cases first.

When there is just 1 equilateral triangle in the base, you need $3$ toothpicks. When there are 3 equilateral triangles in the base, you need $9$ toothpicks in all. When there are 5 equilateral triangles in the base, you need $18$ toothpicks in all. When there are 7 equilateral triangles in the base, you need $30$ toothpicks in all.

Taking the finite differences, we get $6, 9, 12.$ It forms a linear equations. This means the original numbers $(3, 9, 18, 30)$ form a quadratic.

Let the quadratic be $y = ax^2 + bx + c$ where $y = 2* \text{equilateral triangles in base} - 1.$

Then, we have the following points: $(1, 3), (2, 9), (3, 18), (4, 30).$

We can plug these values into $y = ax^2 + bx + c$, giving:

$$a + b + c = 3, 4a + b + c = 9, 9a + 3b + c = 18.$$

Solving gives $a = b = 1.5, c = 0.$ So, $$y = 1.5x^2 + 1.5x.$$

For our problem, we need it when there are $2003$ equilateral triangles in the base. For the quadratic, the corresponding $x$-value would be $\frac{2003 + 1}{2} = 1002.$. So, our answer is simply: $$1.5 * 1002^2 + 1.5*1002 = \boxed{1507509}.$$