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

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==Solution 2==
 
==Solution 2==
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The first row of triangles has <math>1</math> upward-facing triangle, the second row has <math>2</math> upward-facing triangles, the third row has <math>3</math> upward-facing triangles, and so on having <math>n</math> upward-facing triangles in the <math>n^\text{th}</math> row. The last row with <math>2003</math> small triangles has <math>1002^\text{th}</math> upward-facing triangles. By Gauss's formula, the number of the upward-facing triangles in the entire triangle are now <math>\frac{1002\times1003}{2}</math>, meaning that the number of toothpicks are <math>\frac{1002\times1003}{2}\times3</math>, or <math>\boxed{\text{C}}</math>.
+
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 <math>1</math> upward-facing triangle, the second row has <math>2</math> upward-facing triangles, the third row has <math>3</math> upward-facing triangles, and so on having <math>n</math> upward-facing triangles in the <math>n^\text{th}</math> row. The last row with <math>2003</math> small triangles has <math>1002</math> upward-facing triangles. By Gauss's formula, the number of the upward-facing triangles in the entire triangle are now <math>\frac{1002\times1003}{2}</math>, meaning that the number of toothpicks are <math>\frac{1002\times1003}{2}\times3</math>, or <math>\boxed{\text{C}}</math>.
  
 
~mathpro12345
 
~mathpro12345

Revision as of 18:42, 17 December 2020

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 $\boxed{\text{C}}$

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 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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