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

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So, the number of toothpicks on the inside of the large equilateral triangle is <math>\frac{10040004\cdot3-3006}{2}=1504503</math>
 
So, the number of toothpicks on the inside of the large equilateral triangle is <math>\frac{10040004\cdot3-3006}{2}=1504503</math>
  
Therefore the total number of toothpicks is <math>1504503+3006=\boxed{\mathrm{(C)}\ 1,507,509}</math>
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Therefore the total number of toothpicks is <math>1504503+3006=\boxed{\mathrm{(C)}\ 1,507,509}</math>  
 +
~dolphin7
  
 
===Solution 2===
 
===Solution 2===
We see that the bottom row of <math>2003</math> small triangles are formed from <math>1002</math> upward-facing triangles and <math>1001</math> downward-facing triangles. Since each upward-facing triangle uses three distinct toothpicks, and since the total number of upward-facing triangles is <math>1002+1001+...+1=\frac{1003\cdot1002}{2}=502503</math>, we have that the total number of toothpicks is <math>3\cdot 502503=\boxed{\mathrm{(C)}\ 1,507,509}</math>
+
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. By Gauss's formula, the number of the upward-facing triangles in the entire triangle are <math>\frac{1002\times1003}{2}</math>
  
===Solution 3===
+
~mathpro12345
Experimenting a bit we find that the number of toothpicks needs a triangle with <math>1</math>, <math>2</math> and <math>3</math> rows is <math>1\cdot{3}</math>, <math>3\cdot{3}</math> and <math>6\cdot{3}</math> respectively. Since <math>1</math>, <math>3</math> and <math>6</math> are triangular numbers we know that depending on how many rows there are in the triangle, the number we multiply by <math>3</math> to find total no.toothpicks is the corresponding triangular number. Since the triangle in question has <math>2n-1=2003\implies{n=1002}</math> rows, we can use <math>\frac{n(n+1)}{2}</math> to find the triangular number for that row and multiply by <math>3</math>, hence finding the total no.toothpicks; this is just <math>\frac{3\cdot{1002}\cdot{1003}}{2}=3\cdot{501}\cdot{1003}=\boxed{\mathrm{(C)}\ 1,507,509}</math>.
 
===Note===
 
In the final step of the problem, we know that the units digit of the answer must <math>9</math>, so the only answer choice applicable must be <math>\boxed{\mathrm{(C)}\ 1,507,509}</math>. This saves you the time it takes to compute <math>3\cdot{501}\cdot{1003}</math>.
 
 
 
Alternatively, we can note that <math>\frac{3}{2}\cdot{1002}\cdot{1003}\approx 1.5 \cdot 1,000,000 = 1,500,000 \approx \boxed{\mathrm{(C)}\ 1,507,509}</math>.~ dolphin7
 
 
 
===Solution 4===
 
We first see how many small equilateral triangles share a side with the bigger triangle, here we see there are <math>3</math> such triangles in case of the perimeter being <math>3</math>, after some observation, we find that if the perimeter is <math>n</math> the no. of triangles is <math>\frac{n(n-1)}{2}</math>, we know that the perimeter when there are <math>2003</math> base triangles is, <math>\frac{2003+1}{2}</math> which is <math>1002</math>, so the no. of small triangles would be <math>\frac{1001\cdot{1002}}{2}</math>, hence the total no. of toothpicks is, <math>3({501\cdot1001}+1002)</math> which is <math>\boxed{1,507,509}</math>. ~RMOAspirantFaraz
 
  
 
== See Also ==
 
== See Also ==

Revision as of 13:38, 6 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

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

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. By Gauss's formula, the number of the upward-facing triangles in the entire triangle are $\frac{1002\times1003}{2}$

~mathpro12345

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
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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