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

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<math> \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 </math>
 
<math> \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 </math>
  
== Solution 1==
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==Solution==
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=== Solution 1===
 
There are <math>1+3+5+...+2003=1002^{2}=1004004</math> small equilateral triangles.  
 
There are <math>1+3+5+...+2003=1002^{2}=1004004</math> small equilateral triangles.  
  
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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=1,507,509 \Rightarrow \mathrm{(C)}</math>
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Therefore the total number of toothpicks is <math>1504503+3006=\boxed{\mathrm{(C)}\ 1,507,509}</math>
  
==Solution 2==
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===Solution 2===
We see that the bottom row of <math>2003</math> small triangles is formed from <math>1002</math> downward-facing triangles and <math>1001</math> upward-facing triangles. Since each downward-facing triangle uses three distinct toothpicks, and since the total number of downward-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=1507509</math>
+
We see that the bottom row of <math>2003</math> small triangles is formed from <math>1002</math> downward-facing triangles and <math>1001</math> upward-facing triangles. Since each downward-facing triangle uses three distinct toothpicks, and since the total number of downward-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>
  
 
== See Also ==
 
== See Also ==

Revision as of 13:57, 1 August 2011

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?

2003amc10a23.gif

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

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 $3\cdot 502503=\boxed{\mathrm{(C)}\ 1,507,509}$

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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