Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
 
(2 intermediate revisions by 2 users not shown)
Line 31: Line 31:
 
draw(circle(p,0.1));
 
draw(circle(p,0.1));
 
}</asy>
 
}</asy>
 +
 
<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>
  
Line 53: Line 54:
 
===Note===
 
===Note===
 
You don't have to calculate the value of <math>\frac{1002\times1003}{2}\times3</math>, and you can use units digits to find the answer easily. The units digit of <math>1002\times1003</math> is <math>6</math>, and has a unit digit of <math>3</math> after being divided by <math>2</math>. Then this is multiplied by <math>3</math>, now the final number ending with a <math>9</math>. This leaves only one answer choice possible, which is <math>\boxed{\text{C}}</math>
 
You don't have to calculate the value of <math>\frac{1002\times1003}{2}\times3</math>, and you can use units digits to find the answer easily. The units digit of <math>1002\times1003</math> is <math>6</math>, and has a unit digit of <math>3</math> after being divided by <math>2</math>. Then this is multiplied by <math>3</math>, now the final number ending with a <math>9</math>. This leaves only one answer choice possible, which is <math>\boxed{\text{C}}</math>
 +
 +
 +
===Solution 3===
 +
 +
Test out some fewer cases first.
 +
 +
When there is just 1 equilateral triangle in the base, you need <math>3</math> toothpicks.
 +
When there are 3 equilateral triangles in the base, you need <math>9</math> toothpicks in all.
 +
When there are 5 equilateral triangles in the base, you need <math>18</math> toothpicks in all.
 +
When there are 7 equilateral triangles in the base, you need <math>30</math> toothpicks in all.
 +
 +
Taking the finite differences, we get <math>6, 9, 12.</math> It forms a linear equations. This means the original numbers <math>(3, 9, 18, 30)</math> form a quadratic.
 +
 +
Let the quadratic be <math>y = ax^2 + bx + c</math> where <math>y = 2* \text{equilateral triangles in base} - 1.</math>
 +
 +
Then, we have the following points: <math>(1, 3), (2, 9), (3, 18), (4, 30).</math>
 +
 +
We can plug these values into <math>y = ax^2 + bx + c</math>, giving:
 +
 +
<cmath>a + b + c = 3, 4a + b + c = 9, 9a + 3b + c = 18.</cmath>
 +
 +
Solving gives <math>a = b = 1.5, c = 0.</math> So, <cmath>y = 1.5x^2 + 1.5x.</cmath>
 +
 +
For our problem, we need it when there are <math>2003</math> equilateral triangles in the base. For the quadratic, the corresponding <math>x</math>-value would be <math>\frac{2003 + 1}{2} = 1002.</math>. So, our answer is simply: <cmath>1.5 * 1002^2 + 1.5*1002 = \boxed{1507509}.</cmath>
 +
 +
 +
== Video Solution (Meta-Solving Technique) ==
 +
https://youtu.be/GmUWIXXf_uk?t=494
 +
 +
~ pi_is_3.14
  
 
== See Also ==
 
== See Also ==

Latest revision as of 17:01, 28 January 2021

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: \[1.5 * 1002^2 + 1.5*1002 = \boxed{1507509}.\]


Video Solution (Meta-Solving Technique)

https://youtu.be/GmUWIXXf_uk?t=494

~ pi_is_3.14

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_23&oldid=143657"