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== | |
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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~dolphin7 | ~dolphin7 | ||
− | + | ==Solution 2== | |
− | 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 | + | 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 | ||
+ | |||
+ | ===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> | ||
+ | |||
+ | |||
+ | ===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 18:01, 28 January 2021
Contents
Problem
A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have rows of small congruent equilateral triangles, with 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 small equilateral triangles?
Solution 1
There are small equilateral triangles.
Each small equilateral triangle needs toothpicks to make it.
But, each toothpick that isn't one of the toothpicks on the outside of the large equilateral triangle is a side for small equilateral triangles.
So, the number of toothpicks on the inside of the large equilateral triangle is
Therefore the total number of toothpicks is ~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 upward-facing triangle, the second row has upward-facing triangles, the third row has upward-facing triangles, and so on having upward-facing triangles in the row. The last row with small triangles has upward-facing triangles. By Gauss's formula, the number of the upward-facing triangles in the entire triangle are now , meaning that the number of toothpicks are , or .
~mathpro12345
Note
You don't have to calculate the value of , and you can use units digits to find the answer easily. The units digit of is , and has a unit digit of after being divided by . Then this is multiplied by , now the final number ending with a . This leaves only one answer choice possible, which is
Solution 3
Test out some fewer cases first.
When there is just 1 equilateral triangle in the base, you need toothpicks. When there are 3 equilateral triangles in the base, you need toothpicks in all. When there are 5 equilateral triangles in the base, you need toothpicks in all. When there are 7 equilateral triangles in the base, you need toothpicks in all.
Taking the finite differences, we get It forms a linear equations. This means the original numbers form a quadratic.
Let the quadratic be where
Then, we have the following points:
We can plug these values into , giving:
Solving gives So,
For our problem, we need it when there are equilateral triangles in the base. For the quadratic, the corresponding -value would be . So, our answer is simply:
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.