Difference between revisions of "2017 AMC 12A Problems/Problem 6"
(Created page with "Joy has <math>30</math> thin rods, one each of every integer length from <math>1 \text{ cm}</math> through <math>30 \text{ cm}</math>. She places the rods with lengths <math>3...") |
m (Added a video solution) |
||
(11 intermediate revisions by 10 users not shown) | |||
Line 1: | Line 1: | ||
+ | ==Problem== | ||
+ | |||
Joy has <math>30</math> thin rods, one each of every integer length from <math>1 \text{ cm}</math> through <math>30 \text{ cm}</math>. She places the rods with lengths <math>3 \text{ cm}</math>, <math>7 \text{ cm}</math>, and <math>15 \text{cm}</math> on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod? | Joy has <math>30</math> thin rods, one each of every integer length from <math>1 \text{ cm}</math> through <math>30 \text{ cm}</math>. She places the rods with lengths <math>3 \text{ cm}</math>, <math>7 \text{ cm}</math>, and <math>15 \text{cm}</math> on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod? | ||
+ | <math>\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 17 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 19 \qquad\textbf{(E)}\ 20</math> | ||
+ | |||
+ | ==Solution== | ||
− | <math> | + | The quadrilateral cannot be a straight line. Thus, the fourth side must be longer than <math>15 - (3 + 7) = 5</math> and shorter than <math>15 + 3 + 7 = 25</math>. This means Joy can use the <math>19</math> possible integer rod lengths that fall into <math>[6, 24]</math>. However, she has already used the rods of length <math>7</math> cm and <math>15</math> cm so the answer is <math>19 - 2 = 17</math> <math>\boxed{\textbf{(B)}}</math> |
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=1Vi1100kO9o | ||
+ | |||
+ | ~Math4All999 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2017|ab=A|num-b=9|num-a=11}} | ||
+ | {{AMC12 box|year=2017|ab=A|num-b=5|num-a=7}} | ||
+ | {{MAA Notice}} |
Latest revision as of 06:09, 14 September 2024
Contents
Problem
Joy has thin rods, one each of every integer length from through . She places the rods with lengths , , and on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?
Solution
The quadrilateral cannot be a straight line. Thus, the fourth side must be longer than and shorter than . This means Joy can use the possible integer rod lengths that fall into . However, she has already used the rods of length cm and cm so the answer is
Video Solution
https://www.youtube.com/watch?v=1Vi1100kO9o
~Math4All999
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 |
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.