Difference between revisions of "2017 AMC 10A Problems/Problem 10"
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Joy has <math>30</math> thin rods, one each of every integer length from <math>1</math> cm through <math>30</math> cm. She places the rods with lengths <math>3</math> cm, <math>7</math> cm, and <math>15</math> cm 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</math> cm through <math>30</math> cm. She places the rods with lengths <math>3</math> cm, <math>7</math> cm, and <math>15</math> cm 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>\ | + | <math>\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20</math> |
==Solution== | ==Solution== | ||
Revision as of 18:26, 8 February 2017
Problem
Joy has 
thin rods, one each of every integer length from 
cm through cm. She places the rods with lengths 
cm, 
cm, and 
cm 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 triangle inequality generalizes to all polygons, so 
and 
to get 
. Now, we know that there are 
numbers between 
and 
exclusive, but we must subtract 
to account for the 2 lengths already used, which gives 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
