Difference between revisions of "2017 AMC 10A Problems/Problem 10"
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | The triangle inequality generalizes to all polygons, so <math>x < 3+7+15</math> and <math>x+3+7>15</math> to get <math>5<x<25</math>. Now, we know that there are <math>19</math> numbers between <math>5</math> and <math>25</math> exclusive, but we must subtract <math>2</math> to account for the 2 lengths already used, which gives <math>19-2=\boxed{17 | + | The triangle inequality generalizes to all polygons, so <math>x < 3+7+15</math> and <math>x+3+7>15</math> to get <math>5<x<25</math>. Now, we know that there are <math>19</math> numbers between <math>5</math> and <math>25</math> exclusive, but we must subtract <math>2</math> to account for the 2 lengths already used, which gives <math>19-2=\boxed{\textbf{(B)}\ 17}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=A|num-b=9|num-a=11}} | {{AMC10 box|year=2017|ab=A|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:27, 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. 
