Difference between revisions of "2017 AMC 12A Problems/Problem 6"
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==Solution== | ==Solution== | ||
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| + | 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</math> = 30. This means she can use the 19 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> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=A|num-b=5|num-a=7}} | {{AMC12 box|year=2017|ab=A|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:05, 8 February 2017
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 
= 30. This means she can use the 19 possible integer rod lengths that fall into ![$[6, 24]$](http://latex.artofproblemsolving.com/b/9/3/b93d84871b0a3d11e3bfa6cac25df7ac7edc0b5e.png)
. However, she has already used the rods of length 
cm and 
cm so the answer is 
See Also
| 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. 
