2017 AMC 10A Problems/Problem 10
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 1
The triangle inequality generalizes to all polygons, so 
and 
yields 
. Now, we know that there are 
numbers between 
and 
exclusive, but we must subtract 
to account for the 2 lengths already used that are between those numbers, which gives 
Solution 2
The quadrilateral inequality states that 
, where 
and 
and the side lengths, and is the largest one. Now, we can just plug these values in. We have cases. First if this other side length (we will call it 
) is the largest side length we have: ![\[3+7+15 \ge x \rightarrow 25 \ge x\]](http://latex.artofproblemsolving.com/8/1/e/81eef33c4b66635555029469bb622c504b004336.png)
, so there are ![\[25,26,27,28,29,30\]](http://latex.artofproblemsolving.com/f/4/6/f461fd3ab903cb36e9645d66443c427d612b47f4.png)
, so 
values. Now, we have case which is if is the largest side length. We can say that ![\[3+7+x \ge 15 \rightarrow 10+x \ge 15\]](http://latex.artofproblemsolving.com/7/8/c/78c98b7e4bc3f93cf33522828d197ec92be36df1.png)
, so we have the lower bound to be and the upper bound to be , so we have 
values for this case. When adding these cases we get a total of 
Video Solution
~savannahsolver
See Also
| 2017 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
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| All AMC 10 Problems and Solutions | ||
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