Difference between revisions of "2017 AMC 10A Problems/Problem 10"
(→Solution 2) |
Syonaggarwal (talk | contribs) (→Video Solution) |
||
| Line 13: | Line 13: | ||
~savannahsolver | ~savannahsolver | ||
| + | ==EASIEST and FASTEST SOLUTION,== | ||
| + | https://www.youtube.com/watch?v=7u4uE2Fh1NY&feature=youtu.be | ||
==See Also== | ==See Also== | ||
Latest revision as of 11:00, 19 August 2025
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 
Video Solution
~savannahsolver
EASIEST and FASTEST SOLUTION,
https://www.youtube.com/watch?v=7u4uE2Fh1NY&feature=youtu.be
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
