Difference between revisions of "2017 AMC 10A Problems/Problem 10"
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==Solution== | ==Solution== | ||
− | The triangle inequality generalizes to all polygons, so <math>x < 3+7+15</math> and <math>x+3+7 | + | The triangle inequality generalizes to all polygons, so <math>x < 3+7+15</math> and <math>15<x+3+7</math> yields <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 that are between those numbers, which gives <math>19-2=\boxed{\textbf{(B)}\ 17}</math> |
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+ | ==Video Solution== | ||
+ | https://youtu.be/pxg7CroAt20 | ||
+ | |||
+ | https://youtu.be/RFClyXKH49g | ||
+ | |||
+ | ~savannahsolver | ||
==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}} | ||
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+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 15:57, 23 July 2021
Contents
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 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
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.