Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 AMC 10A Problems/Problem 10"
(→‎Solution)
 
(9 intermediate revisions by 9 users not shown)
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{\textbf{(B)}\ 17}</math>
+
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>
 +
 
 +
==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}}
 +
 +
[[Category:Introductory Geometry Problems]]

Latest revision as of 15:57, 23 July 2021

Problem

Joy has $30$ thin rods, one each of every integer length from $1$ cm through $30$ cm. She places the rods with lengths $3$ cm, $7$ cm, and $15$ 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?

$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20$

Solution

The triangle inequality generalizes to all polygons, so $x < 3+7+15$ and $15<x+3+7$ yields $5<x<25$. Now, we know that there are $19$ numbers between $5$ and $25$ exclusive, but we must subtract $2$ to account for the 2 lengths already used that are between those numbers, which gives $19-2=\boxed{\textbf{(B)}\ 17}$

Video Solution

https://youtu.be/pxg7CroAt20

https://youtu.be/RFClyXKH49g

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_10&oldid=159007"