During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

Difference between revisions of "2017 AMC 10A Problems/Problem 10"

m (See Also)
(Solution)
 
(6 intermediate revisions by 6 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 that are between those numbers, 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 (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS