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

Revision as of 17:23, 29 December 2024

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 1

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 $19-2=\boxed{\textbf{(B)}\ 17}$

Solution 2

The quadrilateral inequality states that $a+b+c \ge d$ , where $a,b,c,$ and $d$ and the side lengths, and $d$ is the largest one. Now, we can just plug these values in. We have $2$ cases. First if this other side length (we will call it $x$ ) is the largest side length we have: $3+7+15 \ge x \rightarrow 25 \ge x$ , so there are $25,26,27,28,29,30$ , so $6$ values. Now, we have case $2$ which is if $15$ is the largest side length. We can say that $3+7+x \ge 15 \rightarrow 10+x \ge 15$ , so we have the lower bound to be $5$ and the upper bound to be $15$ , so we have $15-5+1 = 11$ values for this case. When adding these $2$ cases we get a total of $11+6 = \boxed{\textbf{(B)}\ 17}$

Video Solution

https://youtu.be/pxg7CroAt20

https://youtu.be/RFClyXKH49g

~savannahsolver

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.

@@ Line 4: / Line 4: @@
 <math>\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20</math>
-==Solution==
+==Solution 1==
 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>
+==Solution 2==
+The quadrilateral inequality states that <math>a+b+c \ge d</math>, where <math>a,b,c,</math> and <math>d</math> and the side lengths, and <math>d</math> is the largest one. Now, we can just plug these values in. We have <math>2</math> cases. First if this other side length (we will call it <math>x</math>) is the largest side length we have: <cmath>3+7+15 \ge x \rightarrow 25 \ge x</cmath>, so there are <cmath>25,26,27,28,29,30</cmath>, so <math>6</math> values. Now, we have case <math>2</math> which is if <math>15</math> is the largest side length. We can say that <cmath>3+7+x \ge 15 \rightarrow 10+x \ge 15</cmath>, so we have the lower bound to be <math>5</math> and the upper bound to be <math>15</math>, so we have <math>15-5+1 = 11</math> values for this case. When adding these <math>2</math> cases we get a total of <math>11+6 = \boxed{\textbf{(B)}\ 17}</math>
 ==Video Solution==

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