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 "2015 AMC 10B Problems/Problem 13"

m (→‎Solution)
(→‎Solution 2 (very similar to Solution 1))
 
(6 intermediate revisions by 5 users not shown)
Line 5: Line 5:
 
<math>\textbf{(A) } 20 \qquad\textbf{(B) } \dfrac{360}{17} \qquad\textbf{(C) } \dfrac{107}{5} \qquad\textbf{(D) } \dfrac{43}{2} \qquad\textbf{(E) } \dfrac{281}{13} </math>
 
<math>\textbf{(A) } 20 \qquad\textbf{(B) } \dfrac{360}{17} \qquad\textbf{(C) } \dfrac{107}{5} \qquad\textbf{(D) } \dfrac{43}{2} \qquad\textbf{(E) } \dfrac{281}{13} </math>
  
−
==Solution==
+
==Solution 1==
−
We find the x-intercepts and the y-intercepts to find the intersections of the axes and the line. If <math>x=0</math>, then <math>y=12</math>. If <math>y</math> is <math>0</math>, then <math>x=5</math>. Our three vertices are <math>(0,0)</math>, <math>(5,0)</math>, and <math>(0,12)</math>. Two of our altitudes are <math>5</math> and <math>12</math>. Since the area of the triangle is <math>30</math>, letting <math>h</math> be our hypotenuse, our final altitude has to be <math>30(2)/h</math>. The final altitude is <math>\frac{60}{13}</math>, , so the sum of our altitudes is <math>\frac{60+156+65}{13}=\boxed{\textbf{(E)} \dfrac{281}{13}}</math>.
+
We find the <math>x</math>-intercepts and the <math>y</math>-intercepts to find the intersections of the axes and the line. If <math>x=0</math>, then <math>y=12</math>. If <math>y</math> is <math>0</math>, then <math>x=5</math>. Our three vertices are <math>(0,0)</math>, <math>(5,0)</math>, and <math>(0,12)</math>. Two of our altitudes are <math>5</math> and <math>12</math>, and since it is a <math>5</math>-<math>12</math>-<math>13</math> right triangle, the hypotenuse is <math>13</math>. Since the area of the triangle is <math>30</math>, so our final altitude is <math>\frac{30(2)}{13}=\frac{60}{13}</math>. The sum of our altitudes is <math>\frac{60+156+65}{13}=\boxed{\textbf{(E)} \dfrac{281}{13}}</math>. Note that there is no need to calculate the final answer after we know that the third altitude has length <math>\frac{60}{13}</math> since <math>E</math> is the only choice with a denominator of <math>13</math> and <math>13</math> is relatively prime to <math>5</math> and <math>12</math>.
 +
 
 +
==Video Solution 1==
 +
https://youtu.be/zFHtdWEhriQ
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Solution 2 (very similar to Solution 1)==
 +
Noticing that the line has coefficients <math>12</math> and <math>5</math>, we can suspect that we have a <math>5</math>-<math>12</math>-<math>13</math> triangle on our hands. Because we want the sum of the altitudes, the third altitude that is not an axis must have a denominator of <math>13</math>. Since the other altitudes are integers, we choose the option with <math>13</math> as the denominator, namely <math>E</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 16:14, 2 August 2022

Problem

The line $12x+5y=60$ forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?

$\textbf{(A) } 20 \qquad\textbf{(B) } \dfrac{360}{17} \qquad\textbf{(C) } \dfrac{107}{5} \qquad\textbf{(D) } \dfrac{43}{2} \qquad\textbf{(E) } \dfrac{281}{13}$

Solution 1

We find the $x$-intercepts and the $y$-intercepts to find the intersections of the axes and the line. If $x=0$, then $y=12$. If $y$ is $0$, then $x=5$. Our three vertices are $(0,0)$, $(5,0)$, and $(0,12)$. Two of our altitudes are $5$ and $12$, and since it is a $5$-$12$-$13$ right triangle, the hypotenuse is $13$. Since the area of the triangle is $30$, so our final altitude is $\frac{30(2)}{13}=\frac{60}{13}$. The sum of our altitudes is $\frac{60+156+65}{13}=\boxed{\textbf{(E)} \dfrac{281}{13}}$. Note that there is no need to calculate the final answer after we know that the third altitude has length $\frac{60}{13}$ since $E$ is the only choice with a denominator of $13$ and $13$ is relatively prime to $5$ and $12$.

Video Solution 1

https://youtu.be/zFHtdWEhriQ

~Education, the Study of Everything

Solution 2 (very similar to Solution 1)

Noticing that the line has coefficients $12$ and $5$, we can suspect that we have a $5$-$12$-$13$ triangle on our hands. Because we want the sum of the altitudes, the third altitude that is not an axis must have a denominator of $13$. Since the other altitudes are integers, we choose the option with $13$ as the denominator, namely $E$.

See Also

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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=2015_AMC_10B_Problems/Problem_13&oldid=176590"