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Difference between revisions of "2015 AMC 10B Problems/Problem 13"

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==Solution 2 (very similar to Solution 1)==
 
==Solution 2 (very similar to Solution 1)==
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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 a <math>13</math> as the denominator, namely <math>E</math>.
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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==

Revision as of 10:58, 18 July 2021

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 $12$.

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

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