2015 AMC 10B Problems/Problem 13


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


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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AMC 10 Problems and Solutions

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