2015 AMC 10B Problems/Problem 13

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 a $13$ as the denominator, namely $E$ .

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2015 AMC 10B Problems/Problem 13

Contents

Problem

Solution 1

Solution 2 (very similar to Solution 1)

See Also