2015 AMC 10B Problems/Problem 13
Contents
Problem
The line forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?
Solution 1
We find the -intercepts and the -intercepts to find the intersections of the axes and the line. If , then . If is , then . Our three vertices are , , and . Two of our altitudes are and , and since it is a -- right triangle, the hypotenuse is . Since the area of the triangle is , so our final altitude is . The sum of our altitudes is . Note that there is no need to calculate the final answer after we know that the third altitude has length since is the only choice with a denominator of and is relatively prime to and .
Video Solution 1
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Solution 2 (very similar to Solution 1)
Noticing that the line has coefficients and , we can suspect that we have a -- 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 . Since the other altitudes are integers, we choose the option with as the denominator, namely .
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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All AMC 10 Problems and Solutions |
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