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 | |
| 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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