1986 AHSME Problems/Problem 29
Contents
Problem
Two of the altitudes of the scalene triangle have length and . If the length of the third altitude is also an integer, what is the biggest it can be?
Solution 1
Assume we have a scalene triangle . Arbitrarily, let be the height to base and be the height to base . Due to area equivalences, the base must be three times the length of .
Let the base be , thus making . Thus, setting the final height to base to , we note that (by area equivalence) . Thus, . We note that to maximize we must minimize . Using the triangle inequality, , thus or . The minimum value of is , which would output . However, because must be larger than , the minimum integer height must be .
Solution 2
The reciprocals of the altitudes of a triangle themselves form a triangle - this can be easily proven. Let our desired altitude be .
We have , which implies . We also have , which implies . Therefore the maximum integral value of is 5.
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See also
1986 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Problem 30 | |
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