1986 AHSME Problems/Problem 29

Problem

Two of the altitudes of the scalene triangle $ABC$ have length $4$ and $12$ . If the length of the third altitude is also an integer, what is the biggest it can be?

$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ \text{none of these}$

Solution

Assume we have a scalene triangle $ABC$ . Arbitrarily, let 12 be the height to base $AB$ and 4 be the height to base $AC$ . Due to area equivalences, the base $AC$ must be three times the length of $AB$ .

Let the base $AB$ be $x$ , thus making $AC = 3x$ . Thus, setting the final height to base $BC$ to $h$ , we note that $(BC * h)/2 = (3x * 4)/2 = 6x$ . Thus, $h = (12x / BC)$ . We note that to maximize $h$ we must minimize $BC$ . Using the triangle inequality, $BC + AB > AC$ , thus $BC + x > 3x$ or $BC > 2x$ . The minimum value of $BC$ is $2x$ , which would output $h = 6$ . However, because $BC$ must be larger than $2x$ , the minimum integer height must be $5$ .

$\fbox{B}$

1986 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
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 • 26 • 27 • 28 • 29 • 30
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1986 AHSME Problems/Problem 29

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