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 1

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 (by area equivalence) $\frac{BC \cdot h}{2} = \frac{3x \cdot 4}{2} = 6x$ . Thus, $h = \frac{12x}{BC}$ . 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 $5$ .

$\fbox{B}$

Solution 2

The reciprocals of the altitudes of a triangle themselves form a triangle - this can be easily proven. Let our desired altitude be $a$ .

We have $\frac{1}{a}<\frac{1}{4}+\frac{1}{12}=\frac{1}{3}$ , which implies $a>3$ . We also have $\frac{1}{a}>\frac{1}{4}-\frac{1}{12}=\frac{1}{6}$ , which implies $a<6$ . Therefore the maximum integral value of $a$ is 5.

$\fbox{B}$ .

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

Contents

Problem

Solution 1

Solution 2

See also