# 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. $\fbox{B}$.

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