1990 AHSME Problems/Problem 27

Problem

Which of these triples could $\underline{not}$ be the lengths of the three altitudes of a triangle?

$\text{(A) } 1,\sqrt{3},2\quad \text{(B) } 3,4,5\quad \text{(C) } 5,12,13\quad \text{(D) } 7,8,\sqrt{113}\quad \text{(E) } 8,15,17$

Solution

Let $a$ , $b$ , and $c$ be the side lengths of the triangle such that $a<b<c$ . We are given $a+b>c$ by the triangle inequality.

Let $h_a$ , $h_b$ , and $h_c$ be the altitudes to sides $a$ , $b$ , and $c$ respectively. We see that $h_c<h_b<h_a$ . By computing the areas using $a$ , $b$ , and $c$ as bases we get $\frac{1}{2}ah_a=\frac{1}{2}bh_b=\frac{1}{2}ch_c$ Solving for $a$ and $b$ , plugging back into the triangle inequality, and canceling $c$ from both sides leaves us with $\frac{h_c}{h_a}+\frac{h_c}{h_b}>1$ Further manipulation gives $h_c>\frac{h_ah_b}{h_a+h_b}$

Looking at the answer choices and letting $h_c$ be the smallest value each time, we see that $5>\frac{12\cdot13}{12+13}$ is not true. Thus, the answer is $\fbox{C}$ .

1990 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
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1990 AHSME Problems/Problem 27

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