# 1990 AHSME Problems/Problem 27

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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. 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. 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 $\fbox{C}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 