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Difference between revisions of "1990 AHSME Problems/Problem 27"

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== Solution ==
 
== Solution ==
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Let <math>a</math>, <math>b</math>, and <math>c</math> be the lengths of the triangle such that <math>a<b<c</math>. We are given <math>a+b>c</math> by the triangle inequality.
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Let <math>a</math>, <math>b</math>, and <math>c</math> be the side lengths of the triangle such that <math>a<b<c</math>. We are given <math>a+b>c</math> by the triangle inequality.
  
 
Let <math>h_a</math>, <math>h_b</math>, and <math>h_c</math> be the altitudes to sides <math>a</math>, <math>b</math>, and <math>c</math> respectively. We see that <math>h_c<h_b<h_a</math>. By computing the areas using <math>a</math>, <math>b</math>, and <math>c</math> as bases we get <cmath>\frac{1}{2}ah_a=\frac{1}{2}bh_b=\frac{1}{2}ch_c</cmath>Solving for <math>a</math> and <math>b</math>, plugging back into the triangle inequality, and canceling <math>c</math> from both sides leaves us with <cmath>\frac{h_c}{h_a}+\frac{h_c}{h_b}>1</cmath>
 
Let <math>h_a</math>, <math>h_b</math>, and <math>h_c</math> be the altitudes to sides <math>a</math>, <math>b</math>, and <math>c</math> respectively. We see that <math>h_c<h_b<h_a</math>. By computing the areas using <math>a</math>, <math>b</math>, and <math>c</math> as bases we get <cmath>\frac{1}{2}ah_a=\frac{1}{2}bh_b=\frac{1}{2}ch_c</cmath>Solving for <math>a</math> and <math>b</math>, plugging back into the triangle inequality, and canceling <math>c</math> from both sides leaves us with <cmath>\frac{h_c}{h_a}+\frac{h_c}{h_b}>1</cmath>

Latest revision as of 20:26, 31 July 2016

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 $\fbox{C}$.

See also

1990 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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
All AHSME Problems and Solutions

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