Difference between revisions of "1990 AHSME Problems/Problem 27"
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== Solution == | == Solution == | ||
| − | <math>\fbox{C}</math> | + | 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. |
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| + | 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> | ||
| + | Further manipulation gives <cmath>h_c>\frac{h_ah_b}{h_a+h_b}</cmath> | ||
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| + | Looking at the answer choices and letting <math>h_c</math> be the smallest value each time, we see that <math>5>\frac{12\cdot13}{12+13}</math> is not true. Thus, the answer is <math>\fbox{C}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 20:26, 31 July 2016
Problem
Which of these triples could 
be the lengths of the three altitudes of a triangle?
Solution
Let 
, 
, and 
be the side lengths of the triangle such that 
. We are given 
by the triangle inequality.
Let 
, 
, and 
be the altitudes to sides , , and respectively. We see that 
. By computing the areas using , , and as bases we get ![\[\frac{1}{2}ah_a=\frac{1}{2}bh_b=\frac{1}{2}ch_c\]](http://latex.artofproblemsolving.com/7/e/d/7ed2162da72b5bf2f23fac3f3c3a48e40495cd53.png)
Solving for and , plugging back into the triangle inequality, and canceling from both sides leaves us with ![\[\frac{h_c}{h_a}+\frac{h_c}{h_b}>1\]](http://latex.artofproblemsolving.com/a/9/4/a943ffce76fd37937599c3d4c2fd6aa4e4846048.png)
Further manipulation gives ![\[h_c>\frac{h_ah_b}{h_a+h_b}\]](http://latex.artofproblemsolving.com/5/4/6/5460b0370fe414ccd66a4b700d2dc67df152f43a.png)
Looking at the answer choices and letting be the smallest value each time, we see that 
is not true. Thus, the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
