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Difference between revisions of "1986 AHSME Problems/Problem 29"

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\textbf{(E)}\ \text{none of these}  </math>   
 
\textbf{(E)}\ \text{none of these}  </math>   
  
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==Solution==
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==Solution 1==
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Assume we have a scalene triangle <math>ABC</math>. Arbitrarily, let <math>12</math> be the height to base <math>AB</math> and <math>4</math> be the height to base <math>AC</math>. Due to area equivalences, the base <math>AC</math> must be three times the length of <math>AB</math>.
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Let the base <math>AB</math> be <math>x</math>, thus making <math>AC = 3x</math>. Thus, setting the final height to base <math>BC</math> to <math>h</math>, we note that (by area equivalence) <math>\frac{BC \cdot h}{2} = \frac{3x \cdot 4}{2} = 6x</math>. Thus, <math>h = \frac{12x}{BC}</math>. We note that to maximize <math>h</math> we must minimize <math>BC</math>. Using the triangle inequality, <math>BC + AB > AC</math>, thus <math>BC + x > 3x</math> or <math>BC > 2x</math>. The minimum value of <math>BC</math> is <math>2x</math>, which would output <math>h = 6</math>. However, because <math>BC</math> must be larger than <math>2x</math>, the minimum integer height must be <math>5</math>.
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<math>\fbox{B}</math>
 
<math>\fbox{B}</math>
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 +
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==Solution 2==
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 +
The reciprocals of the altitudes of a triangle themselves form a triangle - this can be easily proven. Let our desired altitude be <math>a</math>.
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We have <math>\frac{1}{a}<\frac{1}{4}+\frac{1}{12}=\frac{1}{3}</math>, which implies <math>a>3</math>. We also have <math>\frac{1}{a}>\frac{1}{4}-\frac{1}{12}=\frac{1}{6}</math>, which implies <math>a<6</math>. Therefore the maximum integral value of <math>a</math> is 5.
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<math>\fbox{B}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 13:00, 30 October 2019

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.

$\fbox{B}$.

See also

1986 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
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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