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> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | + | 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> | ||
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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
Contents
Problem
Two of the altitudes of the scalene triangle 
have length 
and 
.
If the length of the third altitude is also an integer, what is the biggest it can be?
Solution 1
Assume we have a scalene triangle . Arbitrarily, let be the height to base 
and be the height to base 
. Due to area equivalences, the base must be three times the length of .
Let the base be 
, thus making 
. Thus, setting the final height to base 
to 
, we note that (by area equivalence) 
. Thus, 
. We note that to maximize we must minimize . Using the triangle inequality, 
, thus 
or 
. The minimum value of is 
, which would output 
. However, because must be larger than , the minimum integer height must be 
.
Solution 2
The reciprocals of the altitudes of a triangle themselves form a triangle - this can be easily proven. Let our desired altitude be 
.
We have 
, which implies 
. We also have 
, which implies 
. Therefore the maximum integral value of is 5.
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
