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>. | 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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<math>\fbox{B}</math> | <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 == |
Revision as of 05:28, 26 June 2018
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 |
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