Difference between revisions of "1986 AHSME Problems/Problem 29"
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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>. | 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>. | ||
− | <math>\fbox{B}</math> | + | <math>\fbox{B}</math> |
== See also == | == See also == |
Revision as of 21:38, 15 April 2018
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
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 .
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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