Difference between revisions of "2015 AIME II Problems/Problem 4"
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==Solution== | ==Solution== | ||
− | Call the trapezoid <math>ABCD</math> with <math>AB</math> as the smaller base and <math>CD</math> as the longer. | + | Call the trapezoid <math>ABCD</math> with <math>AB</math> as the smaller base and <math>CD</math> as the longer. Let the point where an altitude intersects the larger base be <math>E</math>, where <math>E</math> is closer to <math>D</math>. |
Subtract the two bases and divide to find that <math>ED</math> is <math>\log 8</math>. The altitude can be expressed as <math>\frac{4}{3} \log 8</math>. Therefore, the two legs are <math>\frac{5}{3} \log 8</math>, or <math>\log 32</math>. | Subtract the two bases and divide to find that <math>ED</math> is <math>\log 8</math>. The altitude can be expressed as <math>\frac{4}{3} \log 8</math>. Therefore, the two legs are <math>\frac{5}{3} \log 8</math>, or <math>\log 32</math>. |
Latest revision as of 15:16, 28 March 2020
Problem
In an isosceles trapezoid, the parallel bases have lengths and , and the altitude to these bases has length . The perimeter of the trapezoid can be written in the form , where and are positive integers. Find .
Solution
Call the trapezoid with as the smaller base and as the longer. Let the point where an altitude intersects the larger base be , where is closer to .
Subtract the two bases and divide to find that is . The altitude can be expressed as . Therefore, the two legs are , or .
The perimeter is thus which is . So
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.