2015 AIME II Problems/Problem 4
Contents
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
Solution 2 (gratuitous wishful thinking)
Set the base of the log as 2. Then call the trapezoid with as the longer base. Then have the two feet of the altitudes be and , with and in position from left to right respectively. Then, and are (from the log subtraction identity. Then (isosceles trapezoid and being 6. Then the 2 legs of the trapezoid is .
And we have the answer:
-dragoon
Video Solution
https://www.youtube.com/watch?v=9re2qLzOKWk&t=226s
~MathProblemSolvingSkills.com
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 |
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