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
Solution 3
Let be the trapezoid, where and and . Draw altitudes from and to with feet at and , respectively. , so . Now, we attempt to find , or what's left of after we take out . We make use of the two logarithmic rules:
Thus, since , .
Now, why was finding important? Absolutely no reason! Just kidding, lol 🤣 Now, we essentially "glue" triangles and together to get , where is the point where and became one. Note we can do this because and are both right triangles with a common leg length (the altitude of trapezoid ).
Triangle has a base of , which is just equal to . It is equal to because when we brought triangles and together, the length of was not changed except for taking out .
since because the problem tells us we have an isosceles trapezoid. Drop and altitude from to The altitude has length . The altitude also bisects since is isosceles. Let the foot of the altitude be . Then (Remember that C'D' was , and then it got bisected by the altitude). Thus, the hypotenuse, must be from the Pythagorean Theorem or by noticing that you have a 3-4-5 right triangle with a similarity ratio of . Since , .
Now, we have , , and . Thus, their sum is
Thus, . ~Extremelysupercooldude
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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