1987 AIME Problems/Problem 14
Contents
[hide]Problem
Compute
Solution 1 (Sophie Germain Identity)
The Sophie Germain Identity states that can be factored as . Each of the terms is in the form of . Using Sophie Germain, we get that so the original expression becomes
Almost all of the terms cancel out! We are left with .
Solution 2 (Completing the Square and Difference of Squares)
In both the numerator and the denominator, each factor is of the form for some positive integer
We factor by completing the square, then applying the difference of squares: The original expression now becomes ~MRENTHUSIASM
Solution 3 (Complex Numbers)
In both the numerator and the denominator, each factor is of the form for some positive integer
We factor by solving the equation .
SOLUTION IN PROGRESS. WAITING FOR THE API ISSUE TO BE FIXED.
(code testing--this should be rendered correctly).
TEST
Video Solution
https://youtu.be/ZWqHxc0i7ro?t=1023
~ pi_is_3.14
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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