1987 AIME Problems/Problem 14
Contents
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 which simplifies to Almost all of the terms cancel out! We are left with
~Azjps (Solution)
~MRENTHUSIASM (Minor Reformatting)
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 or
Two solutions follow from here:
Solution 3.1 (Polar Form)
We rewrite to the polar form where is the magnitude of such that and is the argument of such that
By De Moivre's Theorem, we have from which
- so
- so
By the Factor Theorem, we get We continue with the last paragraph of Solution 2 to get the answer
~MRENTHUSIASM
Solution 3.2 (Rectangular Form)
We rewrite to the rectangular form for some real numbers and
Note that so there are two cases:
We have We need from which or
We have We need from which or
By the Factor Theorem, we get We continue with the last paragraph of Solution 2 to get the answer
~MRENTHUSIASM
Video Solutions
https://youtu.be/ZWqHxc0i7ro?t=1023
~ pi_is_3.14
https://www.youtube.com/watch?v=yoOWcx2Otcw
~Michael Penn
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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