Problem:
For real numbers w and z,
w1ββz1βw1β+z1ββ=2014
What is wβzw+zβ?
Answer Choices:
A. β2014
B. 2014β1β
C. 20141β
D. 1
E. 2014
Solution:
Note that
2014=w1ββz1βw1β+z1ββ=wzzβwβwzw+zββ=zβww+zβ
Because zβww+zβ=βwβzw+zβ, the requested value is (A)β2014β.
The problems on this page are the property of the MAA's American Mathematics Competitions