Problem:
Let a,b, and c be real numbers such that a−7b+8c=4 and 8a+4b−c=7. Then a2−b2+c2 is
Answer Choices:
A. 0
B. 1
C. 4
D. 7
E. 8
Solution:
We have a+8c=4+7b and 8a−c=7−4b. Squaring both equations and adding the results yields
(a+8c)2+(8a−c)2=(4+7b)2+(7−4b)2.
Expanding gives 65(a2+c2)=65(1+b2). So a2+c2=1+b2, and a2−b2+c2=1.
Answer: B.
The problems on this page are the property of the MAA's American Mathematics Competitions