2002 AMC 10B Problem 20

Question

Let

a,

b,

and

c

be real numbers such that

a - 7b + 8c = 4

and

8a + 4b - c = 7.

What is

a^2 - b^2 + c^2?

0

1

4

7

8

Rearrange to

a + 8c = 4 + 7b

and

8a - c = 7 - 4b

Square both equations and add; the cross terms with

b

cancel nicely

Rewrite the equations as

a + 8c = 4 + 7b

and

8a - c = 7 - 4b.

Squaring both and adding,

\begin{aligned} &(a + 8c)^2 + (8a - c)^2 \\ &= (4 + 7b)^2 \\ &\quad {}+ (7 - 4b)^2. \end{aligned}

The left side expands to

65a^2 + 65c^2

(the

ac

terms cancel), and the right side expands to

65 + 65b^2

(the

b

terms cancel). So

65(a^2 + c^2) = 65(1 + b^2),

giving

a^2 - b^2 + c^2 = 1.

Thus, the correct answer is B.

Problem 20 in other years